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In Chapter 3 "Chemical Reactions", Section 3.3 "Chemical Equations", you learned that applying a small amount of heat to solid ammonium dichromate initiates a vigorous reaction that produces chromium(III) oxide, nitrogen gas, and water vapor. These are not the only products of this reaction that interest chemists, however; the reaction also releases energy in the form of heat and light. So our description of this reaction was incomplete. A complete description of a chemical reaction includes not only the identity, amount, and chemical form of the reactants and products but also the quantity of energy produced or consumed. In combustion reactions, heat is always a product; in other reactions, heat may be produced or consumed.
This chapter introduces you to thermochemistryA branch of chemistry that describes the energy changes that occur during chemical reactions., a branch of chemistry that describes the energy changes that occur during chemical reactions. In some situations, the energy produced by chemical reactions is actually of greater interest to chemists than the material products of the reaction. For example, the controlled combustion of organic molecules, primarily sugars and fats, within our cells provides the energy for physical activity, thought, and other complex chemical transformations that occur in our bodies. Similarly, our energy-intensive society extracts energy from the combustion of fossil fuels, such as coal, petroleum, and natural gas, to manufacture clothing and furniture, heat your home in winter and cool it in summer, and power the car or bus that gets you to class and to the movies. By the end of this chapter, you will know enough about thermochemistry to explain why ice cubes cool a glass of soda, how instant cold packs and hot packs work, and why swimming pools and waterbeds are heated. You will also understand what factors determine the caloric content of your diet and why even “nonpolluting” uses of fossil fuels may be affecting the environment.
Thermodynamic spontaneity. The highly exothermic and dramatic thermite reaction is thermodynamically spontaneous. Reactants of aluminum and a metal oxide, usually iron, which are stable at room temperature, are ignited either in the presence of heat or by the reaction of potassium permanganate and glycerin. The resulting products are aluminum oxide, free and molten elemental metal, and a great deal of heat, which makes this an excellent method for on-site welding. Because this reaction has its own oxygen supply, it can be used for underwater welding as well.
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (). Thermal energyEnergy that results from atomic and molecular motion; the faster the motion, the higher the thermal energy. results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperatureA measure of an object’s thermal energy content. of an object is a measure of its thermal energy content. Radiant energyOne of the five forms of energy, radiant energy is carried by light, microwaves, and radio waves (the other forms of energy are thermal, chemical, nuclear, and electrical). Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energyOne of the five forms of energy, electrical energy results from the flow of electrically charged particles. The other four forms of energy are radiant, thermal, chemical, and nuclear. results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energyOne of the five forms of energy, nuclear energy is stored in the nucleus of an atom. The other four forms of energy are radiant, thermal, chemical, and electrical. is stored in the nucleus of an atom, and chemical energyOne of the five forms of energy, chemical energy is stored within a chemical compound because of a particular arrangement of atoms. The other four forms of energy are radiant, thermal, nuclear, and electrical. is stored within a chemical compound because of a particular arrangement of atoms.
Figure 5.1 Forms of Energy
(a) Thermal energy results from atomic and molecular motion; molten steel at 2000°C has a very high thermal energy content. (b) Radiant energy (e.g., from the sun) is the energy in light, microwaves, and radio waves. (c) Lightning is an example of electrical energy, which is due to the flow of electrically charged particles. (d) Nuclear energy is released when particles in the nucleus of the atom are rearranged. (e) Chemical energy results from the particular arrangement of atoms in a chemical compound; the heat and light produced in this reaction are due to energy released during the breaking and reforming of chemical bonds.
Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE)Energy stored in an object because of its relative position or orientation., which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE)Energy due to the motion of an object: where is the mass of the object and is its velocity. is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other.
Energy can be converted from one form to another () or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to mechanical work to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction.
Figure 5.2 Interconversion of Forms of Energy
When a swimmer steps off the platform to dive into the water, potential energy is converted to kinetic energy. As the swimmer climbs back up to the top of the diving platform, chemical energy is converted to mechanical work.
Although energy can be converted from one form to another, the total amount of energy in the universe remains constant. This is known as the law of conservation of energyThe total amount of energy in the universe remains constant. Energy can be neither created nor destroyed, but it can be converted from one form to another..As you will learn in , the law of conservation of energy is also known as the first law of thermodynamics. Energy cannot be created or destroyed.
One definition of energyThe capacity to do work. is the capacity to do work. The easiest form of work to visualize is mechanical workThe energy required to move an object a distance when opposed by a force : (), which is the energy required to move an object a distance d when opposed by a force F, such as gravity:
Equation 5.1
Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (a), we can also write as follows:Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth.
Equation 5.2
Figure 5.3 An Example of Mechanical Work
One form of energy is mechanical work, the energy required to move an object of mass m a distance d when opposed by a force F, such as gravity.
Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the force of gravity. The amount of work done (w) and thus the energy required depends on three things: (1) the height of the second floor (the distance d); (2) your mass, which must be raised that distance against the downward acceleration due to gravity; and (3) your path, as you will learn in .
In contrast, heat (q)Thermal energy that can be transformed from an object at one temperature to an object at another temperature. is thermal energy that can be transferred from an object at one temperature to an object at another temperature. The net transfer of thermal energy stops when the two objects reach the same temperature.
The energy of an object can be changed only by the transfer of energy to or from another object in the form of heat,As you will learn in , hot objects can also lose energy as radiant energy, such as heat or light. This energy is converted to heat when it is absorbed by another object. Hence radiant energy is equivalent to heat. work performed on or by the object, or some combination of heat and work. Consider, for example, the energy stored in a fully charged battery. As shown in , this energy can be used primarily to perform work (e.g., running an electric fan) or to generate light and heat (e.g., illuminating a light bulb). When the battery is fully discharged in either case, the total change in energy is the same, even though the fraction released as work or heat varies greatly. The sum of the heat produced and the work performed equals the change in energy (ΔE):
Equation 5.3
Energy can be transferred only in the form of heat, work performed on or by an object, or some combination of heat and work.
Figure 5.4 Energy Transfer
Discharging a fully charged battery releases the same amount of energy whether the battery is used to run a fan (a) or illuminate a light bulb (b). In (a), most of the energy is used to perform work, which turns the blades of the fan and thus moves the air; only a small portion of the energy is released as heat by the electric motor. In (b), all the energy is released as heat and light; no work is done.
Energy is an extensive property of matter—for example, the amount of thermal energy in an object is proportional to both its mass and its temperature. (For more information on the properties of matter, see .) A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy.
The kinetic energy of an object is related to its mass m and velocity v:
Equation 5.4
For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is
Equation 5.5
Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J)The SI unit of energy: ,The joule is named after the British physicist James Joule (1818–1889), an early worker in the field of energy. is defined as 1 kilogram·meter2/second2 (kg·m2/s2). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 103 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 105 J or 4.88 × 102 kJ. It is important to remember that the units of energy are the same regardless of the form of energy, whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same.
To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is given by (w = Fd). According to , the force (F) exerted by gravity on any object is equal to its mass (m, in this case, 1360 kg) times the acceleration (a) due to gravity (g, 9.81 m/s2 at Earth’s surface). The distance (d) is the height (h) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows:
Equation 5.6
The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero.
Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage.
A recent and spectacular example of the conversion of potential energy to kinetic energy was seen by the earthquake near the east coast of Honshu, Japan, on March 11, 2011. The magnitude 9.0 earthquake occurred along the Japan Trench subduction zone, the interface boundary between the Pacific and North American geological plates. During its westward movement, the Pacific plate became trapped under the North American plate, and its further movement was prevented. When there was sufficient potential energy to allow the Pacific plate to break free, approximately 7.1 × 1015 kJ of potential energy was released as kinetic energy, the equivalent of 4.75 × 108 tn of TNT (trinitrotoluene) or 25,003 nuclear bombs. The island of Japan experienced the worst devastation in its history from the earthquake, resulting tsunami, and aftershocks. Historical records indicate that an earthquake of such force occurs in some region of the globe approximately every 1000 years. One such earthquake and resulting tsunami is speculated to have caused the destruction of the lost city of Atlantis, referred to by the ancient Greek philosopher Plato.
The units of energy are the same for all forms of energy.
Energy can also be expressed in the non-SI units of calories (cal)A non-SI unit of energy: 1 cal = 4.184 J exactly., where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C.We specify the exact temperatures because the amount of energy needed to raise the temperature of 1 g of water 1°C varies slightly with elevation. To three significant figures, however, this amount is 1.00 cal over the temperature range 0°C–100°C. The name is derived from the Latin calor, meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:
Equation 5.7
In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information, addressed in .
Given: mass and velocity or height
Asked for: kinetic and potential energy
Strategy:
Use to calculate the kinetic energy and to calculate the potential energy, as appropriate.
Solution:
The kinetic energy of an object is given by In this case, we know both the mass and the velocity, but we must convert the velocity to SI units:
The kinetic energy of the baseball is therefore
The increase in potential energy is the same as the amount of work required to raise the ball to its new altitude, which is (250 − 3) = 247 feet above its initial position. Thus
Exercise
Answer:
general definition of work
relationship between energy, heat, and work
kinetic energy
potential energy in a gravitational field
Thermochemistry is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. Energy is the capacity to do work. Mechanical work is the amount of energy required to move an object a given distance when opposed by a force. Thermal energy is due to the random motions of atoms, molecules, or ions in a substance. The temperature of an object is a measure of the amount of thermal energy it contains. Heat (q) is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of potential energy (PE), energy caused by the relative position or orientation of an object. Kinetic energy (KE) is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the law of conservation of energy states that energy can be neither created nor destroyed. The most common units of energy are the joule (J), defined as 1 (kg·m2)/s2, and the calorie, defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J).
What is the relationship between mechanical work and energy?
Does a person with a mass of 50 kg climbing a height of 15 m do work? Explain your answer. Does that same person do work while descending a mountain?
If a person exerts a force on an immovable object, does that person do work? Explain your answer.
Explain the differences between electrical energy, nuclear energy, and chemical energy.
The chapter describes thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy. Which form(s) of energy are represented by each of the following?
Describe the various forms of energy that are interconverted when a flashlight is switched on.
Describe the forms of energy that are interconverted when the space shuttle lifts off.
Categorize each of the following as representing kinetic energy or potential energy.
Are the units for potential energy the same as the units for kinetic energy? Can an absolute value for potential energy be obtained? Explain your answer.
Categorize each of the following as representing kinetic energy or potential energy.
Why does hammering a piece of sheet metal cause the metal to heat up?
Technically, the person is not doing any work, since the object does not move.
The kinetic energy of the hammer is transferred to the metal.
Please be sure you are familiar with the topics discussed in Essential Skills 4 () before proceeding to the Numerical Problems.
Describe the mathematical relationship between (a) the thermal energy stored in an object and that object’s mass and (b) the thermal energy stored in an object and that object’s temperature.
How much energy (in kilojoules) is released or stored when each of the following occurs?
Calculate how much energy (in kilojoules) is released or stored when each of the following occurs:
A car weighing 1438 kg falls off a bridge that is 211 ft high. Ignoring air resistance, how much energy is released when the car hits the water?
A 1 tn roller coaster filled with passengers reaches a height of 28 m before accelerating downhill. How much energy is released when the roller coaster reaches the bottom of the hill? Assume no energy is lost due to friction.
250 kJ released
To study the flow of energy during a chemical reaction, we need to distinguish between a systemThe small, well-defined part of the universe in which we are interested., the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundingsAll the universe that is not the system; that is, system + surroundings = universe., the rest of the universe, including the container in which the reaction is carried out (). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa.
Figure 5.5 A System and Its Surroundings
The system is that part of the universe we are interested in studying, such as a chemical reaction inside a flask. The surroundings are the rest of the universe, including the container in which the reaction is carried out.
Three kinds of systems are important in chemistry. An open systemA system that can exchange both matter and energy with its surroundings. can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed systemA system that can exchange energy but not matter with its surroundings. can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated systemA system that can exchange neither energy nor matter with its suroundings. exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant, which must be true if energy is conserved.
The state of a systemA complete description of the system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state functionA property of a system whose magnitude depends on only the present state of the system, not its previous history. is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent. For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose.
Figure 5.6 Elevation as an Example of a State Function
The change in elevation between state 1 (at the bottom of the parking garage) and state 2 (at the top level of the parking garage) is the same for both paths A and B; it does not depend on which path is taken from the bottom to the top. In contrast, the distance traveled and the work needed to reach the top do depend on which path is taken. Elevation is a state function, but distance and work are not state functions.
The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (see chapter opening image). The balanced chemical equation for the reaction is as follows:
Equation 5.8
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)We can also write this chemical equation as
Equation 5.9
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) + heatto indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat (q) is transferred from a system to its surroundings is described as exothermicA process in which heat is transferred from a system to its surroundings.. By convention, q < 0 for an exothermic reaction.
When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
Equation 5.10
heat + H2O(s) → H2O(l)When heat is transferred to a system from its surroundings, the process is endothermicA process in which heat is transferred to a system from its surroundings.. By convention, q > 0 for an endothermic reaction.
We have stated that the change in energy (ΔE) is equal to the sum of the heat produced and the work performed (). Work done by an expanding gas is called pressure-volume work, also called PV work. Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
Equation 5.11
Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure)
Equation 5.12
w = −PΔVThe negative sign associated with PV work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
Figure 5.7 An Example of Work Performed by a Reaction Carried Out at Constant Pressure
(a) Initially, the system (a copper penny and concentrated nitric acid) is at atmospheric pressure. (b) When the penny is added to the nitric acid, the volume of NO2 gas that is formed causes the piston to move upward to maintain the system at atmospheric pressure. In doing so, the system is performing work on its surroundings.
The symbol E in represents the internal energyThe sum of the kinetic and potential energies of all of a system’s components. Additionally, ΔE = q + w, where q is the heat produced by the system and w is the work performed by the system. Internal energy is a state function. of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H)The sum of a system’s internal energy and the product of its pressure and volume : (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy E plus the product of its pressure P and volume V:
Equation 5.13
H = E + PVBecause internal energy, pressure, and volume are all state functions, enthalpy is also a state function.
If a chemical change occurs at constant pressure (for a given P, ΔP = 0), the change in enthalpy (ΔH)At constant pressure, the amount of heat transferred from the surroundings to the system or vice versa: . is
Equation 5.14
ΔH = Δ(E + PV) = ΔE + ΔPV = ΔE + PΔVSubstituting q + w for ΔE () and −w for PΔV (), we obtain
Equation 5.15
ΔH = ΔE + PΔV = qp + w − w = qpThe subscript p is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From we see that at constant pressure the change in enthalpy, ΔH of the system, defined as Hfinal − Hinitial, is equal to the heat gained or lost.
Equation 5.16
ΔH = Hfinal − Hinitial = qpJust as with ΔE, because enthalpy is a state function, the magnitude of ΔH depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
To find ΔH, measure qp.
When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (ΔHrxn)The change in enthalpy that occurs during a chemical reaction., the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so ΔHrxn is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so ΔHrxn is positive. Thus ΔHrxn < 0 for an exothermic reaction, and ΔHrxn > 0 for an endothermic reaction. In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:
Reaction Type | q | ΔHrxn |
---|---|---|
exothermic | < 0 | < 0 (heat flows from a system to its surroundings) |
endothermic | > 0 | > 0 (heat flows from the surroundings to a system) |
If ΔHrxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (part (a) in ). Conversely, if ΔHrxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (part (b) in ). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.
Bond breaking requires an input of energy; bond making releases energy.
Figure 5.8 The Enthalpy of Reaction
Energy changes in chemical reactions are usually measured as changes in enthalpy. (a) If heat flows from a system to its surroundings, the enthalpy of the system decreases, ΔHrxn is negative, and the reaction is exothermic; it is energetically downhill. (b) Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, ΔHrxn is positive, and the reaction is endothermic; it is energetically uphill.
Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):
Equation 5.17
Equation 5.18
In both cases, the magnitude of the enthalpy change is the same; only the sign is different.
Enthalpy is an extensive property (like mass). The magnitude of ΔH for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider , which describes the reaction of aluminum with iron(III) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al2O3, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe2O3 consumed:
Equation 5.19
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) + 851.5 kJThus ΔH = −851.5 kJ/mol of Fe2O3. We can also describe ΔH for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written after the reaction, as in , it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
Equation 5.20
If 4 mol of Al and 2 mol of Fe2O3 react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:
Equation 5.21
The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example 2.
Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If ΔH is 6.01 kJ/mol for the reaction H2O(s) → H2O(l) at 0°C and constant pressure, how much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 106 metric tons)? (A metric ton is 1000 kg.)
Given: energy per mole of ice and mass of iceberg
Asked for: energy required to melt iceberg
Strategy:
A Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice.
B Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice.
Solution:
A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by ΔH (+6.01 kJ/mol):
B The energy needed to melt the iceberg is thus
Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below.
Possible sources of the approximately 3.34 × 1011 kJ needed to melt a 1.00 × 106 metric ton iceberg |
---|
Combustion of 3.8 × 103 ft3 of natural gas |
Combustion of 68,000 barrels of oil |
Combustion of 15,000 tons of coal |
1.1 × 108 kilowatt-hours of electricity |
Exercise
If 17.3 g of powdered aluminum are allowed to react with excess Fe2O3, how much heat is produced?
Answer: 273 kJ
Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s lawThe enthalpy change for an overall reaction is the sum of the values for the individual reactions., after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.
We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in can be viewed as occurring in three distinct steps with known ΔH values. As shown in , the first reaction produces 1 mol of solid aluminum oxide (Al2O3) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in ); the enthalpy change for this reaction is −732.5 kJ/mol of Fe2O3. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in ); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in ); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in , the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in gives the overall reaction, shown in part (d):
Equation 5.22
The net reaction in part (d) in is identical to . By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in .
Figure 5.9 Energy Changes Accompanying the Thermite Reaction
Because enthalpy is a state function, the overall enthalpy change for the reaction of 2 mol of Al(s) with 1 mol of Fe2O3(s) is −851.1 kJ, whether the reaction occurs in a single step (ΔH4, shown on the left) or in three hypothetical steps (shown on the right) that involve the successive formation of solid Al2O3 and liquid iron (ΔH1), solid iron at 1758°C (ΔH2), and solid iron at 25°C (ΔH3). Thus ΔH4 = ΔH1 + ΔH2 + ΔH3, as stated by Hess’s law.
Comparing parts (a) and (d) in also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in ), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in ). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation.
When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure:
We illustrate how to use this procedure in Example 3.
When carbon is burned with limited amounts of oxygen gas (O2), carbon monoxide (CO) is the main product:
When carbon is burned in excess O2, carbon dioxide (CO2) is produced:
Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O2 to give CO2.
Given: two balanced chemical equations and their ΔH values
Asked for: enthalpy change for a third reaction
Strategy:
A After balancing the chemical equation for the overall reaction, write two equations whose ΔH values are known and that, when added together, give the equation for the overall reaction. (Reverse the direction of one or more of the equations as necessary, making sure to also reverse the sign of ΔH.)
B Multiply the equations by appropriate factors to ensure that they give the desired overall chemical equation when added together. To obtain the enthalpy change per mole of CO, write the resulting equations as a sum, along with the enthalpy change for each.
Solution:
A We begin by writing the balanced chemical equation for the reaction of interest:
There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give
Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO2, is the product we want in Equation 3:
B Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant:
Writing the resulting equations as a sum, along with the enthalpy change for each, gives
Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O2, and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2:
An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms:
The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO2. From Hess’s law, ΔHA + ΔHB = ΔHC, and we are given ΔH for reactions A and C. Substituting the appropriate values gives
This is again the enthalpy change for the conversion of 2 mol of CO to CO2. The enthalpy change for the conversion of 1 mol of CO to CO2 is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem.
Exercise
The reaction of acetylene (C2H2) with hydrogen (H2) can produce either ethylene (C2H4) or ethane (C2H6):
What is ΔH for the reaction of C2H4 with H2 to form C2H6?
Answer: −136.3 kJ/mol of C2H4
, , and presented a wide variety of chemical reactions, and you learned how to write balanced chemical equations that include all the reactants and the products except heat. One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following:
Enthalpy of formation (ΔHf)The enthalpy change for the formation of 1 mol of a compound from its component elements.: The enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The corresponding relationship is
Equation 5.23
For example,
The sign convention for ΔHf is the same as for any enthalpy change: ΔHf < 0 if heat is released when elements combine to form a compound and ΔHf > 0 if heat is absorbed. The values of ΔHvap and ΔHfus for some common substances are listed in . These values are used in enthalpy calculations when any of the substances undergoes a change of physical state during a reaction.
Table 5.1 Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points
Substance | ΔHvap (kJ/mol) | ΔHfus (kJ/mol) |
---|---|---|
argon (Ar) | 6.3 | 1.3 |
methane (CH4) | 9.2 | 0.84 |
ethanol (CH3CH2OH) | 39.3 | 7.6 |
benzene (C6H6) | 31.0 | 10.9 |
water (H2O) | 40.7 | 6.0 |
mercury (Hg) | 59.0 | 2.29 |
iron (Fe) | 340 | 14 |
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane.
The standard conditionsThe conditions under which most thermochemical data are tabulated: 1 atm for all gases and a concentration of 1.0 M for all species in solution. for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard stateThe most stable form of a pure substance at a pressure of 1 atm at a specified temperature.. This is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ()The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition. (which is pronounced “delta H eff naught”). The standard enthalpy of formation of any element in its standard state is zero by definition. For example, although oxygen can exist as ozone (O3), atomic oxygen (O), and molecular oxygen (O2), O2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H2(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (). Therefore, O2(g), H2(g), and graphite have values of zero.
Figure 5.10 Elemental Carbon
Although graphite and diamond are both forms of elemental carbon, graphite is more stable at 1 atm pressure and 25°C than diamond is. Given enough time, diamond will revert to graphite under these conditions. Hence graphite is the standard state of carbon.
The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:
Equation 5.24
It is not possible to measure the value of for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, O2, and H2 and measuring the heat evolved as glucose is formed; the reaction shown in does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of for an extensive list of compounds are given in . Note that values are always reported in kilojoules per mole of the substance of interest. Also notice in that the standard enthalpy of formation of O2(g) is zero because it is the most stable form of oxygen in its standard state.
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
Given: compound
Asked for: balanced chemical equation for its formation from elements in standard states
Strategy:
Use to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made.
Solution:
To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in by a value of 0 kJ/mol.
Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are H2(g) and Cl2(g), respectively, the unbalanced chemical equation is
H2(g) + Cl2(g) → HCl(g)Fractional coefficients are required in this case because values are reported for 1 mol of the product, HCl. Multiplying both H2(g) and Cl2(g) by 1/2 balances the equation:
The standard states of the elements in this compound are Mg(s), C(s, graphite), and O2(g). The unbalanced chemical equation is thus
Mg(s) + C(s, graphite) + O2(g) → MgCO3(s)This equation can be balanced by inspection to give
Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows:
C(s, graphite) + H2(g) + O2(g) → CH3(CH2)14CO2H(s)There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is
16C(s, graphite) + 16H2(g) + O2(g) → CH3(CH2)14CO2H(s)Exercise
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
Answer:
Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose values are known. The standard enthalpy of reaction ()The enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard state. is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction
Equation 5.25
aA + bB → cC + dDwhere A, B, C, and D are chemical substances and a, b, c, and d are their stoichiometric coefficients. The magnitude of is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
Equation 5.26
More generally, we can write
Equation 5.27
where the symbol Σ means “sum of” and m and n are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters.
“Products minus reactants” summations are typical of state functions.
To demonstrate the use of tabulated values, we will use them to calculate for the combustion of glucose, the reaction that provides energy for your brain:
Equation 5.28
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)Equation 5.29
From , the relevant values are , , and . Because O2(g) is a pure element in its standard state, Inserting these values into and changing the subscript to indicate that this is a combustion reaction, we obtain
Equation 5.30
As illustrated in , we can use to calculate for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the values of the reactants. Consequently, the enthalpy changes are
(Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change.) The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O2) to the elements is therefore +1273.3 kJ.
Figure 5.11 A Thermochemical Cycle for the Combustion of Glucose
Two hypothetical pathways are shown from the reactants to the products. The green arrow labeled indicates the combustion reaction. Alternatively, we could first convert the reactants to the elements via the reverse of the equations that define their standard enthalpies of formation (the upward arrow, labeled and ). Then we could convert the elements to the products via the equations used to define their standard enthalpies of formation (the downward arrows, labeled and ). Because enthalpy is a state function, is equal to the sum of the enthalpy changes
The reactions that convert the elements to final products (downward purple arrows in ) are identical to those used to define the values of the products. Consequently, the enthalpy changes (from ) are
The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ):
Equation 5.31
This is the same result we obtained using the “products minus reactants” rule () and values. The two results must be the same because is just a more compact way of describing the thermochemical cycle shown in .
Long-chain fatty acids such as palmitic acid [CH3(CH2)14CO2H] are one of the two major sources of energy in our diet (). Use the data in to calculate for the combustion of palmitic acid. Based on the energy released in combustion per gram, which is the better fuel—glucose or palmitic acid?
Given: compound and values
Asked for: per mole and per gram
Strategy:
A After writing the balanced chemical equation for the reaction, use and the values from to calculate the energy released by the combustion of 1 mol of palmitic acid.
B Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in for the combustion of glucose to determine which is the better fuel.
Solution:
A To determine the energy released by the combustion of palmitic acid, we need to calculate its As always, the first requirement is a balanced chemical equation:
C16H32O2(s) + 23O2(g) → 16CO2(g) + 16H2O(l)Using (“products minus reactants”) with values from (and omitting the physical states of the reactants and products to save space) gives
This is the energy released by the combustion of 1 mol of palmitic acid.
B The energy released by the combustion of 1 g of palmitic acid is
As calculated in , of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore
The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight.
Exercise
Use the data in to calculate for the water–gas shift reaction, which is used industrially on an enormous scale to obtain H2(g):
Answer: −41.2 kJ/mol
We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s which we cannot obtain otherwise. This procedure is illustrated in Example 6.
Beginning in 1923, tetraethyllead [(C2H5)4Pb] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are CO2(g), H2O(l), and red PbO(s). What is the standard enthalpy of formation of tetraethyllead, given that is −19.29 kJ/g for the combustion of tetraethyllead and of red PbO(s) is −219.0 kJ/mol?
Given: reactant, products, and values
Asked for: of reactant
Strategy:
A Write the balanced chemical equation for the combustion of tetraethyllead. Then insert the appropriate quantities into to get the equation for of tetraethyllead.
B Convert per gram given in the problem to per mole by multiplying per gram by the molar mass of tetraethyllead.
C Use to obtain values of for the other reactants and products. Insert these values into the equation for of tetraethyllead and solve the equation.
Solution:
A The balanced chemical equation for the combustion reaction is as follows:
2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l)Solving for gives
The values of all terms other than and are given in .
B The magnitude of is given in the problem in kilojoules per gram of tetraethyllead. We must therefore multiply this value by the molar mass of tetraethyllead (323.44 g/mol) to get for 1 mol of tetraethyllead:
Because the balanced chemical equation contains 2 mol of tetraethyllead, is
C Inserting the appropriate values into the equation for gives
Exercise
Ammonium sulfate [(NH4)2SO4] is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid:
2NH3(g) + H2SO4(aq) → (NH4)2SO4(s)The value of is −2805 kJ/g H2SO4. Use the data in to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole).
Answer: −1181 kJ/mol
Physical changes, such as melting or vaporization, and chemical reactions, in which one substance is converted to another, are accompanied by changes in enthalpy. Two other kinds of changes that are accompanied by changes in enthalpy are the dissolution of solids and the dilution of concentrated solutions.
The dissolution of a solid can be described as follows:
Equation 5.32
solute(s) + solvent(l) → solution(l)The values of ΔHsoln for some common substances are given in . The sign and the magnitude of ΔHsoln depend on specific attractive and repulsive interactions between the solute and the solvent; these factors will be discussed in . When substances dissolve, the process can be either exothermic (ΔHsoln < 0) or endothermic (ΔHsoln > 0), as you can see from the data in .
Table 5.2 Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)
Anion | |||||
Cation | Fluoride | Chloride | Bromide | Iodide | Hydroxide |
lithium | 4.7 | −37.0 | −48.8 | −63.3 | −23.6 |
sodium | 0.9 | 3.9 | −0.6 | −7.5 | −44.5 |
potassium | −17.7 | 17.2 | 19.9 | 20.3 | −57.6 |
ammonium | −1.2 | 14.8 | 16.8 | 13.7 | — |
silver | −22.5 | 65.5 | 84.4 | 112.2 | — |
magnesium | −17.7 | −160.0 | −185.6 | −213.2 | 2.3 |
calcium | 11.5 | −81.3 | −103.1 | −119.7 | −16.7 |
Nitrate | Acetate | Carbonate | Sulfate | ||
lithium | −2.5 | — | −18.2 | −29.8 | |
sodium | 20.5 | −17.3 | −26.7 | 2.4 | |
potassium | 34.9 | −15.3 | −30.9 | 23.8 | |
ammonium | 25.7 | −2.4 | — | 6.6 | |
silver | 22.6 | — | 22.6 | 17.8 | |
magnesium | −90.9 | — | −25.3 | −91.2 | |
calcium | −19.2 | — | −13.1 | −18.0 |
Substances with large positive or negative enthalpies of solution have commercial applications as instant cold or hot packs. Single-use versions of these products are based on the dissolution of either calcium chloride (CaCl2, ΔHsoln = −81.3 kJ/mol) or ammonium nitrate (NH4NO3, ΔHsoln = +25.7 kJ/mol). Both types consist of a plastic bag that contains about 100 mL of water plus a dry chemical (40 g of CaCl2 or 30 g of NH4NO3) in a separate plastic pouch. When the pack is twisted or struck sharply, the inner plastic bag of water ruptures, and the salt dissolves in the water. If the salt is CaCl2, heat is released to produce a solution with a temperature of about 90°C; hence the product is an “instant hot compress.” If the salt is NH4NO3, heat is absorbed when it dissolves, and the temperature drops to about 0° for an “instant cold pack.”
A similar product based on the precipitation of sodium acetate, not its dissolution, is marketed as a reusable hand warmer (). At high temperatures, sodium acetate forms a highly concentrated aqueous solution. With cooling, an unstable supersaturated solution containing excess solute is formed. When the pack is agitated, sodium acetate trihydrate [CH3CO2Na·3H2O] crystallizes, and heat is evolved:
Equation 5.33
A bag of concentrated sodium acetate solution can be carried until heat is needed, at which time vigorous agitation induces crystallization and heat is released. The pack can be reused after it is immersed in hot water until the sodium acetate redissolves.
Figure 5.12 An Instant Hot Pack Based on the Crystallization of Sodium Acetate
The hot pack is at room temperature prior to agitation (left). Because the sodium acetate is in solution, you can see the metal disc inside the pack. After the hot pack has been agitated, the sodium acetate crystallizes (right) to release heat. Because of the mass of white sodium acetate that has crystallized, the metal disc is no longer visible.
The amount of heat released or absorbed when a substance is dissolved is not a constant; it depends on the final concentration of the solute. The ΔHsoln values given previously and in , for example, were obtained by measuring the enthalpy changes at various concentrations and extrapolating the data to infinite dilution.
Because ΔHsoln depends on the concentration of the solute, diluting a solution can produce a change in enthalpy. If the initial dissolution process is exothermic (ΔH < 0), then the dilution process is also exothermic. This phenomenon is particularly relevant for strong acids and bases, which are often sold or stored as concentrated aqueous solutions. If water is added to a concentrated solution of sulfuric acid (which is 98% H2SO4 and 2% H2O) or sodium hydroxide, the heat released by the large negative ΔH can cause the solution to boil. Dangerous spattering of strong acid or base can be avoided if the concentrated acid or base is slowly added to water, so that the heat liberated is largely dissipated by the water. Thus you should never add water to a strong acid or base; a useful way to avoid the danger is to remember: Add water to acid and get blasted!
definition of enthalpy
pressure-volume work
enthalpy change at constant pressure
relationship between and
In chemistry, the small part of the universe that we are studying is the system, and the rest of the universe is the surroundings. Open systems can exchange both matter and energy with their surroundings, closed systems can exchange energy but not matter with their surroundings, and isolated systems can exchange neither matter nor energy with their surroundings. A state function is a property of a system that depends on only its present state, not its history. A reaction or process in which heat is transferred from a system to its surroundings is exothermic. A reaction or process in which heat is transferred to a system from its surroundings is endothermic.
Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy (ΔH) can be measured. A negative ΔH means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. For a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔHrxn. The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion (ΔHfus) and the enthalpy of vaporization (ΔHvap), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is Hess’s law. The enthalpy of combustion (ΔHcomb) is the enthalpy change that occurs when a substance is burned in excess oxygen. The enthalpy of formation (ΔHf) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation () are determined under standard conditions: a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The standard enthalpy of reaction () can be calculated from the sum of the standard enthalpies of formation of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The enthalpy of solution (ΔHsoln) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure.
Please be sure you are familiar with the topics discussed in Essential Skills 4 () before proceeding to the Conceptual Problems.
Heat implies the flow of energy from one object to another. Describe the energy flow in an
Based on the following energy diagram,
Based on the following energy diagram,
When a thermometer is suspended in an insulated thermos that contains a block of ice, the temperature recorded on the thermometer drops. Describe the direction of heat flow.
In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. State whether each process is endothermic or exothermic.
In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. Determine whether each process is endothermic or exothermic.
Is Earth’s environment an isolated system, an open system, or a closed system? Explain your answer.
Why is it impossible to measure the absolute magnitude of the enthalpy of an object or a compound?
Determine whether energy is consumed or released in each scenario. Explain your reasoning.
The chapter states that enthalpy is an extensive property. Why? Describe a situation that illustrates this fact.
The enthalpy of a system is affected by the physical states of the reactants and the products. Explain why.
Is the distance a person travels on a trip a state function? Why or why not?
Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly.
When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state?
What is the difference between and ?
In their elemental form, A2 and B2 exist as diatomic molecules. Given the following reactions, each with an associated ΔH°, describe how you would calculate for the compound AB2.
How can of a compound be determined if the compound cannot be prepared by the reactions used to define its standard enthalpy of formation?
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
Describe the distinction between ΔHsoln and ΔHf.
Does adding water to concentrated acid result in an endothermic or an exothermic process?
The following table lists values for some ionic compounds. If 1 mol of each solute is dissolved in 500 mL of water, rank the resulting solutions from warmest to coldest.
Compound | (kJ/mol) |
---|---|
KOH | −57.61 |
LiNO3 | −2.51 |
KMnO4 | 43.56 |
NaC2H3O2 | −17.32 |
Please be sure you are familiar with the topics discussed in Essential Skills 4 () before proceeding to the Numerical Problems.
Calculate for each chemical equation. If necessary, balance the chemical equations.
Calculate for each reaction. If necessary, balance the chemical equations.
How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas?
How much energy is released or consumed if 10.0 g of N2O5 is completely decomposed to produce gaseous nitrogen dioxide and oxygen?
In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction:
NaCl(s) + H2SO4(l) + MnO2(s) → Na2SO4(s) + MnCl2(s) + H2O(l) + Cl2(g)Calculate for this reaction. Is the reaction exothermic or endothermic?
Would you expect heat to be evolved during each reaction?
How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate?
Methanol is used as a fuel in Indianapolis 500 race cars. Use the following table to determine whether methanol or 2,2,4-trimethylpentane (isooctane) releases more energy per liter during combustion.
Fuel | (kJ/mol) | Density (g/mL) |
---|---|---|
methanol | −726.1 | 0.791 |
2,2,4-trimethylpentane | −5461.4 | 0.692 |
Use the enthalpies of combustion given in the following table to determine which organic compound releases the greatest amount of energy per gram during combustion.
Fuel | (kJ/mol) |
---|---|
methanol | −726.1 |
1-ethyl-2-methylbenzene | −5210.2 |
n-octane | −5470.5 |
Given the enthalpies of combustion, which organic compound is the best fuel per gram?
Fuel | (kJ/mol) |
---|---|
ethanol | −1366.8 |
benzene | −3267.6 |
cyclooctane | −5434.7 |
−174.1 kJ/mol
−20.3 kJ
−34.3 kJ/mol Cl2; exothermic
ΔH = −2.86 kJ CaCl2: −4.6 kJ; K2CO3, −0.65 kJ; mixing, −0.28 kJ
To one decimal place
methanol: ΔH/g = −22.6 kJ C9H12: ΔH/g = −43.3 kJ octane: ΔH/g = −47.9 kJOctane provides the largest amount of heat per gram upon combustion.
ΔHf(C9H17) = −46.1 kJ/mol
Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. CalorimetryA set of techniques used to measure enthalpy changes in chemical processes. describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called calorimeters.
To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object.
We have seen that the temperature of an object changes when it absorbs or loses thermal energy. The magnitude of the temperature change depends on both the amount of thermal energy transferred (q) and the heat capacity of the object. Its heat capacity (C)The amount of energy needed to raise the temperature of an object 1°C. The units of heat capacity are joules per degree Celsius is the amount of energy needed to raise the temperature of the object exactly 1°C; the units of C are joules per degree Celsius (J/°C). The change in temperature (ΔT) is
Equation 5.34
where q is the amount of heat (in joules), C is the heat capacity (in joules per degree Celsius), and ΔT is Tfinal − Tinitial (in degrees Celsius). Note that ΔT is always written as the final temperature minus the initial temperature. The value of C is intrinsically a positive number, but ΔT and q can be either positive or negative, and they both must have the same sign. If ΔT and q are positive, then heat flows from the surroundings into an object. If ΔT and q are negative, then heat flows from an object into its surroundings.
The heat capacity of an object depends on both its mass and its composition. For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity (Cp)The amount of energy needed to increase the temperature of 1 mol of a substance by 1°C. The units of are is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of Cp are thus J/(mol·°C).The subscript p indicates that the value was measured at constant pressure. The specific heat (Cs)The amount of energy needed to increase the temperature of 1 g of a substance by 1°C. The units of are is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g·°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change in two ways:
Equation 5.35
q = nCpΔT, where n = number of moles of substanceEquation 5.36
q = mCsΔT, where m = mass of substance in gramsThe specific heats of some common substances are given in . Note that the specific heat values of most solids are less than 1 J/(g·°C), whereas those of most liquids are about 2 J/(g·°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g·°C), is one of the highest known.
Table 5.3 Specific Heats of Selected Substances at 25°C
Compound | Specific Heat [J/(g·°C)] |
---|---|
H2O(l) | 4.184 |
H2O(g) | 2.062 |
CH3OH (methanol) | 2.531 |
CH3CH2OH (ethanol) | 2.438 |
n-C6H14 (n-hexane) | 2.270 |
C6H6 (benzene) | 1.745 |
C(s) (graphite) | 0.709 |
C(s) (diamond) | 0.509 |
Al(s) | 0.897 |
Fe(s) | 0.449 |
Cu(s) | 0.385 |
Au(s) | 0.129 |
Hg(l) | 0.140 |
NaCl(s) | 0.864 |
MgO(s) | 0.921 |
SiO2(s) (quartz) | 0.742 |
CaCO3(s) (calcite) | 0.915 |
The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1°C. Consequently, the mechanism for maintaining our body temperature at about 37°C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat.
Figure 5.13 The High Specific Heat of Liquid Water Has Major Effects on Climate
Regions that are near very large bodies of water, such as oceans or lakes, tend to have smaller temperature differences between summer and winter months than regions in the center of a continent. The contours on this map show the difference between January and July monthly mean surface temperatures (in degrees Celsius).
A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.)
Passive solar system. During the day (a), sunlight is absorbed by water circulating in the water wall. At night (b), heat stored in the water wall continues to warm the air inside the house.
Given: volume and density of water and initial and final temperatures
Asked for: amount of energy stored
Strategy:
A Use the density of water at 22.0°C to obtain the mass of water (m) that corresponds to 400 L of water. Then compute ΔT for the water.
B Determine the amount of heat absorbed by substituting values for m, Cs, and ΔT into .
Solution:
A The mass of water is
The temperature change (ΔT) is 38.0°C − 22.0°C = +16.0°C.
B From , the specific heat of water is 4.184 J/(g·°C). From , the heat absorbed by the water is thus
Both q and ΔT are positive, consistent with the fact that the water has absorbed energy.
Exercise
Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO2) in .
Answer: 2.7 × 104 kJ (Even though the mass of sandstone is more than six times the mass of the water in Example 7, the amount of thermal energy stored is the same to two significant figures.)
When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process:
Equation 5.37
qcold + qhot = 0The equation implies that the amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite:
Equation 5.38
qcold = −qhotThus heat is conserved in any such process, consistent with the law of conservation of energy.
The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object.
Equation 5.39
[mCsΔT]hot + [mCsΔT]cold = 0which can be rearranged to give
Equation 5.40
[mCsΔT]cold = −[mCsΔT]hotWhen two objects initially at different temperatures are placed in contact, we can use to calculate the final temperature if we know the chemical composition and mass of the objects.
If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings.
Given: mass and initial temperature of two objects
Asked for: final temperature
Strategy:
Using and writing ΔT as Tfinal − Tinitial for both the copper and the water, substitute the appropriate values of m, Cs, and Tinitial into the equation and solve for Tfinal.
Solution:
We can adapt to solve this problem, remembering that ΔT is defined as Tfinal − Tinitial:
Substituting the data provided in the problem and gives
Exercise (a)
If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings?
Answer: 80.0°C
Exercise (b)
A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.)
Answer: 90.6°C
In Example 7, radiant energy from the sun was used to raise the temperature of water. A calorimetric experiment uses essentially the same procedure, except that the thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (qrxn < 0), then heat is absorbed by the calorimeter (qcalorimeter > 0) and its temperature increases. Conversely, if the reaction absorbs heat (qrxn > 0), then heat is transferred from the calorimeter to the system (qcalorimeter < 0) and the temperature of the calorimeter decreases. In both cases, the amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction. The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants.
Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeterA device used to measure enthalpy changes in chemical processes at constant pressure. give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10−6°C). Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and ΔHrxn is
Equation 5.41
ΔHrxn = qrxn = −qcalorimeter = −mCsΔTThe use of a constant-pressure calorimeter is illustrated in Example 9.
Figure 5.14 A Coffee-Cup Calorimeter
This simplified version of a constant-pressure calorimeter consists of two Styrofoam cups nested and sealed with an insulated stopper to thermally isolate the system (the solution being studied) from the surroundings (the air and the laboratory bench). Two holes in the stopper allow the use of a thermometer to measure the temperature and a stirrer to mix the reactants.
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is ΔHsoln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.
Given: mass of substance, volume of solvent, and initial and final temperatures
Asked for: ΔHsoln
Strategy:
A Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.
B Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into .
C Use the molar mass of KOH to calculate ΔHsoln.
Solution:
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
The temperature change is (34.7°C − 23.0°C) = +11.7°C.
B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From , we see that
ΔHrxn = −qcalorimeter = −5.13 kJThis experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.
C The last step is to use the molar mass of KOH to calculate ΔHsoln—the heat released when dissolving 1 mol of KOH:
Exercise
A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example 9, find ΔHsoln for NH4Br (in kilojoules per mole).
Answer: 16.6 kJ/mol
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeterA device used to measure energy changes in chemical processes. shown schematically in . The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.
Figure 5.15 A Bomb Calorimeter
After the temperature of the water in the insulated container has reached a constant value, the combustion reaction is initiated by passing an electric current through a wire embedded in the sample. Because this calorimeter operates at constant volume, the heat released is not precisely the same as the enthalpy change for the reaction.
Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. As you will learn in , the heat released by a reaction carried out at constant volume is identical to the change in internal energy (ΔE) rather than the enthalpy change (ΔH); ΔE is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that ΔE < ΔH, the relationship between the measured temperature change and ΔHcomb is given in , where Cbomb is the total heat capacity of the steel bomb and the water surrounding it:
Equation 5.42
ΔHcomb < qcomb = −qcalorimeter = −CbombΔTTo measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C6H5CO2H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔHcomb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in to determine Cbomb. The use of a bomb calorimeter to measure the ΔHcomb of a substance is illustrated in Example 10.
The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔHcomb of glucose?
Given: mass and ΔT for combustion of standard and sample
Asked for: ΔHcomb of glucose
Strategy:
A Calculate the value of qrxn for benzoic acid by multiplying the mass of benzoic acid by its ΔHcomb. Then use to determine the heat capacity of the calorimeter (Cbomb) from qcomb and ΔT.
B Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔHcomb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose.
Solution:
The first step is to use and the information obtained from the combustion of benzoic acid to calculate Cbomb. We are given ΔT, and we can calculate qcomb from the mass of benzoic acid:
B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:
Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
This result is in good agreement (< 1% error) with the value of ΔHcomb = −2803 kJ/mol that we calculated in using enthalpies of formation.
Exercise
When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH3NHNH2) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔHcomb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.
Answer: −1.30 × 103 kJ/mol
relationship of quantity of a substance, heat capacity, heat flow, and temperature change
constant-pressure calorimetry
: ΔHrxn = qrxn = −qcalorimeter = −mCsΔT
constant-volume calorimetry
Calorimetry is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called calorimeters, which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The heat capacity (C) of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The specific heat (Cs) of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the molar heat capacity (Cp) is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a constant-pressure calorimeter, which gives ΔH values directly, or a bomb calorimeter, which operates at constant volume and is particularly useful for measuring enthalpies of combustion.
Can an object have a negative heat capacity? Why or why not?
What two factors determine the heat capacity of an object? Does the specific heat also depend on these two factors? Explain your answer.
Explain why regions along seacoasts have a more moderate climate than inland regions do.
Although soapstone is more expensive than brick, soapstone is frequently the building material of choice for fireplaces, particularly in northern climates with harsh winters. Propose an explanation for this.
Please be sure you are familiar with the topics discussed in Essential Skills 4 () before proceeding to the Numerical Problems.
Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C.
Element | q (J) | Cp [J/(mol·K)] | Final T (°C) |
---|---|---|---|
nickel | 137 | 26.07 | |
silicon | 19.789 | 3.0 | |
zinc | 603 | 77.5 | |
mercury | 137 | 57 |
Using , how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal.
Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K).
In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy.
How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change?
A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in and solubility rules to identify the salt.
Using the data in , calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)?
A 3.00 g sample of TNT (trinitrotoluene, C7H5N3O6) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the ΔHcomb of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the ΔHf of TNT?
Cp = Cs × (molar mass)
For Cu: q = 58 J; For Al: q = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: Cs = 0.385 and 0.897 J/(g·K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al.
4.7 kJ
Tfinal = 43.1°C; the combustion reaction is 4C7H5N3O6(s) + 21O2(g) → 28CO2(g) + 10H2O(g) + 6N2(g); (TNT) = −65.5 kJ/mol
The thermochemical quantities that you probably encounter most often are the caloric values of food. Food supplies the raw materials that your body needs to replace cells and the energy that keeps those cells functioning. About 80% of this energy is released as heat to maintain your body temperature at a sustainable level to keep you alive.
The nutritional CalorieA unit used to indicate the caloric content of food. It is equal to 1 kilocalorie (1 kcal). (with a capital C) that you see on food labels is equal to 1 kcal (kilocalorie). The caloric content of food is determined from its enthalpy of combustion (ΔHcomb) per gram, as measured in a bomb calorimeter, using the general reaction
Equation 5.43
food + excess O2(g) → CO2(g) + H2O(l) + N2(g)There are two important differences, however, between the caloric values reported for foods and the ΔHcomb of the same foods burned in a calorimeter. First, the ΔHcomb described in joules (or kilojoules) are negative for all substances that can be burned. In contrast, the caloric content of a food is always expressed as a positive number because it is stored energy. Therefore,
Equation 5.44
caloric content = −ΔHcombSecond, when foods are burned in a calorimeter, any nitrogen they contain (largely from proteins, which are rich in nitrogen) is transformed to N2. In the body, however, nitrogen from foods is converted to urea [(H2N)2C=O], rather than N2 before it is excreted. The ΔHcomb of urea measured by bomb calorimetry is −632.0 kJ/mol. Consequently, the enthalpy change measured by calorimetry for any nitrogen-containing food is greater than the amount of energy the body would obtain from it. The difference in the values is equal to the ΔHcomb of urea multiplied by the number of moles of urea formed when the food is broken down. This point is illustrated schematically in the following equations:
Equation 5.45
All three ΔH values are negative, and, by Hess’s law, ΔH3 = ΔH1 + ΔH2. The magnitude of ΔH1 must be less than ΔH3, the calorimetrically measured ΔHcomb for a food. By producing urea rather than N2, therefore, humans are excreting some of the energy that was stored in their food.
Because of their different chemical compositions, foods vary widely in caloric content. As we saw in Example 5, for instance, a fatty acid such as palmitic acid produces about 39 kJ/g during combustion, while a sugar such as glucose produces 15.6 kJ/g. Fatty acids and sugars are the building blocks of fats and carbohydrates, respectively, two of the major sources of energy in the diet. Nutritionists typically assign average values of 38 kJ/g (about 9 Cal/g) and 17 kJ/g (about 4 Cal/g) for fats and carbohydrates, respectively, although the actual values for specific foods vary because of differences in composition. Proteins, the third major source of calories in the diet, vary as well. Proteins are composed of amino acids, which have the following general structure:
General structure of an amino acid. An amino acid contains an amine group (−NH2) and a carboxylic acid group (−CO2H).
In addition to their amine and carboxylic acid components, amino acids may contain a wide range of other functional groups: R can be hydrogen (–H); an alkyl group (e.g., –CH3); an aryl group (e.g., –CH2C6H5); or a substituted alkyl group that contains an amine, an alcohol, or a carboxylic acid (). Of the 20 naturally occurring amino acids, 10 are required in the human diet; these 10 are called essential amino acids because our bodies are unable to synthesize them from other compounds. Because R can be any of several different groups, each amino acid has a different value of ΔHcomb. Proteins are usually estimated to have an average ΔHcomb of 17 kJ/g (about 4 Cal/g).
Figure 5.16 The Structures of 10 Amino Acids
Essential amino acids in this group are indicated with an asterisk.
Calculate the amount of available energy obtained from the biological oxidation of 1.000 g of alanine (an amino acid). Remember that the nitrogen-containing product is urea, not N2, so biological oxidation of alanine will yield less energy than will combustion. The value of ΔHcomb for alanine is −1577 kJ/mol.
Given: amino acid and ΔHcomb per mole
Asked for: caloric content per gram
Strategy:
A Write balanced chemical equations for the oxidation of alanine to CO2, H2O, and urea; the combustion of urea; and the combustion of alanine. Multiply both sides of the equations by appropriate factors and then rearrange them to cancel urea from both sides when the equations are added.
B Use Hess’s law to obtain an expression for ΔH for the oxidation of alanine to urea in terms of the ΔHcomb of alanine and urea. Substitute the appropriate values of ΔHcomb into the equation and solve for ΔH for the oxidation of alanine to CO2, H2O, and urea.
C Calculate the amount of energy released per gram by dividing the value of ΔH by the molar mass of alanine.
Solution:
The actual energy available biologically from alanine is less than its ΔHcomb because of the production of urea rather than N2. We know the ΔHcomb values for alanine and urea, so we can use Hess’s law to calculate ΔH for the oxidation of alanine to CO2, H2O, and urea.
A We begin by writing balanced chemical equations for (1) the oxidation of alanine to CO2, H2O, and urea; (2) the combustion of urea; and (3) the combustion of alanine. Because alanine contains only a single nitrogen atom, whereas urea and N2 each contain two nitrogen atoms, it is easier to balance Equations 1 and 3 if we write them for the oxidation of 2 mol of alanine:
Adding Equations 1 and 2 and canceling urea from both sides give the overall chemical equation directly:
B By Hess’s law, ΔH3 = ΔH1 + ΔH2. We know that ΔH3 = 2ΔHcomb (alanine), ΔH2 = ΔHcomb (urea), and ΔH1 = 2ΔH (alanine → urea). Rearranging and substituting the appropriate values gives
Thus ΔH (alanine → urea) = −2522 kJ/(2 mol of alanine) = −1261 kJ/mol of alanine. Oxidation of alanine to urea rather than to nitrogen therefore results in about a 20% decrease in the amount of energy released (−1261 kJ/mol versus −1577 kJ/mol).
C The energy released per gram by the biological oxidation of alanine is
This is equal to −3.382 Cal/g.
Exercise
Calculate the energy released per gram from the oxidation of valine (an amino acid) to CO2, H2O, and urea. Report your answer to three significant figures. The value of ΔHcomb for valine is −2922 kJ/mol.
Answer: −22.2 kJ/g (−5.31 Cal/g)
The reported caloric content of foods does not include ΔHcomb for those components that are not digested, such as fiber. Moreover, meats and fruits are 50%−70% water, which cannot be oxidized by O2 to obtain energy. So water contains no calories. Some foods contain large amounts of fiber, which is primarily composed of sugars. Although fiber can be burned in a calorimeter just like glucose to give carbon dioxide, water, and heat, humans lack the enzymes needed to break fiber down into smaller molecules that can be oxidized. Hence fiber also does not contribute to the caloric content of food.
We can determine the caloric content of foods in two ways. The most precise method is to dry a carefully weighed sample and carry out a combustion reaction in a bomb calorimeter. The more typical approach, however, is to analyze the food for protein, carbohydrate, fat, water, and “minerals” (everything that doesn’t burn) and then calculate the caloric content using the average values for each component that produces energy (9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals). An example of this approach is shown in for a slice of roast beef. The compositions and caloric contents of some common foods are given in .
Table 5.4 Approximate Composition and Fuel Value of an 8 oz Slice of Roast Beef
Composition | Calories |
---|---|
97.5 g of water | × 0 Cal/g = 0 |
58.7 g of protein | × 4 Cal/g = 235 |
69.3 g of fat | × 9 Cal/g = 624 |
0 g of carbohydrates | × 4 Cal/g = 0 |
1.5 g of minerals | × 0 Cal/g = 0 |
Total mass: 227.0 g | Total calories: about 900 Cal |
Table 5.5 Approximate Compositions and Fuel Values of Some Common Foods
Food (quantity) | Approximate Composition (%) | Food Value (Cal/g) | Calories | |||
---|---|---|---|---|---|---|
Water | Carbohydrate | Protein | Fat | |||
beer (12 oz) | 92 | 3.6 | 0.3 | 0 | 0.4 | 150 |
coffee (6 oz) | 100 | ~0 | ~0 | ~0 | ~0 | ~0 |
milk (1 cup) | 88 | 4.5 | 3.3 | 3.3 | 0.6 | 150 |
egg (1 large) | 75 | 2 | 12 | 12 | 1.6 | 80 |
butter (1 tbsp) | 16 | ~0 | ~0 | 79 | 7.1 | 100 |
apple (8 oz) | 84 | 15 | ~0 | 0.5 | 0.6 | 125 |
bread, white (2 slices) | 37 | 48 | 8 | 4 | 2.6 | 130 |
brownie (40 g) | 10 | 55 | 5 | 30 | 4.8 | 190 |
hamburger (4 oz) | 54 | 0 | 24 | 21 | 2.9 | 326 |
fried chicken (1 drumstick) | 53 | 8.3 | 22 | 15 | 2.7 | 195 |
carrots (1 cup) | 87 | 10 | 1.3 | ~0 | 0.4 | 70 |
Because the Calorie represents such a large amount of energy, a few of them go a long way. An average 73 kg (160 lb) person needs about 67 Cal/h (1600 Cal/day) to fuel the basic biochemical processes that keep that person alive. This energy is required to maintain body temperature, keep the heart beating, power the muscles used for breathing, carry out chemical reactions in cells, and send the nerve impulses that control those automatic functions. Physical activity increases the amount of energy required but not by as much as many of us hope (). A moderately active individual requires about 2500−3000 Cal/day; athletes or others engaged in strenuous activity can burn 4000 Cal/day. Any excess caloric intake is stored by the body for future use, usually in the form of fat, which is the most compact way to store energy. When more energy is needed than the diet supplies, stored fuels are mobilized and oxidized. We usually exhaust the supply of stored carbohydrates before turning to fats, which accounts in part for the popularity of low-carbohydrate diets.
Table 5.6 Approximate Energy Expenditure by a 160 lb Person Engaged in Various Activities
Activity | Cal/h |
---|---|
sleeping | 80 |
driving a car | 120 |
standing | 140 |
eating | 150 |
walking 2.5 mph | 210 |
mowing lawn | 250 |
swimming 0.25 mph | 300 |
roller skating | 350 |
tennis | 420 |
bicycling 13 mph | 660 |
running 10 mph | 900 |
What is the minimum number of Calories expended by a 160 lb person who climbs a 30-story building? (Assume each flight of stairs is 14 ft high.) How many grams of glucose are required to supply this amount of energy? (The energy released during the combustion of glucose was calculated in Example 5.)
Given: mass, height, and energy released by combustion of glucose
Asked for: calories expended and mass of glucose needed
Strategy:
A Convert mass and height to SI units and then substitute these values into to calculate the change in potential energy (in kilojoules). Divide the calculated energy by 4.184 Cal/kJ to convert the potential energy change to Calories.
B Use the value obtained in Example 5 for the combustion of glucose to calculate the mass of glucose needed to supply this amount of energy.
Solution:
The energy needed to climb the stairs equals the difference between the person’s potential energy (PE) at the top of the building and at ground level.
A Recall from that PE = mgh. Because m and h are given in non-SI units, we must convert them to kilograms and meters, respectively:
Thus
PE = (72.6 kg)(9.81 m/s2)(128 m) = 8.55 × 104 (kg·m2)/s2 = 91.2 kJTo convert to Calories, we divide by 4.184 kJ/kcal:
B Because the combustion of glucose produces 15.6 kJ/g (Example 5), the mass of glucose needed to supply 85.5 kJ of energy is
This mass corresponds to only about a teaspoonful of sugar! Because the body is only about 30% efficient in using the energy in glucose, the actual amount of glucose required would be higher: (100%/30%) × 5.85 g = 19.5 g. Nonetheless, this calculation illustrates the difficulty many people have in trying to lose weight by exercise alone.
Exercise
Calculate how many times a 160 lb person would have to climb the tallest building in the United States, the 110-story Willis Tower in Chicago, to burn off 1.0 lb of stored fat. Assume that each story of the building is 14 ft high and use a calorie content of 9.0 kcal/g of fat.
Answer: About 55 times
The calculations in Example 12 ignore various factors, such as how fast the person is climbing. Although the rate is irrelevant in calculating the change in potential energy, it is very relevant to the amount of energy actually required to ascend the stairs. The calculations also ignore the fact that the body’s conversion of chemical energy to mechanical work is significantly less than 100% efficient. According to the average energy expended for various activities listed in , a person must run more than 4.5 h at 10 mph or bicycle for 6 h at 13 mph to burn off 1 lb of fat (1.0 lb × 454 g/lb × 9.0 Cal/g = 4100 Cal). But if a person rides a bicycle at 13 mph for only 1 h per day 6 days a week, that person will burn off 50 lb of fat in the course of a year (assuming, of course, the cyclist doesn’t increase his or her intake of calories to compensate for the exercise).
The nutritional Calorie is equivalent to 1 kcal (4.184 kJ). The caloric content of a food is its ΔHcomb per gram. The combustion of nitrogen-containing substances produces N2(g), but the biological oxidation of such substances produces urea. Hence the actual energy available from nitrogen-containing substances, such as proteins, is less than the ΔHcomb of urea multiplied by the number of moles of urea produced. The typical caloric contents for food are 9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals.
Can water be considered a food? Explain your answer.
Describe how you would determine the caloric content of a bag of popcorn using a calorimeter.
Why do some people initially feel cold after eating a meal and then begin to feel warm?
In humans, one of the biochemical products of the combustion/digestion of amino acids is urea. What effect does this have on the energy available from these reactions? Speculate why conversion to urea is preferable to the generation of N2.
Please be sure you are familiar with the topics discussed in Essential Skills 4 () before proceeding to the Numerical Problems.
Determine the amount of energy available from the biological oxidation of 1.50 g of leucine (an amino acid, ΔHcomb = −3581.7 kJ/mol).
Calculate the energy released (in kilojoules) from the metabolism of 1.5 oz of vodka that is 62% water and 38% ethanol by volume, assuming that the total volume is equal to the sum of the volume of the two components. The density of ethanol is 0.824 g/mL. What is this enthalpy change in nutritional Calories?
While exercising, a person lifts an 80 lb barbell 7 ft off the ground. Assuming that the transformation of chemical energy to mechanical energy is only 35% efficient, how many Calories would the person use to accomplish this task? From , how many grams of glucose would be needed to provide the energy to accomplish this task?
A 30 g sample of potato chips is placed in a bomb calorimeter with a heat capacity of 1.80 kJ/°C, and the bomb calorimeter is immersed in 1.5 L of water. Calculate the energy contained in the food per gram if, after combustion of the chips, the temperature of the calorimeter increases to 58.6°C from an initial temperature of 22.1°C.
Our contemporary society requires the constant expenditure of huge amounts of energy to heat our homes, provide telephone and cable service, transport us from one location to another, provide light when it is dark outside, and run the machinery that manufactures material goods. The United States alone consumes almost 106 kJ per person per day, which is about 100 times the normal required energy content of the human diet. This figure is about 30% of the world’s total energy usage, although only about 5% of the total population of the world lives in the United States. In contrast, the average energy consumption elsewhere in the world is about 105 kJ per person per day, although actual values vary widely depending on a country’s level of industrialization. In this section, we describe various sources of energy and their impact on the environment.
According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another. What is consumed on a huge scale, however, are resources that can be readily converted to a form of energy that is useful for doing work. As you will see in Chapter 18 "Chemical Thermodynamics", energy that is not used to perform work is either stored as potential energy for future use or transferred to the surroundings as heat.
A major reason for the huge consumption of energy by our society is the low efficiency of most machines in transforming stored energy into work. Efficiency can be defined as the ratio of useful work accomplished to energy expended. Automobiles, for example, are only about 20% efficient in converting the energy stored in gasoline to mechanical work; the rest of the energy is released as heat, either emitted in the exhaust or produced by friction in bearings and tires. The production of electricity by coal- or oil-powered steam turbines is significantly more efficient (Figure 5.17 "Electricity from Coal"): about 38% of the energy released from combustion is converted to electricity. In comparison, modern nuclear power plants can be more than 50% efficient.
Figure 5.17 Electricity from Coal
A coal-powered electric power plant uses the combustion of coal to produce steam, which drives a turbine to produce electricity.
In general, it is more efficient to use primary sources of energy directly (such as natural gas or oil) than to transform them to a secondary source such as electricity prior to their use. For example, if a furnace is well maintained, heating a house with natural gas is about 70% efficient. In contrast, burning the natural gas in a remote power plant, converting it to electricity, transmitting it long distances through wires, and heating the house by electric baseboard heaters have an overall efficiency of less than 35%.
The total expenditure of energy in the world each year is about 3 × 1017 kJ. More than 80% of this energy is provided by the combustion of fossil fuels: oil, coal, and natural gas. (The sources of the energy consumed in the United States in 2009 are shown in Figure 5.18 "Energy Consumption in the United States by Source, 2009".) Natural gas and petroleum, whose compositions were described in Chapter 2 "Molecules, Ions, and Chemical Formulas", are the preferred fuels because they or products derived from them are gases or liquids that are readily transported, stored, and burned. Natural gas and petroleum are derived from the remains of marine creatures that died hundreds of millions of years ago and were buried beneath layers of sediment. As the sediment turned to rock, the tremendous heat and pressure inside Earth transformed the organic components of the buried sea creatures to petroleum and natural gas.
Figure 5.18 Energy Consumption in the United States by Source, 2009
More than 80% of the total energy expended is provided by the combustion of fossil fuels, such as oil, coal, and natural gas.
CoalA complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. It is used as a fuel. is a complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. Because plants contain large amounts of cellulose, derived from linked glucose units, the structure of coal is more complex than that of petroleum (Figure 5.19 "The Structures of Cellulose and Coal"). In particular, coal contains a large number of oxygen atoms that link parts of the structure together, in addition to the basic framework of carbon–carbon bonds. It is impossible to draw a single structure for coal; however, because of the prevalence of rings of carbon atoms (due to the original high cellulose content), coal is more similar to an aromatic hydrocarbon than an aliphatic one.
Figure 5.19 The Structures of Cellulose and Coal
(a) Cellulose consists of long chains of cyclic glucose molecules linked by hydrogen bonds. (b) When cellulose is subjected to high pressures and temperatures for long periods of time, water is eliminated, and bonds are formed between the rings, eventually producing coal. This drawing shows some of the common structural features of coal; note the presence of many different kinds of ring structures.
There are four distinct classes of coal (Table 5.7 "Properties of Different Types of Coal"); their hydrogen and oxygen contents depend on the length of time the coal has been buried and the pressures and temperatures to which it has been subjected. Lignite, with a hydrogen:carbon ratio of about 1.0 and a high oxygen content, has the lowest ΔHcomb. Anthracite, in contrast, with a hydrogen:carbon ratio of about 0.5 and the lowest oxygen content, has the highest ΔHcomb and is the highest grade of coal. The most abundant form in the United States is bituminous coal, which has a high sulfur content because of the presence of small particles of pyrite (FeS2). As discussed in Chapter 4 "Reactions in Aqueous Solution", the combustion of coal releases the sulfur in FeS2 as SO2, which is a major contributor to acid rain. Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds" compares the ΔHcomb per gram of oil, natural gas, and coal with those of selected organic compounds.
Table 5.7 Properties of Different Types of Coal
Type | % Carbon | Hydrogen:Carbon Mole Ratio | % Oxygen | % Sulfur | Heat Content | US Deposits |
---|---|---|---|---|---|---|
anthracite | 92 | 0.5 | 3 | 1 | high | Pennsylvania, New York |
bituminous | 80 | 0.6 | 8 | 5 | medium | Appalachia, Midwest, Utah |
subbituminous | 77 | 0.9 | 16 | 1 | medium | Rocky Mountains |
lignite | 71 | 1.0 | 23 | 1 | low | Montana |
Table 5.8 Enthalpies of Combustion of Common Fuels and Selected Organic Compounds
Fuel | ΔHcomb (kJ/g) |
---|---|
dry wood | −15 |
peat | −20.8 |
bituminous coal | −28.3 |
charcoal | −35 |
kerosene | −37 |
C6H6 (benzene) | −41.8 |
crude oil | −43 |
natural gas | −50 |
C2H2 (acetylene) | −50.0 |
CH4 (methane) | −55.5 |
gasoline | −84 |
hydrogen | −143 |
Peat, a precursor to coal, is the partially decayed remains of plants that grow in swampy areas. It is removed from the ground in the form of soggy bricks of mud that will not burn until they have been dried. Even though peat is a smoky, poor-burning fuel that gives off relatively little heat, humans have burned it since ancient times (Figure 5.20 "A Peat Bog"). If a peat bog were buried under many layers of sediment for a few million years, the peat could eventually be compressed and heated enough to become lignite, the lowest grade of coal; given enough time and heat, lignite would eventually become anthracite, a much better fuel.
Figure 5.20 A Peat Bog
Peat is a smoky fuel that burns poorly and produces little heat, but it has been used as a fuel since ancient times.
Oil and natural gas resources are limited. Current estimates suggest that the known reserves of petroleum will be exhausted in about 60 years, and supplies of natural gas are estimated to run out in about 120 years. Coal, on the other hand, is relatively abundant, making up more than 90% of the world’s fossil fuel reserves. As a solid, coal is much more difficult to mine and ship than petroleum (a liquid) or natural gas. Consequently, more than 75% of the coal produced each year is simply burned in power plants to produce electricity. A great deal of current research focuses on developing methods to convert coal to gaseous fuels (coal gasification) or liquid fuels (coal liquefaction). In the most common approach to coal gasification, coal reacts with steam to produce a mixture of CO and H2 known as synthesis gas, or syngas:Because coal is 70%–90% carbon by mass, it is approximated as C in Equation 5.46.
Equation 5.46
Converting coal to syngas removes any sulfur present and produces a clean-burning mixture of gases.
Syngas is also used as a reactant to produce methane and methanol. A promising approach is to convert coal directly to methane through a series of reactions:
Equation 5.47
Burning a small amount of coal or methane provides the energy consumed by these reactions. Unfortunately, methane produced by this process is currently significantly more expensive than natural gas. As supplies of natural gas become depleted, however, this coal-based process may well become competitive in cost.
Measuring crude oil. The standard industrial unit of measure for crude oil is the 42 gal barrel.
Similarly, the techniques available for converting coal to liquid fuels are not yet economically competitive with the production of liquid fuels from petroleum. Current approaches to coal liquefaction use a catalyst to break the complex network structure of coal into more manageable fragments. The products are then treated with hydrogen (from syngas or other sources) under high pressure to produce a liquid more like petroleum. Subsequent distillation, cracking, and reforming can be used to create products similar to those obtained from petroleum. (For more information about cracking, see Chapter 2 "Molecules, Ions, and Chemical Formulas", Section 2.6 "Industrially Important Chemicals".) The total yield of liquid fuels is about 5.5 bbl of crude liquid per ton of coal (1 bbl is 42 gal or 160 L). Although the economics of coal liquefaction are currently even less attractive than for coal gasification, liquid fuels based on coal are likely to become economically competitive as supplies of petroleum are consumed.
If bituminous coal is converted to methane by the process in Equation 5.46, what is the ratio of the ΔHcomb of the methane produced to the enthalpy of the coal consumed to produce the methane? (Note that 1 mol of CH4 is produced for every 2 mol of carbon in coal.)
Given: chemical reaction and ΔHcomb (Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds")
Asked for: ratio of ΔHcomb of methane produced to coal consumed
Strategy:
A Write a balanced chemical equation for the conversion of coal to methane. Referring to Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds", calculate the ΔHcomb of methane and carbon.
B Calculate the ratio of the energy released by combustion of the methane to the energy released by combustion of the carbon.
Solution:
A The balanced chemical equation for the conversion of coal to methane is as follows:
2C(s) + 2H2O(g) → CH4(g) + CO2(g)Thus 1 mol of methane is produced for every 2 mol of carbon consumed. The ΔHcomb of 1 mol of methane is
The ΔHcomb of 2 mol of carbon (as coal) is
B The ratio of the energy released from the combustion of methane to the energy released from the combustion of carbon is
The energy released from the combustion of the product (methane) is 131% of that of the reactant (coal). The fuel value of coal is actually increased by the process!
How is this possible when the law of conservation of energy states that energy cannot be created? The reaction consumes 2 mol of water () but produces only 1 mol of CO2 (). Part of the difference in potential energy between the two (approximately 180 kJ/mol) is stored in CH4 and can be released during combustion.
Exercise
Using the data in Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds", calculate the mass of hydrogen necessary to provide as much energy during combustion as 1 bbl of crude oil (density approximately 0.75 g/mL).
Answer: 36 kg
Even if carbon-based fuels could be burned with 100% efficiency, producing only CO2(g) and H2O(g), doing so could still potentially damage the environment when carried out on the vast scale required by an industrial society. The amount of CO2 released is so large and is increasing so rapidly that it is apparently overwhelming the natural ability of the planet to remove CO2 from the atmosphere. In turn, the elevated levels of CO2 are thought to be affecting the temperature of the planet through a mechanism known as the greenhouse effectThe phenomenon in which substances absorb thermal energy radiated by Earth, thus trapping thermal energy in the atmosphere.. As you will see, there is little doubt that atmospheric CO2 levels are increasing, and the major reason for this increase is the combustion of fossil fuels. There is substantially less agreement, however, on whether the increased CO2 levels are responsible for a significant increase in temperature.
Figure 5.21 "The Global Carbon Cycle" illustrates the global carbon cycleThe distribution and flow of carbon throughout the planet., the distribution and flow of carbon on Earth. Normally, the fate of atmospheric CO2 is to either (1) dissolve in the oceans and eventually precipitate as carbonate rocks or (2) be taken up by plants. The rate of uptake of CO2 by the ocean is limited by its surface area and the rate at which gases dissolve, which are approximately constant. The rate of uptake of CO2 by plants, representing about 60 billion metric tons of carbon per year, partly depends on how much of Earth’s surface is covered by vegetation. Unfortunately, the rapid deforestation for agriculture is reducing the overall amount of vegetation, and about 60 billion metric tons of carbon are released annually as CO2 from animal respiration and plant decay. The amount of carbon released as CO2 every year by fossil fuel combustion is estimated to be about 5.5 billion metric tons. The net result is a system that is slightly out of balance, experiencing a slow but steady increase in atmospheric CO2 levels (Figure 5.22 "Changes in Atmospheric CO"). As a result, average CO2 levels have increased by about 30% since 1850.
Figure 5.21 The Global Carbon Cycle
Most of Earth’s carbon is found in the crust, where it is stored as calcium and magnesium carbonate in sedimentary rocks. The oceans also contain a large reservoir of carbon, primarily as the bicarbonate ion (HCO3−). Green plants consume about 60 billion metric tons of carbon per year as CO2 during photosynthesis, and about the same amount of carbon is released as CO2 annually from animal and plant respiration and decay. The combustion of fossil fuels releases about 5.5 billion metric tons of carbon per year as CO2.
Figure 5.22 Changes in Atmospheric CO2 Levels
(a) Average worldwide CO2 levels have increased by about 30% since 1850. (b) Atmospheric CO2 concentrations measured at Mauna Loa in Hawaii show seasonal variations caused by the removal of CO2 from the atmosphere by green plants during the growing season along with a general increase in CO2 levels.
The increasing levels of atmospheric CO2 are of concern because CO2 absorbs thermal energy radiated by the Earth, as do other gases such as water vapor, methane, and chlorofluorocarbons. Collectively, these substances are called greenhouse gasesA substance that absorbs thermal energy radiated by Earth, thus trapping thermal energy in the atmosphere.; they mimic the effect of a greenhouse by trapping thermal energy in the Earth’s atmosphere, a phenomenon known as the greenhouse effect (Figure 5.23 "The Greenhouse Effect").
Figure 5.23 The Greenhouse Effect
Thermal energy can be trapped in Earth’s atmosphere by gases such as CO2, water vapor, methane, and chlorofluorocarbons before it can be radiated into space—like the effect of a greenhouse. It is not yet clear how large an increase in the temperature of Earth’s surface can be attributed to this phenomenon.
Venus is an example of a planet that has a runaway greenhouse effect. The atmosphere of Venus is about 95 times denser than that of Earth and contains about 95% CO2. Because Venus is closer to the sun, it also receives more solar radiation than Earth does. The result of increased solar radiation and high CO2 levels is an average surface temperature of about 450°C, which is hot enough to melt lead.
Data such as those in Figure 5.22 "Changes in Atmospheric CO" indicate that atmospheric levels of greenhouse gases have increased dramatically over the past 100 years, and it seems clear that the heavy use of fossil fuels by industry is largely responsible. It is not clear, however, how large an increase in temperature (global warming) may result from a continued increase in the levels of these gases. Estimates of the effects of doubling the preindustrial levels of CO2 range from a 0°C to a 4.5°C increase in the average temperature of Earth’s surface, which is currently about 14.4°C. Even small increases, however, could cause major perturbations in our planet’s delicately balanced systems. For example, an increase of 5°C in Earth’s average surface temperature could cause extensive melting of glaciers and the Antarctic ice cap. It has been suggested that the resulting rise in sea levels could flood highly populated coastal areas, such as New York City, Calcutta, Tokyo, Rio de Janeiro, and Sydney. An analysis conducted in 2009 by leading climate researchers from the US National Oceanic and Atmospheric Administration, Switzerland, and France shows that CO2 in the atmosphere will remain near peak levels far longer than other greenhouse gases, which dissipate more quickly. The study predicts a rise in sea levels of approximately 3 ft by the year 3000, excluding the rise from melting glaciers and polar ice caps. According to the analysis, southwestern North America, the Mediterranean, and southern Africa are projected to face droughts comparable to that of the Dust Bowl of the 1930s as a result of global climate changes.
The increase in CO2 levels is only one of many trends that can affect Earth’s temperature. In fact, geologic evidence shows that the average temperature of Earth has fluctuated significantly over the past 400,000 years, with a series of glacial periods (during which the temperature was 10°C–15°C lower than it is now and large glaciers covered much of the globe) interspersed with relatively short, warm interglacial periods (Figure 5.24 "Average Surface Temperature of Earth over the Past 400,000 Years"). Although average temperatures appear to have increased by 0.5°C in the last century, the statistical significance of this increase is open to question, as is the existence of a cause-and-effect relationship between the temperature change and CO2 levels. Despite the lack of incontrovertible scientific evidence, however, many people believe that we should take steps now to limit CO2 emissions and explore alternative sources of energy, such as solar energy, geothermal energy from volcanic steam, and nuclear energy, to avoid even the possibility of creating major perturbations in Earth’s environment. In 2010, international delegates met in Cancún, Mexico, and agreed on a broad array of measures that would advance climate protection. These included the development of low-carbon technologies, providing a framework to reduce deforestation, and aiding countries in assessing their own vulnerabilities. They avoided, however, contentious issues of assigning emissions reductions commitments.
Figure 5.24 Average Surface Temperature of Earth over the Past 400,000 Years
The dips correspond to glacial periods, and the peaks correspond to relatively short, warm interglacial periods. Because of these fluctuations, the statistical significance of the 0.5°C increase in average temperatures observed in the last century is open to question.
A student at UCLA decided to fly home to New York for Christmas. The round trip was 4500 air miles, and part of the cost of her ticket went to buy the 100 gal of jet fuel necessary to transport her and her baggage. Assuming that jet fuel is primarily n-dodecane (C12H26) with a density of 0.75 g/mL, how much energy was expended and how many tons of CO2 were emitted into the upper atmosphere to get her home and back?
Given: volume and density of reactant in combustion reaction
Asked for: energy expended and mass of CO2 emitted
Strategy:
A After writing a balanced chemical equation for the reaction, calculate using Equation 5.27 and the values given in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C".
B Determine the number of moles of dodecane in 100 gal by using the density and molar mass of dodecane and the appropriate conversion factors.
C Obtain the amount of energy expended by multiplying by the number of moles of dodecane. Calculate the amount of CO2 emitted in tons by using mole ratios from the balanced chemical equation and the appropriate conversion factors.
Solution:
A We first need to write a balanced chemical equation for the reaction:
2C12H26(l) + 37O2(g) → 24CO2(g) + 26H2O(l)We can calculate for this reaction from Equation 5.27 and the values in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" corresponding to each substance in the specified phase (phases are not shown for simplicity):
According to the balanced chemical equation for the reaction, this value is for the combustion of 2 mol of n-dodecane. So we must divide by 2 to obtain per mole of n-dodecane:
B The number of moles of dodecane in 100 gal can be calculated as follows, using density, molar mass, and appropriate conversion factors:
C The total energy released is
From the balanced chemical equation for the reaction, we see that each mole of dodecane forms 12 mol of CO2 upon combustion. Hence the amount of CO2 emitted is
Exercise
Suppose the student in Example 14 couldn’t afford the plane fare, so she decided to drive home instead. Assume that the round-trip distance by road was 5572 miles, her fuel consumption averaged 31 mpg, and her fuel was pure isooctane (C8H18, density = 0.6919 g/mL). How much energy was expended and how many tons of CO2 were produced during her trip?
Answer: 2.2 × 107 kJ; 1.6 tons of CO2 (about twice as much as is released by flying)
More than 80% of the energy used by modern society (about 3 × 1017 kJ/yr) is from the combustion of fossil fuels. Because of their availability, ease of transport, and facile conversion to convenient fuels, natural gas and petroleum are currently the preferred fuels. Supplies of coal, a complex solid material derived from plants that lived long ago, are much greater, but the difficulty in transporting and burning a solid makes it less attractive as a fuel. Coal releases the smallest amount of energy per gram of any fossil fuel, and natural gas the greatest amount. The combustion of fossil fuels releases large amounts of CO2 that upset the balance of the carbon cycle and result in a steady increase in atmospheric CO2 levels. Because CO2 is a greenhouse gas, which absorbs heat before it can be radiated from Earth into space, CO2 in the atmosphere can result in increased surface temperatures (the greenhouse effect). The temperature increases caused by increased CO2 levels because of human activities are, however, superimposed on much larger variations in Earth’s temperature that have produced phenomena such as the ice ages and are still poorly understood.
Why is it preferable to convert coal to syngas before use rather than burning coal as a solid fuel?
What is meant by the term greenhouse gases? List three greenhouse gases that have been implicated in global warming.
Name three factors that determine the rate of planetary CO2 uptake.
The structure of coal is quite different from the structure of gasoline. How do their structural differences affect their enthalpies of combustion? Explain your answer.
Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems.
One of the side reactions that occurs during the burning of fossil fuels is
4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g)How many kilograms of CO2 are released during the combustion of 16 gal of gasoline? Assume that gasoline is pure isooctane with a density of 0.6919 g/mL. If this combustion was used to heat 4.5 × 103 L of water from an initial temperature of 11.0°C, what would be the final temperature of the water assuming 42% efficiency in the energy transfer?
A 60 W light bulb is burned for 6 hours. If we assume an efficiency of 38% in the conversion of energy from oil to electricity, how much oil must be consumed to supply the electrical energy needed to light the bulb? (1 W = 1 J/s)
How many liters of cyclohexane must be burned to release as much energy as burning 10.0 lb of pine logs? The density of cyclohexane is 0.7785 g/mL, and its ΔHcomb = −46.6 kJ/g.
The previous Essential Skills sections introduced some fundamental operations that you need to successfully manipulate mathematical equations in chemistry. This section describes how to convert between temperature scales and further develops the topic of unit conversions started in Essential Skills 2 in , .
The concept of temperature may seem familiar to you, but many people confuse temperature with heat. Temperature is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas heat is the flow of thermal energy between objects with different temperatures.
Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes.
The Fahrenheit temperature scale was developed in 1717 by the German physicist Gabriel Fahrenheit, who designated the temperature of a bath of ice melting in a solution of salt as the zero point on his scale. Such a solution was commonly used in the 18th century to carry out low-temperature reactions in the laboratory. The scale was measured in increments of 12; its upper end, designated as 96°, was based on the armpit temperature of a healthy person—in this case, Fahrenheit’s wife. Later, the number of increments shown on a thermometer increased as measurements became more precise. The upper point is based on the boiling point of water, designated as 212° to maintain the original magnitude of a Fahrenheit degree, whereas the melting point of ice is designated as 32°.
The Celsius scale was developed in 1742 by the Swedish astronomer Anders Celsius. It is based on the melting and boiling points of water under normal atmospheric conditions. The current scale is an inverted form of the original scale, which was divided into 100 increments. Because of these 100 divisions, the Celsius scale is also called the centigrade scale.
Lord Kelvin, working in Scotland, developed the Kelvin scale in 1848. His scale uses molecular energy to define the extremes of hot and cold. Absolute zero, or 0 K, corresponds to the point at which molecular energy is at a minimum. The Kelvin scale is preferred in scientific work, although the Celsius scale is also commonly used. Temperatures measured on the Kelvin scale are reported simply as K, not °K. compares the three scales.
Figure 5.25 A Comparison of the Fahrenheit, Celsius, and Kelvin Temperature Scales
Because the difference between the freezing point of water and the boiling point of water is 100° on both the Celsius and Kelvin scales, the size of a degree Celsius (°C) and a kelvin (K) are precisely the same. In contrast, both a degree Celsius and a kelvin are 9/5 the size of a degree Fahrenheit (°F).
The kelvin is the same size as the Celsius degree, so measurements are easily converted from one to the other. The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K. The Kelvin and Celsius scales are related as follows:
T (in °C) + 273.15 = T (in K) T (in K) − 273.15 = T (in °C)Degrees on the Fahrenheit scale, however, are based on an English tradition of using 12 divisions, just as 1 ft = 12 in. The relationship between degrees Fahrenheit and degrees Celsius is as follows:
where the coefficient for degrees Fahrenheit is exact. (Some calculators have a function that allows you to convert directly between °F and °C.) There is only one temperature for which the numerical value is the same on both the Fahrenheit and Celsius scales: −40°C = −40°F.
Solution:
In Essential Skills 2, you learned a convenient way of converting between units of measure, such as from grams to kilograms or seconds to hours. The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis. For example, if we observe experimentally that an object’s potential energy is related to its mass, its height from the ground, and to a gravitational force, then when multiplied, the units of mass, height, and the force of gravity must give us units corresponding to those of energy.
Energy is typically measured in joules, calories, or electron volts (eV), defined by the following expressions:
1 J = 1 (kg·m2)/s2 = 1 coulomb·volt 1 cal = 4.184 J 1 eV = 1.602 × 10−19 JTo illustrate the use of dimensional analysis to solve energy problems, let us calculate the kinetic energy in joules of a 320 g object traveling at 123 cm/s. To obtain an answer in joules, we must convert grams to kilograms and centimeters to meters. Using , the calculation may be set up as follows:
Alternatively, the conversions may be carried out in a stepwise manner:
However, this second method involves an additional step.
Now suppose you wish to report the number of kilocalories of energy contained in a 7.00 oz piece of chocolate in units of kilojoules per gram. To obtain an answer in kilojoules, we must convert 7.00 oz to grams and kilocalories to kilojoules. Food reported to contain a value in Calories actually contains that same value in kilocalories. If the chocolate wrapper lists the caloric content as 120 Calories, the chocolate contains 120 kcal of energy. If we choose to use multiple steps to obtain our answer, we can begin with the conversion of kilocalories to kilojoules:
We next convert the 7.00 oz of chocolate to grams:
The number of kilojoules per gram is therefore
Alternatively, we could solve the problem in one step:
The discrepancy between the two answers is attributable to rounding to the correct number of significant figures for each step when carrying out the calculation in a stepwise manner. Recall that all digits in the calculator should be carried forward when carrying out a calculation using multiple steps. In this problem, we first converted kilocalories to kilojoules and then converted ounces to grams. Skill Builder ES2 allows you to practice making multiple conversions between units in a single step.
Solution:
Our goal is to convert 1.5 kJ/g to Calories in 8 oz:
Our goal is to use the energy content, 48 kJ/g, and the density, 0.70 g/mL, to obtain the number of joules in 16.3 gal of gasoline:
Please be sure you are familiar with the topics discussed in Essential Skills 4 () before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts.
Palm trees grow on the coast of southern England even though the latitude is the same as that of Winnipeg, Canada. What is a plausible explanation for this phenomenon? (Hint: the Gulf Stream current is a factor.)
During intense exercise, your body cannot provide enough oxygen to allow the complete combustion of glucose to carbon dioxide. Under these conditions, an alternative means of obtaining energy from glucose is used in which glucose (C6H12O6) is converted to lactic acid (C3H5O3H). The equation for this reaction is as follows:
C6H12O6 → 2C3H5O3H♦ During the late spring, icebergs in the North Atlantic pose a hazard to shipping. To avoid them, ships travel routes that are about 30% longer. Many attempts have been made to destroy icebergs, including using explosives, torpedoes, and bombs. How much heat must be generated to melt 15% of a 1.9 × 108 kg iceberg? How many kilograms of TNT (trinitrotoluene, C7H5N3O6) would be needed to provide enough energy to melt the ice? (The ΔH for explosive decomposition of TNT is −1035.8 kJ/mol.)
Many biochemical processes occur through sequences of reactions called pathways. The total energy released by many of these pathways is much more than the energy a cell could handle if it were all released in a single step. For example, the combustion of glucose in a single step would release enough energy to kill a cell. By using a series of smaller steps that release less energy per reaction, however, the cell can extract the maximum energy from glucose without being destroyed. Referring to , calculate how many grams of glucose would need to be metabolized to raise the temperature of a liver cell from an average body temperature of 37°C to 100°C, if the cell has a volume of 5000 μm3. Although the cell is only 69% water, assume that the density of the cell is 1.00 and that its Cs is the same as that of water.
♦ During smelting, naturally occurring metal oxides are reduced by carbon at high temperature. For copper(II) oxide, this process includes the following series of chemical equations:
CuO(s) + C(s) → Cu(l) + CO(g) 2C(s) + O2(g) → 2CO(g) CuO(l) + CO(g) → Cu(s) + CO2(g)The final products are CO2 and Cu. The discovery of this process led to the increasing use of ores as sources of metals in ancient cultures. In fact, between 3000 BC and 2000 BC, the smelting of copper was well established, and beads made from copper are some of the earliest known metal artifacts.
♦ The earliest known Egyptian artifacts made from tin metal date back to approximately 1400 BC. If the smelting process for SnO2 occurs via the reaction sequence
SnO2(s) + 2C(s) → Sn(s) + 2CO(g) SnO2(s) + 2CO(g) → Sn(s) + 2CO2(g)what is ΔHrxn for the conversion of SnO2 to Sn (s, white) by this smelting process? How much heat was released or required if 28 g of Sn metal was produced from its ore, assuming complete reaction?
♦ An average American consumes approximately 106 kJ of energy per day. The average life expectancy of an American is 77.9 years.
Several theories propose that life on Earth evolved in the absence of oxygen. One theory is that primitive organisms used fermentation processes, in which sugars are decomposed in an oxygen-free environment, to obtain energy. Many kinds of fermentation processes are possible, including the conversion of glucose to lactic acid (a), to CO2 and ethanol (b), and to ethanol and acetic acid (c):
Reaction (a) occurs in rapidly exercising muscle cells, reaction (b) occurs in yeast, and reaction (c) occurs in intestinal bacteria. Using , calculate which reaction gives the greatest energy yield (most negative) per mole of glucose.
A 70 kg person expends 85 Cal/h watching television. If the person eats 8 cups of popcorn that contains 55 Cal per cup, how many kilojoules of energy from the popcorn will have been burned during a 2 h movie? After the movie, the person goes outside to play tennis and burns approximately 500 Cal/h. How long will that person have to play tennis to work off all the residual energy from the popcorn?
♦ Photosynthesis in higher plants is a complex process in which glucose is synthesized from atmospheric carbon dioxide and water in a sequence of reactions that uses light as an energy source. The overall reaction is as follows:
Glucose may then be used to produce the complex carbohydrates, such as cellulose, that constitute plant tissues.
Adipose (fat) tissue consists of cellular protoplasm, which is mostly water, and fat globules. Nearly all the energy stored in adipose tissue comes from the chemical energy of its fat globules, totaling approximately 3500 Cal per pound of tissue. How many kilojoules of energy are stored in 10 g of adipose tissue? How many 50 g brownies would you need to consume to generate 10 lb of fat? (Refer to for the necessary caloric data.)
Proteins contain approximately 4 Cal/g, carbohydrates approximately 4 Cal/g, and fat approximately 9 Cal/g. How many kilojoules of energy are available from the consumption of one serving (8 oz) of each food in the table? (Data are shown per serving.)
Food | Protein (g) | Fat (g) | Carbohydrates (g) |
---|---|---|---|
sour cream | 7 | 48 | 10 |
banana | 2 | 1 | 35 |
cheeseburger | 60 | 31 | 40 |
green peas | 8 | 1 | 21 |
When you eat a bowl of cereal with 500 g of milk, how many Calories must your body burn to warm the milk from 4°C to a normal body temperature of 37°C? (Assume milk has the same specific heat as water.) How many Calories are burned warming the same amount of milk in a 32°C bowl of oatmeal from 32°C to normal body temperature? In some countries that experience starvation conditions, it has been found that infants don’t starve even though the milk from their mothers doesn’t contain the number of Calories thought necessary to sustain them. Propose an explanation for this.
If a person’s fever is caused by an increase in the temperature of water inside the body, how much additional energy is needed if a 70 kg person with a normal body temperature of 37°C runs a temperature of 39.5°C? (A person’s body is approximately 79% water.) The old adage “feed a fever” may contain some truth in this case. How many 4 oz hamburgers would the person need to consume to cause this change? (Refer to for the necessary caloric data.)
Approximately 810 kJ of energy is needed to evaporate water from the leaves of a 9.2 m tree in one day. What mass of water is evaporated from the tree?
through introduced you to a wide variety of chemical substances and some of the most fundamental concepts in chemistry, including general descriptions of chemical bonding, mass relationships in chemical equations, and the energy changes associated with chemical reactions. You learned that the atoms of each element contain a unique number of positively charged protons in the nucleus and that a neutral atom has the same number of electrons as protons. You also learned that protons and neutrons constitute most of the mass of the atom and that electrons occupy most of the volume of an atom. These facts do not, however, explain the stoichiometries and the structures of chemical compounds; a deeper understanding of the electronic structure of atoms is required.
The noble gases. Helium, neon, argon, krypton, and xenon, all monatomic elements, are five of the six known noble gases. When atoms of these gases are excited by the flow of electrons in the discharge tubes, they emit photons of visible light in their characteristic spectral colors. All noble gases have the maximum number of electrons possible in their outer shell (2 for helium, 8 for all others), so they do not form chemical compounds easily.
In this chapter, we describe how electrons are arranged in atoms and how the spatial arrangements of electrons are related to their energies. We also explain how knowing the arrangement of electrons in an atom enables chemists to predict and explain the chemistry of an element. As you study the material presented in this chapter, you will discover how the shape of the periodic table reflects the electronic arrangements of elements. In this and subsequent chapters, we build on this information to explain why certain chemical changes occur and others do not.
After reading this chapter, you will know enough about the theory of the electronic structure of atoms to explain what causes the characteristic colors of neon signs, how laser beams are created, and why gemstones and fireworks have such brilliant colors. In later chapters, we will develop the concepts introduced here to explain why the only compound formed by sodium and chlorine is NaCl, an ionic compound, whereas neon and argon do not form any stable compounds, and why carbon and hydrogen combine to form an almost endless array of covalent compounds, such as CH4, C2H2, C2H4, and C2H6. You will discover that knowing how to use the periodic table is the single most important skill you can acquire to understand the incredible chemical diversity of the elements.
Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation.
Figure 6.1 A Wave in Water
When a drop of water falls onto a smooth water surface, it generates a set of waves that travel outward in a circular direction.
A waveA periodic oscillation that transmits energy through space. is a periodic oscillation that transmits energy through space. Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down.
Figure 6.2 Important Properties of Waves
(a) Wavelength (λ), frequency (ν, labeled in Hz), and amplitude are indicated on this drawing of a wave. (b) The wave with the shortest wavelength has the greatest number of wavelengths per unit time (i.e., the highest frequency). If two waves have the same frequency and speed, the one with the greater amplitude has the higher energy.
Waves have characteristic properties (). As you may have noticed in , waves are periodicPhenomena, such as waves, that repeat regularly in both space and time.; that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength (λ)The distance between two corresponding points in a wave—between the midpoints of two peaks or two troughs..λ is the lowercase Greek lambda, and ν is the lowercase Greek nu. Wavelengths are described by a unit of distance, typically meters. The frequency (ν)The number of oscillations (i.e., of a wave) that pass a particular point in a given period of time. of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations per second (1/s = s−1), which in the SI system is called the hertz (Hz).Named after German physicist Heinrich Hertz (1857–1894), a pioneer in the field of electromagnetic radiation. The amplitudeThe vertical height of a wave, which is defined as half the peak-to-trough height., or vertical height, of a wave is defined as half the peak-to-trough height; as the amplitude of a wave with a given frequency increases, so does its energy. As you can see in , two waves can have the same amplitude but different wavelengths and vice versa. The distance traveled by a wave per unit time is its speed (v)The distance traveled by a wave per unit time., which is typically measured in meters per second (m/s). The speed of a wave is equal to the product of its wavelength and frequency:
Equation 6.1
Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs.
Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiationEnergy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields. (). Some forms of electromagnetic radiation are shown in . In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (c)The speed with which all forms of electromagnetic radiation travel in a vacuum., a fundamental physical constant with a value of 2.99792458 × 108 m/s (which is about 3.00 ×108 m/s or 1.86 × 105 mi/s). This is about a million times faster than the speed of sound.
Figure 6.3 The Nature of Electromagnetic Radiation
All forms of electromagnetic radiation consist of perpendicular oscillating electric and magnetic fields.
Because the various kinds of electromagnetic radiation all have the same speed (c), they differ in only wavelength and frequency. As shown in and , the wavelengths of familiar electromagnetic radiation range from 101 m for radio waves to 10−12 m for gamma rays, which are emitted by nuclear reactions. By replacing v with c in , we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength:
Equation 6.2
For example, the frequency of radio waves is about 108 Hz, whereas the frequency of gamma rays is about 1020 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10−7 m (700 nm, or 4.3 × 1014 Hz) and 4 × 10−7 m (400 nm, or 7.5 × 1014 Hz). Within this range, the eye perceives radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism, as shown in part (b) in . A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms.
Figure 6.4 The Electromagnetic Spectrum
(a) This diagram shows the wavelength and frequency ranges of electromagnetic radiation. The visible portion of the electromagnetic spectrum is the narrow region with wavelengths between about 400 and 700 nm. (b) When white light is passed through a prism, it is split into light of different wavelengths, whose colors correspond to the visible spectrum.
Table 6.1 Common Wavelength Units for Electromagnetic Radiation
Unit | Symbol | Wavelength (m) | Type of Radiation |
---|---|---|---|
picometer | pm | 10−12 | gamma ray |
angstrom | Å | 10−10 | x-ray |
nanometer | nm | 10−9 | x-ray |
micrometer | μm | 10−6 | infrared |
millimeter | mm | 10−3 | infrared |
centimeter | cm | 10−2 | microwave |
meter | m | 100 | radio |
As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength:
Equation 6.3
Equation 6.4
Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer described in absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation.
The energy of electromagnetic radiation increases with increasing frequency and decreasing wavelength.
Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation?
Given: frequency
Asked for: wavelength
Strategy:
Substitute the value for the speed of light in meters per second into to calculate the wavelength in meters.
Solution:
From , we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 108 m/s:
λν = c = 2.998 × 108 m/sThus the wavelength λ is given by
Exercise
As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun?
Answer: 8.45 mm
In and , we describe how scientists developed our current understanding of the structure of atoms using the scientific method described in . You will discover why scientists had to rethink their classical understanding of the nature of electromagnetic energy, which clearly distinguished between the particulate behavior of matter and the wavelike nature of energy.
relationship between wavelength, frequency, and speed of a wave
relationship between wavelength, frequency, and speed of electromagnetic radiation
A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties: wavelength (λ), the distance between successive waves; frequency (ν), the number of waves that pass a fixed point per unit time; speed (v), the rate at which the wave propagates through space; and amplitude, the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. Electromagnetic radiation consists of two perpendicular waves, one electric and one magnetic, propagating at the speed of light (c). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ only in their frequencies and wavelengths.
What are the characteristics of a wave? What is the relationship between electromagnetic radiation and wave energy?
At constant wavelength, what effect does increasing the frequency of a wave have on its speed? its amplitude?
List the following forms of electromagnetic radiation in order of increasing wavelength: x-rays, radio waves, infrared waves, microwaves, ultraviolet waves, visible waves, and gamma rays. List them in order of increasing frequency. Which has the highest energy?
A large industry is centered on developing skin-care products, such as suntan lotions and cosmetics, that cannot be penetrated by ultraviolet radiation. How does the wavelength of visible light compare with the wavelength of ultraviolet light? How does the energy of visible light compare with the energy of ultraviolet light? Why is this industry focused on blocking ultraviolet light rather than visible light?
The human eye is sensitive to what fraction of the electromagnetic spectrum, assuming a typical spectral range of 104 to 1020 Hz? If we came from the planet Krypton and had x-ray vision (i.e., if our eyes were sensitive to x-rays in addition to visible light), how would this fraction be changed?
What is the frequency in megahertz corresponding to each wavelength?
What is the frequency in megahertz corresponding to each wavelength?
Line spectra are also observed for molecular species. Given the following characteristic wavelengths for each species, identify the spectral region (ultraviolet, visible, etc.) in which the following line spectra will occur. Given 1.00 mol of each compound and the wavelength of absorbed or emitted light, how much energy does this correspond to?
What is the speed of a wave in meters per second that has a wavelength of 1250 m and a frequency of 2.36 × 105 s−1?
A wave travels at 3.70 m/s with a frequency of 4.599 × 107 Hz and an amplitude of 1.0 m. What is its wavelength in nanometers?
An AM radio station broadcasts with a wavelength of 248.0 m. What is the broadcast frequency of the station in kilohertz? An AM station has a broadcast range of 92.6 MHz. What is the corresponding wavelength range in meters for this reception?
An FM radio station broadcasts with a wavelength of 3.21 m. What is the broadcast frequency of the station in megahertz? An FM radio typically has a broadcast range of 82–112 MHz. What is the corresponding wavelength range in meters for this reception?
A microwave oven operates at a frequency of approximately 2450 MHz. What is the corresponding wavelength? Water, with its polar molecules, absorbs electromagnetic radiation primarily in the infrared portion of the spectrum. Given this fact, why are microwave ovens used for cooking food?
By the late 19th century, many physicists thought their discipline was well on the way to explaining most natural phenomena. They could calculate the motions of material objects using Newton’s laws of classical mechanics, and they could describe the properties of radiant energy using mathematical relationships known as Maxwell’s equations, developed in 1873 by James Clerk Maxwell, a Scottish physicist. The universe appeared to be a simple and orderly place, containing matter, which consisted of particles that had mass and whose location and motion could be accurately described, and electromagnetic radiation, which was viewed as having no mass and whose exact position in space could not be fixed. Thus matter and energy were considered distinct and unrelated phenomena. Soon, however, scientists began to look more closely at a few inconvenient phenomena that could not be explained by the theories available at the time.
One phenomenon that seemed to contradict the theories of classical physics was blackbody radiationElectromagnetic radiation whose wavelength and color depends on the temperature of the object., the energy emitted by an object when it is heated. The wavelength of energy emitted by an object depends on only its temperature, not its surface or composition. Hence an electric stove burner or the filament of a space heater glows dull red or orange when heated, whereas the much hotter tungsten wire in an incandescent light bulb gives off a yellowish light ().
Figure 6.5 Blackbody Radiation
When heated, all objects emit electromagnetic radiation whose wavelength (and color) depends on the temperature of the object. A relatively low-temperature object, such as an electric stove element, on a low setting appears red, whereas a higher-temperature object, such as the filament of an incandescent light bulb, appears yellow or white.
The intensity of radiation is a measure of the energy emitted per unit area. A plot of the intensity of blackbody radiation as a function of wavelength for an object at various temperatures is shown in . One of the major assumptions of classical physics was that energy increased or decreased in a smooth, continuous manner. For example, classical physics predicted that as wavelength decreased, the intensity of the radiation an object emits should increase in a smooth curve without limit at all temperatures, as shown by the broken line for 6000 K in . Thus classical physics could not explain the sharp decrease in the intensity of radiation emitted at shorter wavelengths (primarily in the ultraviolet region of the spectrum), which we now refer to as the “ultraviolet catastrophe.” In 1900, however, the German physicist Max Planck (1858–1947) explained the ultraviolet catastrophe by proposing that the energy of electromagnetic waves is quantized rather than continuous. This means that for each temperature, there is a maximum intensity of radiation that is emitted in a blackbody object, corresponding to the peaks in , so the intensity does not follow a smooth curve as the temperature increases, as predicted by classical physics. Thus energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantumThe smallest possible unit of energy. Energy can be gained or lost only in integral multiples of a quantum..
Figure 6.6 Relationship between the Temperature of an Object and the Spectrum of Blackbody Radiation It Emits
At relatively low temperatures, most radiation is emitted at wavelengths longer than 700 nm, which is in the infrared portion of the spectrum. The dull red glow of the electric stove element in is due to the small amount of radiation emitted at wavelengths less than 700 nm, which the eye can detect. As the temperature of the object increases, the maximum intensity shifts to shorter wavelengths, successively resulting in orange, yellow, and finally white light. At high temperatures, all wavelengths of visible light are emitted with approximately equal intensities. The white light spectrum shown for an object at 6000 K closely approximates the spectrum of light emitted by the sun (). Note the sharp decrease in the intensity of radiation emitted at wavelengths below 400 nm, which constituted the ultraviolet catastrophe. The classical prediction fails to fit the experimental curves entirely and does not have a maximum intensity.
In addition to being a physicist, Planck was a gifted pianist, who at one time considered music as a career. During the 1930s, Planck felt it was his duty to remain in Germany, despite his open opposition to the policies of the Nazi government. One of his sons was executed in 1944 for his part in an unsuccessful attempt to assassinate Hitler, and bombing during the last weeks of World War II destroyed Planck’s home.
Although quantization may seem to be an unfamiliar concept, we encounter it frequently. For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. Even electrical charge is quantized: an ion may have a charge of −1 or −2 but not −1.33.
Planck postulated that the energy of a particular quantum of radiant energy could be described explicitly by the equation
Equation 6.5
E = hνwhere the proportionality constant h is called Planck’s constant, one of the most accurately known fundamental constants in science. For our purposes, its value to four significant figures is generally sufficient:
h = 6.626 × 10−34 J·s (joule-seconds)As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. By assuming that energy can be emitted by an object only in integral multiples of hν, Planck devised an equation that fit the experimental data shown in . We can understand Planck’s explanation of the ultraviolet catastrophe qualitatively as follows: At low temperatures, radiation with only relatively low frequencies is emitted, corresponding to low-energy quanta. As the temperature of an object increases, there is an increased probability of emitting radiation with higher frequencies, corresponding to higher-energy quanta. At any temperature, however, it is simply more probable for an object to lose energy by emitting n lower-energy quanta than a single very high-energy quantum that corresponds to ultraviolet radiation. The result is a maximum in the plot of intensity of emitted radiation versus wavelength, as shown in , and a shift in the position of the maximum to lower wavelength (higher frequency) with increasing temperature.
At the time he proposed his radical hypothesis, Planck could not explain why energies should be quantized. Initially, his hypothesis explained only one set of experimental data—blackbody radiation. If quantization were observed for a large number of different phenomena, then quantization would become a law (as defined in ). In time, a theory might be developed to explain that law. As things turned out, Planck’s hypothesis was the seed from which modern physics grew.
Only five years after he proposed it, Planck’s quantization hypothesis was used to explain a second phenomenon that conflicted with the accepted laws of classical physics. When certain metals are exposed to light, electrons are ejected from their surface (). Classical physics predicted that the number of electrons emitted and their kinetic energy should depend on only the intensity of the light, not its frequency. In fact, however, each metal was found to have a characteristic threshold frequency of light; below that frequency, no electrons are emitted regardless of the light’s intensity. Above the threshold frequency, the number of electrons emitted was found to be proportional to the intensity of the light, and their kinetic energy was proportional to the frequency. This phenomenon was called the photoelectric effectA phenomenon in which electrons are ejected from the surface of a metal that has been exposed to light..
Figure 6.7 The Photoelectric Effect
(a) Irradiating a metal surface with photons of sufficiently high energy causes electrons to be ejected from the metal. (b) A photocell that uses the photoelectric effect, similar to those found in automatic door openers. When light strikes the metal cathode, electrons are emitted and attracted to the anode, resulting in a flow of electrical current. If the incoming light is interrupted by, for example, a passing person, the current drops to zero. (c) In contrast to predictions using classical physics, no electrons are emitted when photons of light with energy less than Eo, such as red light, strike the cathode. The energy of violet light is above the threshold frequency, so the number of emitted photons is proportional to the light’s intensity.
Albert Einstein (1879–1955; Nobel Prize in Physics, 1921) quickly realized that Planck’s hypothesis about the quantization of radiant energy could also explain the photoelectric effect. The key feature of Einstein’s hypothesis was the assumption that radiant energy arrives at the metal surface in particles that we now call photonsA quantum of radiant energy, each of which possesses a particular energy given by , each possessing a particular energy E given by . Einstein postulated that each metal has a particular electrostatic attraction for its electrons that must be overcome before an electron can be emitted from its surface (Eo = hνo). If photons of light with energy less than Eo strike a metal surface, no single photon has enough energy to eject an electron, so no electrons are emitted regardless of the intensity of the light. If a photon with energy greater than Eo strikes the metal, then part of its energy is used to overcome the forces that hold the electron to the metal surface, and the excess energy appears as the kinetic energy of the ejected electron:
Equation 6.6
kinetic energy of ejected electron = E − Eo = hν − hνo = h(ν − νo)When a metal is struck by light with energy above the threshold energy Eo, the number of emitted electrons is proportional to the intensity of the light beam, which corresponds to the number of photons per square centimeter, but the kinetic energy of the emitted electrons is proportional to the frequency of the light. Thus Einstein showed that the energy of the emitted electrons depended on the frequency of the light, contrary to the prediction of classical physics.
In 1900, Einstein was working in the Swiss patent office in Bern. He was born in Germany and throughout his childhood his parents and teachers had worried that he might be developmentally disabled. The patent office job was a low-level civil service position that was not very demanding, but it did allow Einstein to spend a great deal of time reading and thinking about physics. In 1905, he published his paper on the photoelectric effect, for which he received the Nobel Prize in 1921.
Planck’s and Einstein’s postulate that energy is quantized is in many ways similar to Dalton’s description of atoms. Both theories are based on the existence of simple building blocks, atoms in one case and quanta of energy in the other. The work of Planck and Einstein thus suggested a connection between the quantized nature of energy and the properties of individual atoms. In fact, Einstein’s Nobel Prize was awarded for his work on the photoelectric effect (not for his more famous equation E = mc2), demonstrating its fundamental importance.
Recently, scientists have theorized that even our sense of smell has its basis in quantum physics. Preliminary research indicates that odorant molecules absorb a quantum of energy that causes their bonds to vibrate at a specific frequency. Because different assemblages of molecules have different characteristic frequencies, these vibrations seem to act as a molecular signature that can be detected as an odor. Studies with flies show that they can distinguish between similar molecules with different vibrational frequencies. This vibrational theory of smell could serve as a discriminatory process in nature in other ways and is an active area of research.
Figure 6.8 A Beam of Red Light Emitted by a Ruby Laser
Ruby lasers, which emit red light at a wavelength of 694.3 nm, are used to read bar codes. When used for commercial applications, such lasers are generally designed to emit radiation over a narrow range of wavelengths to reduce their cost.
A ruby laser, a device that produces light in a narrow range of wavelengths (), emits red light at a wavelength of 694.3 nm (). What is the energy in joules of a
Given: wavelength
Asked for: energy of single photon and mole of photons
Strategy:
A Use and to calculate the energy in joules.
B Multiply the energy of a single photon by Avogadro’s number to obtain the energy in a mole of photons.
Solution:
The energy of a single photon is given by E = hν = hc/λ.
B To calculate the energy in a mole of photons, we multiply the energy of a single photon by the number of photons in a mole (Avogadro’s number). If we write the energy of a photon as 2.861 × 10−19 J/photon, we obtain the energy of a mole of photons with wavelength 694.3 nm:
This energy is of the same magnitude as some of the enthalpies of reaction in , and, as you will see in , it is comparable to the strength of many chemical bonds. As a result, light can be used to initiate chemical reactions. In fact, an entire area of chemistry called photochemistry is devoted to studying such processes. In the phenomenon of photosynthesis, green plants use the energy of visible light to convert carbon dioxide and water into sugars such as glucose.
Exercise
An x-ray generator, such as those used in hospitals, emits radiation with a wavelength of 1.544 Å.
Answer:
The properties of blackbody radiation, the radiation emitted by hot objects, could not be explained with classical physics. Max Planck postulated that energy was quantized and could be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used Planck’s concept of the quantization of energy to explain the photoelectric effect, the ejection of electrons from certain metals when exposed to light. Einstein postulated the existence of what today we call photons, particles of light with a particular energy, E = hν. Both energy and matter have fundamental building blocks: quanta and atoms, respectively.
Describe the relationship between the energy of a photon and its frequency.
How was the ultraviolet catastrophe explained?
If electromagnetic radiation with a continuous range of frequencies above the threshold frequency of a metal is allowed to strike a metal surface, is the kinetic energy of the ejected electrons continuous or quantized? Explain your answer.
The vibrational energy of a plucked guitar string is said to be quantized. What do we mean by this? Are the sounds emitted from the 88 keys on a piano also quantized?
Which of the following exhibit quantized behavior: a human voice, the speed of a car, a harp, the colors of light, automobile tire sizes, waves from a speedboat?
The energy of a photon is directly proportional to the frequency of the electromagnetic radiation.
Quantized: harp, tire size, speedboat waves; continuous: human voice, colors of light, car speed.
What is the energy of a photon of light with each wavelength? To which region of the electromagnetic spectrum does each wavelength belong?
How much energy is contained in each of the following? To which region of the electromagnetic spectrum does each wavelength belong?
A mole of photons is found to have an energy of 225 kJ. What is the wavelength of the radiation?
Use the data in to calculate how much more energetic a single gamma-ray photon is than a radio-wave photon. How many photons from a radio source operating at a frequency of 8 × 105 Hz would be required to provide the same amount of energy as a single gamma-ray photon with a frequency of 3 × 1019 Hz?
A radio station has a transmitter that broadcasts at a frequency of 100.7 MHz with a power output of 50 kW. Given that 1 W = 1 J/s, how many photons are emitted by the transmitter each second?
532 nm
The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. The concept of the photon, however, emerged from experimentation with thermal radiation, electromagnetic radiation emitted as the result of a source’s temperature, which produces a continuous spectrum of energies. More direct evidence was needed to verify the quantized nature of electromagnetic radiation. In this section, we describe how experimentation with visible light provided this evidence.
Although objects at high temperature emit a continuous spectrum of electromagnetic radiation (Figure 6.6 "Relationship between the Temperature of an Object and the Spectrum of Blackbody Radiation It Emits"), a different kind of spectrum is observed when pure samples of individual elements are heated. For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H2 emit a red light. Unlike blackbody radiation, the color of the light emitted by the hydrogen atoms does not depend greatly on the temperature of the gas in the tube. When the emitted light is passed through a prism, only a few narrow lines, called a line spectrumA spectrum in which light of only a certain wavelength is emitted or absorbed, rather than a continuous range of wavelengths., are seen (Figure 6.9 "The Emission of Light by Hydrogen Atoms"), rather than a continuous range of colors. The light emitted by hydrogen atoms is red because, of its four characteristic lines, the most intense line in its spectrum is in the red portion of the visible spectrum, at 656 nm. With sodium, however, we observe a yellow color because the most intense lines in its spectrum are in the yellow portion of the spectrum, at about 589 nm.
Figure 6.9 The Emission of Light by Hydrogen Atoms
(a) A sample of excited hydrogen atoms emits a characteristic red light. (b) When the light emitted by a sample of excited hydrogen atoms is split into its component wavelengths by a prism, four characteristic violet, blue, green, and red emission lines can be observed, the most intense of which is at 656 nm.
Such emission spectra were observed for many other elements in the late 19th century, which presented a major challenge because classical physics was unable to explain them. Part of the explanation is provided by Planck’s equation (Equation 6.5): the observation of only a few values of λ (or ν) in the line spectrum meant that only a few values of E were possible. Thus the energy levels of a hydrogen atom had to be quantized; in other words, only states that had certain values of energy were possible, or allowed. If a hydrogen atom could have any value of energy, then a continuous spectrum would have been observed, similar to blackbody radiation.
In 1885, a Swiss mathematics teacher, Johann Balmer (1825–1898), showed that the frequencies of the lines observed in the visible region of the spectrum of hydrogen fit a simple equation that can be expressed as follows:
Equation 6.7
where n = 3, 4, 5, 6. As a result, these lines are known as the Balmer series. The Swedish physicist Johannes Rydberg (1854–1919) subsequently restated and expanded Balmer’s result in the Rydberg equation:
Equation 6.8
where n1 and n2 are positive integers, n2 > n1, and the Rydberg constant, has a value of 1.09737 × 107 m−1.
A mathematics teacher at a secondary school for girls in Switzerland, Balmer was 60 years old when he wrote the paper on the spectral lines of hydrogen that made him famous. He published only one other paper on the topic, which appeared when he was 72 years old.
Like Balmer’s equation, Rydberg’s simple equation described the wavelengths of the visible lines in the emission spectrum of hydrogen (with n1 = 2, n2 = 3, 4, 5,…). More important, Rydberg’s equation also described the wavelengths of other series of lines that would be observed in the emission spectrum of hydrogen: one in the ultraviolet (n1 = 1, n2 = 2, 3, 4,…) and one in the infrared (n1 = 3, n2 = 4, 5, 6). Unfortunately, scientists had not yet developed any theoretical justification for an equation of this form.
In 1913, a Danish physicist, Niels Bohr (1885–1962; Nobel Prize in Physics, 1922), proposed a theoretical model for the hydrogen atom that explained its emission spectrum. Bohr’s model required only one assumption: The electron moves around the nucleus in circular orbits that can have only certain allowed radii. As discussed in Chapter 1 "Introduction to Chemistry", Rutherford’s earlier model of the atom had also assumed that electrons moved in circular orbits around the nucleus and that the atom was held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electron. Although we now know that the assumption of circular orbits was incorrect, Bohr’s insight was to propose that the electron could occupy only certain regions of space.
During the Nazi occupation of Denmark in World War II, Bohr escaped to the United States, where he became associated with the Atomic Energy Project. In his final years, he devoted himself to the peaceful application of atomic physics and to resolving political problems arising from the development of atomic weapons.
Using classical physics, Bohr showed that the energy of an electron in a particular orbit is given by
Equation 6.9
where is the Rydberg constant, h is Planck’s constant, c is the speed of light, and n is a positive integer corresponding to the number assigned to the orbit, with n = 1 corresponding to the orbit closest to the nucleus.The negative sign in Equation 6.9 is a convention indicating that the electron-nucleus pair has a lower energy when they are near each other than when they are infinitely far apart, corresponding to n = ∞. The latter condition is arbitrarily assigned an energy of zero. Thus the orbit with n = 1 is the lowest in energy. Because a hydrogen atom with its one electron in this orbit has the lowest possible energy, this is the ground stateThe most stable arrangement of electrons for an element or a compound., the most stable arrangement for a hydrogen atom. As n increases, the radius of the orbit increases; the electron is farther from the proton, which results in a less stable arrangement with higher potential energy (Figure 6.10 "The Bohr Model of the Hydrogen Atom"). A hydrogen atom with an electron in an orbit with n > 1 is therefore in an excited stateAny arrangement of electrons that is higher in energy than the ground state.: its energy is higher than the energy of the ground state. When an atom in an excited state undergoes a transition to the ground state in a process called decay, it loses energy by emitting a photon whose energy corresponds to the difference in energy between the two states (Figure 6.11 "The Emission of Light by a Hydrogen Atom in an Excited State").
Figure 6.10 The Bohr Model of the Hydrogen Atom
(a) The distance of the orbit from the nucleus increases with increasing n. (b) The energy of the orbit becomes increasingly less negative with increasing n.
Figure 6.11 The Emission of Light by a Hydrogen Atom in an Excited State
(a) Light is emitted when the electron undergoes a transition from an orbit with a higher value of n (at a higher energy) to an orbit with a lower value of n (at lower energy). (b) The Balmer series of emission lines is due to transitions from orbits with n ≥ 3 to the orbit with n = 2. The differences in energy between these levels corresponds to light in the visible portion of the electromagnetic spectrum.
So the difference in energy (ΔE) between any two orbits or energy levels is given by where n1 is the final orbit and n2 the initial orbit. Substituting from Bohr’s equation (Equation 6.9) for each energy value gives
Equation 6.10
If n2 > n1, the transition is from a higher energy state (larger-radius orbit) to a lower energy state (smaller-radius orbit), as shown by the dashed arrow in part (a) in Figure 6.11 "The Emission of Light by a Hydrogen Atom in an Excited State". Substituting hc/λ for ΔE gives
Equation 6.11
Canceling hc on both sides gives
Equation 6.12
Except for the negative sign, this is the same equation that Rydberg obtained experimentally. The negative sign in Equation 6.11 and Equation 6.12 indicates that energy is released as the electron moves from orbit n2 to orbit n1 because orbit n2 is at a higher energy than orbit n1. Bohr calculated the value of independently and obtained a value of 1.0974 × 107 m−1, the same number Rydberg had obtained by analyzing the emission spectra.
We can now understand the physical basis for the Balmer series of lines in the emission spectrum of hydrogen (part (b) in Figure 6.9 "The Emission of Light by Hydrogen Atoms"). As shown in part (b) in Figure 6.11 "The Emission of Light by a Hydrogen Atom in an Excited State", the lines in this series correspond to transitions from higher-energy orbits (n > 2) to the second orbit (n = 2). Thus the hydrogen atoms in the sample have absorbed energy from the electrical discharge and decayed from a higher-energy excited state (n > 2) to a lower-energy state (n = 2) by emitting a photon of electromagnetic radiation whose energy corresponds exactly to the difference in energy between the two states (part (a) in Figure 6.11 "The Emission of Light by a Hydrogen Atom in an Excited State"). The n = 3 to n = 2 transition gives rise to the line at 656 nm (red), the n = 4 to n = 2 transition to the line at 486 nm (green), the n = 5 to n = 2 transition to the line at 434 nm (blue), and the n = 6 to n = 2 transition to the line at 410 nm (violet). Because a sample of hydrogen contains a large number of atoms, the intensity of the various lines in a line spectrum depends on the number of atoms in each excited state. At the temperature in the gas discharge tube, more atoms are in the n = 3 than the n ≥ 4 levels. Consequently, the n = 3 to n = 2 transition is the most intense line, producing the characteristic red color of a hydrogen discharge (part (a) in Figure 6.9 "The Emission of Light by Hydrogen Atoms"). Other families of lines are produced by transitions from excited states with n > 1 to the orbit with n = 1 or to orbits with n ≥ 3. These transitions are shown schematically in Figure 6.12 "Electron Transitions Responsible for the Various Series of Lines Observed in the Emission Spectrum of Hydrogen".
Figure 6.12 Electron Transitions Responsible for the Various Series of Lines Observed in the Emission Spectrum of Hydrogen
The Lyman series of lines is due to transitions from higher-energy orbits to the lowest-energy orbit (n = 1); these transitions release a great deal of energy, corresponding to radiation in the ultraviolet portion of the electromagnetic spectrum. The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation. (Orbits are not drawn to scale.)
In contemporary applications, electron transitions are used in timekeeping that needs to be exact. Telecommunications systems, such as cell phones, depend on timing signals that are accurate to within a millionth of a second per day, as are the devices that control the US power grid. Global positioning system (GPS) signals must be accurate to within a billionth of a second per day, which is equivalent to gaining or losing no more than one second in 1,400,000 years. Quantifying time requires finding an event with an interval that repeats on a regular basis. To achieve the accuracy required for modern purposes, physicists have turned to the atom. The current standard used to calibrate clocks is the cesium atom. Supercooled cesium atoms are placed in a vacuum chamber and bombarded with microwaves whose frequencies are carefully controlled. When the frequency is exactly right, the atoms absorb enough energy to undergo an electronic transition to a higher-energy state. Decay to a lower-energy state emits radiation. The microwave frequency is continually adjusted, serving as the clock’s pendulum. In 1967, the second was defined as the duration of 9,192,631,770 oscillations of the resonant frequency of a cesium atom, called the cesium clock. Research is currently under way to develop the next generation of atomic clocks that promise to be even more accurate. Such devices would allow scientists to monitor vanishingly faint electromagnetic signals produced by nerve pathways in the brain and geologists to measure variations in gravitational fields, which cause fluctuations in time, that would aid in the discovery of oil or minerals.
The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur?
Given: lowest-energy orbit in the Lyman series
Asked for: wavelength of the lowest-energy Lyman line and corresponding region of the spectrum
Strategy:
A Substitute the appropriate values into Equation 6.8 (the Rydberg equation) and solve for λ.
B Use Figure 6.4 "The Electromagnetic Spectrum" to locate the region of the electromagnetic spectrum corresponding to the calculated wavelength.
Solution:
We can use the Rydberg equation to calculate the wavelength:
A For the Lyman series, n1 = 1. The lowest-energy line is due to a transition from the n = 2 to n = 1 orbit because they are the closest in energy.
and
λ = 1.215 × 10−7 m = 122 nmB This wavelength is in the ultraviolet region of the spectrum.
Exercise
The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the n = 5 orbit. Calculate the wavelength of the second line in the Pfund series to three significant figures. In which region of the spectrum does it lie?
Answer: 4.65 × 103 nm; infrared
Bohr’s model of the hydrogen atom gave an exact explanation for its observed emission spectrum. The following are his key contributions to our understanding of atomic structure:
Unfortunately, Bohr could not explain why the electron should be restricted to particular orbits. Also, despite a great deal of tinkering, such as assuming that orbits could be ellipses rather than circles, his model could not quantitatively explain the emission spectra of any element other than hydrogen (Figure 6.13 "The Emission Spectra of Elements Compared with Hydrogen"). In fact, Bohr’s model worked only for species that contained just one electron: H, He+, Li2+, and so forth. Scientists needed a fundamental change in their way of thinking about the electronic structure of atoms to advance beyond the Bohr model.
Figure 6.13 The Emission Spectra of Elements Compared with Hydrogen
These images show (a) hydrogen gas, which is atomized to hydrogen atoms in the discharge tube; (b) neon; and (c) mercury.
Thus far we have explicitly considered only the emission of light by atoms in excited states, which produces an emission spectrumA spectrum produced by the emission of light by atoms in excited states.. The converse, absorption of light by ground-state atoms to produce an excited state, can also occur, producing an absorption spectrumA spectrum produced by the absorption of light by ground-state atoms.. Because each element has characteristic emission and absorption spectra, scientists can use such spectra to analyze the composition of matter, as we describe in Section 6.4 "The Relationship between Energy and Mass".
When an atom emits light, it decays to a lower energy state; when an atom absorbs light, it is excited to a higher energy state.
If white light is passed through a sample of hydrogen, hydrogen atoms absorb energy as an electron is excited to higher energy levels (orbits with n ≥ 2). If the light that emerges is passed through a prism, it forms a continuous spectrum with black lines (corresponding to no light passing through the sample) at 656, 468, 434, and 410 nm. These wavelengths correspond to the n = 2 to n = 3, n = 2 to n = 4, n = 2 to n = 5, and n = 2 to n = 6 transitions. Any given element therefore has both a characteristic emission spectrum and a characteristic absorption spectrum, which are essentially complementary images.
Absorption of light by a hydrogen atom. (a) When a hydrogen atom absorbs a photon of light, an electron is excited to an orbit that has a higher energy and larger value of n. (b) Images of the emission and absorption spectra of hydrogen are shown here.
Emission and absorption spectra form the basis of spectroscopy, which uses spectra to provide information about the structure and the composition of a substance or an object. In particular, astronomers use emission and absorption spectra to determine the composition of stars and interstellar matter. As an example, consider the spectrum of sunlight shown in Figure 6.14 "The Visible Spectrum of Sunlight". Because the sun is very hot, the light it emits is in the form of a continuous emission spectrum. Superimposed on it, however, is a series of dark lines due primarily to the absorption of specific frequencies of light by cooler atoms in the outer atmosphere of the sun. By comparing these lines with the spectra of elements measured on Earth, we now know that the sun contains large amounts of hydrogen, iron, and carbon, along with smaller amounts of other elements. During the solar eclipse of 1868, the French astronomer Pierre Janssen (1824–1907) observed a set of lines that did not match those of any known element. He suggested that they were due to the presence of a new element, which he named helium, from the Greek helios, meaning “sun.” Helium was finally discovered in uranium ores on Earth in 1895.
Figure 6.14 The Visible Spectrum of Sunlight
The characteristic dark lines are mostly due to the absorption of light by elements that are present in the cooler outer part of the sun’s atmosphere; specific elements are indicated by the labels. The lines at 628 and 687 nm, however, are due to the absorption of light by oxygen molecules in Earth’s atmosphere.
The familiar red color of “neon” signs used in advertising is due to the emission spectrum of neon shown in part (b) in Figure 6.13 "The Emission Spectra of Elements Compared with Hydrogen". Similarly, the blue and yellow colors of certain street lights are caused, respectively, by mercury and sodium discharges. In all these cases, an electrical discharge excites neutral atoms to a higher energy state, and light is emitted when the atoms decay to the ground state. In the case of mercury, most of the emission lines are below 450 nm, which produces a blue light (part (c) in Figure 6.13 "The Emission Spectra of Elements Compared with Hydrogen"). In the case of sodium, the most intense emission lines are at 589 nm, which produces an intense yellow light.
Sodium and mercury spectra. Many street lights use bulbs that contain sodium or mercury vapor. Due to the very different emission spectra of these elements, they emit light of different colors.
Figure 6.15 The Chemistry of Fireworks
(a) In the “multibreak” shell used for fireworks, the chambers contain mixtures of fuels and oxidizers plus compounds for special effects (“stars”) connected by time-delay fuses so that the chambers explode in stages. (b) The finale of a fireworks display usually consists of many shells fired simultaneously to give a dazzling multicolor display. The labels indicate the substances that are responsible for the colors of some of the fireworks shown.
The colors of fireworks are also due to atomic emission spectra. As shown in part (a) in Figure 6.15 "The Chemistry of Fireworks", a typical shell used in a fireworks display contains gunpowder to propel the shell into the air and a fuse to initiate a variety of redox reactions that produce heat and small explosions. Thermal energy excites the atoms to higher energy states; as they decay to lower energy states, the atoms emit light that gives the familiar colors. When oxidant/reductant mixtures listed in Table 6.2 "Common Chemicals Used in the Manufacture of Fireworks*" are ignited, a flash of white or yellow light is produced along with a loud bang. Achieving the colors shown in part (b) in Figure 6.15 "The Chemistry of Fireworks" requires adding a small amount of a substance that has an emission spectrum in the desired portion of the visible spectrum. For example, sodium is used for yellow because of its 589 nm emission lines. The intense yellow color of sodium would mask most other colors, so potassium and ammonium salts, rather than sodium salts, are usually used as oxidants to produce other colors, which explains the preponderance of such salts in Table 6.2 "Common Chemicals Used in the Manufacture of Fireworks*". Strontium salts, which are also used in highway flares, emit red light, whereas barium gives a green color. Blue is one of the most difficult colors to achieve. Copper(II) salts emit a pale blue light, but copper is dangerous to use because it forms highly unstable explosive compounds with anions such as chlorate. As you might guess, preparing fireworks with the desired properties is a complex, challenging, and potentially hazardous process.
Table 6.2 Common Chemicals Used in the Manufacture of Fireworks*
Oxidizers | Fuels (reductants) | Special effects |
---|---|---|
ammonium perchlorate | aluminum | blue flame: copper carbonate, copper sulfate, or copper oxide |
barium chlorate | antimony sulfide | red flame: strontium nitrate or strontium carbonate |
barium nitrate | charcoal | white flame: magnesium or aluminum |
potassium chlorate | magnesium | yellow flame: sodium oxalate or cryolite (Na3AlF6) |
potassium nitrate | sulfur | green flame: barium nitrate or barium chlorate |
potassium perchlorate | titanium | white smoke: potassium nitrate plus sulfur |
strontium nitrate | colored smoke: potassium chlorate and sulfur, plus organic dye | |
whistling noise: potassium benzoate or sodium salicylate | ||
white sparks: aluminum, magnesium, or titanium | ||
gold sparks: iron fillings or charcoal | ||
*Almost any combination of an oxidizer and a fuel may be used along with the compounds needed to produce a desired special effect. |
Most light emitted by atoms is polychromatic—containing more than one wavelength. In contrast, lasers (from light amplification by stimulated emission of radiation) emit monochromatic light—a single wavelength only. Lasers have many applications in fiber-optic telecommunications, the reading and recording of compact discs (CDs) and digital video discs (DVDs), steel cutting, and supermarket checkout scanners. Laser beams are generated by the same general phenomenon that gives rise to emission spectra, with one difference: only a single excited state is produced, which in principle results in only a single frequency of emitted light. In practice, however, inexpensive commercial lasers actually emit light with a very narrow range of wavelengths.
How a CD player uses a laser to read a CD. Inside a CD is a flat, light-reflecting layer called “land.” On the land are many “pits” recorded in a spiral-shaped track. (From the laser’s point of view, pits are actually the “bumps” shown here because the master disc with pits is duplicated negatively, turning the pits into bumps.) Pits have the same light-reflecting surface as land, but there are differences in the frequencies of the reflected light in the pit and the land, making light reflected by pits relatively dark compared with light reflected by land.
The operation of a ruby laser, the first type of laser used commercially, is shown schematically in Figure 6.16 "A Ruby Laser". Ruby is an impure form of aluminum oxide (Al2O3) in which Cr3+ replaces some of the Al3+ ions. The red color of the gem is caused by the absorption of light in the blue region of the visible spectrum by Cr3+ ions, which leaves only the longer wavelengths to be reflected back to the eye. One end of a ruby bar is coated with a fully reflecting mirror, and the mirror on the other end is only partially reflecting. When flashes of white light from a flash lamp excite the Cr3+ ions, they initially decay to a relatively long-lived excited state and can subsequently decay to the ground state by emitting a photon of red light. Some of these photons are reflected back and forth by the mirrored surfaces. As shown in part (b) in Figure 6.16 "A Ruby Laser", each time a photon interacts with an excited Cr3+ ion, it can stimulate that ion to emit another photon that has the same wavelength and is synchronized (in phase) with the first wave. This process produces a cascade of photons traveling back and forth, until the intense beam emerges through the partially reflecting mirror. Ruby is only one substance that is used to produce a laser; the choice of material determines the wavelength of light emitted, from infrared to ultraviolet, and the light output can be either continuous or pulsed.
Figure 6.16 A Ruby Laser
(a) This cutaway view of a ruby laser shows the ruby rod, the flash lamp used to excite the Cr3+ ions in the ruby, and the totally and partially reflective mirrors. (b) This schematic drawing illustrates how light from the flash lamp excites the Cr3+ ions to a short-lived excited state, which is followed by decay to a longer-lived excited state that is responsible for the stimulated in-phase emission of light by the laser.
When used in a DVD player or a CD player, light emitted by a laser passes through a transparent layer of plastic on the CD and is reflected by an underlying aluminum layer, which contains pits or flat regions that were created when the CD was recorded. Differences in the frequencies of the transmitted and reflected light are detected by light-sensitive equipment that converts these differences into binary code, a series of 1s and 0s, which is translated electronically into recognizable sounds and images.
Atoms of individual elements emit light at only specific wavelengths, producing a line spectrum rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states. Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Bohr’s model could not, however, explain the spectra of atoms heavier than hydrogen.
Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy excited state to a higher-energy excited state. This produces an absorption spectrum, which has dark lines in the same position as the bright lines in the emission spectrum of an element.
Is the spectrum of the light emitted by isolated atoms of an element discrete or continuous? How do these spectra differ from those obtained by heating a bulk sample of a solid element? Explain your answers.
Explain why each element has a characteristic emission and absorption spectra. If spectral emissions had been found to be continuous rather than discrete, what would have been the implications for Bohr’s model of the atom?
Explain the differences between a ground state and an excited state. Describe what happens in the spectrum of a species when an electron moves from a ground state to an excited state. What happens in the spectrum when the electron falls from an excited state to a ground state?
What phenomenon causes a neon sign to have a characteristic color? If the emission spectrum of an element is constant, why do some neon signs have more than one color?
How is light from a laser different from the light emitted by a light source such as a light bulb? Describe how a laser produces light.
Using a Bohr model and the transition from n = 2 to n = 3 in an atom with a single electron, describe the mathematical relationship between an emission spectrum and an absorption spectrum. What is the energy of this transition? What does the sign of the energy value represent in this case? What range of light is associated with this transition?
If a hydrogen atom is excited from an n = 1 state to an n = 3 state, how much energy does this correspond to? Is this an absorption or an emission? What is the wavelength of the photon involved in this process? To what region of the electromagnetic spectrum does this correspond?
The hydrogen atom emits a photon with a 486 nm wavelength, corresponding to an electron decaying from the n = 4 level to which level? What is the color of the emission?
An electron in a hydrogen atom can decay from the n = 3 level to n = 2 level. What is the color of the emitted light? What is the energy of this transition?
Calculate the wavelength and energy of the photon that gives rise to the third line in order of increasing energy in the Lyman series in the emission spectrum of hydrogen. In what region of the spectrum does this wavelength occur? Describe qualitatively what the absorption spectrum looks like.
The wavelength of one of the lines in the Lyman series of hydrogen is 121 nm. In what region of the spectrum does this occur? To which electronic transition does this correspond?
The emission spectrum of helium is shown. What change in energy (ΔE) in kilojoules per mole gives rise to each line?
Removing an electron from solid potassium requires 222 kJ/mol. Would you expect to observe a photoelectric effect for potassium using a photon of blue light (λ = 485 nm)? What is the longest wavelength of energy capable of ejecting an electron from potassium? What is the corresponding color of light of this wavelength?
The binding energy of an electron is the energy needed to remove an electron from its lowest energy state. According to Bohr’s postulates, calculate the binding energy of an electron in a hydrogen atom in kilojoules per mole. What wavelength in nanometers is required to remove such an electron?
As a radio astronomer, you have observed spectral lines for hydrogen corresponding to a state with n = 320, and you would like to produce these lines in the laboratory. Is this feasible? Why or why not?
656 nm; red light
n = 2, blue-green light
97.2 nm, 2.04 × 10−18 J/photon, ultraviolet light, absorption spectrum is a single dark line at a wavelength of 97.2 nm
Violet: 390 nm, 307 kJ/mol photons; Blue-purple: 440 nm, 272 kJ/mol photons; Blue-green: 500 nm, 239 kJ/mol photons; Orange: 580 nm, 206 kJ/mol photons; Red: 650 nm, 184 kJ/mol photons
1313 kJ/mol, λ ≤ 91.1 nm
Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a photoelectron from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom.
Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the following equation:
Equation 6.13
According to this theory, a photon of wavelength λ and frequency ν has a nonzero mass, which is given as follows:
Equation 6.14
That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle dualityA principle that matter and energy have properties typical of both waves and particles.. Depending on conditions, light could be viewed as either a wave or a particle.
In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie proposed that a particle such as an electron could be described by a wave whose wavelength is given by
Equation 6.15
where h is Planck’s constant, m is the mass of the particle, and v is the velocity of the particle. This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. For his work, de Broglie received the Nobel Prize in Physics in 1929.
If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation, which is an extremely small number. As you will calculate in Example 4, Planck’s constant (6.63 × 10−34 J·s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus) to be noticeable.
Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h.
Given: mass and speed of object
Asked for: wavelength
Strategy:
A Convert the speed of the baseball to the appropriate SI units: meters per second.
B Substitute values into and solve for the wavelength.
Solution:
The wavelength of a particle is given by λ = h/mv. We know that m = 0.149 kg, so all we need to find is the speed of the baseball:
B Recall that the joule is a derived unit, whose units are (kg·m2)/s2. Thus the wavelength of the baseball is
(You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10−14 m, the wavelength of the baseball is almost unimaginably small.
Exercise
Calculate the wavelength of a neutron that is moving at 3.00 × 103 m/s.
Answer: 1.32 Å, or 132 pm
As you calculated in Example 4, objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, such as electrons, exhibit the properties of both particles and waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an electron microscope, which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can ().
Figure 6.17 A Comparison of Images Obtained Using a Light Microscope and an Electron Microscope
Because of their shorter wavelength, high-energy electrons have a higher resolving power than visible light. Consequently, an electron microscope (b) is able to resolve finer details than a light microscope (a). (Radiolaria, which are shown here, are unicellular planktonic organisms.)
De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing waveA wave that does not travel in space., a wave that does not travel in space. An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (). If the length of the string is L, then the lowest-energy vibration (the fundamentalThe lowest-energy standing wave.) has wavelength
Equation 6.16
Higher-energy vibrations (overtonesThe vibration of a standing wave that is higher in energy than the fundamental vibration.) are produced when the string is plucked more strongly; they have wavelengths given by
Equation 6.17
where n has any integral value. Thus the vibrational energy of the string is quantized, and only certain wavelengths and frequencies are possible. Notice in that all overtones have one or more nodesThe point where the amplitude of a wave is zero., points where the string does not move. The amplitude of the wave at a node is zero.
Figure 6.18 Standing Waves on a Vibrating String
The vibration with n = 1 is the fundamental and contains no nodes. Vibrations with higher values of n are called overtones; they contain n − 1 nodes.
Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength
Equation 6.18
2πr = nλwhere r is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a standing circular wave (). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing constructive interference. Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing destructive interference. De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to n = 1 in , one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.
Standing waves are often observed on rivers, reservoirs, ponds, and lakes when seismic waves from an earthquake travel through the area. The waves are called seismic seiches, a term first used in 1955 when lake levels in England and Norway oscillated from side to side as a result of the Assam earthquake of 1950 in Tibet. They were first described in the Proceedings of the Royal Society in 1755 when they were seen in English harbors and ponds after a large earthquake in Lisbon, Portugal. Seismic seiches were also observed in many places in North America after the Alaska earthquake of March 28, 1964. Those occurring in western reservoirs lasted for two hours or longer, and amplitudes reached as high as nearly 6 ft along the Gulf Coast. The height of seiches is approximately proportional to the thickness of surface sediments; a deeper channel will produce a higher seiche.
Figure 6.19 Standing Circular Wave and Destructive Interference
(a) In a standing circular wave with n = 5, the circumference of the circle corresponds to exactly five wavelengths, which results in constructive interference of the wave with itself when overlapping occurs. (b) If the circumference of the circle is not equal to an integral multiple of wavelengths, then the wave does not overlap exactly with itself, and the resulting destructive interference will result in cancellation of the wave. Consequently, a standing wave cannot exist under these conditions.
As you will see, several of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron, the concept of standing waves, and the presence of nodes that increase in number as the energy level increases. Unfortunately, his explanation also contains one major feature that we know to be incorrect: in the currently accepted model, the electron in a given orbit is not always at the same distance from the nucleus.
Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a particle that exhibits wavelike behavior. This situation was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the Heisenberg uncertainty principleA principle stating that the uncertainty in the position of a particle multiplied by the uncertainty in its momentum is greater than or equal to Planck’s constant divided by 4π: states that the uncertainty in the position of a particle (Δx) multiplied by the uncertainty in its momentum [Δ(mv)] is greater than or equal to Planck’s constant divided by 4π:
Equation 6.19
Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths.
If the precise position x of a particle is known absolutely (Δx = 0), then the uncertainty in its momentum must be infinite:
Equation 6.20
Because the mass of the electron at rest (m) is both constant and accurately known, the uncertainty in Δ(mv) must be due to the Δv term, which would have to be infinitely large for Δ(mv) to equal infinity. That is, according to , the more accurately we know the exact position of the electron (as Δx → 0), the less accurately we know the speed and the kinetic energy of the electron (1/2 mv2) because Δ(mv) → ∞. Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as Δ(mv) → 0], then Δx → ∞ and we have no idea where the electron is.
Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify simultaneously both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the most probable radius of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model.
Calculate the minimum uncertainty in the position of the pitched baseball from Example 4 that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h.
Given: mass and speed of object
Asked for: minimum uncertainty in its position
Strategy:
A Rearrange the inequality that describes the Heisenberg uncertainty principle () to solve for the minimum uncertainty in the position of an object (Δx).
B Find Δv by converting the velocity of the baseball to the appropriate SI units: meters per second.
C Substitute the appropriate values into the expression for the inequality and solve for Δx.
Solution:
A The Heisenberg uncertainty principle tells us that (Δx)[Δ(mv)] = h/4π. Rearranging the inequality gives
B We know that h = 6.626 × 10−34 J·s and m = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ(mv) = mΔv and Δv = ±1 mi/h. We have
C Therefore,
Inserting the definition of a joule (1 J = 1 kg·m2/s2) gives
This is equal to 3.12 × 10−32 inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out.
Exercise
Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest.
Answer: 6 × 10−10 m, or 0.6 nm (about the diameter of a benzene molecule)
Einstein’s relationship between mass and energy
De Broglie’s relationship between mass, speed, and wavelength
Heisenberg’s uncertainty principle
The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes, points where the amplitude of the wave is always zero. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
Explain what is meant by each term and illustrate with a sketch:
How does Einstein’s theory of relativity illustrate the wave–particle duality of light? What properties of light can be explained by a wave model? What properties can be explained by a particle model?
In the modern theory of the electronic structure of the atom, which of de Broglie’s ideas have been retained? Which proved to be incorrect?
According to Bohr, what is the relationship between an atomic orbit and the energy of an electron in that orbit? Is Bohr’s model of the atom consistent with Heisenberg’s uncertainty principle? Explain your answer.
The development of ideas frequently builds on the work of predecessors. Complete the following chart by filling in the names of those responsible for each theory shown.
How much heat is generated by shining a carbon dioxide laser with a wavelength of 1.065 μm on a 68.95 kg sample of water if 1.000 mol of photons is absorbed and converted to heat? Is this enough heat to raise the temperature of the water 4°C?
Show the mathematical relationship between energy and mass and between wavelength and mass. What is the effect of doubling the
What is the de Broglie wavelength of a 39 g bullet traveling at 1020 m/s ± 10 m/s? What is the minimum uncertainty in the bullet’s position?
What is the de Broglie wavelength of a 6800 tn aircraft carrier traveling at 18 ± 0.1 knots (1 knot = 1.15 mi/h)? What is the minimum uncertainty in its position?
Calculate the mass of a particle if it is traveling at 2.2 × 106 m/s and has a frequency of 6.67 × 107 Hz. If the uncertainty in the velocity is known to be 0.1%, what is the minimum uncertainty in the position of the particle?
Determine the wavelength of a 2800 lb automobile traveling at 80 mi/h ± 3%. How does this compare with the diameter of the nucleus of an atom? You are standing 3 in. from the edge of the highway. What is the minimum uncertainty in the position of the automobile in inches?
E = 112.3 kJ, ΔT = 0.3893°C, over ten times more light is needed for a 4.0°C increase in temperature
1.7 × 10−35 m, uncertainty in position is ≥ 1.4 × 10−34 m
9.1 × 10−39 kg, uncertainty in position ≥ 2.6 m
The paradox described by Heisenberg’s uncertainty principle and the wavelike nature of subatomic particles such as the electron made it impossible to use the equations of classical physics to describe the motion of electrons in atoms. Scientists needed a new approach that took the wave behavior of the electron into account. In 1926, an Austrian physicist, Erwin Schrödinger (1887–1961; Nobel Prize in Physics, 1933), developed wave mechanics, a mathematical technique that describes the relationship between the motion of a particle that exhibits wavelike properties (such as an electron) and its allowed energies. In doing so, Schrödinger developed the theory of quantum mechanicsA theory developed by Erwin Schrödinger that describes the energies and spatial distributions of electrons in atoms and molecules., which is used today to describe the energies and spatial distributions of electrons in atoms and molecules.
Schrödinger’s unconventional approach to atomic theory was typical of his unconventional approach to life. He was notorious for his intense dislike of memorizing data and learning from books. When Hitler came to power in Germany, Schrödinger escaped to Italy. He then worked at Princeton University in the United States but eventually moved to the Institute for Advanced Studies in Dublin, Ireland, where he remained until his retirement in 1955.
Although quantum mechanics uses sophisticated mathematics, you do not need to understand the mathematical details to follow our discussion of its general conclusions. We focus on the properties of the wave functions that are the solutions of Schrödinger’s equations.
A wave function (Ψ)A mathematical function that relates the location of an electron at a given point in space to the amplitude of its wave, which corresponds to its energy.Ψ is the uppercase Greek psi. is a mathematical function that relates the location of an electron at a given point in space (identified by x, y, and z coordinates) to the amplitude of its wave, which corresponds to its energy. Thus each wave function is associated with a particular energy E. The properties of wave functions derived from quantum mechanics are summarized here:
Figure 6.20 The Four Variables (Latitude, Longitude, Depth, and Time) Required to Precisely Locate an Object
If you are the captain of a ship trying to intercept an enemy submarine, you need to deliver your depth charge to the right location at the right time.
Figure 6.21 Probability of Finding the Electron in the Ground State of the Hydrogen Atom at Different Points in Space
(a) The density of the dots shows electron probability. (b) In this plot of Ψ2 versus r for the ground state of the hydrogen atom, the electron probability density is greatest at r = 0 (the nucleus) and falls off with increasing r. Because the line never actually reaches the horizontal axis, the probability of finding the electron at very large values of r is very small but not zero.
Schrödinger’s approach uses three quantum numbers (n, l, and ml) to specify any wave function. The quantum numbers provide information about the spatial distribution of an electron. Although n can be any positive integer, only certain values of l and ml are allowed for a given value of n.
The principal quantum number (n)One of three quantum numbers that tells the average relative distance of an electron from the nucleus. tells the average relative distance of an electron from the nucleus:
Equation 6.21
n = 1, 2, 3, 4,…As n increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of n are easier to remove from an atom. All wave functions that have the same value of n are said to constitute a principal shellAll the wave functions that have the same value of because those electrons have similar average distances from the nucleus. because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number n corresponds to the n used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels.
The second quantum number is often called the azimuthal quantum number (l)One of three quantum numbers that discribes the shape of the region of space occupied by an electron.. The value of l describes the shape of the region of space occupied by the electron. The allowed values of l depend on the value of n and can range from 0 to n − 1:
Equation 6.22
l = 0, 1, 2,…, n − 1For example, if n = 1, l can be only 0; if n = 2, l can be 0 or 1; and so forth. For a given atom, all wave functions that have the same values of both n and l form a subshellA group of wave functions that have the same values of and . The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space.
The third quantum number is the magnetic quantum number (ml)One of three quantum numbers that describes the orientation of the region of space occupied by an electron with respect to an applied magnetic field.. The value of ml describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of ml depend on the value of l: ml can range from −l to l in integral steps:
Equation 6.23
ml = −l, −l + 1,…, 0,…, l − 1, lFor example, if l = 0, ml can be only 0; if l = 1, ml can be −1, 0, or +1; and if l = 2, ml can be −2, −1, 0, +1, or +2.
Each wave function with an allowed combination of n, l, and ml values describes an atomic orbitalA wave function with an allowed combination of , , and quantum numbers., a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals.
How many subshells and orbitals are contained within the principal shell with n = 4?
Given: value of n
Asked for: number of subshells and orbitals in the principal shell
Strategy:
A Given n = 4, calculate the allowed values of l. From these allowed values, count the number of subshells.
B For each allowed value of l, calculate the allowed values of ml. The sum of the number of orbitals in each subshell is the number of orbitals in the principal shell.
Solution:
A We know that l can have all integral values from 0 to n − 1. If n = 4, then l can equal 0, 1, 2, or 3. Because the shell has four values of l, it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of ml.
B For l = 0, ml can be only 0, and thus the l = 0 subshell has only one orbital. For l = 1, ml can be 0 or ±1; thus the l = 1 subshell has three orbitals. For l = 2, ml can be 0, ±1, or ±2, so there are five orbitals in the l = 2 subshell. The last allowed value of l is l = 3, for which ml can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the l = 3 subshell. The total number of orbitals in the n = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to n2:
Exercise
How many subshells and orbitals are in the principal shell with n = 3?
Answer: three subshells; nine orbitals
Rather than specifying all the values of n and l every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of l for a particular subshell or orbital:
l = | 0 | 1 | 2 | 3 |
Designation | s | p | d | f |
The principal quantum number is named first, followed by the letter s, p, d, or f as appropriate. These orbital designations are derived from corresponding spectroscopic characteristics: sharp, principle, diffuse, and fundamental. A 1s orbital has n = 1 and l = 0; a 2p subshell has n = 2 and l = 1 (and has three 2p orbitals, corresponding to ml = −1, 0, and +1); a 3d subshell has n = 3 and l = 2 (and has five 3d orbitals, corresponding to ml = −2, −1, 0, +1, and +2); and so forth.
We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows ():
Each principal shell has n subshells, and each subshell has 2l + 1 orbitals.
Table 6.3 Values of n, l, and ml through n = 4
n | l | Subshell Designation | ml | Number of Orbitals in Subshell | Number of Orbitals in Shell |
---|---|---|---|---|---|
1 | 0 | 1s | 0 | 1 | 1 |
2 | 0 | 2s | 0 | 1 | 4 |
1 | 2p | −1, 0, 1 | 3 | ||
3 | 0 | 3s | 0 | 1 | 9 |
1 | 3p | −1, 0, 1 | 3 | ||
2 | 3d | −2, −1, 0, 1, 2 | 5 | ||
4 | 0 | 4s | 0 | 1 | 16 |
1 | 4p | −1, 0, 1 | 3 | ||
2 | 4d | −2, −1, 0, 1, 2 | 5 | ||
3 | 4f | −3, −2, −1, 0, 1, 2, 3 | 7 |
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
One way of representing electron probability distributions was illustrated in for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at r = 0 (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx− 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (part (a) in ), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (part (b) in ), so the density of dots is greatest for the smallest spherical shells in part (a) in . In contrast, the surface area of each spherical shell is equal to 4πr2, which increases very rapidly with increasing r (part (c) in ). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (part (d) in ). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (part (d) in ).
Figure 6.22 Most Probable Radius for the Electron in the Ground State of the Hydrogen Atom
(a) Imagine dividing the atom’s total volume into very thin concentric shells as shown in the onion drawing. (b) A plot of electron probability density Ψ2 versus r shows that the electron probability density is greatest at r = 0 and falls off smoothly with increasing r. The density of the dots is therefore greatest in the innermost shells of the onion. (c) The surface area of each shell, given by 4πr2, increases rapidly with increasing r. (d) If we count the number of dots in each spherical shell, we obtain the total probability of finding the electron at a given value of r. Because the surface area of each shell increases more rapidly with increasing r than the electron probability density decreases, a plot of electron probability versus r (the radial probability) shows a peak. This peak corresponds to the most probable radius for the electron, 52.9 pm, which is exactly the radius predicted by Bohr’s model of the hydrogen atom.
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in ). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability.
Figure 6.23 Probability Densities for the 1s, 2s, and 3s Orbitals of the Hydrogen Atom
(a) The electron probability density in any plane that contains the nucleus is shown. Note the presence of circular regions, or nodes, where the probability density is zero. (b) Contour surfaces enclose 90% of the electron probability, which illustrates the different sizes of the 1s, 2s, and 3s orbitals. The cutaway drawings give partial views of the internal spherical nodes. The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative. (c) In these plots of electron probability as a function of distance from the nucleus (r) in all directions (radial probability), the most probable radius increases as n increases, but the 2s and 3s orbitals have regions of significant electron probability at small values of r.
Three things happen to s orbitals as n increases ():
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron densityElectron distributions that are represented as standing waves., as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in . Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
Figure 6.24 Electron Probability Distribution for a Hydrogen 2p Orbital
The nodal plane of zero electron density separates the two lobes of the 2p orbital. As in , the colors correspond to regions of space where the phase of the wave function is positive (orange) and negative (blue).
The electron probability distribution for one of the hydrogen 2p orbitals is shown in . Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a 2pz orbital. As shown in , the other two 2p orbitals have identical shapes, but they lie along the x axis (2px) and y axis (2py), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
Figure 6.25 The Three Equivalent 2p Orbitals of the Hydrogen Atom
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative.
Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in .
Subshells with l = 2 have five d orbitals; the first principal shell to have a d subshell corresponds to n = 3. The five d orbitals have ml values of −2, −1, 0, +1, and +2.
Figure 6.26 The Five Equivalent 3d Orbitals of the Hydrogen Atom
The surfaces shown enclose 90% of the total electron probability for the five hydrogen 3d orbitals. Four of the five 3d orbitals consist of four lobes arranged in a plane that is intersected by two perpendicular nodal planes. These four orbitals have the same shape but different orientations. The fifth 3d orbital, , has a distinct shape even though it is mathematically equivalent to the others. The phase of the wave function for the different lobes is indicated by color: orange for positive and blue for negative.
The hydrogen 3d orbitals, shown in , have more complex shapes than the 2p orbitals. All five 3d orbitals contain two nodal surfaces, as compared to one for each p orbital and zero for each s orbital. In three of the d orbitals, the lobes of electron density are oriented between the x and y, x and z, and y and z planes; these orbitals are referred to as the 3dxy, 3dxz, and 3dyz orbitals, respectively. A fourth d orbital has lobes lying along the x and y axes; this is the orbital. The fifth 3d orbital, called the orbital, has a unique shape: it looks like a 2pz orbital combined with an additional doughnut of electron probability lying in the xy plane. Despite its peculiar shape, the orbital is mathematically equivalent to the other four and has the same energy. In contrast to p orbitals, the phase of the wave function for d orbitals is the same for opposite pairs of lobes. As shown in , the phase of the wave function is positive for the two lobes of the orbital that lie along the z axis, whereas the phase of the wave function is negative for the doughnut of electron density in the xy plane. Like the s and p orbitals, as n increases, the size of the d orbitals increases, but the overall shapes remain similar to those depicted in .
Principal shells with n = 4 can have subshells with l = 3 and ml values of −3, −2, −1, 0, +1, +2, and +3. These subshells consist of seven f orbitals. Each f orbital has three nodal surfaces, so their shapes are complex. Because f orbitals are not particularly important for our purposes, we do not discuss them further, and orbitals with higher values of l are not discussed at all.
Although we have discussed the shapes of orbitals, we have said little about their comparative energies. We begin our discussion of orbital energiesA particular energy associated with a given set of quantum numbers. by considering atoms or ions with only a single electron (such as H or He+).
The relative energies of the atomic orbitals with n ≤ 4 for a hydrogen atom are plotted in ; note that the orbital energies depend on only the principal quantum number n. Consequently, the energies of the 2s and 2p orbitals of hydrogen are the same; the energies of the 3s, 3p, and 3d orbitals are the same; and so forth. The orbital energies obtained for hydrogen using quantum mechanics are exactly the same as the allowed energies calculated by Bohr. In contrast to Bohr’s model, however, which allowed only one orbit for each energy level, quantum mechanics predicts that there are 4 orbitals with different electron density distributions in the n = 2 principal shell (one 2s and three 2p orbitals), 9 in the n = 3 principal shell, and 16 in the n = 4 principal shell.The different values of l and ml for the individual orbitals within a given principal shell are not important for understanding the emission or absorption spectra of the hydrogen atom under most conditions, but they do explain the splittings of the main lines that are observed when hydrogen atoms are placed in a magnetic field. As we have just seen, however, quantum mechanics also predicts that in the hydrogen atom, all orbitals with the same value of n (e.g., the three 2p orbitals) are degenerateHaving the same energy., meaning that they have the same energy. shows that the energy levels become closer and closer together as the value of n increases, as expected because of the 1/n2 dependence of orbital energies.
Figure 6.27 Orbital Energy Level Diagram for the Hydrogen Atom
Each box corresponds to one orbital. Note that the difference in energy between orbitals decreases rapidly with increasing values of n.
The energies of the orbitals in any species with only one electron can be calculated by a minor variation of Bohr’s equation (), which can be extended to other single-electron species by incorporating the nuclear charge Z (the number of protons in the nucleus):
Equation 6.24
In general, both energy and radius decrease as the nuclear charge increases. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. For example, in the ground state of the hydrogen atom, the single electron is in the 1s orbital, whereas in the first excited state, the atom has absorbed energy and the electron has been promoted to one of the n = 2 orbitals. In ions with only a single electron, the energy of a given orbital depends on only n, and all subshells within a principal shell, such as the px, py, and pz orbitals, are degenerate.
For an atom or an ion with only a single electron, we can calculate the potential energy by considering only the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When more than one electron is present, however, the total energy of the atom or the ion depends not only on attractive electron-nucleus interactions but also on repulsive electron-electron interactions. When there are two electrons, the repulsive interactions depend on the positions of both electrons at a given instant, but because we cannot specify the exact positions of the electrons, it is impossible to exactly calculate the repulsive interactions. Consequently, we must use approximate methods to deal with the effect of electron-electron repulsions on orbital energies.
If an electron is far from the nucleus (i.e., if the distance r between the nucleus and the electron is large), then at any given moment, most of the other electrons will be between that electron and the nucleus. Hence the electrons will cancel a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between it and the electron farther away. As a result, the electron farther away experiences an effective nuclear charge (Zeff)The nuclear charge an electron actually experiences because of shielding from other electrons closer to the nucleus. that is less than the actual nuclear charge Z. This effect is called electron shieldingThe effect by which electrons closer to the nucleus neutralize a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between the nucleus and an electron father away.. As the distance between an electron and the nucleus approaches infinity, Zeff approaches a value of 1 because all the other (Z − 1) electrons in the neutral atom are, on the average, between it and the nucleus. If, on the other hand, an electron is very close to the nucleus, then at any given moment most of the other electrons are farther from the nucleus and do not shield the nuclear charge. At r ≈ 0, the positive charge experienced by an electron is approximately the full nuclear charge, or Zeff ≈ Z. At intermediate values of r, the effective nuclear charge is somewhere between 1 and Z: 1 ≤ Zeff ≤ Z. Thus the actual Zeff experienced by an electron in a given orbital depends not only on the spatial distribution of the electron in that orbital but also on the distribution of all the other electrons present. This leads to large differences in Zeff for different elements, as shown in for the elements of the first three rows of the periodic table. Notice that only for hydrogen does Zeff = Z, and only for helium are Zeff and Z comparable in magnitude.
Figure 6.28 Relationship between the Effective Nuclear Charge Zeff and the Atomic Number Z for the Outer Electrons of the Elements of the First Three Rows of the Periodic Table
Except for hydrogen, Zeff is always less than Z, and Zeff increases from left to right as you go across a row.
The energies of the different orbitals for a typical multielectron atom are shown in . Within a given principal shell of a multielectron atom, the orbital energies increase with increasing l. An ns orbital always lies below the corresponding np orbital, which in turn lies below the nd orbital. These energy differences are caused by the effects of shielding and penetration, the extent to which a given orbital lies inside other filled orbitals. As shown in , for example, an electron in the 2s orbital penetrates inside a filled 1s orbital more than an electron in a 2p orbital does. Hence in an atom with a filled 1s orbital, the Zeff experienced by a 2s electron is greater than the Zeff experienced by a 2p electron. Consequently, the 2s electron is more tightly bound to the nucleus and has a lower energy, consistent with the order of energies shown in .
Due to electron shielding, Zeff increases more rapidly going across a row of the periodic table than going down a column.
Figure 6.29 Orbital Energy Level Diagram for a Typical Multielectron Atom
Because of the effects of shielding and the different radial distributions of orbitals with the same value of n but different values of l, the different subshells are not degenerate in a multielectron atom. (Compare this with .) For a given value of n, the ns orbital is always lower in energy than the np orbitals, which are lower in energy than the nd orbitals, and so forth. As a result, some subshells with higher principal quantum numbers are actually lower in energy than subshells with a lower value of n; for example, the 4s orbital is lower in energy than the 3d orbitals for most atoms.
Figure 6.30 Orbital Penetration
A comparison of the radial probability distribution of the 2s and 2p orbitals for various states of the hydrogen atom shows that the 2s orbital penetrates inside the 1s orbital more than the 2p orbital does. Consequently, when an electron is in the small inner lobe of the 2s orbital, it experiences a relatively large value of Zeff, which causes the energy of the 2s orbital to be lower than the energy of the 2p orbital.
Notice in that the difference in energies between subshells can be so large that the energies of orbitals from different principal shells can become approximately equal. For example, the energy of the 3d orbitals in most atoms is actually between the energies of the 4s and the 4p orbitals.
Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wave functions (Ψ) to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Wave functions have five important properties: (1) the wave function uses three variables (Cartesian axes x, y, and z) to describe the position of an electron; (2) the magnitude of the wave function is proportional to the intensity of the wave; (3) the probability of finding an electron at a given point is proportional to the square of the wave function at that point, leading to a distribution of probabilities in space that is often portrayed as an electron density plot; (4) describing electron distributions as standing waves leads naturally to the existence of sets of quantum numbers characteristic of each wave function; and (5) each spatial distribution of the electron described by a wave function with a given set of quantum numbers has a particular energy.
Quantum numbers provide important information about the energy and spatial distribution of an electron. The principal quantum number n can be any positive integer; as n increases for an atom, the average distance of the electron from the nucleus also increases. All wave functions with the same value of n constitute a principal shell in which the electrons have similar average distances from the nucleus. The azimuthal quantum number l can have integral values between 0 and n − 1; it describes the shape of the electron distribution. Wave functions that have the same values of both n and l constitute a subshell, corresponding to electron distributions that usually differ in orientation rather than in shape or average distance from the nucleus. The magnetic quantum number ml can have 2l + 1 integral values, ranging from −l to +l, and describes the orientation of the electron distribution. Each wave function with a given set of values of n, l, and ml describes a particular spatial distribution of an electron in an atom, an atomic orbital.
The four chemically important types of atomic orbital correspond to values of l = 0, 1, 2, and 3. Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. All orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. Orbitals with l = 3 are f orbitals, which are still more complex.
Because its average distance from the nucleus determines the energy of an electron, each atomic orbital with a given set of quantum numbers has a particular energy associated with it, the orbital energy. In atoms or ions with only a single electron, all orbitals with the same value of n have the same energy (they are degenerate), and the energies of the principal shells increase smoothly as n increases. An atom or ion with the electron(s) in the lowest-energy orbital(s) is said to be in its ground state, whereas an atom or ion in which one or more electrons occupy higher-energy orbitals is said to be in an excited state. The calculation of orbital energies in atoms or ions with more than one electron (multielectron atoms or ions) is complicated by repulsive interactions between the electrons. The concept of electron shielding, in which intervening electrons act to reduce the positive nuclear charge experienced by an electron, allows the use of hydrogen-like orbitals and an effective nuclear charge (Zeff) to describe electron distributions in more complex atoms or ions. The degree to which orbitals with different values of l and the same value of n overlap or penetrate filled inner shells results in slightly different energies for different subshells in the same principal shell in most atoms.
Why does an electron in an orbital with n = 1 in a hydrogen atom have a lower energy than a free electron (n = ∞)?
What four variables are required to fully describe the position of any object in space? In quantum mechanics, one of these variables is not explicitly considered. Which one and why?
Chemists generally refer to the square of the wave function rather than to the wave function itself. Why?
Orbital energies of species with only one electron are defined by only one quantum number. Which one? In such a species, is the energy of an orbital with n = 2 greater than, less than, or equal to the energy of an orbital with n = 4? Justify your answer.
In each pair of subshells for a hydrogen atom, which has the higher energy? Give the principal and the azimuthal quantum number for each pair.
What is the relationship between the energy of an orbital and its average radius? If an electron made a transition from an orbital with an average radius of 846.4 pm to an orbital with an average radius of 476.1 pm, would an emission spectrum or an absorption spectrum be produced? Why?
In making a transition from an orbital with a principal quantum number of 4 to an orbital with a principal quantum number of 7, does the electron of a hydrogen atom emit or absorb a photon of energy? What would be the energy of the photon? To what region of the electromagnetic spectrum does this energy correspond?
What quantum number defines each of the following?
In an attempt to explain the properties of the elements, Niels Bohr initially proposed electronic structures for several elements with orbits holding a certain number of electrons, some of which are in the following table:
Element | Number of Electrons | Electrons in orbits with n = | |||
---|---|---|---|---|---|
4 | 3 | 2 | 1 | ||
H | 1 | 1 | |||
He | 2 | 2 | |||
Ne | 10 | 8 | 2 | ||
Ar | 18 | 8 | 8 | 2 | |
Li | 3 | 1 | 2 | ||
Na | 11 | 1 | 8 | 2 | |
K | 19 | 1 | 8 | 8 | 2 |
Be | 4 | 2 | 2 |
What happens to the energy of a given orbital as the nuclear charge Z of a species increases? In a multielectron atom and for a given nuclear charge, the Zeff experienced by an electron depends on its value of l. Why?
The electron density of a particular atom is divided into two general regions. Name these two regions and describe what each represents.
As the principal quantum number increases, the energy difference between successive energy levels decreases. Why? What would happen to the electron configurations of the transition metals if this decrease did not occur?
Describe the relationship between electron shielding and Zeff on the outermost electrons of an atom. Predict how chemical reactivity is affected by a decreased effective nuclear charge.
If a given atom or ion has a single electron in each of the following subshells, which electron is easier to remove?
How many subshells are possible for n = 3? What are they?
How many subshells are possible for n = 5? What are they?
What value of l corresponds to a d subshell? How many orbitals are in this subshell?
What value of l corresponds to an f subshell? How many orbitals are in this subshell?
State the number of orbitals and electrons that can occupy each subshell.
State the number of orbitals and electrons that can occupy each subshell.
How many orbitals and subshells are found within the principal shell n = 6? How do these orbital energies compare with those for n = 4?
How many nodes would you expect a 4p orbital to have? A 5s orbital?
A p orbital is found to have one node in addition to the nodal plane that bisects the lobes. What would you predict to be the value of n? If an s orbital has two nodes, what is the value of n?
Three subshells, with l = 0 (s), l = 1 (p), and l = 2 (d).
A d subshell has l = 2 and contains 5 orbitals.
A principal shell with n = 6 contains six subshells, with l = 0, 1, 2, 3, 4, and 5, respectively. These subshells contain 1, 3, 5, 7, 9, and 11 orbitals, respectively, for a total of 36 orbitals. The energies of the orbitals with n = 6 are higher than those of the corresponding orbitals with the same value of l for n = 4.
Now you can use the information you learned in to determine the electronic structure of every element in the periodic table. The process of describing each atom’s electronic structure consists, essentially, of beginning with hydrogen and adding one proton and one electron at a time to create the next heavier element in the table.All stable nuclei other than hydrogen also contain one or more neutrons. Because neutrons have no electrical charge, however, they can be ignored in the following discussion. Before demonstrating how to do this, however, we must introduce the concept of electron spin and the Pauli principle.
When scientists analyzed the emission and absorption spectra of the elements more closely, they saw that for elements having more than one electron, nearly all the lines in the spectra were actually pairs of very closely spaced lines. Because each line represents an energy level available to electrons in the atom, there are twice as many energy levels available as would be predicted solely based on the quantum numbers n, l, and ml. Scientists also discovered that applying a magnetic field caused the lines in the pairs to split farther apart. In 1925, two graduate students in physics in the Netherlands, George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spinThe magnetic moment that results when an electron spins. Electrons have two possible orientations (spin up and spin down), which are described by a fourth quantum number (ms)..
Figure 6.31 Electron Spin
In a magnetic field, an electron has two possible orientations with different energies, one with spin up, aligned with the magnetic field, and one with spin down, aligned against it. All other orientations are forbidden.
In an external magnetic field, the electron has two possible orientations (). These are described by a fourth quantum number (ms), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript s is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it.
The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist,Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons. He developed the Pauli exclusion principleA principle stating that no two electrons in an atom can have the same value of all four quantum numbers.: No two electrons in an atom can have the same values of all four quantum numbers (n, l, ml, ms).
By giving the values of n, l, and ml, we also specify a particular orbital (e.g., 1s with n = 1, l = 0, ml = 0). Because ms has only two possible values (+½ or −½), two electrons, and only two electrons, can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which, as you learned in , was originally based on the physical and chemical properties of the known elements.
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 2p orbital and predict the maximum number of electrons the 2p subshell can accommodate.
Given: orbital
Asked for: allowed quantum numbers and maximum number of electrons in orbital
Strategy:
A List the quantum numbers (n, l, ml) that correspond to an n = 2p orbital. List all allowed combinations of (n, l, ml).
B Build on these combinations to list all the allowed combinations of (n, l, ml, ms).
C Add together the number of combinations to predict the maximum number of electrons the 2p subshell can accommodate.
Solution:
A For a 2p orbital, we know that n = 2, l = n − 1 = 1, and ml = −l, (−l +1),…, (l − 1), l. There are only three possible combinations of (n, l, ml): (2, 1, 1), (2, 1, 0), and (2, 1, −1).
B Because ms is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (n, l, ml, ms): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½).
C Hence the 2p subshell, which consists of three 2p orbitals (2px, 2py, and 2pz), can contain a total of six electrons, two in each orbital.
Exercise
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for a 6s orbital, and predict the total number of electrons it can contain.
Answer: (6, 0, 0, +½), (6, 0, 0, −½); two electrons
The electron configurationThe arrangement of an element’s electrons in its atomic orbitals. of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry.
We construct the periodic table by following the aufbau principleThe process used to build up the periodic table by adding protons one by one to the nucleus and adding the corresponding electrons to the lowest-energy orbital available without violating the Pauli exclusion principle. (from German, meaning “building up”). First we determine the number of electrons in the atom; then we add electrons one at a time to the lowest-energy orbital available without violating the Pauli principle. We use the orbital energy diagram of , recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to ms = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to ms = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be paired. Here is a schematic orbital diagram for a hydrogen atom in its ground state:
From the orbital diagram, we can write the electron configuration in an abbreviated form in which the occupied orbitals are identified by their principal quantum number n and their value of l (s, p, d, or f), with the number of electrons in the subshell indicated by a superscript. For hydrogen, therefore, the single electron is placed in the 1s orbital, which is the orbital lowest in energy (), and the electron configuration is written as 1s1 and read as “one-s-one.”
A neutral helium atom, with an atomic number of 2 (Z = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1s orbital. From the Pauli exclusion principle, we know that an orbital can contain two electrons with opposite spin, so we place the second electron in the same orbital as the first but pointing down, so that the electrons are paired. The orbital diagram for the helium atom is therefore
written as 1s2, where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle.
The next element is lithium, with Z = 3 and three electrons in the neutral atom. We know that the 1s orbital can hold two of the electrons with their spins paired. tells us that the next lowest energy orbital is 2s, so the orbital diagram for lithium is
This electron configuration is written as 1s22s1.
The next element is beryllium, with Z = 4 and four electrons. We fill both the 1s and 2s orbitals to achieve a 1s22s2 electron configuration:
When we reach boron, with Z = 5 and five electrons, we must place the fifth electron in one of the 2p orbitals. Because all three 2p orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1s22s22p1:
At carbon, with Z = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2p orbital that already has an electron, or should it go in one of the empty 2p orbitals? If it goes in an empty 2p orbital, will the sixth electron have its spin aligned with or be opposite to the spin of the fifth? In short, which of the following three orbital diagrams is correct for carbon, remembering that the 2p orbitals are degenerate?
Because of electron-electron repulsions, it is more favorable energetically for an electron to be in an unoccupied orbital than in one that is already occupied; hence we can eliminate choice a. Similarly, experiments have shown that choice b is slightly higher in energy (less stable) than choice c because electrons in degenerate orbitals prefer to line up with their spins parallel; thus, we can eliminate choice b. Choice c illustrates Hund’s ruleA rule stating that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. (named after the German physicist Friedrich H. Hund, 1896–1997), which today says that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. By Hund’s rule, the electron configuration of carbon, which is 1s22s22p2, is understood to correspond to the orbital diagram shown in c. Experimentally, it is found that the ground state of a neutral carbon atom does indeed contain two unpaired electrons.
When we get to nitrogen (Z = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is
with three unpaired electrons. The electron configuration of nitrogen is thus 1s22s22p3.
At oxygen, with Z = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2p orbitals, which gives us two unpaired electrons and a 1s22s22p4 electron configuration. Because all the 2p orbitals are degenerate, it doesn’t matter which one has the pair of electrons.
Similarly, fluorine has the electron configuration 1s22s22p5:
When we reach neon, with Z = 10, we have filled the 2p subshell, giving a 1s22s22p6 electron configuration:
Notice that for neon, as for helium, all the orbitals through the 2p level are completely filled. This fact is very important in dictating both the chemical reactivity and the bonding of helium and neon, as you will see.
As we continue through the periodic table in this way, writing the electron configurations of larger and larger atoms, it becomes tedious to keep copying the configurations of the filled inner subshells. In practice, chemists simplify the notation by using a bracketed noble gas symbol to represent the configuration of the noble gas from the preceding row because all the orbitals in a noble gas are filled. For example, [Ne] represents the 1s22s22p6 electron configuration of neon (Z = 10), so the electron configuration of sodium, with Z = 11, which is 1s22s22p63s1, is written as [Ne]3s1:
Neon | Z = 10 | 1s22s22p6 |
Sodium | Z = 11 | 1s22s22p63s1 = [Ne]3s1 |
Because electrons in filled inner orbitals are closer to the nucleus and more tightly bound to it, they are rarely involved in chemical reactions. This means that the chemistry of an atom depends mostly on the electrons in its outermost shell, which are called the valence electronsElectrons in the outermost shell of an atom.. The simplified notation allows us to see the valence-electron configuration more easily. Using this notation to compare the electron configurations of sodium and lithium, we have:
Sodium | 1s22s22p63s1 = [Ne]3s1 |
Lithium | 1s22s1 = [He]2s1 |
It is readily apparent that both sodium and lithium have one s electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case.
As we continue to build the eight elements of period 3, the 3s and 3p orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3s23p6, corresponding to a filled valence shell.
Draw an orbital diagram and use it to derive the electron configuration of phosphorus, Z = 15. What is its valence electron configuration?
Given: atomic number
Asked for: orbital diagram and valence electron configuration for phosphorus
Strategy:
A Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
B Referring to , draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
C Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.
Solution:
A Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus.
B The additional five electrons are placed in the next available orbitals, which tells us are the 3s and 3p orbitals:
Because the 3s orbital is lower in energy than the 3p orbitals, we fill it first:
Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3p orbitals separately but with their spins aligned:
The electron configuration is [Ne]3s23p3.
C We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3s23p3.
Exercise
Draw an orbital diagram and use it to derive the electron configuration of chlorine, Z = 17. What is its valence electron configuration?
Answer: [Ne]3s23p5; 3s23p5
The general order in which orbitals are filled is depicted in . Subshells corresponding to each value of n are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4s orbital is filled prior to the 3d orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4s1, and the configuration of calcium is [Ar]4s2. Five 3d orbitals are filled by the next 10 elements, the transition metals, followed by three 4p orbitals. Notice that the last member of this row is the noble gas krypton (Z = 36), [Ar]4s23d104p6 = [Kr], which has filled 4s, 3d, and 4p orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5s, 4d, and 5p orbitals are filled sequentially.
Figure 6.32 Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms
If you write the subshells for each value of the principal quantum number on successive lines, the observed order in which they are filled is indicated by a series of diagonal lines running from the upper right to the lower left.
The sixth row of the periodic table will be different from the preceding two because the 4f orbitals, which can hold 14 electrons, are filled between the 6s and the 5d orbitals. The elements that contain 4f orbitals in their valence shell are the lanthanides. When the 6p orbitals are finally filled, we have reached the next (and last known) noble gas, radon (Z = 86), [Xe]6s24f145d106p6 = [Rn]. In the last row, the 5f orbitals are filled between the 7s and the 6d orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.
Write the electron configuration of mercury (Z = 80), showing all the inner orbitals.
Given: atomic number
Asked for: complete electron configuration
Strategy:
Using the orbital diagram in and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.
Solution:
By placing the electrons in orbitals following the order shown in and using the periodic table as a guide, we obtain
1s2 | row 1 | 2 electrons |
2s22p6 | row 2 | 8 electrons |
3s23p6 | row 3 | 8 electrons |
4s23d104p6 | row 4 | 18 electrons |
5s24d105p6 | row 5 | 18 electrons |
row 1–5 | 54 electrons |
After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to , we need to fill the 6s (2 electrons), 4f (14 electrons), and 5d (10 electrons) orbitals. The result is mercury’s electron configuration:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 = Hg = [Xe]6s24f145d10with a filled 5d subshell, a 6s24f145d10 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.)
Exercise
Although element 114 is not stable enough to occur in nature, two isotopes of element 114 were created for the first time in a nuclear reactor in 1999 by a team of Russian and American scientists. Write the complete electron configuration for element 114.
Answer: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p2
The electron configurations of the elements are presented in , which lists the orbitals in the order in which they are filled. In several cases, the ground state electron configurations are different from those predicted by . Some of these anomalies occur as the 3d orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4s13d5 rather than the predicted [Ar]4s23d4. Similarly, the observed electron configuration of copper is [Ar]4s13d10 instead of [Ar]s23d9. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (ns1, np3, nd5, nf7) or filled (ns2, np6, nd10, nf14) subshell. Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. In heavier elements, other more complex effects can also be important, leading to some of the additional anomalies indicated in . For example, cerium has an electron configuration of [Xe]6s24f15d1, which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences.
Additional stability is associated with half-filled or filled subshells.
As you have learned, the electron configurations of the elements explain the otherwise peculiar shape of the periodic table. Although the table was originally organized on the basis of physical and chemical similarities between the elements within groups, these similarities are ultimately attributable to orbital energy levels and the Pauli principle, which cause the individual subshells to be filled in a particular order. As a result, the periodic table can be divided into “blocks” corresponding to the type of subshell that is being filled, as illustrated in . For example, the two columns on the left, known as the s blockThe elements in the left two columns of the periodic table in which the ns orbital is being filled., consist of elements in which the ns orbitals are being filled. The six columns on the right, elements in which the np orbitals are being filled, constitute the p blockThe elements in the six columns on the right of the periodic table in which the np orbitals are being filled.. In between are the 10 columns of the d blockThe elements in the periodic table in which the (n − 1)d orbitals are being filled., elements in which the (n − 1)d orbitals are filled. At the bottom lie the 14 columns of the f blockThe elements in the periodic table in which the (n − 2)f orbitals are being filled., elements in which the (n − 2)f orbitals are filled. Because two electrons can be accommodated per orbital, the number of columns in each block is the same as the maximum electron capacity of the subshell: 2 for ns, 6 for np, 10 for (n − 1)d, and 14 for (n − 2)f. Within each column, each element has the same valence electron configuration—for example, ns1 (group 1) or ns2np1 (group 13). As you will see, this is reflected in important similarities in the chemical reactivity and the bonding for the elements in each column.
Because each orbital can have a maximum of 2 electrons, there are 2 columns in the s block, 6 columns in the p block, 10 columns in the d block, and 14 columns in the f block.
Figure 6.33 The Periodic Table, Showing How the Elements Are Grouped According to the Kind of Subshell (s, p, d, f) Being Filled with Electrons in the Valence Shell of Each Element
Hydrogen and helium are placed somewhat arbitrarily. Although hydrogen is not an alkali metal, its 1s1 electron configuration suggests a similarity to lithium ([He]2s1) and the other elements in the first column. Although helium, with a filled ns subshell, should be similar chemically to other elements with an ns2 electron configuration, the closed principal shell dominates its chemistry, justifying its placement above neon on the right. In , we will examine how electron configurations affect the properties and reactivity of the elements.
Figure 6.34 Electron Configurations of the Elements
The electron configurations of elements indicated in red are exceptions due to the added stability associated with half-filled and filled subshells. The electron configurations of the elements indicated in blue are also anomalous, but the reasons for the observed configurations are more complex. For elements after No, the electron configurations are tentative.
Use the periodic table to predict the valence electron configuration of all the elements of group 2 (beryllium, magnesium, calcium, strontium, barium, and radium).
Given: series of elements
Asked for: valence electron configurations
Strategy:
A Identify the block in the periodic table to which the group 2 elements belong. Locate the nearest noble gas preceding each element and identify the principal quantum number of the valence shell of each element.
B Write the valence electron configuration of each element by first indicating the filled inner shells using the symbol for the nearest preceding noble gas and then listing the principal quantum number of its valence shell, its valence orbitals, and the number of valence electrons in each orbital as superscripts.
Solution:
A The group 2 elements are in the s block of the periodic table, and as group 2 elements, they all have two valence electrons. Beginning with beryllium, we see that its nearest preceding noble gas is helium and that the principal quantum number of its valence shell is n = 2.
B Thus beryllium has an [He]s2 electron configuration. The next element down, magnesium, is expected to have exactly the same arrangement of electrons in the n = 3 principal shell: [Ne]s2. By extrapolation, we expect all the group 2 elements to have an ns2 electron configuration.
Exercise
Use the periodic table to predict the characteristic valence electron configuration of the halogens in group 17.
Answer: All have an ns2np5 electron configuration, one electron short of a noble gas electron configuration. (Note that the heavier halogens also have filled (n − 1)d10 subshells, as well as an (n − 2)f14 subshell for Rn; these do not, however, affect their chemistry in any significant way.
In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. The concept of electron spin has important consequences for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin). Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the valence electrons. The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively.
A set of four quantum numbers specifies each wave function. What information is given by each quantum number? What does the specified wave function describe?
List two pieces of evidence to support the statement that electrons have a spin.
The periodic table is divided into blocks. Identify each block and explain the principle behind the divisions. Which quantum number distinguishes the horizontal rows?
Identify the element with each ground state electron configuration.
Identify the element with each ground state electron configuration.
Propose an explanation as to why the noble gases are inert.
How many magnetic quantum numbers are possible for a 4p subshell? A 3d subshell? How many orbitals are in these subshells?
How many magnetic quantum numbers are possible for a 6s subshell? A 4f subshell? How many orbitals does each subshell contain?
If l = 2 and ml = 2, give all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in the corresponding 3d subshell.
Give all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 4d subshell. How many electrons can the 4d orbital accommodate? How would this differ from a situation in which there were only three quantum numbers (n, l, m)?
Given the following sets of quantum numbers (n, l, ml, ms), identify each principal shell and subshell.
Is each set of quantum numbers allowed? Explain your answers.
List the set of quantum numbers for each electron in the valence shell of each element.
List the set of quantum numbers for each electron in the valence shell of each element.
Sketch the shape of the periodic table if there were three possible values of ms for each electron (+½, −½, and 0); assume that the Pauli principle is still valid.
Predict the shape of the periodic table if eight electrons could occupy the p subshell.
If the electron could only have spin +½, what would the periodic table look like?
If three electrons could occupy each s orbital, what would be the electron configuration of each species?
If Hund’s rule were not followed and maximum pairing occurred, how many unpaired electrons would each species have? How do these numbers compare with the number found using Hund’s rule?
Write the electron configuration for each element in the ground state.
Write the electron configuration for each element in the ground state.
Give the complete electron configuration for each element.
Give the complete electron configuration for each element.
Write the valence electron configuration for each element:
Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element.
Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element.
How many unpaired electrons does each species contain?
How many unpaired electrons does each species contain?
For each element, give the complete electron configuration, draw the valence electron configuration, and give the number of unpaired electrons present.
Use an orbital diagram to illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule for each element.
For a 4p subshell, n = 4 and l = 1. The allowed values of the magnetic quantum number, ml, are therefore +1, 0, −1, corresponding to three 4p orbitals. For a 3d subshell, n = 3 and l = 2. The allowed values of the magnetic quantum number, ml, are therefore +2, +1, 0, −1, −2, corresponding to five 3d orbitals.
n = 3, l = 2, ml = 2, ms = ;n = 3, l = 2, ml = 2, ms =
The lamps in street lights use emission of light from excited states of atoms to produce a characteristic glow. Light is generated by electron bombardment of a metal vapor. Of calcium and strontium, which metal vapor would you use to produce yellow light? Which metal would you use to produce red light? Calculate the energy associated with each transition and propose an explanation for the colors of the emitted light.
Lasers have useful medical applications because their light is directional (permitting tight focus of the laser beam for precise cutting), monochromatic, and intense. Carbon dioxide lasers, emitting at a wavelength of 1.06 × 104 nm, are typically used in surgery.
An excimer (meaning “excited dimer”) laser emits light in the ultraviolet region of the spectrum. An example of such a laser is krypton fluoride (KrF), which emits light at a wavelength of 248 nm. What is the energy in joules of a mole of photons emitted from this laser? How much more energetic is a single photon of this wavelength than a photon from a carbon dioxide laser used in surgery (10,600 nm)?
Wavelengths less than 10 nm are needed to “see” objects on an atomic or molecular scale. Such imaging can be accomplished with an electron microscope, which uses electric and magnetic fields to focus and accelerate a beam of electrons to a high velocity. Electron microscopy is now a powerful tool in chemical research. What electron velocity is needed to produce electrons with a wavelength of 4 × 10−3 nm, which is sufficient to produce an image of an atom? If electromagnetic radiation were used, what region of the electromagnetic spectrum would this correspond to?
Microwave ovens operate by emitting microwave radiation, which is primarily absorbed by water molecules in food. The absorbed radiation is converted to heat through rapid oscillations of polar water molecules, which cooks the food and warms beverages. If 7.2 × 1028 photons are needed to heat 150.0 g of water from 20.0°C to 100.0°C in a microwave oven, what is the frequency of the microwaves? Metal objects should not be placed in a microwave oven because they cause sparks. Why does this cause sparks?
The magnitude of the energy gap between an excited state and a ground state determines the color of visible light that is absorbed. The observed color of an object is not the color of the light it absorbs but rather the complement of that color. The accompanying rosette, first developed by Isaac Newton, shows the colors increasing in energy from red to violet. Any two colors that are opposite each other are said to be complementary (e.g., red and green are complementary).
Given the absorption spectra and following table, what are the colors of the objects that produce spectra A, B, and C?
Wavelength (nm) | Color of Light |
---|---|
390–453 | violet |
453–492 | blue |
492–577 | green |
577–597 | yellow |
597–622 | orange |
622–780 | red |
Photodegradation of atmospheric ozone occurs via the reaction O3 + hν → O2 + O; the maximum absorption occurs at approximately 255 nm. In what region of the electromagnetic spectrum does this occur? Based on this information, what would be the effect of depleting the ozone layer of Earth’s atmosphere?
A microscope’s resolution (its ability to distinguish between two points separated by a given distance) depends on the wavelength of light used to illuminate an object. The resolution R is given by the equation R = λ/2N, where N is a constant related to the aperture. If a microscope has an aperture constant of 0.25, what is the smallest distance between two objects that can be resolved using the following light sources?
Silver bromide is the photosensitive material in 35 mm photographic film. When monochromatic light falls on film, the photons are recorded if they contain sufficient energy to react with silver bromide in the film. Given that the minimum energy needed to do this is approximately 57.9 kJ/mol, explain why red light is used to light a darkroom. What happens when the door to the darkroom is opened, allowing yellow light to enter?
A lighting system has recently been developed that uses a quartz bulb the size of a golf ball filled with an inert gas and a small amount of sulfur. When irradiated by microwaves, the bulb puts out as much light as hundreds of high-intensity mercury vapor lamps. Because 1000 kJ/mol is needed to ionize sulfur, can this process occur simply by irradiating sulfur atoms with microwaves? Explain your answer.
The following table lists the ionization energies of some common atmospheric species:
Species | Ionization Energy (kJ/mol) |
---|---|
NO | 897 |
CO2 | 1330 |
O2 | 1170 |
An artist used a pigment that has a significant absorption peak at 450 nm, with a trace absorption at 530 nm. Based on the color chart and table in Problem 6, what was the color of the paint? Draw the absorption pattern. What would the absorption spectrum have looked like if the artist had wanted green? Using absorption spectra, explain why an equal combination of red and yellow paints produces orange.
You live in a universe where an electron has four different spins (ms = +½, +¼, −½, −¼) and the periodic table has only 36 elements. Which elements would be noble gases? What would the periodic table look like? (Assume that the Pauli exclusion principle is still valid.)
If you were living on a planet where there were three quantum numbers (n, l, m) instead of four, what would be the allowed combinations for an electron in a 3p orbital? How many electrons would this orbital contain assuming the Pauli exclusion principle were still in effect? How does this compare with the actual number of allowed combinations found on Earth?
X-rays are frequently used to project images of the human body. Recently, however, a superior technique called magnetic resonance imaging (MRI) has been developed that uses proton spin to image tissues in spectacular detail. In MRI, spinning hydrogen nuclei in an organic material are irradiated with photons that contain enough energy to flip the protons to the opposite orientation. If 33.121 kJ/mol of energy is needed to flip a proton, what is the resonance frequency required to produce an MRI spectrum? Suggest why this frequency of electromagnetic radiation would be preferred over x-rays.
Vanadium has been found to be a key component in a biological catalyst that reduces nitrogen to ammonia. What is the valence electron configuration of vanadium? What are the quantum numbers for each valence electron? How many unpaired electrons does vanadium have?
Tellurium, a metal used in semiconductor devices, is also used as a coloring agent in porcelains and enamels. Illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule using tellurium metal.
A new element is believed to have been discovered by a team of Russian and American scientists, although its existence is yet to be independently confirmed. Six atoms of element 117, temporarily named ununseptium, were created by smashing together isotopes of calcium with the element berkelium. Give the following:
In Chapter 6 "The Structure of Atoms", we presented the contemporary quantum mechanical model of the atom. In using this model to describe the electronic structures of the elements in order of increasing atomic number, we saw that periodic similarities in electron configuration correlate with periodic similarities in properties, which is the basis for the structure of the periodic table. For example, the noble gases have what is often called filled or closed-shell valence electron configurations. These closed shells are actually filled s and p subshells with a total of eight electrons, which are called octets; helium is an exception, with a closed 1s shell that has only two electrons. Because of their filled valence shells, the noble gases are generally unreactive. In contrast, the alkali metals have a single valence electron outside a closed shell and readily lose this electron to elements that require electrons to achieve an octet, such as the halogens. Thus because of their periodic similarities in electron configuration, atoms in the same column of the periodic table tend to form compounds with the same oxidation states and stoichiometries. Chapter 6 "The Structure of Atoms" ended with the observation that, because all the elements in a column have the same valence electron configuration, the periodic table can be used to find the electron configuration of most of the elements at a glance.
Crookes’s Spiral Periodic Table, 1888. Created by Sir William Crookes (1832–1919), the spiral represents the relationships between the elements and the order of evolution of the elements from what he believed to be primal matter.
In this chapter, we explore the relationship between the electron configurations of the elements, as reflected in their arrangement in the periodic table, and their physical and chemical properties. In particular, we focus on the similarities between elements in the same column and on the trends in properties that are observed across horizontal rows or down vertical columns. By the end of this chapter, your understanding of these trends and relationships will provide you with clues as to why argon is used in incandescent light bulbs, why coal and wood burst into flames when they come in contact with pure F2, why aluminum was discovered so late despite being the third most abundant element in Earth’s crust, and why lithium is commonly used in batteries. We begin by expanding on the brief discussion of the history of the periodic table presented in Chapter 1 "Introduction to Chemistry" and describing how it was created many years before electrons had even been discovered, much less discussed in terms of shells, subshells, orbitals, and electron spin.
The modern periodic table has evolved through a long history of attempts by chemists to arrange the elements according to their properties as an aid in predicting chemical behavior. One of the first to suggest such an arrangement was the German chemist Johannes Dobereiner (1780–1849), who noticed that many of the known elements could be grouped in triadsA set of three elements that have similar properties., sets of three elements that have similar properties—for example, chlorine, bromine, and iodine; or copper, silver, and gold. Dobereiner proposed that all elements could be grouped in such triads, but subsequent attempts to expand his concept were unsuccessful. We now know that portions of the periodic table—the d block in particular—contain triads of elements with substantial similarities. The middle three members of most of the other columns, such as sulfur, selenium, and tellurium in group 16 or aluminum, gallium, and indium in group 13, also have remarkably similar chemistry.
By the mid-19th century, the atomic masses of many of the elements had been determined. The English chemist John Newlands (1838–1898), hypothesizing that the chemistry of the elements might be related to their masses, arranged the known elements in order of increasing atomic mass and discovered that every seventh element had similar properties (). (The noble gases were still unknown.) Newlands therefore suggested that the elements could be classified into octavesA group of seven elements, corresponding to the horizontal rows in the main group elements (not counting the noble gases, which were unknown at the time)., corresponding to the horizontal rows in the main group elements. Unfortunately, Newlands’s “law of octaves” did not seem to work for elements heavier than calcium, and his idea was publicly ridiculed. At one scientific meeting, Newlands was asked why he didn’t arrange the elements in alphabetical order instead of by atomic mass, since that would make just as much sense! Actually, Newlands was on the right track—with only a few exceptions, atomic mass does increase with atomic number, and similar properties occur every time a set of ns2np6 subshells is filled. Despite the fact that Newlands’s table had no logical place for the d-block elements, he was honored for his idea by the Royal Society of London in 1887.
Newlands noticed that elemental properties repeated every seventh (or multiple of seven) element, as musical notes repeat every eighth note.
Figure 7.1 The Arrangement of the Elements into Octaves as Proposed by Newlands
The table shown here accompanied a letter from a 27-year-old Newlands to the editor of the journal Chemical News in which he wrote: “If the elements are arranged in the order of their equivalents, with a few slight transpositions, as in the accompanying table, it will be observed that elements belonging to the same group usually appear on the same horizontal line. It will also be seen that the numbers of analogous elements generally differ either by 7 or by some multiple of seven; in other words, members of the same group stand to each other in the same relation as the extremities of one or more octaves in music. Thus, in the nitrogen group, between nitrogen and phosphorus there are 7 elements; between phosphorus and arsenic, 14; between arsenic and antimony, 14; and lastly, between antimony and bismuth, 14 also. This peculiar relationship I propose to provisionally term the Law of Octaves. I am, &c. John A. R. Newlands, F.C.S. Laboratory, 19, Great St. Helen’s, E.C., August 8, 1865.”
The periodic table achieved its modern form through the work of the German chemist Julius Lothar Meyer (1830–1895) and the Russian chemist Dimitri Mendeleev (1834–1907), both of whom focused on the relationships between atomic mass and various physical and chemical properties. In 1869, they independently proposed essentially identical arrangements of the elements. Meyer aligned the elements in his table according to periodic variations in simple atomic properties, such as “atomic volume” (), which he obtained by dividing the atomic mass (molar mass) in grams per mole by the density of the element in grams per cubic centimeter. This property is equivalent to what is today defined as molar volumeThe molar mass of an element divided by its density. (measured in cubic centimeters per mole):
Equation 7.1
As shown in , the alkali metals have the highest molar volumes of the solid elements. In Meyer’s plot of atomic volume versus atomic mass, the nonmetals occur on the rising portion of the graph, and metals occur at the peaks, in the valleys, and on the downslopes.
When his family’s glass factory was destroyed by fire, Mendeleev moved to St. Petersburg, Russia, to study science. He became ill and was not expected to recover, but he finished his PhD with the help of his professors and fellow students. In addition to the periodic table, another of Mendeleev’s contributions to science was an outstanding textbook, The Principles of Chemistry, which was used for many years.
Figure 7.2 Variation of Atomic Volume with Atomic Number, Adapted from Meyer’s Plot of 1870
Note the periodic increase and decrease in atomic volume. Because the noble gases had not yet been discovered at the time this graph was formulated, the peaks correspond to the alkali metals (group 1).
Mendeleev, who first published his periodic table in 1869 (), is usually credited with the origin of the modern periodic table. The key difference between his arrangement of the elements and that of Meyer and others is that Mendeleev did not assume that all the elements had been discovered (actually, only about two-thirds of the naturally occurring elements were known at the time). Instead, he deliberately left blanks in his table at atomic masses 44, 68, 72, and 100, in the expectation that elements with those atomic masses would be discovered. Those blanks correspond to the elements we now know as scandium, gallium, germanium, and technetium.
Figure 7.3 Mendeleev’s Periodic Table, as Published in the German Journal Annalen der Chemie und Pharmacie in 1872
The column headings “Reihen” and “Gruppe” are German for “row” and “group.” Formulas indicate the type of compounds formed by each group, with “R” standing for “any element” and superscripts used where we now use subscripts. Atomic masses are shown after equal signs and increase across each row from left to right.
The most convincing evidence in support of Mendeleev’s arrangement of the elements was the discovery of two previously unknown elements whose properties closely corresponded with his predictions (). Two of the blanks Mendeleev had left in his original table were below aluminum and silicon, awaiting the discovery of two as-yet-unknown elements, eka-aluminum and eka-silicon (from the Sanskrit eka, meaning “one,” as in “one beyond aluminum”). The observed properties of gallium and germanium matched those of eka-aluminum and eka-silicon so well that once they were discovered, Mendeleev’s periodic table rapidly gained acceptance.
Table 7.1 Comparison of the Properties Predicted by Mendeleev in 1869 for eka-Aluminum and eka-Silicon with the Properties of Gallium (Discovered in 1875) and Germanium (Discovered in 1886)
Property | eka-Aluminum (predicted) | Gallium (observed) | eka-Silicon (predicted) | Germanium (observed) |
---|---|---|---|---|
atomic mass | 68 | 69.723 | 72 | 72.64 |
element | metal | metal | dirty-gray metal | gray-white metal |
low mp* | mp = 29.8°C | high mp | mp = 938°C | |
d = 5.9 g/cm3 | d = 5.91 g/cm3 | d = 5.5 g/cm3 | d = 5.323 g/cm3 | |
oxide | E2O3 | Ga2O3 | EO2 | GeO2 |
d = 5.5 g/cm3 | d = 6.0 g/cm3 | d = 4.7 g/cm3 | d = 4.25 g/cm3 | |
chloride | ECl3 | GaCl3 | ECl4 | GeCl4 |
volatile |
mp = 78°C bp* = 201°C |
bp < 100°C | bp = 87°C | |
*mp = melting point; bp = boiling point. |
When the chemical properties of an element suggested that it might have been assigned the wrong place in earlier tables, Mendeleev carefully reexamined its atomic mass. He discovered, for example, that the atomic masses previously reported for beryllium, indium, and uranium were incorrect. The atomic mass of indium had originally been reported as 75.6, based on an assumed stoichiometry of InO for its oxide. If this atomic mass were correct, then indium would have to be placed in the middle of the nonmetals, between arsenic (atomic mass 75) and selenium (atomic mass 78). Because elemental indium is a silvery-white metal, however, Mendeleev postulated that the stoichiometry of its oxide was really In2O3 rather than InO. This would mean that indium’s atomic mass was actually 113, placing the element between two other metals, cadmium and tin.
One group of elements that is absent from Mendeleev’s table is the noble gases, all of which were discovered more than 20 years later, between 1894 and 1898, by Sir William Ramsay (1852–1916; Nobel Prize in Chemistry 1904). Initially, Ramsay did not know where to place these elements in the periodic table. Argon, the first to be discovered, had an atomic mass of 40. This was greater than chlorine’s and comparable to that of potassium, so Ramsay, using the same kind of reasoning as Mendeleev, decided to place the noble gases between the halogens and the alkali metals.
Despite its usefulness, Mendeleev’s periodic table was based entirely on empirical observation supported by very little understanding. It was not until 1913, when a young British physicist, H. G. J. Moseley (1887–1915), while analyzing the frequencies of x-rays emitted by the elements, discovered that the underlying foundation of the order of the elements was by the atomic number, not the atomic mass. Moseley hypothesized that the placement of each element in his series corresponded to its atomic number Z, which is the number of positive charges (protons) in its nucleus. Argon, for example, although having an atomic mass greater than that of potassium (39.9 amu versus 39.1 amu, respectively), was placed before potassium in the periodic table. While analyzing the frequencies of the emitted x-rays, Moseley noticed that the atomic number of argon is 18, whereas that of potassium is 19, which indicated that they were indeed placed correctly. Moseley also noticed three gaps in his table of x-ray frequencies, so he predicted the existence of three unknown elements: technetium (Z = 43), discovered in 1937; promethium (Z = 61), discovered in 1945; and rhenium (Z = 75), discovered in 1925.
Moseley left his research work at the University of Oxford to join the British army as a telecommunications officer during World War I. He was killed during the Battle of Gallipoli in Turkey.
Before its discovery in 1999, some theoreticians believed that an element with a Z of 114 existed in nature. Use Mendeleev’s reasoning to name element 114 as eka-______; then identify the known element whose chemistry you predict would be most similar to that of element 114.
Given: atomic number
Asked for: name using prefix eka-
Strategy:
A Using the periodic table (see ), locate the n = 7 row. Identify the location of the unknown element with Z = 114; then identify the known element that is directly above this location.
B Name the unknown element by using the prefix eka- before the name of the known element.
Solution:
A The n = 7 row can be filled in by assuming the existence of elements with atomic numbers greater than 112, which is underneath mercury (Hg). Counting three boxes to the right gives element 114, which lies directly below lead (Pb). B If Mendeleev were alive today, he would call element 114 eka-lead.
Exercise
Use Mendeleev’s reasoning to name element 112 as eka-______; then identify the known element whose chemistry you predict would be most similar to that of element 112.
Answer: eka-mercury
The periodic table arranges the elements according to their electron configurations, such that elements in the same column have the same valence electron configurations. Periodic variations in size and chemical properties are important factors in dictating the types of chemical reactions the elements undergo and the kinds of chemical compounds they form. The modern periodic table was based on empirical correlations of properties such as atomic mass; early models using limited data noted the existence of triads and octaves of elements with similar properties. The periodic table achieved its current form through the work of Dimitri Mendeleev and Julius Lothar Meyer, who both focused on the relationship between atomic mass and chemical properties. Meyer arranged the elements by their atomic volume, which today is equivalent to the molar volume, defined as molar mass divided by molar density. The correlation with the electronic structure of atoms was made when H. G. J. Moseley showed that the periodic arrangement of the elements was determined by atomic number, not atomic mass.
Johannes Dobereiner is credited with developing the concept of chemical triads. Which of the group 15 elements would you expect to compose a triad? Would you expect B, Al, and Ga to act as a triad? Justify your answers.
Despite the fact that Dobereiner, Newlands, Meyer, and Mendeleev all contributed to the development of the modern periodic table, Mendeleev is credited with its origin. Why was Mendeleev’s periodic table accepted so rapidly?
How did Moseley’s contribution to the development of the periodic table explain the location of the noble gases?
The eka- naming scheme devised by Mendeleev was used to describe undiscovered elements.
Based on the data given, complete the table.
Species | Molar Mass (g/mol) | Density (g/cm3) | Molar Volume (cm3/mol) |
---|---|---|---|
A | 40.078 | 25.85 | |
B | 39.09 | 0.856 | |
C | 32.065 | 16.35 | |
D | 1.823 | 16.98 | |
E | 26.98 | 9.992 | |
F | 22.98 | 0.968 |
Plot molar volume versus molar mass for these substances. According to Meyer, which would be considered metals and which would be considered nonmetals?
Species | Molar Mass (g/mol) | Density (g/cm3) | Molar Volume (cm3/mol) |
---|---|---|---|
A | 40.078 | 1.550 | 25.85 |
B | 39.09 | 0.856 | 45.67 |
C | 32.065 | 1.961 | 16.35 |
D | 30.95 | 1.823 | 16.98 |
E | 26.98 | 2.700 | 9.992 |
F | 22.98 | 0.968 | 23.7 |
Meyer found that the alkali metals had the highest molar volumes, and that molar volumes decreased steadily with increasing atomic mass, then leveled off, and finally rose again. The elements located on the rising portion of a plot of molar volume versus molar mass were typically nonmetals. If we look at the plot of the data in the table, we can immediately identify those elements with the largest molar volumes (A, B, F) as metals located on the left side of the periodic table. The element with the smallest molar volume (E) is aluminum. The plot shows that the subsequent elements (C, D) have molar volumes that are larger than that of E, but smaller than those of A and B. Thus, C and D are most likely to be nonmetals (which is the case: C = sulfur, D = phosphorus).
Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes. In this section, we discuss how atomic and ion “sizes” are defined and obtained.
Recall from Chapter 6 "The Structure of Atoms" that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure 7.4 "Plots of Radial Probability as a Function of Distance from the Nucleus for He, Ne, and Ar", which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom.
Figure 7.4 Plots of Radial Probability as a Function of Distance from the Nucleus for He, Ne, and Ar
In He, the 1s electrons have a maximum radial probability at ≈30 pm from the nucleus. In Ne, the 1s electrons have a maximum at ≈8 pm, and the 2s and 2p electrons combine to form another maximum at ≈35 pm (the n = 2 shell). In Ar, the 1s electrons have a maximum at ≈2 pm, the 2s and 2p electrons combine to form a maximum at ≈18 pm, and the 3s and 3p electrons combine to form a maximum at ≈70 pm.
Figure 7.4 "Plots of Radial Probability as a Function of Distance from the Nucleus for He, Ne, and Ar" also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell (n = 1), it shows only a single peak. In contrast, neon, with filled n = 1 and 2 principal shells, has two peaks. Argon, with filled n = 1, 2, and 3 principal shells, has three peaks. The peak for the filled n = 1 shell occurs at successively shorter distances for neon (Z = 10) and argon (Z = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1s2 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of n. Consequently, the two electrons in the n = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the n = 1 shell also decreases tremendously (the filled 1s orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled n = 2 shell in argon is located closer to the nucleus and has a lower energy than the n = 2 shell in neon.
Figure 7.4 "Plots of Radial Probability as a Function of Distance from the Nucleus for He, Ne, and Ar" illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl2 molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a covalent atomic radius (rcov)Half the distance between the nuclei of two like atoms joined by a covalent bond in the same molecule. of 99 pm or 0.99 Å (part (a) in Figure 7.5 "Definitions of the Atomic Radius").Atomic radii are often measured in angstroms (Å), a non-SI unit: 1 Å = 1 × 10−10 m = 100 pm.
Figure 7.5 Definitions of the Atomic Radius
(a) The covalent atomic radius, rcov, is half the distance between the nuclei of two like atoms joined by a covalent bond in the same molecule, such as Cl2. (b) The metallic atomic radius, rmet, is half the distance between the nuclei of two adjacent atoms in a pure solid metal, such as aluminum. (c) The van der Waals atomic radius, rvdW, is half the distance between the nuclei of two like atoms, such as argon, that are closely packed but not bonded. (d) This is a depiction of covalent versus van der Waals radii of chlorine.
In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds.A similar approach for measuring the size of ions is discussed later in this section.
Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the metallic atomic radius(rmet)Half the distance between the nuclei of two adjacent metal atoms. is defined as half the distance between the nuclei of two adjacent metal atoms (part (b) in Figure 7.5 "Definitions of the Atomic Radius"). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the van der Waals atomic radius(rvdW)Half the internuclear distance between two nonbonded atoms in the solid., which is half the internuclear distance between two nonbonded atoms in the solid (part (c) in Figure 7.5 "Definitions of the Atomic Radius"). An atom such as chlorine has both a covalent radius (the distance between the two atoms in a Cl2 molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, Cl2(s) at low temperatures). These radii are generally not the same (part (d) in Figure 7.5 "Definitions of the Atomic Radius").
Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions described in Chapter 6 "The Structure of Atoms". Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure 7.6 "A Plot of Periodic Variation of Atomic Radius with Atomic Number for the First Six Rows of the Periodic Table"). In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure 7.7 "Calculated Atomic Radii (in Picometers) of the ").
Figure 7.6 A Plot of Periodic Variation of Atomic Radius with Atomic Number for the First Six Rows of the Periodic Table
There is a similarity to the plot of atomic volume versus atomic number (Figure 7.2 "Variation of Atomic Volume with Atomic Number, Adapted from Meyer’s Plot of 1870")—a variation of Meyer’s early plot.
Figure 7.7 Calculated Atomic Radii (in Picometers) of the s-, p-, and d-Block Elements
The sizes of the circles illustrate the relative sizes of the atoms. The calculated values are based on quantum mechanical wave functions.
Source: http://www.webelements.com.
Atomic radii decrease from left to right across a row and increase from top to bottom down a column.
Trends in atomic size result from differences in the effective nuclear charges (Zeff) experienced by electrons in the outermost orbitals of the elements. As we described in Chapter 6 "The Structure of Atoms", for all elements except H, the effective nuclear charge is always less than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius.
The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. (For more information on electron shielding, see Chapter 6 "The Structure of Atoms", Section 6.5 "Atomic Orbitals and Their Energies", and Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom".) All have a filled 1s2 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2s and 2p orbitals, electrons in the same principal shell are not very effective at shielding one another from the nuclear charge. Thus the single 2s electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1s2 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of Zeff = +1.26 for Li.) In contrast, the two 2s electrons in beryllium do not shield each other very well, although the filled 1s2 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2s electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2s and 2p orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size.
The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number n is not constant. As we saw in Chapter 6 "The Structure of Atoms", the size of the orbitals increases as n increases, provided the nuclear charge remains the same. In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to ns orbitals with increasing values of n. However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number n increases from 2 to 6, but the nuclear charge increases from +3 to +55! If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6s1 valence electron configuration, is much larger than lithium, with a 2s1 valence electron configuration. The effective nuclear charge changes relatively little from lithium to cesium because electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1s22s22p63s23p64s23d104p65s24d105p6 shells, abbreviated as [Xe]5s24d105p6, which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge.
Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge.
On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon.
Given: three elements
Asked for: arrange in order of increasing atomic radius
Strategy:
A Identify the location of the elements in the periodic table. Determine the relative sizes of elements located in the same column from their principal quantum number n. Then determine the order of elements in the same row from their effective nuclear charges. If the elements are not in the same column or row, use pairwise comparisons.
B List the elements in order of increasing atomic radius.
Solution:
A These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater. B Combining the two inequalities gives the overall order: C < Si < Al.
Exercise
On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur.
Answer: O < S < P < K
As you learned in Chapter 2 "Molecules, Ions, and Chemical Formulas", ionic compounds consist of regular repeating arrays of alternating cations and anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it is possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radiusThe radius of a cation or anion. of one or both. As illustrated in Figure 7.8 "Definition of Ionic Radius", the internuclear distance corresponds to the sum of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na+ ion is essentially the same in NaCl and Na2S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed.
Figure 7.8 Definition of Ionic Radius
(a) The internuclear distance is apportioned between adjacent cations and anions in the ionic structure, as shown here for Na+ and Cl− in sodium chloride. (b) This depiction of electron density contours for a single plane of atoms in the NaCl structure shows how the lines connect points of equal electron density. Note the relative sizes of the electron density contour lines around Cl− and Na+.
A comparison of ionic radii with atomic radii (Figure 7.9 "Ionic Radii (in Picometers) of the Most Common Oxidation States of the ") shows that a cation is always smaller than its parent neutral atom, and an anion is always larger than the parent neutral atom. When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases (compare Li at 167 pm with Li+ at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe2+ at 78 pm with Fe3+ at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases (compare F at 42 pm with F− at 133 pm).
Figure 7.9 Ionic Radii (in Picometers) of the Most Common Oxidation States of the s-, p-, and d-Block Elements
Gray circles indicate the sizes of the ions shown; colored circles indicate the sizes of the neutral atoms, previously shown in Figure 7.7 "Calculated Atomic Radii (in Picometers) of the ".
Source: Ionic radius data from R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767.
Cations are always smaller than the neutral atom, and anions are always larger.
Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na− ion, allowing comparison of its size with that of the far more familiar Na+ ion, which is found in many compounds. The radius of sodium in each of its three known oxidation states is given in Table 7.2 "Experimentally Measured Values for the Radius of Sodium in Its Three Known Oxidation States". All three species have a nuclear charge of +11, but they contain 10 (Na+), 11 (Na0), and 12 (Na−) electrons. The Na+ ion is significantly smaller than the neutral Na atom because the 3s1 electron has been removed to give a closed shell with n = 2. The Na− ion is larger than the parent Na atom because the additional electron produces a 3s2 valence electron configuration, while the nuclear charge remains the same.
Table 7.2 Experimentally Measured Values for the Radius of Sodium in Its Three Known Oxidation States
Na+ | Na0 | Na− | |
---|---|---|---|
Electron Configuration | 1s22s22p6 | 1s22s22p63s1 | 1s22s22p63s2 |
Radius (pm) | 102 | 154* | 202† |
*The metallic radius measured for Na(s). | |||
†Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253. |
Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of n lie at successively greater distances from the nucleus.
Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an isoelectronic seriesA group of ions or atoms and ions that have the same number of electrons and thus the same ground-state electron configuration.. For example, the isoelectronic series of species with the neon closed-shell configuration (1s22s22p6) is shown in Table 7.3 "Radius of Ions with the Neon Closed-Shell Electron Configuration". The sizes of the ions in this series decrease smoothly from N3− to Al3+. All six of the ions contain 10 electrons in the 1s, 2s, and 2p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al3+) is the smallest, and the ion with the smallest nuclear charge (N3−) is the largest. One member of this isoelectronic series is not listed in Table 7.3 "Radius of Ions with the Neon Closed-Shell Electron Configuration": the neon atom. Because neon forms no covalent or ionic compounds, its radius is difficult to measure.
Table 7.3 Radius of Ions with the Neon Closed-Shell Electron Configuration
Ion | Radius (pm) | Atomic Number |
---|---|---|
N3− | 146 | 7 |
O2− | 140 | 8 |
F− | 133 | 9 |
Na+ | 102 | 11 |
Mg2+ | 72 | 12 |
Al3+ | 53.5 | 13 |
Source: R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767.
Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl−, K+, S2−, and Se2−.
Given: four ions
Asked for: order by increasing radius
Strategy:
A Determine which ions form an isoelectronic series. Of those ions, predict their relative sizes based on their nuclear charges. For ions that do not form an isoelectronic series, locate their positions in the periodic table.
B Determine the relative sizes of the ions based on their principal quantum numbers n and their locations within a row.
Solution:
A We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K+, Cl−, and S2− form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K+ has the greatest nuclear charge (Z = 19), its radius is smallest, and S2− with Z = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se2− ion to be even larger than S2−. B The order must therefore be K+ < Cl− < S2− < Se2−.
Exercise
Based on their positions in the periodic table, arrange these ions in order of increasing size: Br−, Ca2+, Rb+, and Sr2+.
Answer: Ca2+ < Sr2+ < Rb+ < Br−
A variety of methods have been established to measure the size of a single atom or ion. The covalent atomic radius (rcov) is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the metallic atomic radius (rmet) is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The van der Waals radius (rvdW) of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The ionic radii of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series, shows a clear correlation between increasing nuclear charge and decreasing size.
The electrons of the 1s shell have a stronger electrostatic attraction to the nucleus than electrons in the 2s shell. Give two reasons for this.
Predict whether Na or Cl has the more stable 1s2 shell and explain your rationale.
Arrange K, F, Ba, Pb, B, and I in order of decreasing atomic radius.
Arrange Ag, Pt, Mg, C, Cu, and Si in order of increasing atomic radius.
Using the periodic table, arrange Li, Ga, Ba, Cl, and Ni in order of increasing atomic radius.
Element M is a metal that forms compounds of the type MX2, MX3, and MX4, where X is a halogen. What is the expected trend in the ionic radius of M in these compounds? Arrange these compounds in order of decreasing ionic radius of M.
The atomic radii of Na and Cl are 190 and 79 pm, respectively, but the distance between sodium and chlorine in NaCl is 282 pm. Explain this discrepancy.
Are shielding effects on the atomic radius more pronounced across a row or down a group? Why?
What two factors influence the size of an ion relative to the size of its parent atom? Would you expect the ionic radius of S2− to be the same in both MgS and Na2S? Why or why not?
Arrange Br−, Al3+, Sr2+, F−, O2−, and I− in order of increasing ionic radius.
Arrange P3−, N3−, Cl−, In3+, and S2− in order of decreasing ionic radius.
How is an isoelectronic series different from a series of ions with the same charge? Do the cations in magnesium, strontium, and potassium sulfate form an isoelectronic series? Why or why not?
What isoelectronic series arises from fluorine, nitrogen, magnesium, and carbon? Arrange the ions in this series by
What would be the charge and electron configuration of an ion formed from calcium that is isoelectronic with
The 1s shell is closer to the nucleus and therefore experiences a greater electrostatic attraction. In addition, the electrons in the 2s subshell are shielded by the filled 1s2 shell, which further decreases the electrostatic attraction to the nucleus.
Ba > K > Pb > I > B > F
The sum of the calculated atomic radii of sodium and chlorine atoms is 253 pm. The sodium cation is significantly smaller than a neutral sodium atom (102 versus 154 pm), due to the loss of the single electron in the 3s orbital. Conversely, the chloride ion is much larger than a neutral chlorine atom (181 versus 99 pm), because the added electron results in greatly increased electron–electron repulsions within the filled n = 3 principal shell. Thus, transferring an electron from sodium to chlorine decreases the radius of sodium by about 50%, but causes the radius of chlorine to almost double. The net effect is that the distance between a sodium ion and a chloride ion in NaCl is greater than the sum of the atomic radii of the neutral atoms.
Plot the ionic charge versus ionic radius using the following data for Mo: Mo3+, 69 pm; Mo4+, 65 pm; and Mo5+, 61 pm. Then use this plot to predict the ionic radius of Mo6+. Is the observed trend consistent with the general trends discussed in the chapter? Why or why not?
Internuclear distances for selected ionic compounds are given in the following table.
If the ionic radius of Li+ is 76 pm, what is the ionic radius of each of the anions?
LiF | LiCl | LiBr | LiI | |
---|---|---|---|---|
Distance (pm) | 209 | 257 | 272 | 296 |
What is the ionic radius of Na+?
NaF | NaCl | NaBr | NaI | |
---|---|---|---|---|
Distance (pm) | 235 | 282 | 298 | 322 |
Arrange the gaseous species Mg2+, P3−, Br−, S2−, F−, and N3− in order of increasing radius and justify your decisions.
We have seen that when elements react, they often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation.
Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy(I)The minimum amount of energy needed to remove an electron from the gaseous atom in its ground state: of an element as the amount of energy needed to remove an electron from the gaseous atom E in its ground state. I is therefore the energy required for the reaction
Equation 7.2
Because an input of energy is required, the ionization energy is always positive (I > 0) for the reaction as written in . Larger values of I mean that the electron is more tightly bound to the atom and harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV):
1 eV/atom = 96.49 kJ/molIf an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define a first ionization energy (I1), a second ionization energy (I2), and in general an nth ionization energy (In) according to the following reactions:
Equation 7.3
Equation 7.4
Equation 7.5
Values for the ionization energies of Li and Be listed in show that successive ionization energies for an element increase steadily; that is, it takes more energy to remove the second electron from an atom than the first, and so forth. There are two reasons for this trend. First, the second electron is being removed from a positively charged species rather than a neutral one, so in accordance with Coulomb’s law, more energy is required. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger.
Successive ionization energies for an element increase steadily.
Table 7.4 Ionization Energies (in kJ/mol) for Removing Successive Electrons from Li and Be
Reaction | I | Reaction | I |
---|---|---|---|
I1 = 520.2 | I1 = 899.5 | ||
I2 = 7298.2 | I2 = 1757.1 | ||
I3 = 11,815.0 | I3 = 14,848.8 | ||
I4 = 21,006.6 |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
The most important consequence of the values listed in is that the chemistry of Li is dominated by the Li+ ion, while the chemistry of Be is dominated by the +2 oxidation state. The energy required to remove the second electron from Li
Li+(g) → Li2+(g) + e−is more than 10 times greater than the energy needed to remove the first electron. Similarly, the energy required to remove the third electron from Be
Be2+(g) → Be3+(g) + e−is about 15 times greater than the energy needed to remove the first electron and around 8 times greater than the energy required to remove the second electron. Both Li+ and Be2+ have 1s2 closed-shell configurations, and much more energy is required to remove an electron from the 1s2 core than from the 2s valence orbital of the same element. The chemical consequences are enormous: lithium (and all the alkali metals) forms compounds with the 1+ ion but not the 2+ or 3+ ions. Similarly, beryllium (and all the alkaline earth metals) forms compounds with the 2+ ion but not the 3+ or 4+ ions. The energy required to remove electrons from a filled core is prohibitively large and simply cannot be achieved in normal chemical reactions.
The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions.
Ionization energies of the elements in the third row of the periodic table exhibit the same pattern as those of Li and Be (): successive ionization energies increase steadily as electrons are removed from the valence orbitals (3s or 3p, in this case), followed by an especially large increase in ionization energy when electrons are removed from filled core levels as indicated by the bold diagonal line in . Thus in the third row of the periodic table, the largest increase in ionization energy corresponds to removing the fourth electron from Al, the fifth electron from Si, and so forth—that is, removing an electron from an ion that has the valence electron configuration of the preceding noble gas. This pattern explains why the chemistry of the elements normally involves only valence electrons. Too much energy is required to either remove or share the inner electrons.
Table 7.5 Successive Ionization Energies (in kJ/mol) for the Elements in the Third Row of the Periodic Table
Element | I 1 | I 2 | I 3 | I 4 | I 5 | I 6 | I 7 |
---|---|---|---|---|---|---|---|
Na | 495.8 | 4562.4* | — | — | — | — | — |
Mg | 737.7 | 1450.7 | 7732.7 | — | — | — | — |
Al | 577.5 | 1816.7 | 2744.8 | 11,577.5 | — | — | — |
Si | 786.5 | 1577.1 | 3231.6 | 4355.5 | 16,090.6 | — | — |
P | 1011.8 | 1907.5 | 2914.1 | 4963.6 | 6274.0 | 21,267.4 | — |
S | 999.6 | 2251.8 | 3357 | 4556.2 | 7004.3 | 8495.8 | 27,107.4 |
Cl | 1251.2 | 2297.7 | 3822 | 5158.6 | 6540 | 9362 | 11,018.2 |
Ar | 1520.6 | 2665.9 | 3931 | 5771 | 7238 | 8781.0 | 11,995.3 |
*Inner-shell electron |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
From their locations in the periodic table, predict which of these elements has the highest fourth ionization energy: B, C, or N.
Given: three elements
Asked for: element with highest fourth ionization energy
Strategy:
A List the electron configuration of each element.
B Determine whether electrons are being removed from a filled or partially filled valence shell. Predict which element has the highest fourth ionization energy, recognizing that the highest energy corresponds to the removal of electrons from a filled electron core.
Solution:
A These elements all lie in the second row of the periodic table and have the following electron configurations:
B: [He]2s22p1 C: [He]2s22p2 N: [He]2s22p3B The fourth ionization energy of an element (I4) is defined as the energy required to remove the fourth electron:
E3+(g) → E4+(g) + e−Because carbon and nitrogen have four and five valence electrons, respectively, their fourth ionization energies correspond to removing an electron from a partially filled valence shell. The fourth ionization energy for boron, however, corresponds to removing an electron from the filled 1s2 subshell. This should require much more energy. The actual values are as follows: B, 25,026 kJ/mol; C, 6223 kJ/mol; and N, 7475 kJ/mol.
Exercise
From their locations in the periodic table, predict which of these elements has the lowest second ionization energy: Sr, Rb, or Ar.
Answer: Sr
The first column of data in shows that first ionization energies tend to increase across the third row of the periodic table. This is because the valence electrons do not screen each other very well, allowing the effective nuclear charge to increase steadily across the row. The valence electrons are therefore attracted more strongly to the nucleus, so atomic sizes decrease and ionization energies increase. These effects represent two sides of the same coin: stronger electrostatic interactions between the electrons and the nucleus further increase the energy required to remove the electrons.
However, the first ionization energy decreases at Al ([Ne]3s23p1) and at S ([Ne]3s23p4). The electrons in aluminum’s filled 3s2 subshell are better at screening the 3p1 electron than they are at screening each other from the nuclear charge, so the s electrons penetrate closer to the nucleus than the p electron does. The decrease at S occurs because the two electrons in the same p orbital repel each other. This makes the S atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements.
The first ionization energies of the elements in the first six rows of the periodic table are plotted in . They are presented numerically and graphically in . These figures illustrate three important trends:
Generally, I1 increases diagonally from the lower left of the periodic table to the upper right.
Figure 7.10 A Plot of Periodic Variation of First Ionization Energy with Atomic Number for the First Six Rows of the Periodic Table
There is a decrease in ionization energy within a group (most easily seen here for groups 1 and 18).
Figure 7.11 First Ionization Energies of the s-, p-, d-, and f-Block Elements
The darkness of the shading inside the cells of the table indicates the relative magnitudes of the ionization energies. Elements in gray have undetermined first ionization energies.
Source: Data from CRC Handbook of Chemistry and Physics (2004).
Gallium (Ga), which is the first element following the first row of transition metals, has the following electron configuration: [Ar]4s23d104p1. Its first ionization energy is significantly lower than that of the immediately preceding element, zinc, because the filled 3d10 subshell of gallium lies inside the 4p subshell, screening the single 4p electron from the nucleus. Experiments have revealed something of even greater interest: the second and third electrons that are removed when gallium is ionized come from the 4s2 orbital, not the 3d10 subshell. The chemistry of gallium is dominated by the resulting Ga3+ ion, with its [Ar]3d10 electron configuration. This and similar electron configurations are particularly stable and are often encountered in the heavier p-block elements. They are sometimes referred to as pseudo noble gas configurationsThe and similar electron configurations that are particularly stable and are often encountered in the heavier -block elements.. In fact, for elements that exhibit these configurations, no chemical compounds are known in which electrons are removed from the (n − 1)d10filled subshell.
As we noted, the first ionization energies of the transition metals and the lanthanides change very little across each row. Differences in their second and third ionization energies are also rather small, in sharp contrast to the pattern seen with the s- and p-block elements. The reason for these similarities is that the transition metals and the lanthanides form cations by losing the ns electrons before the (n − 1)d or (n − 2)f electrons, respectively. This means that transition metal cations have (n − 1)dn valence electron configurations, and lanthanide cations have (n − 2)fn valence electron configurations. Because the (n − 1)d and (n − 2)f shells are closer to the nucleus than the ns shell, the (n − 1)d and (n − 2)f electrons screen the ns electrons quite effectively, reducing the effective nuclear charge felt by the ns electrons. As Z increases, the increasing positive charge is largely canceled by the electrons added to the (n − 1)d or (n − 2)f orbitals.
That the ns electrons are removed before the (n − 1)d or (n − 2)f electrons may surprise you because the orbitals were filled in the reverse order. (For more information on shell filling order, see , .) In fact, the ns, the (n − 1)d, and the (n − 2)f orbitals are so close to one another in energy, and interpenetrate one another so extensively, that very small changes in the effective nuclear charge can change the order of their energy levels. As the d orbitals are filled, the effective nuclear charge causes the 3d orbitals to be slightly lower in energy than the 4s orbitals. The [Ar]3d2 electron configuration of Ti2+ tells us that the 4s electrons of titanium are lost before the 3d electrons; this is confirmed by experiment. A similar pattern is seen with the lanthanides, producing cations with an (n − 2)fn valence electron configuration.
Because their first, second, and third ionization energies change so little across a row, these elements have important horizontal similarities in chemical properties in addition to the expected vertical similarities. For example, all the first-row transition metals except scandium form stable compounds as M2+ ions, whereas the lanthanides primarily form compounds in which they exist as M3+ ions.
Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr.
Given: six elements
Asked for: element with lowest first ionization energy
Strategy:
Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy.
Solution:
These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb.
Exercise
Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn.
Answer: As
The electron affinity(EA)The energy change that occurs when an electron is added to a gaseous atom: of an element E is defined as the energy change that occurs when an electron is added to a gaseous atom:
Equation 7.6
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with our discussion of energy changes in , where a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released.
Chlorine has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element:
Equation 7.7
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
Equation 7.8
Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy:
Equation 7.9
Electron affinities for the first six rows of the periodic table are plotted in and presented numerically and graphically in . Both figures show that the halogens, with their ns2np5 valence electron configuration, have the most negative electron affinities. In general, electron affinities become more negative as we go across a row of the periodic table. This pattern corresponds to the increased effective nuclear charge felt by the valence electrons across a row, which leads to increased electrostatic attraction between the added electron and the nucleus (a more negative electron affinity). The trend, however, is not as uniform as the one observed for ionization energies. Some of the alkaline earths (group 2), the elements of group 12, and all the noble gases (group 18) have effective electron affinities that are greater than or equal to zero, while the electron affinities for the elements of group 15 are usually less negative than those for the group 14 elements. These exceptions can be explained by the groups’ electron configurations. Both the alkaline earth metals and the noble gases have valence electron shells with filled subshells (ns2 and ns2np6, respectively). In each case, the added electron must enter a higher-energy orbital, requiring an input of energy. All the group 15 elements have an ns2np3 valence electron configuration, in which each of the three p orbitals has a single electron, in accordance with Hund’s rule; hence the added electron must enter an already occupied p orbital. The resulting electron–electron repulsions destabilize the anion, causing the electron affinities of these elements to be less negative than we would otherwise expect. In the case of nitrogen, the 2p orbital is quite small, and the electron–electron repulsions are so strong that nitrogen has approximately zero affinity for an extra electron. In the heavier elements, however, the effect is relatively small because they have larger valence p orbitals.
Generally, electron affinities become more negative across a row of the periodic table.
Figure 7.12 A Plot of Periodic Variation of Electron Affinity with Atomic Number for the First Six Rows of the Periodic Table
Figure 7.13 Electron Affinities (in kJ/mol) of the s-, p-, and d-Block Elements
There are many more exceptions to the trends across rows and down columns than with first ionization energies. Elements that do not form stable ions, such as the noble gases, are assigned an effective electron affinity that is greater than or equal to zero. Elements for which no data are available are shown in gray.
Source: Data from Journal of Physical and Chemical Reference Data 28, no. 6 (1999).
In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as n increases, the extra electrons enter orbitals that are increasingly far from the nucleus. Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend:
In general, electron affinities become more negative across a row and less negative down a column.
The equations for second and higher electron affinities are analogous to those for second and higher ionization energies:
Equation 7.10
Equation 7.11
As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol:
Equation 7.12
Equation 7.13
Thus the formation of a gaseous oxide (O2−) ion is energetically quite unfavorable:
Equation 7.14
Similarly, the formation of all common dianions (such as S2−) or trianions (such as P3−) is energetically unfavorable in the gas phase.
While first electron affinities can be negative, positive, or zero, second electron affinities are always positive.
If energy is required to form both positively charged ions and monatomic polyanions, why do ionic compounds such as MgO, Na2S, and Na3P form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions in the crystalline salt. We will describe the energetics of ionic compounds in more detail in .
Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity?
Given: three elements
Asked for: element with most negative electron affinity
Strategy:
A Locate the elements in the periodic table. Use the trends in electron affinities going down a column for elements in the same group. Similarly, use the trends in electron affinities from left to right for elements in the same row.
B Place the elements in order, listing the element with the most negative electron affinity first.
Solution:
A We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements.
Exercise
Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion?
Answer: Rb
The elements with the highest ionization energies are generally those with the most negative electron affinities, which are located toward the upper right corner of the periodic table (compare and ). Conversely, the elements with the lowest ionization energies are generally those with the least negative electron affinities and are located in the lower left corner of the periodic table.
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativityThe relative ability of an atom to attract electrons to itself in a chemical compound. (represented by the Greek letter chi, χ, pronounced “ky” as in “sky”), defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property.
The original electronegativity scale, developed in the 1930s by Linus Pauling (1901– 1994) was based on measurements of the strengths of covalent bonds between different elements. Pauling arbitrarily set the electronegativity of fluorine at 4.0 (although today it has been refined to 3.98), thereby creating a scale in which all elements have values between 0 and 4.0.
Periodic variations in Pauling’s electronegativity values are illustrated in and . If we ignore the inert gases and elements for which no stable isotopes are known, we see that fluorine (χ = 3.98) is the most electronegative element and cesium is the least electronegative nonradioactive element (χ = 0.79). Because electronegativities generally increase diagonally from the lower left to the upper right of the periodic table, elements lying on diagonal lines running from upper left to lower right tend to have comparable values (e.g., O and Cl and N, S, and Br).
Figure 7.14 A Plot of Periodic Variation of Electronegativity with Atomic Number for the First Six Rows of the Periodic Table
Pauling won two Nobel Prizes, one for chemistry in 1954 and one for peace in 1962. When he was nine, Pauling’s father died, and his mother tried to convince him to quit school to support the family. He did not quit school but was denied a high school degree because of his refusal to take a civics class.
Figure 7.15 Pauling Electronegativity Values of the s-, p-, d-, and f-Block Elements
Values for most of the actinides are approximate. Elements for which no data are available are shown in gray.
Source: Data from L. Pauling, The Nature of the Chemical Bond, 3rd ed. (1960).
Pauling’s method is limited by the fact that many elements do not form stable covalent compounds with other elements; hence their electronegativities cannot be measured by his method. Other definitions have since been developed that address this problem.
An alternative method for measuring electronegativity was developed by Robert Mulliken (1896–1986; Nobel Prize in Chemistry 1966). Mulliken noticed that elements with large first ionization energies tend to have very negative electron affinities and gain electrons in chemical reactions. Conversely, elements with small first ionization energies tend to have slightly negative (or even positive) electron affinities and lose electrons in chemical reactions. Mulliken recognized that an atom’s tendency to gain or lose electrons could therefore be described quantitatively by the average of the values of its first ionization energy and the absolute value of its electron affinity. Using our definition of electron affinity, we can write Mulliken’s original expression for electronegativity as follows:Mulliken’s definition used the magnitude of the ionization energy and the electron affinity. By definition, the magnitude of a quantity is a positive number. Our definition of electron affinity produces negative values for the electron affinity for most elements, so vertical lines indicating absolute value are needed in to make sure that we are adding two positive numbers in the numerator.
Equation 7.15
Elements with a large first ionization energy and a very negative electron affinity have a large positive value in the numerator of , so their electronegativity is high. Elements with a small first ionization energy and a small electron affinity have a small positive value for the numerator in , so they have a low electronegativity.Inserting the appropriate data from and into gives a Mulliken electronegativity value for fluorine of 1004.6 kJ/mol. To compare Mulliken’s electronegativity values with those obtained by Pauling, Mulliken’s values are divided by 252.4 kJ/mol, which gives Pauling’s value (3.98).
As noted previously, all electronegativity scales give essentially the same results for one element relative to another. Even though the Mulliken scale is based on the properties of individual atoms and the Pauling scale is based on the properties of atoms in molecules, they both apparently measure the same basic property of an element. In the following discussion, we will focus on the relationship between electronegativity and the tendency of atoms to form positive or negative ions. We will therefore be implicitly using the Mulliken definition of electronegativity. Because of the parallels between the Mulliken and Pauling definitions, however, the conclusions are likely to apply to atoms in molecules as well.
An element’s electronegativity provides us with a single value that we can use to characterize the chemistry of an element. Elements with a high electronegativity (χ ≥ 2.2 in ) have very negative affinities and large ionization potentials, so they are generally nonmetals and electrical insulators that tend to gain electrons in chemical reactions (i.e., they are oxidants). In contrast, elements with a low electronegativity (χ ≤ 1.8) have electron affinities that have either positive or small negative values and small ionization potentials, so they are generally metals and good electrical conductors that tend to lose their valence electrons in chemical reactions (i.e., they are reductants). In between the metals and nonmetals, along the heavy diagonal line running from B to At in , is a group of elements with intermediate electronegativities (χ ~ 2.0). These are the semimetals, elements that have some of the chemical properties of both nonmetals and metals. The distinction between metals and nonmetals is one of the most fundamental we can make in categorizing the elements and predicting their chemical behavior. shows the strong correlation between electronegativity values, metallic versus nonmetallic character, and location in the periodic table.
Figure 7.16 Three-Dimensional Plots Demonstrating the Relationship between Electronegativity and the Metallic/Nonmetallic Character of the Elements
(a) A plot of electrical resistivity (measured resistivity to electron flow) at or near room temperature shows that substances with high resistivity (little to no measured electron flow) are electrical insulators, whereas substances with low resistivity (high measured electron flow) are metals. (b) A plot of Pauling electronegativities for a like set of elements shows that high electronegativity values (≥ about 2.2) correlate with high electrical resistivities (insulators). Low electronegativity values (≤ about 2.2) correlate with low resistivities (metals). Because electrical resistivity is typically measured only for solids and liquids, the gaseous elements do not appear in part (a).
The rules for assigning oxidation states that were introduced in are based on the relative electronegativities of the elements; the more electronegative element in a binary compound is assigned a negative oxidation state. As we shall see, electronegativity values are also used to predict bond energies, bond polarities, and the kinds of reactions that compounds undergo.
On the basis of their positions in the periodic table, arrange Cl, Se, Si, and Sr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a semimetal.
Given: four elements
Asked for: order by increasing electronegativity and classification
Strategy:
A Locate the elements in the periodic table. From their diagonal positions from lower left to upper right, predict their relative electronegativities.
B Arrange the elements in order of increasing electronegativity.
C Classify each element as a metal, a nonmetal, or a semimetal according to its location about the diagonal belt of semimetals running from B to At.
Solution:
A Electronegativity increases from lower left to upper right in the periodic table (). Because Sr lies far to the left of the other elements given, we can predict that it will have the lowest electronegativity. Because Cl lies above and to the right of Se, we can predict that χCl > χSe. Because Si is located farther from the upper right corner than Se or Cl, its electronegativity should be lower than those of Se and Cl but greater than that of Sr. B The overall order is therefore χSr < χSi < χSe < χCl.
C To classify the elements, we note that Sr lies well to the left of the diagonal belt of semimetals running from B to At; while Se and Cl lie to the right and Si lies in the middle. We can predict that Sr is a metal, Si is a semimetal, and Se and Cl are nonmetals.
Exercise
On the basis of their positions in the periodic table, arrange Ge, N, O, Rb, and Zr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a semimetal.
Answer: Rb < Zr < Ge < N < O; metals (Rb, Zr); semimetal (Ge); nonmetal (N, O)
Electronegativity values increase from lower left to upper right in the periodic table.
The trends in periodic properties are summarized in . As discussed, atomic radii decrease from lower left to upper right in the periodic table; ionization energies become more positive, electron affinities become more negative, and electronegativities increase from the lower left to the upper right.
Figure 7.17 Summary of Major Periodic Trends
The general trends for the first ionization energy, electron affinity, and electronegativity are opposite to the general trend for covalent atomic radius.
The tendency of an element to lose or gain electrons is one of the most important factors in determining the kind of compounds it forms. Periodic behavior is most evident for ionization energy (I), the energy required to remove an electron from a gaseous atom. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell. Consequently, only valence electrons can be removed in chemical reactions, leaving the filled inner shell intact. Ionization energies explain the common oxidation states observed for the elements. Ionization energies increase diagonally from the lower left of the periodic table to the upper right. Minor deviations from this trend can be explained in terms of particularly stable electronic configurations, called pseudo noble gas configurations, in either the parent atom or the resulting ion. The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity. Elements with a high electronegativity are generally nonmetals and electrical insulators and tend to behave as oxidants in chemical reactions. Conversely, elements with a low electronegativity are generally metals and good electrical conductors and tend to behave as reductants in chemical reactions.
Identify each statement as either true or false and explain your reasoning.
Based on electronic configurations, explain why the first ionization energies of the group 16 elements are lower than those of the group 15 elements, which is contrary to the general trend.
The first through third ionization energies do not vary greatly across the lanthanides. Why? How does the effective nuclear charge experienced by the ns electron change when going from left to right (with increasing atomic number) in this series?
Most of the first row transition metals can form at least two stable cations, for example iron(II) and iron(III). In contrast, scandium and zinc each form only a single cation, the Sc3+ and Zn2+ ions, respectively. Use the electron configuration of these elements to provide an explanation.
Of the elements Nd, Al, and Ar, which will readily form(s) +3 ions? Why?
Orbital energies can reverse when an element is ionized. Of the ions B3+, Ga3+, Pr3+, Cr3+, and As3+, in which would you expect this reversal to occur? Explain your reasoning.
The periodic trends in electron affinities are not as regular as periodic trends in ionization energies, even though the processes are essentially the converse of one another. Why are there so many more exceptions to the trends in electron affinities compared to ionization energies?
Elements lying on a lower right to upper left diagonal line cannot be arranged in order of increasing electronegativity according to where they occur in the periodic table. Why?
Why do ionic compounds form, if energy is required to form gaseous cations?
Why is Pauling’s definition of electronegativity considered to be somewhat limited?
Based on their positions in the periodic table, arrange Sb, O, P, Mo, K, and H in order of increasing electronegativity.
Based on their positions in the periodic table, arrange V, F, B, In, Na, and S in order of decreasing electronegativity.
Both Al and Nd will form a cation with a +3 charge. Aluminum is in Group 13, and loss of all three valence electrons will produce the Al3+ ion with a noble gas configuration. Neodymium is a lanthanide, and all of the lanthanides tend to form +3 ions because the ionization potentials do not vary greatly across the row, and a +3 charge can be achieved with many oxidants.
K < Mo ≈ Sb < P ≈ H < O
The following table gives values of the first and third ionization energies for selected elements:
Number of Electrons | Element | I1 (E → E+ + e−, kJ/mol) | Element | I3 (E2+ → E3+ + e−, kJ/mol) |
---|---|---|---|---|
11 | Na | 495.9 | Al | 2744.8 |
12 | Mg | 737.8 | Si | 3231.6 |
13 | Al | 577.6 | P | 2914.1 |
14 | Si | 786.6 | S | 3357 |
15 | P | 1011.9 | Cl | 3822 |
16 | S | 999.6 | Ar | 3931 |
17 | Cl | 1251.2 | K | 4419.6 |
18 | Ar | 1520.6 | Ca | 4912.4 |
Plot the ionization energies versus number of electrons. Explain why the slopes of the I1 and I3 plots are different, even though the species in each row of the table have the same electron configurations.
Would you expect the third ionization energy of iron, corresponding to the removal of an electron from a gaseous Fe2+ ion, to be larger or smaller than the fourth ionization energy, corresponding to removal of an electron from a gaseous Fe3+ ion? Why? How would these ionization energies compare to the first ionization energy of Ca?
Which would you expect to have the highest first ionization energy: Mg, Al, or Si? Which would you expect to have the highest third ionization energy. Why?
Use the values of the first ionization energies given in to construct plots of first ionization energy versus atomic number for (a) boron through oxygen in the second period; and (b) oxygen through tellurium in group 16. Which plot shows more variation? Explain the reason for the variation in first ionization energies for this group of elements.
Arrange Ga, In, and Zn in order of increasing first ionization energies. Would the order be the same for second and third ionization energies? Explain your reasoning.
Arrange each set of elements in order of increasing magnitude of electron affinity.
Arrange each set of elements in order of decreasing magnitude of electron affinity.
Of the species F, O−, Al3+, and Li+, which has the highest electron affinity? Explain your reasoning.
Of the species O−, N2−, Hg2+, and H+, which has the highest electron affinity? Which has the lowest electron affinity? Justify your answers.
The Mulliken electronegativity of element A is 542 kJ/mol. If the electron affinity of A is −72 kJ/mol, what is the first ionization energy of element A? Use the data in the following table as a guideline to decide if A is a metal, a nonmetal, or a semimetal. If 1 g of A contains 4.85 × 1021 molecules, what is the identity of element A?
Na | Al | Si | S | Cl | |
---|---|---|---|---|---|
EA (kJ/mol) | −59.6 | −41.8 | −134.1 | −200.4 | −348.6 |
I (kJ/mol) | 495.8 | 577.5 | 786.5 | 999.6 | 1251.2 |
Based on their valence electron configurations, classify the following elements as either electrical insulators, electrical conductors, or substances with intermediate conductivity: S, Ba, Fe, Al, Te, Be, O, C, P, Sc, W, Na, B, and Rb.
Using the data in Problem 10, what conclusions can you draw with regard to the relationship between electronegativity and electrical properties? Estimate the approximate electronegativity of a pure element that is very dense, lustrous, and malleable.
Of the elements Al, Mg, O2, Ti, I2, and H2, which, if any, would you expect to be a good reductant? Explain your reasoning.
Of the elements Zn, B, Li, Se, Co, and Br2, which if any, would you expect to be a good oxidant? Explain your reasoning.
Determine whether each species is a good oxidant, a good reductant, or neither.
Determine whether each species is a good oxidant, a good reductant, or neither.
Of the species I2, O−, Zn, Sn2+, and K+, choose which you would expect to be a good oxidant. Then justify your answer.
Based on the valence electron configuration of the noble gases, would you expect them to have positive or negative electron affinities? What does this imply about their most likely oxidation states? their reactivity?
The general features of both plots are roughly the same, with a small peak at 12 electrons and an essentially level region from 15–16 electrons. The slope of the I3 plot is about twice as large as the slope of the I1 plot, however, because the I3 values correspond to removing an electron from an ion with a +2 charge rather than a neutral atom. The greater charge increases the effect of the steady rise in effective nuclear charge across the row.
Electron configurations: Mg, 1s22s22p63s2; Al, 1s22s22p63s23p1; Si, 1s22s22p63s23p2; First ionization energies increase across the row due to a steady increase in effective nuclear charge; thus, Si has the highest first ionization energy. The third ionization energy corresponds to removal of a 3s electron for Al and Si, but for Mg it involves removing a 2p electron from a filled inner shell; consequently, the third ionization energy of Mg is the highest.
Hg2+ > H+ > O− > N2−; Hg2+ has the highest positive charge plus a relatively low energy vacant set of orbitals (the 6p subshell) to accommodate an added electron, giving it the greatest electron affinity; N2− has a greater negative charge than O−, so electron–electron repulsions will cause its electron affinity to be even lower (more negative) than that of O−.
insulators: S, O, C (diamond), P; conductors: Ba, Fe, Al, C (graphite), Be, Sc, W, Na, Rb; Te and B are semimetals and semiconductors.
Mg, Al, Ti, and H2
I2 is the best oxidant, with a moderately strong tendency to accept an electron to form the I− ion, with a closed shell electron configuration. O− would probably also be an oxidant, with a tendency to add an electron to form salts containing the oxide ion, O2−. Zn and Sn2+ are all reductants, while K+ has no tendency to act as an oxidant or a reductant.
Periodic trends in properties such as atomic size and ionic size, ionization energy, electron affinity, and electronegativity illustrate the strong connection between the chemical properties and the reactivity of the elements and their positions in the periodic table. In this section, we explore that connection by focusing on two periodic properties that correlate strongly with the chemical behavior of the elements: valence electron configurations and Mulliken electronegativities.
We have said that elements with the same valence electron configuration (i.e., elements in the same column of the periodic table) often have similar chemistry. This correlation is particularly evident for the elements of groups 1, 2, 3, 13, 16, 17, and 18. The intervening families in the p block (groups 14 and 15) straddle the diagonal line separating metals from nonmetals. The lightest members of these two families are nonmetals, so they react differently compared to the heaviest members, which are metals. We begin our survey with the alkali metals (group 1), which contain only a single electron outside a noble gas electron configuration, and end with the noble gases (group 18), which have full valence electron shells.
The elements of group 1 are called the alkali metals. Alkali (from the Arabic al-qili, meaning “ashes of the saltwort plant from salt marshes”) was a general term for substances derived from wood ashes, all of which possessed a bitter taste and were able to neutralize acids. Although oxides of both group 1 and group 2 elements were obtained from wood ashes, the alkali metals had lower melting points.
Potassium and sodium were first isolated in 1807 by the British chemist Sir Humphry Davy (1778–1829) by passing an electrical current through molten samples of potash (K2CO3) and soda ash (Na2CO3). The potassium burst into flames as soon as it was produced because it reacts readily with oxygen at the higher temperature. However, the group 1 elements, like the group 2 elements, become less reactive with air or water as their atomic number decreases. The heaviest element (francium) was not discovered until 1939. It is so radioactive that studying its chemistry is very difficult.
The alkali metals have ns1 valence electron configurations and the lowest electronegativity of any group; hence they are often referred to as being electropositive elements. As a result, they have a strong tendency to lose their single valence electron to form compounds in the +1 oxidation state, producing the EX monohalides and the E2O oxides.
Because they are so reactive, pure group 1 elements are powerful reducing agents that are used in lithium batteries and cardiac pacemakers. Sodium salts such as common table salt (NaCl), baking soda (NaHCO3), soda ash (Na2CO3), and caustic soda (NaOH) are important industrial chemicals. Other compounds of the alkali metals are important in biology. For example, because potassium is required for plant growth, its compounds are used in fertilizers, and lithium salts are used to treat manic-depressive, or bipolar, disorders.
Potassium burning. A piece of potassium dropped in a beaker of water will burn as it skips across the top of the water.
The elements of group 2 are collectively referred to as the alkaline earth metals, a name that originated in the Middle Ages, when an “earth” was defined as a substance that did not melt and was not transformed by fire. Alkalis that did not melt easily were called “alkaline earths.”
Recall that the trend in most groups is for the lightest member to have properties that are quite different from those of the heavier members. Consistent with this trend, the properties of the lightest element—in this case, beryllium—tend to be different from those of its heavier congeners, the other members of the group. Beryllium is relatively unreactive but forms many covalent compounds, whereas the other group members are much more reactive metals and form ionic compounds. As is the case with the alkali metals, the heaviest element, radium, is highly radioactive, making its size difficult to measure. Radium was discovered in 1902 by Marie Curie (1867–1934; Nobel Prize in Chemistry 1903 and Nobel Prize in Chemistry 1911), who, with her husband, Pierre, isolated 120 mg of radium chloride from tons of residues from uranium mining. (For more information about radioactivity, see , .)
All the alkaline earth metals have ns2 valence electron configurations, and all have electronegativities less than 1.6. This means that they behave chemically as metals (although beryllium compounds are covalent) and lose the two valence electrons to form compounds in the +2 oxidation state. Examples include the dihalides (EX2) and the oxides (EO).
Compounds of the group 2 elements have been commercially important since Egyptian and Roman times, when blocks of limestone or marble, which are both CaCO3, were used as building materials, and gypsum (CaSO4·2 H2O) or lime (CaO) was used as mortar. Calcium sulfate is still used in Portland cement and plaster of Paris. Magnesium and beryllium form lightweight, high-strength alloys that are used in the aerospace, automotive, and other high-tech industries. As you learned in , one of the most impressive uses of these elements is in fireworks; strontium and barium salts, for example, give red or green colors, respectively. Except for beryllium, which is highly toxic, the group 2 elements are also important biologically. Bone is largely hydroxyapatite [Ca5(PO4)3OH], mollusk shells are calcium carbonate, magnesium is part of the chlorophyll molecule in green plants, and calcium is important in hormonal and nerve signal transmission. Because BaSO4 is so insoluble, it is used in “barium milk shakes” to obtain x-rays of the gastrointestinal tract.
Of the group 13 elements, only the lightest, boron, lies on the diagonal line that separates nonmetals and metals. Thus boron is a semimetal, whereas the rest of the group 13 elements are metals. Elemental boron has an unusual structure consisting of B12 icosahedra covalently bonded to one another; the other elements are typical metallic solids.
No group 13 elements were known in ancient times, not because they are scarce—Al is the third most abundant element in Earth’s crust—but because they are highly reactive and form extremely stable compounds with oxygen. To isolate the pure elements, potent reducing agents and careful handling were needed.
The elements of group 13 have ns2np1 valence electron configurations. Consequently, two oxidation states are important: +3, from losing three valence electrons to give the closed-shell electron configuration of the preceding noble gas; and +1, from losing the single electron in the np subshell. Because these elements have small, negative electron affinities (boron’s is only −27.0 kJ/mol), they are unlikely to acquire five electrons to reach the next noble gas configuration. In fact, the chemistry of these elements is almost exclusively characterized by +3. Only the heaviest element (Tl) has extensive chemistry in the +1 oxidation state. It loses the single 6p electron to produce TlX monohalides and the oxide Tl2O.
In the 19th century, aluminum was considered a precious metal. In fact, it was considered so precious that aluminum knives and forks were reserved for the French Emperor Louis Napoleon III, while his less important guests had to be content with gold or silver cutlery. Because of the metal’s rarity the dedication of the Washington Monument in 1885 was celebrated by placing a 100 oz chunk of pure aluminum at the top. In contrast, today aluminum is used on an enormous scale in aircraft, automobile engines, armor, cookware, and beverage containers. It is valued for its combination of low density, high strength, and corrosion resistance. Aluminum is also found in compounds that are the active ingredients in most antiperspirant deodorants.
Compounds of boron, such as one form of BN, are hard, have a high melting point, and are resistant to corrosion. They are particularly useful in materials that are exposed to extreme conditions, such as aircraft turbines, brake linings, and polishing compounds. Boron is also a major component of many kinds of glasses, and sodium perborate [Na2B2O4(OH)4] is the active ingredient in many so-called color-safe laundry bleaches.
Gallium, indium, and thallium are less widely used, but gallium arsenide is the red light-emitting diode (LED) in digital readouts in electronics, and MgGa2O4 produces the green light emitted in many xerographic machines. Compounds of thallium(I) are extremely toxic. Although Tl2SO4 is an excellent rat or ant poison, it is so toxic to humans that it is no longer used for this purpose.
The group 14 elements straddle the diagonal line that divides nonmetals from metals. Of the elements in this group, carbon is a nonmetal, silicon and germanium are semimetals, and tin and lead are metals. As a result of this diversity, the structures of the pure elements vary greatly.
The ns2np2 valence electron configurations of group 14 gives rise to three oxidation states: −4, in which four electrons are added to achieve the closed-shell electron configuration of the next noble gas; +4, in which all four valence electrons are lost to give the closed-shell electron configuration of the preceding noble gas; and +2, in which the loss of two np2 electrons gives a filled ns2 subshell.
The electronegativity of carbon is only 2.5, placing it in the middle of the electronegativity range, so carbon forms covalent compounds with a wide variety of elements and is the basis of all organic compounds. All of the group 14 elements form compounds in the +4 oxidation state, so all of them are able to form dioxides (from CO2 to PbO2) and tetrachlorides (CCl4 and PbCl4). Only the two metallic elements, Sn and Pb, form an extensive series of compounds in the +2 oxidation state. Tin salts are sprayed onto glass to make an electrically conductive coating, and then the glass is used in the manufacture of frost-free windshields. Lead sulfate is formed when your car battery discharges.
Carbon has at least four allotropes (forms or crystal structures) that are stable at room temperature: graphite; diamond; a group of related cage structures called fullerenesOne of at least four allotropes of carbon comprising a group of related cage structures. (such as C60); and nanotubesOne of at least four allotropes of carbon that are cylinders of carbon atoms and are intermediate in structure between graphite and the fullerenes., which are cylinders of carbon atoms (). Graphite consists of extended planes of covalently bonded hexagonal rings. Because the planes are not linked by covalent bonds, they can slide across one another easily. This makes graphite ideally suited as a lubricant and as the “lead” in lead pencils. Graphite also provides the black color in inks and tires, and graphite fibers are used in high-tech items such as golf clubs, tennis rackets, airplanes, and sailboats because of their lightweight, strength, and stiffness.
Figure 7.18 Four Allotropes of Carbon
Diamond consists of a rigid three-dimensional array of carbon atoms, making it one of the hardest substances known. In contrast, graphite forms from extended planes of covalently bonded hexagonal rings of carbon atoms that can slide across one another easily. Fullerenes are spherical or ellipsoidal molecules with six- and five-membered rings of carbon atoms, and nanotubes are sheets of graphite rolled up into a cylinder.
In contrast to the layered structure of graphite, each carbon atom in diamond is bonded to four others to form a rigid three-dimensional array, making diamond one of the hardest substances known; consequently, it is used in industry as a cutting tool. Fullerenes, on the other hand, are spherical or ellipsoidal molecules with six- and five-membered rings of carbon atoms; they are volatile substances that dissolve in organic solvents. Fullerenes of extraterrestrial origin have been found in meteorites and have been discovered in a cloud of cosmic dust surrounding a distant star, which makes them the largest molecules ever seen in space. Carbon nanotubes, intermediate in structure between graphite and the fullerenes, can be described as sheets of graphite that have been rolled up into a cylinder or, alternatively, fullerene cages that have been stretched in one direction. Carbon nanotubes are being studied for use in the construction of molecular electronic devices and computers. For example, fabrics that are dipped in an ink of nanotubes and then pressed to thin out the coating are turned into batteries that maintain their flexibility. This creates “wearable electronics” and allows for the possibility of incorporating electronics into flexible surfaces. When applied to a t-shirt, for example, the t-shirt is converted into an “e-shirt.”
Silicon is the second must abundant element in Earth’s crust. Both silicon and germanium have strong, three-dimensional network structures similar to that of diamond. Sand is primarily SiO2, which is used commercially to make glass and prevent caking in food products. Complex compounds of silicon and oxygen with elements such as aluminum are used in detergents and talcum powder and as industrial catalysts. Because silicon-chip technology laid the foundation for the modern electronics industry, the San Jose region of California, where many of the most important advances in electronics and computers were developed, has been nicknamed “Silicon Valley.”
Elemental tin and lead are metallic solids. Tin is primarily used to make alloys such as bronze, which consists of tin and copper; solder, which is tin and lead; and pewter, which is tin, antimony, and copper.
In ancient times, lead was used for everything from pipes to cooking pots because it is easily hammered into different shapes. In fact, the term plumbing is derived from plumbum, the Latin name for lead. Lead compounds were used as pigments in paints, and tetraethyllead was an important antiknock agent in gasoline. Now, however, lead has been banned from many uses because of its toxicity, although it is still widely used in lead storage batteries for automobiles. In previous centuries, lead salts were frequently used as medicines. Evidence suggests, for example, that Beethoven’s death was caused by the application of various lead-containing medicines by his physician. Beethoven contracted pneumonia and was treated with lead salts, but in addition, he suffered from a serious liver ailment. His physician treated the ailment by repeatedly puncturing his abdominal cavity and then sealing the wound with a lead-laced poultice. It seems that the repeated doses of lead compounds contributed to Beethoven’s death.
The group 15 elements are called the pnicogensThe elements in group 15 of the periodic table.—from the Greek pnigein, meaning “to choke,” and genes, meaning “producing”—ostensibly because of the noxious fumes that many nitrogen and phosphorus compounds produce. This family has five stable elements; one isotope of bismuth (209Bi) is nonradioactive and is the heaviest nonradioactive isotope of any element. Once again, the lightest member of the family has unique properties. Although both nitrogen and phosphorus are nonmetals, nitrogen under standard conditions is a diatomic gas (N2), whereas phosphorus consists of three allotropes: white, a volatile, low-melting solid consisting of P4 tetrahedra; a red solid comprised of P8, P9, and P10 cages linked by P2 units; and black layers of corrugated phosphorus sheets. The next two elements, arsenic and antimony, are semimetals with extended three-dimensional network structures, and bismuth is a silvery metal with a pink tint.
All of the pnicogens have ns2np3 valence electron configurations, leading to three common oxidation states: −3, in which three electrons are added to give the closed-shell electron configuration of the next noble gas; +5, in which all five valence electrons are lost to give the closed-shell electron configuration of the preceding noble gas; and +3, in which only the three np electrons are lost to give a filled ns2 subshell. Because the electronegativity of nitrogen is similar to that of chlorine, nitrogen accepts electrons from most elements to form compounds in the −3 oxidation state (such as in NH3). Nitrogen has only positive oxidation states when combined with highly electronegative elements, such as oxygen and the halogens (e.g., HNO3, NF3). Although phosphorus and arsenic can combine with active metals and hydrogen to produce compounds in which they have a −3 oxidation state (PH3, for example), they typically attain oxidation states of +3 and +5 when combined with more electronegative elements, such as PCl3 and H3PO4. Antimony and bismuth are relatively unreactive metals, but form compounds with oxygen and the halogens in which their oxidation states are +3 and +5 (as in Bi2O3 and SbF5).
Although it is present in most biological molecules, nitrogen was the last pnicogen to be discovered. Nitrogen compounds such as ammonia, nitric acid, and their salts are used agriculturally in huge quantities; nitrates and nitrites are used as preservatives in meat products such as ham and bacon, and nitrogen is a component of nearly all explosives.
Phosphorus, too, is essential for life, and phosphate salts are used in fertilizers, toothpaste, and baking powder. One, phosphorus sulfide, P4S3, is used to ignite modern safety matches. Arsenic, in contrast, is toxic; its compounds are used as pesticides and poisons. Antimony and bismuth are primarily used in metal alloys, but a bismuth compound is the active ingredient in the popular antacid medication Pepto-Bismol.
The group 16 elements are often referred to as the chalcogensThe elements in group 16 of the periodic table.—from the Greek chalk, meaning “copper,” and genes, meaning “producing”—because the most ancient copper ore, copper sulfide, is also rich in two other group 16 elements: selenium and tellurium. Once again, the lightest member of the family has unique properties. In its most common pure form, oxygen is a diatomic gas (O2), whereas sulfur is a volatile solid with S8 rings, selenium and tellurium are gray or silver solids that have chains of atoms, and polonium is a silvery metal with a regular array of atoms. Like astatine and radon, polonium is a highly radioactive metallic element.
All of the chalcogens have ns2np4 valence electron configurations. Their chemistry is dominated by three oxidation states: −2, in which two electrons are added to achieve the closed-shell electron configuration of the next noble gas; +6, in which all six valence electrons are lost to give the closed-shell electron configuration of the preceding noble gas; and +4, in which only the four np electrons are lost to give a filled ns2 subshell. Oxygen has the second highest electronegativity of any element; its chemistry is dominated by the −2 oxidation state (as in MgO and H2O). No compounds of oxygen in the +4 or +6 oxidation state are known. In contrast, sulfur can form compounds in all three oxidation states. Sulfur accepts electrons from less electronegative elements to give H2S and Na2S, for example, and it donates electrons to more electronegative elements to give compounds such as SO2, SO3, and SF6. Selenium and tellurium, near the diagonal line in the periodic table, behave similarly to sulfur but are somewhat more likely to be found in positive oxidation states.
Oxygen, the second most electronegative element in the periodic table, was not discovered until the late 18th century, even though it constitutes 20% of the atmosphere and is the most abundant element in Earth’s crust. Oxygen is essential for life; our metabolism is based on the oxidation of organic compounds by O2 to produce CO2 and H2O. Commercially, oxygen is used in the conversion of pig iron to steel, as the oxidant in oxyacetylene torches for cutting steel, as a fuel for the US space shuttle, and in hospital respirators.
Sulfur is the brimstone in “fire and brimstone” from ancient times. Partly as a result of its long history, it is employed in a wide variety of commercial products and processes. In fact, as you learned in , more sulfuric acid is produced worldwide than any other compound. Sulfur is used to cross-link the polymers in rubber in a process called vulcanization, which was discovered by Charles Goodyear in the 1830s and commercialized by Benjamin Goodrich in the 1870s. Vulcanization gives rubber its unique combination of strength, elasticity, and stability.
Selenium, the only other commercially important chalcogen, was discovered in 1817, and today it is widely used in light-sensitive applications. For example, photocopying, or xerography, from the Greek xèrós, meaning “dry,” and graphia, meaning “writing,” uses selenium films to transfer an image from one piece of paper to another, while compounds such as cadmium selenide are used to measure light in photographic light meters and automatic streetlights.
The term halogen, derived from the Greek háls, meaning “salt,” and genes, meaning “producing,” was first applied to chlorine because of its tendency to react with metals to form salts. All of the halogens have an ns2np5 valence electron configuration, and all but astatine are diatomic molecules in which the two halogen atoms share a pair of electrons. Diatomic F2 and Cl2 are pale yellow-green and pale green gases, respectively, while Br2 is a red liquid, and I2 is a purple solid. The halogens were not isolated until the 18th and 19th centuries.
Because of their relatively high electronegativities, the halogens are nonmetallic and generally react by gaining one electron per atom to attain a noble gas electron configuration and an oxidation state of −1. Halides are produced according to the following equation, in which X denotes a halogen:
Equation 7.16
2 E + nX2 → 2 EXnIf the element E has a low electronegativity (as does Na), the product is typically an ionic halide (NaCl). If the element E is highly electronegative (as P is), the product is typically a covalent halide (PCl5). Ionic halides tend to be nonvolatile substances with high melting points, whereas covalent halides tend to be volatile substances with low melting points. Fluorine is the most reactive of the halogens, and iodine the least, which is consistent with their relative electronegativities ().As we shall see in subsequent chapters, however, factors such as bond strengths are also important in dictating the reactivities of these elements. In fact, fluorine reacts with nearly all elements at room temperature. Under more extreme conditions, it combines with all elements except helium, neon, and argon.
The halogens react with hydrogen to form the hydrogen halides (HX):
Equation 7.17
H2(g) + X2(g,l,s) → 2 HX(g)Fluorine is so reactive that any substance containing hydrogen, including coal, wood, and even water, will burst into flames if it comes into contact with pure F2.
Because it is the most electronegative element known, fluorine never has a positive oxidation state in any compound. In contrast, the other halogens (Cl, Br, I) form compounds in which their oxidation states are +1, +3, +5, and +7, as in the oxoanions, XOn−, where n = 1–4. Because oxygen has the second highest electronegativity of any element, it stabilizes the positive oxidation states of the halogens in these ions.
All of the halogens except astatine (which is radioactive) are commercially important. NaCl in salt water is purified for use as table salt. Chlorine and hypochlorite (OCl−) salts are used to sanitize public water supplies, swimming pools, and wastewater, and hypochlorite salts are also used as bleaches because they oxidize colored organic molecules. Organochlorine compounds are used as drugs and pesticides. Fluoride (usually in the form of NaF) is added to many municipal water supplies to help prevent tooth decay, and bromine (in AgBr) is a component of the light-sensitive coating on photographic film. Because iodine is essential to life—it is a key component of the hormone produced by the thyroid gland—small amounts of KI are added to table salt to produce “iodized salt,” which prevents thyroid hormone deficiencies.
The noble gases are helium, neon, argon, krypton, xenon, and radon. All have filled valence electron configurations and therefore are unreactive elements found in nature as monatomic gases. The noble gases were long referred to as either “rare gases” or “inert gases,” but they are neither rare nor inert. Argon constitutes about 1% of the atmosphere, which also contains small amounts of the lighter group 18 elements, and helium is found in large amounts in many natural gas deposits. The group’s perceived “rarity” stems in part from the fact that the noble gases were the last major family of elements to be discovered.
The noble gases have EA ≥ 0, so they do not form compounds in which they have negative oxidation states. Because ionization energies decrease down the column, the only noble gases that form compounds in which they have positive oxidation states are Kr, Xe, and Rn. Of these three elements, only xenon forms an extensive series of compounds. The chemistry of radon is severely limited by its extreme radioactivity, and the chemistry of krypton is limited by its high ionization energy (1350.8 kJ/mol versus 1170.4 kJ/mol for xenon). In essentially all its compounds, xenon is bonded to highly electronegative atoms such as fluorine or oxygen. In fact, the only significant reaction of xenon is with elemental fluorine, which can give XeF2, XeF4, or XeF6. Oxides such as XeO3 are produced when xenon fluorides react with water, and oxidation with ozone produces the perxenate ion [XeO64−], in which xenon acquires a +8 oxidation state by formally donating all eight of its valence electrons to the more electronegative oxygen atoms. In all of its stable compounds, xenon has a positive, even-numbered oxidation state: +2, +4, +6, or +8. The actual stability of these compounds varies greatly. For example, XeO3 is a shock-sensitive, white crystalline solid with explosive power comparable to that of TNT (trinitrotoluene), whereas another compound, Na2XeF8, is stable up to 300°C.
Although none of the noble gas compounds is commercially significant, the elements themselves have important applications. For example, argon is used in incandescent light bulbs, where it provides an inert atmosphere that protects the tungsten filament from oxidation, and in compact fluorescent light bulbs (CFLs). It is also used in arc welding and in the manufacture of reactive elements, such as titanium, or of ultrapure products, such as the silicon used by the electronics industry. Helium, with a boiling point of only 4.2 K, is used as a liquid for studying the properties of substances at very low temperatures. It is also combined in an 80:20 mixture with oxygen used by scuba divers, rather than compressed air, when they descend to great depths. Because helium is less soluble in water than N2—a component of compressed air—replacing N2 with He prevents the formation of bubbles in blood vessels, a condition called “the bends” that can occur during rapid ascents. Neon is familiar to all of us as the gas responsible for the red glow in neon lights.
As expected for elements with the same valence electron configuration, the elements in each column of the d block have vertical similarities in chemical behavior. In contrast to the s- and p-block elements, however, elements in the d block also display strong horizontal similarities. The horizontal trends compete with the vertical trends. In further contrast to the p-block elements, which tend to have stable oxidation states that are separated by two electrons, the transition metalsAny element in groups 3–12 in the periodic table. All of the transition elements are metals. have multiple oxidation states that are separated by only one electron.
The p-block elements form stable compounds in oxidation states that tend to be separated by two electrons, whereas the transition metals have multiple oxidation states that are separated by one electron.
The group 6 elements, chromium, molybdenum, and tungsten, illustrate the competition that occurs between these horizontal and vertical trends. For example, the maximum oxidation state for all elements in group 6 is +6, achieved by losing all six valence electrons (recall that Cr has a 4s13d5 valence electron configuration), yet nearly all the elements in the first row of the transition metals, including chromium, form compounds with the dication M2+, and many also form the trication M3+. As a result, the transition metals in group 6 have very different tendencies to achieve their maximum oxidation state. The most common oxidation state for chromium is +3, whereas the most common oxidation state for molybdenum and tungsten is +6.
The d-block elements display both strong vertical and horizontal similarities.
Groups 3 (scandium, lanthanum, actinium), 11 (copper, silver, gold), and 12 (zinc, cadmium, mercury) are the only transition metal groups in which the oxidation state predicted by the valence electron configuration dominates the chemistry of the group. The elements of group 3 have three valence electrons outside an inner closed shell, so their chemistry is almost exclusively that of the M3+ ions produced by losing all three valence electrons. The elements of group 11 have 11 valence electrons in an ns1(n − 1)d10 valence electron configuration, and so all three lose a single electron to form the monocation M+ with a closed (n − 1)d10 electron configuration. Consequently, compounds of Cu+, Ag+, and Au+ are very common, although there is also a great deal of chemistry involving Cu2+. Similarly, the elements of group 12 all have an ns2(n − 1)d10 valence electron configuration, so they lose two electrons to form M2+ ions with an (n − 1)d10 electron configuration; indeed, the most important ions for these elements are Zn2+, Cd2+, and Hg2+. Mercury, however, also forms the dimeric mercurous ion (Hg22+) because of a subtle balance between the energies needed to remove additional electrons and the energy released when bonds are formed. The +3 oxidation state is the most important for the lanthanidesAny of the 14 elements between (cerium) and (lutetium). and for most of the actinidesAny of the 14 elements between (thorium) and (lawrencium)..
Based on the following information, determine the most likely identities for elements D and E.
Given: physical and chemical properties of two elements
Asked for: identities
Strategy:
A Based on the conductivity of the elements, determine whether each is a metal, a nonmetal, or a semimetal. Confirm your prediction from its physical appearance.
B From the compounds each element forms, determine its common oxidation states.
C If the element is a nonmetal, it must be located in the p block of the periodic table. If a semimetal, it must lie along the diagonal line of semimetals from B to At. Transition metals can have two oxidation states separated by one electron.
D From your classification, the oxidation states of the element, and its physical appearance, deduce its identity.
Solution:
Exercise
Based on the following information, determine the most likely identities for elements G and J.
Answer:
The chemical families consist of elements that have the same valence electron configuration and tend to have similar chemistry. The alkali metals (group 1) have ns1 valence electron configurations and form M+ ions, while the alkaline earth metals (group 2) have ns2 valence electron configurations and form M2+ ions. Group 13 elements have ns2np1 valence electron configurations and have an overwhelming tendency to form compounds in the +3 oxidation state. Elements in group 14 have ns2np2 valence electron configurations but exhibit a variety of chemical behaviors because they range from a nonmetal (carbon) to metals (tin/lead). Carbon, the basis of organic compounds, has at least four allotropes with distinct structures: diamond, graphite, fullerenes, and carbon nanotubes. The pnicogens (group 15) all have ns2np3 valence electron configurations; they form compounds in oxidation states ranging from −3 to +5. The chalcogens (group 16) have ns2np4 valence electron configurations and react chemically by either gaining two electrons or by formally losing four or six electrons. The halogens (group 17) all have ns2np5 valence electron configurations and are diatomic molecules that tend to react chemically by accepting a single electron. The noble gases (group 18) are monatomic gases that are chemically quite unreactive due to the presence of a filled shell of electrons. The transition metals (groups 3–10) contain partially filled sets of d orbitals, and the lanthanides and the actinides are those groups in which f orbitals are being filled. These groups exhibit strong horizontal similarities in behavior. Many of the transition metals form M2+ ions, whereas the chemistry of the lanthanides and actinides is dominated by M3+ ions.
Of the group 1 elements, which would you expect to be the best reductant? Why? Would you expect boron to be a good reductant? Why or why not?
Classify each element as a metal, a nonmetal, or a semimetal: Hf, I, Tl, S, Si, He, Ti, Li, and Sb. Which would you expect to be good electrical conductors? Why?
Classify each element as a metal, a nonmetal, or a semimetal: Au, Bi, P, Kr, V, Na, and Po. Which would you expect to be good electrical insulators? Why?
Of the elements Kr, Xe, and Ar, why does only xenon form an extensive series of compounds? Would you expect Xe2+ to be a good oxidant? Why or why not?
Identify each statement about the halogens as either true or false and explain your reasoning.
Nitrogen forms compounds in the +5, +4, +3, +2, and −3 oxidation states, whereas Bi forms ions only in the +5 and +3 oxidation states. Propose an explanation for the differences in behavior.
Of the elements Mg, Al, O, P, and Ne, which would you expect to form covalent halides? Why? How do the melting points of covalent halides compare with those of ionic halides?
Of the elements Li, Ga, As, and Xe, would you expect to form ionic chlorides? Explain your reasoning. Which are usually more volatile—ionic or covalent halides? Why?
Predict the relationship between the oxidative strength of the oxoanions of bromine—BrOn− (n = 1–4)—and the number of oxygen atoms present (n). Explain your reasoning.
The stability of the binary hydrides of the chalcogens decreases in the order H2O > H2S > H2Se > H2Te. Why?
Of the elements O, Al, H, and Cl, which will form a compound with nitrogen in a positive oxidation state? Write a reasonable chemical formula for an example of a binary compound with each element.
How do you explain the differences in chemistry observed for the group 14 elements as you go down the column? Classify each group 14 element as a metal, a nonmetal, or a semimetal. Do you expect the group 14 elements to form covalent or ionic compounds? Explain your reasoning.
Why is the chemistry of the group 13 elements less varied than the chemistry of the group 15 elements? Would you expect the chemistry of the group 13 elements to be more or less varied than that of the group 17 elements? Explain your reasoning.
If you needed to design a substitute for BaSO4, the barium milkshake used to examine the large and small intestine by x-rays, would BeSO4 be an inappropriate substitute? Explain your reasoning.
The alkali metals have an ns1 valence electron configuration, and consequently they tend to lose an electron to form ions with +1 charge. Based on their valence electron configuration, what other kind of ion can the alkali metals form? Explain your answer.
Would Mo or W be the more appropriate biological substitute for Cr? Explain your reasoning.
Nitrogen will have a positive oxidation state in its compounds with O and Cl, because both O and Cl are more electronegative than N. Reasonable formulas for binary compounds are: N2O5 or N2O3 and NCl3.
Write a balanced equation for formation of XeO3 from elemental Xe and O2. What is the oxidation state of Xe in XeO3? Would you expect Ar to undergo an analogous reaction? Why or why not?
Which of the p-block elements exhibit the greatest variation in oxidation states? Why? Based on their valence electron configurations, identify these oxidation states.
Based on its valence electron configuration, what are the three common oxidation states of selenium? In a binary compound, what atoms bonded to Se will stabilize the highest oxidation state? the lowest oxidation state?
Would you expect sulfur to be readily oxidized by HCl? Why or why not? Would you expect phosphorus to be readily oxidized by sulfur? Why or why not?
What are the most common oxidation states for the pnicogens? What factors determine the relative stabilities of these oxidation states for the lighter and the heavier pnicogens? What is likely to be the most common oxidation state for phosphorus and arsenic? Why?
Of the compounds NF3, NCl3, and NI3, which would be the least stable? Explain your answer. Of the ions BrO−, ClO−, or FO−, which would be the least stable? Explain your answer.
In an attempt to explore the chemistry of the superheavy element ununquadium, Z = 114, you isolated two distinct salts by exhaustively oxidizing metal samples with chlorine gas. These salts are found to have the formulas MCl2 and MCl4. What would be the name of ununquadium using Mendeleev’s eka-notation?
Would you expect the compound CCl2 to be stable? SnCl2? Why or why not?
A newly discovered element (Z) is a good conductor of electricity and reacts only slowly with oxygen. Reaction of 1 g of Z with oxygen under three different sets of conditions gives products with masses of 1.333 g, 1.668 g, and 1.501 g, respectively. To what family of elements does Z belong? What is the atomic mass of the element?
An unknown element (Z) is a dull, brittle powder that reacts with oxygen at high temperatures. Reaction of 0.665 gram of Z with oxygen under two different sets of conditions forms gaseous products with masses of 1.328 g and 1.660 g. To which family of elements does Z belong? What is the atomic mass of the element?
Why are the alkali metals such powerful reductants? Would you expect Li to be able to reduce H2? Would Li reduce V? Why or why not?
What do you predict to be the most common oxidation state for Au, Sc, Ag, and Zn? Give the valence electron configuration for each element in its most stable oxidation state.
Complete the following table.
Mg | C | Ne | Fe | Br | |
---|---|---|---|---|---|
Valence Electron Configuration | |||||
Common Oxidation States | |||||
Oxidizing Strength |
Use the following information to identify elements T, X, D, and Z. Element T reacts with oxygen to form at least three compounds: TO, T2O3, and TO2. Element X reacts with oxygen to form XO2, but X is also known to form compounds in the +2 oxidation state. Element D forms D2O3, and element Z reacts vigorously and forms Z2O. Electrical conductivity measurements showed that element X exhibited electrical conductivity intermediate between metals and insulators, while elements T, D, and Z were good conductors of electricity. Element T is a hard, lustrous, silvery metal, element X is a blue-gray metal, element D is a light, silvery metal, and element Z is a soft, low-melting metal.
Predict whether Cs, F2, Al, and He will react with oxygen. If a reaction will occur, identify the products.
Predict whether K, Ar, O, and Al will react with Cl2. If a reaction will occur, identify the products.
Use the following information to identify elements X, T, and Z.
Adding a reactive metal to water in the presence of oxygen results in a fire. In the absence of oxygen, the addition of 551 mg of the metal to water produces 6.4 mg of hydrogen gas. Treatment of 2.00 g of this metal with 6.3 g of Br2 results in the formation of 3.86 g of an ionic solid. To which chemical family does this element belong? What is the identity of the element? Write and balance the chemical equation for the reaction of water with the metal to form hydrogen gas.
2 Xe + 3 O2 → 2 XeO3
The oxidation state of xenon in XeO3 is +6. No, Ar is much more difficult to oxidize than Xe.
The valence electron configuration of Se is [Ar]4s23d104p4. Its common oxidation states are: +6, due to loss of all six electrons in the 4s and 4p subshells; +4, due to loss of only the four 4p electrons; and −2, due to addition of two electrons to give an [Ar]4s23d104p6 electron configuration, which is isoelectronic with the following noble gas, Kr. The highest oxidation state (+6) will be stabilized by bonds to highly electronegative atoms such as F (SeF6) and O (SeO3), while the lowest oxidation state will be stabilized in covalent compounds by bonds to less electronegative atoms such as H (H2Se) or C [(CH3)2Se], or in ionic compounds with cations of electropositive metals (Na2Se).
All of the pnicogens have ns2np3 valence electron configurations. The pnicogens therefore tend to form compounds in three oxidation states: +5, due to loss of all five valence electrons; +3, due to loss of the three np3 electrons; and −3, due to addition of three electrons to give a closed shell electron configuration. Bonds to highly electronegative atoms such as F and O will stabilize the higher oxidation states, while bonds to less electronegative atoms such as H and C will stabilize the lowest oxidation state, as will formation of an ionic compound with the cations of electropositive metals. The most common oxidation state for phosphorus and arsenic is +5.
Uuq = eka-lead
The ratios of the masses of the element to the mass of oxygen give empirical formulas of ZO, Z2O3, and ZO2. The high electrical conductivity of the element immediately identifies it as a metal, and the existence of three oxides of the element with oxidation states separated by only one electron identifies it as a transition metal. If 1 g of Z reacts with 0.33 g O2 to give ZO, the balanced equation for the reaction must be 2 Z + O2 → 2 ZO. Using M to represent molar mass, the ratio of the molar masses of ZO and Z is therefore:
MZO:MZ = (MZ + MO): MZ = (MZ + 16.0): MZ = 1.33:1 = 1.33.Solving for MZ gives a molar mass of 48 g/mol and an atomic mass of 48 amu for Z, which identifies it as titanium.
Alkali metals are powerful reductants because they have a strong tendency to lose their ns1 valence electron, as reflected in their low first ionization energies and electronegativities. Lithium has a more positive electron affinity than hydrogen and a substantially lower first ionization energy, so we expect lithium to reduce hydrogen. Transition metals have low electron affinities and do not normally form compounds in negative oxidation states. Therefore, we do not expect lithium to reduce vanadium.
Mg | C | Ne | Fe | Br | |
---|---|---|---|---|---|
Valence Electron Configuration | 3s2 | 2s22p2 | 2s22p6 | 4s23d6 | 4s24p5 |
Common Oxidation States | +2 | −4, +4 | 0 | +2, +3 | −1, +1, +3, +5, +7 |
Oxidizing Strength | None | Weak | None | None | Strong |
4 Cs(s) + O2(g) → 2 Cs2O(s) 2 F2(g) + O2(g) → OF2(g) 4 Al(s) + 3 O2(g) → 2 Al2O3(s) He + O2(g) → no reaction
Of the more than 100 known elements, approximately 28 are known to be essential for the growth of at least one biological species, and only 19 are essential to humans. (For more information on essential elements, see , , and .) What makes some elements essential to an organism and the rest nonessential? There are at least two reasons:
As you can see in , many of the elements that are abundant in Earth’s crust are nevertheless not found in an easily accessible form (e.g., as ions dissolved in seawater). Instead, they tend to form insoluble oxides, hydroxides, or carbonate salts. Although silicon is the second most abundant element in Earth’s crust, SiO2 and many silicate minerals are insoluble, so they are not easily absorbed by living tissues. This is also the case for iron and aluminum, which form insoluble hydroxides. Many organisms have therefore developed elaborate strategies to obtain iron from their environment. In contrast, molybdenum and iodine, though not particularly abundant, are highly soluble—molybdenum as molybdate (MoO42−) and iodine as iodide (I−) and iodate (IO3−)—and thus are more abundant in seawater than iron. Not surprisingly, both molybdenum and iodine are used by many organisms.
Table 7.6 Relative Abundance of Some Essential Elements in Earth’s Crust and Oceans
Element* | Crust (ppm; average) | Seawater (mg/L = ppm) |
---|---|---|
O | 461,000 | 857,000 |
Si | 282,000 | 2.2 |
Al | 82,300 | 0.002 |
Fe | 56,300 | 0.002 |
Ca | 41,500 | 412 |
Na | 23,600 | 10,800 |
Mg | 23,300 | 1290 |
K | 20,900 | 399 |
H | 1400 | 108,000 |
P | 1050 | 0.06 |
Mn | 950 | 0.0002 |
F | 585 | 1.3 |
S | 350 | 905 |
C | 200 | 28 |
Cl | 145 | 19,400 |
V | 120 | 0.0025 |
Cr | 102 | 0.0003 |
Ni | 84 | 0.00056 |
Zn | 70 | 0.0049 |
Cu | 60 | 0.00025 |
Co | 25 | 0.00002 |
Li | 20 | 0.18 |
N | 19 | 0.5 |
Br | 2.4 | 67.3 |
Mo | 1.2 | 0.01 |
I | 0.45 | 0.06 |
Se | 0.05 | 0.0002 |
*Elements in boldface are known to be essential to humans. |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
Fortunately, many of the elements essential to life are necessary in only small amounts. ( lists trace elements in humans.) Even so, elements that are present in trace amounts can exert large effects on the health of an organism. Such elements function as part of an amplification mechanismA process by which elements that are present in trace amouts can exert large effects on the health of an organism., in which a molecule containing a trace element is an essential part of a larger molecule that acts in turn to regulate the concentrations of other molecules, and so forth. The amplification mechanism enables small variations in the concentration of the trace element to have large biochemical effects.
Essential trace elementsElements that are required for the growth of most organisms. in mammals can have four general roles: (1) they can behave as macrominerals, (2) they can participate in the catalysis of group-transfer reactions, (3) they can participate in oxidation–reduction reactions, or (4) they can serve as structural components.
The macromineralsAny of the six essential elements (Na, Mg, K, Ca, Cl, and P) that provide essential ions in body fluids and form the major structural components of the body.—Na, Mg, K, Ca, Cl, and P—are found in large amounts in biological tissues and are present as inorganic compounds, either dissolved or precipitated. All form monatomic ions (Na+, Mg2+, K+, Ca2+, Cl−) except for phosphorus, which is found as the phosphate ion (PO43−). Recall that calcium salts are used by many organisms as structural materials, such as in bone [hydroxyapatite, Ca5(PO4)3OH]; calcium salts are also in sea shells and egg shells (CaCO3), and they serve as a repository for Ca2+ in plants (calcium oxalate).
The body fluids of all multicellular organisms contain relatively high concentrations of these ions. Some ions (Na+, Ca2+, and Cl−) are located primarily in extracellular fluids such as blood plasma, whereas K+, Mg2+, and phosphate are located primarily in intracellular fluids. Substantial amounts of energy are required to selectively transport these ions across cell membranes. The selectivity of these ion pumpsA complex assembly of proteins that selectively transports ions across cell membranes toward the side with the higher concentration. is based on differences in ionic radius () and ionic charge.
Maintaining optimum levels of macrominerals is important because temporary changes in their concentration within a cell affect biological functions. For example, nerve impulse transmission requires a sudden, reversible increase in the amount of Na+ that flows into the nerve cell. Similarly, when hormones bind to a cell, they can cause Ca2+ ions to enter that cell. In a complex series of reactions, the Ca2+ ions trigger events such as muscle contraction, the release of neurotransmitters, or the secretion of hormones. When people who exercise vigorously for long periods of time overhydrate with water, low blood salt levels can result in a condition known as hyponatremia, which causes nausea, fatigue, weakness, seizures, and even death. For this reason, athletes should hydrate with a sports drink containing salts, not just water.
Trace metal ions also play crucial roles in many biological group-transfer reactionsA reaction in which a recognizable functional group is transferred from one molecule to another.. In these reactions, a recognizable functional group, such as a phosphoryl unit (−PO3−), is transferred from one molecule to another. In this example,
Equation 7.18
a unit is transferred from an alkoxide (RO−) to hydroxide (OH−). To neutralize the negative charge on the molecule that is undergoing the reaction, many biological reactions of this type require the presence of metal ions, such as Zn2+, Mn2+, Ca2+, or Mg2+ and occasionally Ni2+ or Fe3+. The effectiveness of the metal ion depends largely on its charge and radius.
Zinc is an important component of enzymes that catalyze the hydrolysis of proteins, the addition of water to CO2 to produce HCO3− and H+, and most of the reactions involved in DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) synthesis, repair, and replication. Consequently, zinc deficiency has serious adverse effects, including abnormal growth and sexual development and a loss of the sense of taste.
A third important role of trace elements is to transfer electrons in biological oxidation–reduction reactions. Iron and copper, for example, are found in proteins and enzymes that participate in O2 transport, the reduction of O2, the oxidation of organic molecules, and the conversion of atmospheric N2 to NH3. These metals usually transfer one electron per metal ion by alternating between oxidation states, such as 3+/2+ (Fe) or 2+/1+ (Cu).
Because most transition metals have multiple oxidation states separated by only one electron, they are uniquely suited to transfer multiple electrons one at a time. Examples include molybdenum (+6/+5/+4), which is widely used for two-electron oxidation–reduction reactions, and cobalt (+3/+2/+1), which is found in vitamin B12. In contrast, many of the p-block elements are well suited for transferring two electrons at once. Selenium (+4/+2), for example, is found in the enzyme that catalyzes the oxidation of glutathione (GSH) to its disulfide form (GSSG): 2 GSH + H2O2 → 2 H2O + GSSG. (For more information about glutathione and its disulfide form, see , , Example 20.)
Trace elements also act as essential structural components of biological tissues or molecules. In many systems where trace elements do not change oxidation states or otherwise participate directly in biochemical reactions, it is often assumed, though frequently with no direct evidence, that the element stabilizes a particular three-dimensional structure of the biomolecule in which it is found. One example is a sugar-binding protein containing Mn2+ and Ca2+ that is a part of the biological defense system of certain plants. Other examples include enzymes that require Zn2+ at one site for activity to occur at a different site on the molecule. Some nonmetallic elements, such as F−, also appear to have structural roles. Fluoride, for example, displaces the hydroxide ion from hydroxyapatite in bone and teeth to form fluoroapatite [Ca5(PO4)3F]. Fluoroapatite is less soluble in acid and provides increased resistance to tooth decay. Another example of a nonmetal that plays a structural role is iodine, which in humans is found in only one molecule, the thyroid hormone thyroxine. When a person’s diet does not contain sufficient iodine, the thyroid glands in their neck become greatly enlarged, leading to a condition called goiter. Because iodine is found primarily in ocean fish and seaweed, many of the original settlers of the American Midwest developed goiter due to the lack of seafood in their diet. Today most table salt contains small amounts of iodine [actually potassium iodide (KI)] to prevent this problem.
An individual with goiter. In the United States, “iodized salt” prevents the occurrence of goiter.
There is some evidence that tin is an essential element in mammals. Based solely on what you know about the chemistry of tin and its position in the periodic table, predict a likely biological function for tin.
Asked for: likely biological function
Strategy:
From the position of tin in the periodic table, its common oxidation states, and the data in , predict a likely biological function for the element.
Solution:
From its position in the lower part of group 14, we know that tin is a metallic element whose most common oxidation states are +4 and +2. Given the low levels of tin in mammals (140 mg/70 kg human), tin is unlikely to function as a macromineral. Although a role in catalyzing group-transfer reactions or as an essential structural component cannot be ruled out, the most likely role for tin would be in catalyzing oxidation–reduction reactions that involve two-electron transfers. This would take advantage of the ability of tin to have two oxidation states separated by two electrons.
Exercise
Based solely on what you know about the chemistry of vanadium and its position in the periodic table, predict a likely biological function for vanadium.
Answer: Vanadium likely catalyzes oxidation–reduction reactions because it is a first-row transition metal and is likely to have multiple oxidation states.
Many of the elements in the periodic table are essential trace elements that are required for the growth of most organisms. Although they are present in only small quantities, they have important biological effects because of their participation in an amplification mechanism. Macrominerals are present in larger amounts and play structural roles or act as electrolytes whose distribution in cells is tightly controlled. These ions are selectively transported across cell membranes by ion pumps. Other trace elements catalyze group-transfer reactions or biological oxidation–reduction reactions, while others yet are essential structural components of biological molecules.
Give at least one criterion for essential elements involved in biological oxidation–reduction reactions. Which region of the periodic table contains elements that are very well suited for this role? Explain your reasoning.
What are the general biological roles of trace elements that do not have two or more accessible oxidation states?
Problems marked with a ♦ involve multiple concepts.
♦ Most plants, animals, and bacteria use oxygen as the terminal oxidant in their respiration process. In a few locations, however, whole ecosystems have developed in the absence of oxygen, containing creatures that can use sulfur compounds instead of oxygen. List the common oxidation states of sulfur, provide an example of any oxoanions containing sulfur in these oxidation states, and name the ions. Which ion would be the best oxidant?
Titanium is currently used in the aircraft industry and is now used in ships, which operate in a highly corrosive environment. Interest in this metal is due to the fact that titanium is strong, light, and corrosion resistant. The densities of selected elements are given in the following table. Why can an element with an even lower density such as calcium not be used to produce an even lighter structural material?
Element | Density (g/cm3) | Element | Density (g/cm3) |
---|---|---|---|
K | 0.865 | Cr | 7.140 |
Ca | 1.550 | Mn | 7.470 |
Sc | 2.985 | Fe | 7.874 |
Ti | 4.507 | Co | 8.900 |
V | 6.110 | Ni | 8.908 |
♦ The compound Fe3O4 was called lodestone in ancient times because it responds to Earth’s magnetic field and can be used to construct a primitive compass. Today Fe3O4 is commonly called magnetite because it contains both Fe2+ and Fe3+, and the unpaired electrons on these ions align to form tiny magnets. How many unpaired electrons does each ion have? Would you expect to observe magnetic behavior in compounds containing Zn2+? Why or why not? Would you expect Fe or Zn to have the lower third ionization energy? Why?
♦ Understanding trends in periodic properties allows us to predict the properties of individual elements. For example, if we need to know whether francium is a liquid at room temperature (approximately 20°C), we could obtain this information by plotting the melting points of the other alkali metals versus atomic number. Based on the data in the following table, would you predict francium to be a solid, a liquid, or a gas at 20°C?
Li | Na | K | Rb | Cs | |
---|---|---|---|---|---|
Melting Point (°C) | 180 | 97.8 | 63.7 | 39.0 | 28.5 |
Francium is found in minute traces in uranium ores. Is this consistent with your conclusion? Why or why not? Why would francium be found in these ores, but only in small quantities?
Due to its 3s2 3p4 electron configuration, sulfur has three common oxidation states: +6, +4, and −2. Examples of each are: −2 oxidation state, the sulfide anion, S2− or hydrogen sulfide, H2S; +4 oxidation state, the sulfite ion, SO32−; +6 oxidation state, the sulfate ion, SO42−. The sulfate ion would be the best biological oxidant, because it can accept the greatest number of electrons.
Iron(II) has four unpaired electrons, and iron(III) has five unpaired electrons. Compounds of Zn2+ do not exhibit magnetic behavior, because the Zn2+ ion has no unpaired electrons. The third ionization potential of zinc is larger than that of iron, because removing a third electron from zinc requires breaking into the closed 3d10 subshell.
In Chapter 7 "The Periodic Table and Periodic Trends", we described the relationship between the chemical properties and reactivity of an element and its position in the periodic table. In this chapter and Chapter 9 "Molecular Geometry and Covalent Bonding Models", we describe the interactions that hold atoms together in chemical substances, and we examine the factors that determine how the atoms of a substance are arranged in space. Our goal is to understand how the properties of the component atoms in a chemical compound determine the structure and reactivity of the compound.
The properties described in Chapter 6 "The Structure of Atoms" and Chapter 7 "The Periodic Table and Periodic Trends" were properties of isolated atoms, yet most of the substances in our world consist of atoms held together in molecules, ionic compounds, or metallic solids. The properties of these substances depend on not only the characteristics of the component atoms but also how those atoms are bonded to one another.
Carbon and silicon bonding. Both the group 14 elements carbon and silicon form bonds with oxygen, but how they form those bonds results in a vast difference in physical properties. Because of its simple molecular bond, carbon dioxide is a gas that exists as a volatile molecular solid, known as “dry ice,” at temperatures of −78°C and below. Silicon dioxide is a giant covalent structure, whose strong bonds in three dimensions make it a hard, high-melting-point solid, such as quartz.
What you learn in this chapter about chemical bonding and molecular structure will help you understand how different substances with the same atoms can have vastly different physical and chemical properties. For example, oxygen gas (O2) is essential for life, yet ozone (O3) is toxic to cells, although as you learned in Chapter 3 "Chemical Reactions", ozone in the upper atmosphere shields us from harmful ultraviolet light. Moreover, you saw in Chapter 7 "The Periodic Table and Periodic Trends" that diamond is a hard, transparent solid that is a gemstone; graphite is a soft, black solid that is a lubricant; and fullerenes are molecular species with carbon cage structures—yet all of these are composed of carbon. As you learn about bonding, you will also discover why, although carbon and silicon both have ns2np2 valence electron configurations and form dioxides, CO2 is normally a gas that condenses into the volatile molecular solid known as dry ice, whereas SiO2 is a nonvolatile solid with a network structure that can take several forms, including beach sand and quartz crystals.
In Chapter 2 "Molecules, Ions, and Chemical Formulas", we defined a chemical bond as the force that holds atoms together in a chemical compound. We also introduced two idealized types of bonding: covalent bondingA type of chemical bonding in which electrons are shared between atoms in a molecule or polyatomic ion., in which electrons are shared between atoms in a molecule or polyatomic ion, and ionic bondingA type of chemical bonding in which positively and negatively charged ions are held together by electrostatic forces., in which positively and negatively charged ions are held together by electrostatic forces. The concepts of covalent and ionic bonding were developed to explain the properties of different kinds of chemical substances. Ionic compounds, for example, typically dissolve in water to form aqueous solutions that conduct electricity. (For more information about solution conductivity, see Chapter 4 "Reactions in Aqueous Solution", Section 4.1 "Aqueous Solutions".) In contrast, most covalent compounds that dissolve in water form solutions that do not conduct electricity. Furthermore, many covalent compounds are volatile, whereas ionic compounds are not.
Despite the differences in the distribution of electrons between these two idealized types of bonding, all models of chemical bonding have three features in common:
Energy is required to dissociate bonded atoms or ions.
We explore these characteristics further, after briefly describing the energetic factors involved in the formation of an ionic bond.
Chemical bonding is the general term used to describe the forces that hold atoms together in molecules and ions. Two idealized types of bonding are ionic bonding, in which positively and negatively charged ions are held together by electrostatic forces, and covalent bonding, in which electron pairs are shared between atoms. All models of chemical bonding have three common features: atoms form bonds because the products are more stable than the isolated atoms; bonding interactions are characterized by a particular energy (the bond energy or lattice energy), which is the amount of energy required to dissociate the substance into its components; and bonding interactions have an optimal internuclear distance, the bond distance.
Describe the differences between covalent bonding and ionic bonding. Which best describes the bonding in MgCl2 and PF5?
What three features do all chemical bonds have in common?
explained that ionic bonds are formed when positively and negatively charged ions are held together by electrostatic forces. You learned that the energy of the electrostatic attraction (E), a measure of the force’s strength, is inversely proportional to the internuclear distance between the charged particles (r):
Equation 8.1
where each ion’s charge is represented by the symbol Q. The proportionality constant k is equal to 2.31 × 10−28 J·m.This value of k includes the charge of a single electron (1.6022 × 10−19 C) for each ion. The equation can also be written using the charge of each ion, expressed in coulombs (C), incorporated in the constant. In this case, the proportionality constant, k, equals 8.999 × 109 J·m/C2. In the example given, Q1 = +1(1.6022 × 10−19 C) and Q2 = −1(1.6022 × 10−19 C). If Q1 and Q2 have opposite signs (as in NaCl, for example, where Q1 is +1 for Na+ and Q2 is −1 for Cl−), then E is negative, which means that energy is released when oppositely charged ions are brought together from an infinite distance to form an isolated ion pair. As shown by the green curve in the lower half of , predicts that the maximum energy is released when the ions are infinitely close to each other, at r = 0. Because ions occupy space, however, they cannot be infinitely close together. At very short distances, repulsive electron–electron interactions between electrons on adjacent ions become stronger than the attractive interactions between ions with opposite charges, as shown by the red curve in the upper half of . The total energy of the system is a balance between the attractive and repulsive interactions. The purple curve in shows that the total energy of the system reaches a minimum at r0, the point where the electrostatic repulsions and attractions are exactly balanced. This distance is the same as the experimentally measured bond distance.
Energy is released when a bond is formed.
Figure 8.1 A Plot of Potential Energy versus Internuclear Distance for the Interaction between a Gaseous Na+ Ion and a Gaseous Cl− Ion
The energy of the system reaches a minimum at a particular distance (r0) when the attractive and repulsive interactions are balanced.
Let’s consider the energy released when a gaseous Na+ ion and a gaseous Cl− ion are brought together from r = ∞ to r = r0. Given that the observed gas-phase internuclear distance is 236 pm, the energy change associated with the formation of an ion pair from an Na+(g) ion and a Cl−(g) ion is as follows:
Equation 8.2
The negative value indicates that energy is released. To calculate the energy change in the formation of a mole of NaCl pairs, we need to multiply the energy per ion pair by Avogadro’s number:
Equation 8.3
This is the energy released when 1 mol of gaseous ion pairs is formed, not when 1 mol of positive and negative ions condenses to form a crystalline lattice. Because of long-range interactions in the lattice structure, this energy does not correspond directly to the lattice energy of the crystalline solid. However, the large negative value indicates that bringing positive and negative ions together is energetically very favorable, whether an ion pair or a crystalline lattice is formed.
We summarize the important points about ionic bonding:
Calculate the amount of energy released when 1 mol of gaseous Li+F− ion pairs is formed from the separated ions. The observed internuclear distance in the gas phase is 156 pm.
Given: cation and anion, amount, and internuclear distance
Asked for: energy released from formation of gaseous ion pairs
Strategy:
Substitute the appropriate values into to obtain the energy released in the formation of a single ion pair and then multiply this value by Avogadro’s number to obtain the energy released per mole.
Solution:
Inserting the values for Li+F− into (where Q1 = +1, Q2 = −1, and r = 156 pm), we find that the energy associated with the formation of a single pair of Li+F− ions is
Then the energy released per mole of Li+F− ion pairs is
Because Li+ and F− are smaller than Na+ and Cl− (see ), the internuclear distance in LiF is shorter than in NaCl. Consequently, in accordance with , much more energy is released when 1 mol of gaseous Li+F− ion pairs is formed (−891 kJ/mol) than when 1 mol of gaseous Na+Cl− ion pairs is formed (−589 kJ/mol).
Exercise
Calculate the amount of energy released when 1 mol of gaseous MgO ion pairs is formed from the separated ions. The internuclear distance in the gas phase is 175 pm.
Answer: −3180 kJ/mol = −3.18 × 103 kJ/mol
The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance. The total energy of the system is a balance between the repulsive interactions between electrons on adjacent ions and the attractive interactions between ions with opposite charges.
Describe the differences in behavior between NaOH and CH3OH in aqueous solution. Which solution would be a better conductor of electricity? Explain your reasoning.
What is the relationship between the strength of the electrostatic attraction between oppositely charged ions and the distance between the ions? How does the strength of the electrostatic interactions change as the size of the ions increases?
Which will result in the release of more energy: the interaction of a gaseous sodium ion with a gaseous oxide ion or the interaction of a gaseous sodium ion with a gaseous bromide ion? Why?
Which will result in the release of more energy: the interaction of a gaseous chloride ion with a gaseous sodium ion or a gaseous potassium ion? Explain your answer.
What are the predominant interactions when oppositely charged ions are
Several factors contribute to the stability of ionic compounds. Describe one type of interaction that destabilizes ionic compounds. Describe the interactions that stabilize ionic compounds.
What is the relationship between the electrostatic attractive energy between charged particles and the distance between the particles?
The interaction of a sodium ion and an oxide ion. The electrostatic attraction energy between ions of opposite charge is directly proportional to the charge on each ion (Q1 and Q2 in ). Thus, more energy is released as the charge on the ions increases (assuming the internuclear distance does not increase substantially). A sodium ion has a +1 charge; an oxide ion, a −2 charge; and a bromide ion, a −1 charge. For the interaction of a sodium ion with an oxide ion, Q1 = +1 and Q2 = −2, whereas for the interaction of a sodium ion with a bromide ion, Q1 = +1 and Q2 = −1. The larger value of Q1 × Q2 for the sodium ion–oxide ion interaction means it will release more energy.
How does the energy of the electrostatic interaction between ions with charges +1 and −1 compare to the interaction between ions with charges +3 and −1 if the distance between the ions is the same in both cases? How does this compare with the magnitude of the interaction between ions with +3 and −3 charges?
How many grams of gaseous MgCl2 are needed to give the same electrostatic attractive energy as 0.5 mol of gaseous LiCl? The ionic radii are Li+ = 76 pm, Mg+2 = 72 pm, and Cl− = 181 pm.
Sketch a diagram showing the relationship between potential energy and internuclear distance (from r = ∞ to r = 0) for the interaction of a bromide ion and a potassium ion to form gaseous KBr. Explain why the energy of the system increases as the distance between the ions decreases from r = r0 to r = 0.
Calculate the magnitude of the electrostatic attractive energy (E, in kilojoules) for 85.0 g of gaseous SrS ion pairs. The observed internuclear distance in the gas phase is 244.05 pm.
What is the electrostatic attractive energy (E, in kilojoules) for 130 g of gaseous HgI2? The internuclear distance is 255.3 pm.
At r < r0, the energy of the system increases due to electron–electron repulsions between the overlapping electron distributions on adjacent ions. At very short internuclear distances, electrostatic repulsions between adjacent nuclei also become important.
Recall from that the reaction of a metal with a nonmetal usually produces an ionic compound; that is, electrons are transferred from the metal (the reductant) to the nonmetal (the oxidant). Ionic compounds are usually rigid, brittle, crystalline substances with flat surfaces that intersect at characteristic angles. They are not easily deformed, and they melt at relatively high temperatures. NaCl, for example, melts at 801°C. These properties result from the regular arrangement of the ions in the crystalline lattice and from the strong electrostatic attractive forces between ions with opposite charges.
While has demonstrated that the formation of ion pairs from isolated ions releases large amounts of energy, even more energy is released when these ion pairs condense to form an ordered three-dimensional array (). In such an arrangement each cation in the lattice is surrounded by more than one anion (typically four, six, or eight) and vice versa, so it is more stable than a system consisting of separate pairs of ions, in which there is only one cation–anion interaction in each pair. Note that r0 may differ between the gas-phase dimer and the lattice.
An ionic lattice is more stable than a system consisting of separate ion pairs.
The lattice energy of nearly any ionic solid can be calculated rather accurately using a modified form of :
Equation 8.4
U, which is always a positive number, represents the amount of energy required to dissociate 1 mol of an ionic solid into the gaseous ions. If we assume that ΔV = 0, then the lattice energy, U, is approximately equal to the change in enthalpy, ΔH (see , ):
Equation 8.5
As before, Q1 and Q2 are the charges on the ions and r0 is the internuclear distance. We see from that lattice energy is directly related to the product of the ion charges and inversely related to the internuclear distance. The value of the constant k′ depends on the specific arrangement of ions in the solid lattice and their valence electron configurations, topics that will be discussed in more detail in . Representative values for calculated lattice energies, which range from about 600 to 10,000 kJ/mol, are listed in . Energies of this magnitude can be decisive in determining the chemistry of the elements.
Table 8.1 Representative Calculated Lattice Energies
Substance | U (kJ/mol) |
---|---|
NaI | 682 |
CaI2 | 1971 |
MgI2 | 2293 |
NaOH | 887 |
Na2O | 2481 |
NaNO3 | 755 |
Ca3(PO4)2 | 10,602 |
CaCO3 | 2804 |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
Because the lattice energy depends on the product of the charges of the ions, a salt having a metal cation with a +2 charge (M2+) and a nonmetal anion with a −2 charge (X2−) will have a lattice energy four times greater than one with M+ and X−, assuming the ions are of comparable size (and have similar internuclear distances). For example, the calculated value of U for NaF is 910 kJ/mol, whereas U for MgO (containing Mg2+ and O2− ions) is 3795 kJ/mol.
Because lattice energy is inversely related to the internuclear distance, it is also inversely proportional to the size of the ions. This effect is illustrated in , which shows that lattice energy decreases for the series LiX, NaX, and KX as the radius of X− increases. Because r0 in is the sum of the ionic radii of the cation and the anion (r0 = r+ + r−), r0 increases as the cation becomes larger in the series, so the magnitude of U decreases. A similar effect is seen when the anion becomes larger in a series of compounds with the same cation.
Figure 8.2 A Plot of Lattice Energy versus the Identity of the Halide for the Lithium, Sodium, and Potassium Halides
Because the ionic radii of the cations decrease in the order K+ > Na+ > Li+ for a given halide ion, the lattice energy decreases smoothly from Li+ to K+. Conversely, for a given alkali metal ion, the fluoride salt always has the highest lattice energy and the iodide salt the lowest.
Lattice energies are highest for substances with small, highly charged ions.
Arrange GaP, BaS, CaO, and RbCl in order of increasing lattice energy.
Given: four compounds
Asked for: order of increasing lattice energy
Strategy:
Using , predict the order of the lattice energies based on the charges on the ions. For compounds with ions with the same charge, use the relative sizes of the ions to make this prediction.
Solution:
The compound GaP, which is used in semiconductor electronics, contains Ga3+ and P3− ions; the compound BaS contains Ba2+ and S2− ions; the compound CaO contains Ca2+ and O2− ions; and the compound RbCl has Rb+ and Cl− ions. We know from that lattice energy is directly proportional to the product of the ionic charges. Consequently, we expect RbCl, with a (−1)(+1) term in the numerator, to have the lowest lattice energy, and GaP, with a (+3)(−3) term, the highest. To decide whether BaS or CaO has the greater lattice energy, we need to consider the relative sizes of the ions because both compounds contain a +2 metal ion and a −2 chalcogenide ion. Because Ba2+ lies below Ca2+ in the periodic table, Ba2+ is larger than Ca2+. Similarly, S2− is larger than O2−. Because the cation and the anion in BaS are both larger than the corresponding ions in CaO, the internuclear distance is greater in BaS and its lattice energy will be lower than that of CaO. The order of increasing lattice energy is RbCl < BaS < CaO < GaP.
Exercise
Arrange InAs, KBr, LiCl, SrSe, and ZnS in order of decreasing lattice energy.
Answer: InAs > ZnS > SrSe > LiCl > KBr
The magnitude of the forces that hold an ionic substance together has a dramatic effect on many of its properties. The melting pointThe temperature at which the individual ions in a lattice or the individual molecules in a covalent compound have enough kinetic energy to overcome the attractive forces that hold them together in the solid., for example, is the temperature at which the individual ions have enough kinetic energy to overcome the attractive forces that hold them in place. At the melting point, the ions can move freely, and the substance becomes a liquid. Thus melting points vary with lattice energies for ionic substances that have similar structures. The melting points of the sodium halides (), for example, decrease smoothly from NaF to NaI, following the same trend as seen for their lattice energies (). Similarly, the melting point of MgO is 2825°C, compared with 996°C for NaF, reflecting the higher lattice energies associated with higher charges on the ions. In fact, because of its high melting point, MgO is used as an electrical insulator in heating elements for electric stoves.
Figure 8.3 A Plot of Melting Point versus the Identity of the Halide for the Sodium Halides
The melting points follow the same trend as the magnitude of the lattice energies in .
The hardnessThe resistance of ionic materials to scratching or abrasion. of ionic materials—that is, their resistance to scratching or abrasion—is also related to their lattice energies. Hardness is directly related to how tightly the ions are held together electrostatically, which, as we saw, is also reflected in the lattice energy. As an example, MgO is harder than NaF, which is consistent with its higher lattice energy.
In addition to determining melting point and hardness, lattice energies affect the solubilities of ionic substances in water. In general, the higher the lattice energy, the less soluble a compound is in water. For example, the solubility of NaF in water at 25°C is 4.13 g/100 mL, but under the same conditions, the solubility of MgO is only 0.65 mg/100 mL, meaning that it is essentially insoluble.
High lattice energies lead to hard, insoluble compounds with high melting points.
In principle, lattice energies could be measured by combining gaseous cations and anions to form an ionic solid and then measuring the heat evolved. Unfortunately, measurable quantities of gaseous ions have never been obtained under conditions where heat flow can be measured. Instead, lattice energies are found using the experimentally determined enthalpy changes for other chemical processes, Hess’s law, and a thermochemical cycle called the Born–Haber cycleA thermochemical cycle that decribes the process in which an ionic solid is conceptually formed from its component elements in a stepwise manner., similar to those introduced in . Developed by Max Born and Fritz Haber in 1919, the Born–Haber cycle describes a process in which an ionic solid is conceptually formed from its component elements in a stepwise manner.
Let’s use the Born–Haber cycle to determine the lattice energy of CsF(s). CsF is a nearly ideal ionic compound because Cs is the least electronegative element that is not radioactive and F is the most electronegative element. To construct a thermochemical cycle for the formation of CsF, we need to know its enthalpy of formation, ΔHf, which is defined by the following chemical reaction:
Equation 8.6
Because enthalpy is a state function, the overall ΔH for a series of reactions is the sum of the values of ΔH for the individual reactions. (For more information about state functions and Hess’s law, see , .) We can therefore use a thermochemical cycle to determine the enthalpy change that accompanies the formation of solid CsF from the parent elements (not ions).
The Born–Haber cycle for calculating the lattice energy of cesium fluoride is shown in . This particular cycle consists of six reactions, plus the following five reactions:
Figure 8.4 The Born–Haber Cycle Illustrating the Enthalpy Changes Involved in the Formation of Solid Cesium Fluoride from Its Elements
Reaction 1
This equation describes the sublimationThe conversion of a solid directly to a gas (without an intervening liquid phase). of elemental cesium, the conversion of the solid directly to a gas. The accompanying enthalpy change is called the enthalpy of sublimation(ΔHsub)The enthalpy change that accompanies the conversion of a solid directly to a gas. () and is always positive because energy is required to sublime a solid.
Table 8.2 Selected Enthalpies of Sublimation at 298 K
Substance | ΔHsub (kJ/mol) |
---|---|
Li | 159.3 |
Na | 107.5 |
K | 89.0 |
Rb | 80.9 |
Cs | 76.5 |
Be | 324.0 |
Mg | 147.1 |
Ca | 177.8 |
Sr | 164.4 |
Ba | 180.0 |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
Reaction 2
This equation describes the ionization of cesium, so the enthalpy change is the first ionization energy of cesium. Recall from that energy is needed to ionize any neutral atom. Hence, regardless of the compound, the enthalpy change for this portion of the Born–Haber cycle is always positive.
Reaction 3
This equation describes the dissociation of fluorine molecules into fluorine atoms, where D is the energy required for dissociation to occur (). We need to dissociate only mol of F2(g) molecules to obtain 1 mol of F(g) atoms. The ΔH for this reaction, too, is always positive because energy is required to dissociate any stable diatomic molecule into the component atoms.
Table 8.3 Selected Bond Dissociation Enthalpies at 298 K
Substance | D (kJ/mol) |
---|---|
H2(g) | 436.0 |
N2(g) | 945.3 |
O2(g) | 498.4 |
F2(g) | 158.8 |
Cl2(g) | 242.6 |
Br2(g) | 192.8 |
I2(g) | 151.1 |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
Reaction 4
This equation describes the formation of a gaseous fluoride ion from a fluorine atom; the enthalpy change is the electron affinity of fluorine. Recall from that electron affinities can be positive, negative, or zero. In this case, ΔH is negative because of the highly negative electron affinity of fluorine.
Reaction 5
This equation describes the formation of the ionic solid from the gaseous ions. Because Reaction 5 is the reverse of the equation used to define lattice energy and U is defined to be a positive number, ΔH5 is always negative, as it should be in a step that forms bonds.
If the enthalpy of formation of CsF from the elements is known (ΔHf = −553.5 kJ/mol at 298 K), then the thermochemical cycle shown in has only one unknown, the quantity ΔH5 = −U. From Hess’s law, we can write
Equation 8.7
ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5Equation 8.8
−ΔH5 = ΔH1 + ΔH2 + ΔH3 + ΔH4 − ΔHfSubstituting for the individual ΔHs, we obtain
Substituting the appropriate values into this equation gives
Equation 8.9
U = 76.5 kJ/mol + 375.7 kJ/mol + 79.4 kJ/mol + (−328.2 kJ/mole) − (−553.5 kJ/mol) = 756.9 kJ/molU is larger in magnitude than any of the other quantities in . The process we have used to arrive at this value is summarized in .
Table 8.4 Summary of Reactions in the Born–Haber Cycle for the Formation of CsF(s)
Reaction | Enthalpy Change (kJ/mol) | |
---|---|---|
(1) | Cs(s) → Cs(g) | ΔHsub = 76.5 |
(2) | Cs(g) → Cs + (g) + e− | I1 = 375.7 |
(3) | ½F2(g) → F(g) | ½D = 79.4 |
(4) | F(g) + e− → F−(g) | EA = −328.2 |
(5) | Cs + (g) + F−(g) → CsF(s) | −U = −756.9 |
Cs(s) + ½F2(g) → CsF(s) | ΔHf = −553.5 |
may be used as a tool for predicting which ionic compounds are likely to form from particular elements. As we have noted, ΔH1 (ΔHsub), ΔH2 (I), and ΔH3 (D) are always positive numbers, and ΔH2 can be quite large. In contrast, ΔH4 (EA) is comparatively small and can be positive, negative, or zero. Thus the first three terms in make the formation of an ionic substance energetically unfavorable, and the fourth term contributes little either way. The formation of an ionic compound will be exothermic (ΔHf < 0) if and only if ΔH5 (−U) is a large negative number. This means that lattice energy is the most important factor in determining the stability of an ionic compound. Another example is the formation of BaO:
Equation 8.10
whose Born–Haber cycle is compared with that for the formation of CsF in .
Figure 8.5 Comparison of the Enthalpy Changes Involved in the Formation of Solid CsF and BaO from Their Elements
The lattice energy of BaO, with a dipositive cation and a dinegative anion, dominates the Born–Haber cycle.
Reaction 1
More than twice as much energy is required to sublime barium metal (180.0 kJ/mol) as is required to sublime cesium (76.5 kJ/mol).
Reaction 2
Nearly four times the energy is needed to form Ba2+ ions (I1 = 502.9 kJ/mol, I2 = 965.2 kJ/mol, I1 + I2 = 1468.1 kJ/mol) as Cs+ ions (I1 = 375.7 kJ/mol).
Reaction 3
Because the bond energy of O2(g) is 498.4 kJ/mol compared with 158.8 kJ/mol for F2(g), more than three times the energy is needed to form oxygen atoms from O2 molecules as is required to form fluorine atoms from F2.
Reaction 4
Forming gaseous oxide (O2−) ions is energetically unfavorable. Even though adding one electron to an oxygen atom is exothermic (EA1 = −141 kJ/mol), adding a second electron to an O−(g) ion is energetically unfavorable (EA2 = +744 kJ/mol)—so much so that the overall cost of forming O2−(g) from O(g) is energetically prohibitive (EA1 + EA2 = +603 kJ/mol).
If the first four terms in the Born–Haber cycle are all substantially more positive for BaO than for CsF, why does BaO even form? The answer is the formation of the ionic solid from the gaseous ions (Reaction 5):
Reaction 5
Remember from that lattice energies are directly proportional to the product of the charges on the ions and inversely proportional to the internuclear distance. Although the internuclear distances are not significantly different for BaO and CsF (275 and 300 pm, respectively), the larger ionic charges in BaO produce a much higher lattice energy. Substituting values for BaO (ΔHf = −548.0 kJ/mol) into the equation and solving for U gives
Thus U for BaO is slightly more than four times greater than U for CsF. The extra energy released when BaO forms from its ions more than compensates for the additional energy required to form Ba2+(g) and O2−(g) ions from Ba(s) and
If the formation of ionic lattices containing multiply charged ions is so energetically favorable, why does CsF contain Cs+ and F− ions rather than Cs2+ and F2− ions? If we assume that U for a Cs2+F2− salt would be approximately the same as U for BaO, the formation of a lattice containing Cs2+ and F2− ions would release 2291 kJ/mol (3048 kJ/mol − 756.9 kJ/mol) more energy than one containing Cs+ and F− ions. To form the Cs2+ ion from Cs+, however, would require removing a 5p electron from a filled inner shell, which calls for a great deal of energy: I2 = 2234.4 kJ/mol for Cs. Furthermore, forming an F2− ion is expected to be even more energetically unfavorable than forming an O2− ion. Not only is an electron being added to an already negatively charged ion, but because the F− ion has a filled 2p subshell, the added electron would have to occupy an empty high-energy 3s orbital. Cesium fluoride, therefore, is not Cs2+F2− because the energy cost of forming the doubly charged ions would be greater than the additional lattice energy that would be gained.
Lattice energy is usually the most important energy factor in determining the stability of an ionic compound.
Use data from , , , , and to calculate the lattice energy of MgH2.
Given: chemical compound and data from figures and tables
Asked for: lattice energy
Strategy:
A Write a series of stepwise reactions for forming MgH2 from its elements via the gaseous ions.
B Use Hess’s law and data from the specified figures and tables to calculate the lattice energy.
Solution:
A Hess’s law allows us to use a thermochemical cycle (the Born–Haber cycle) to calculate the lattice energy for a given compound. We begin by writing reactions in which we form the component ions from the elements in a stepwise manner and then assemble the ionic solid:
B lists the first and second ionization energies for the period 3 elements [I1(Mg) = 737.7 kJ/mol, I2(Mg) = 1450.7 kJ/mol]. First electron affinities for all elements are given in [EA(H) = −72.8 kJ/mol]. lists selected enthalpies of sublimation [ΔHsub(Mg) = 147.1 kJ/mol]. lists selected bond dissociation energies [D(H2) = 436.0 kJ/mol]. Enthalpies of formation (ΔHf = −75.3 kJ/mol for MgH2) are listed in . From Hess’s law, ΔHf is equal to the sum of the enthalpy changes for Reactions 1–5:
For MgH2, U = 2701.2 kJ/mol. Once again, lattice energy provides the driving force for forming this compound because ΔH1, ΔH2, ΔH3 > 0. When solving this type of problem, be sure to write the chemical equation for each step and double-check that the enthalpy value used for each step has the correct sign for the reaction in the direction it is written.
Exercise
Use data from , , , , and to calculate the lattice energy of Li2O. Remember that the second electron affinity for oxygen [O−(g) + e− → O2−(g)] is positive (+744 kJ/mol; see ).
Answer: 2809 kJ/mol
Ionic compounds have strong electrostatic attractions between oppositely charged ions in a regular array. The lattice energy (U) of an ionic substance is defined as the energy required to dissociate the solid into gaseous ions; U can be calculated from the charges on the ions, the arrangement of the ions in the solid, and the internuclear distance. Because U depends on the product of the ionic charges, substances with di- or tripositive cations and/or di- or trinegative anions tend to have higher lattice energies than their singly charged counterparts. Higher lattice energies typically result in higher melting points and increased hardness because more thermal energy is needed to overcome the forces that hold the ions together. Lattice energies cannot be measured directly but are obtained from a thermochemical cycle called the Born–Haber cycle, in which Hess’s law is used to calculate the lattice energy from the measured enthalpy of formation of the ionic compound, along with other thermochemical data. The Born–Haber cycle can be used to predict which ionic compounds are likely to form. Sublimation, the conversion of a solid directly to a gas, has an accompanying enthalpy change called the enthalpy of sublimation.
If a great deal of energy is required to form gaseous ions, why do ionic compounds form at all?
What are the general physical characteristics of ionic compounds?
Ionic compounds consist of crystalline lattices rather than discrete ion pairs. Why?
What factors affect the magnitude of the lattice energy of an ionic compound? What is the relationship between ionic size and lattice energy?
Which would have the larger lattice energy—an ionic compound consisting of a large cation and a large anion or one consisting of a large anion and a small cation? Explain your answer and any assumptions you made.
How would the lattice energy of an ionic compound consisting of a monovalent cation and a divalent anion compare with the lattice energy of an ionic compound containing a monovalent cation and a monovalent anion, if the internuclear distance was the same in both compounds? Explain your answer.
Which would have the larger lattice energy—CrCl2 or CrCl3—assuming similar arrangements of ions in the lattice? Explain your answer.
Which cation in each pair would be expected to form a chloride salt with the larger lattice energy, assuming similar arrangements of ions in the lattice? Explain your reasoning.
Which cation in each pair would be expected to form an oxide with the higher melting point, assuming similar arrangements of ions in the lattice? Explain your reasoning.
How can a thermochemical cycle be used to determine lattice energies? Which steps in such a cycle require an input of energy?
Although NaOH and CH3OH have similar formulas and molecular masses, the compounds have radically different properties. One has a high melting point, and the other is a liquid at room temperature. Which compound is which and why?
Arrange SrO, PbS, and PrI3 in order of decreasing lattice energy.
Compare BaO and MgO with respect to each of the following properties.
Would you expect the formation of SrO from its component elements to be exothermic or endothermic? Why or why not? How does the valence electron configuration of the component elements help you determine this?
Using the information in Problem 4 and Problem 5, predict whether CaO or MgCl2 will have the higher melting point.
Lattice energy is directly proportional to the product of the ionic charges and inversely proportional to the internuclear distance. Therefore, PrI3 > SrO > PbS.
U = 2522.2 kJ/mol
Despite the fact that Mg2+ is smaller than Ca2+, the higher charge of O2− versus Cl− gives CaO a larger lattice energy than MgCl2. Consequently, we expect CaO to have the higher melting point.
At the beginning of the 20th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbolsA system that can be used to predict the number of bonds formed by most elements in their compounds., often shortened to Lewis dot symbols—that can be used for predicting the number of bonds formed by most elements in their compounds (). Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons. Cesium, for example, has the electron configuration [Xe]6s1, which indicates one valence electron outside a closed shell. In the Lewis dot symbol, this single electron is represented as a single dot:
Figure 8.6 G. N. Lewis and the Octet Rule
(a) Lewis is working in the laboratory. (b) In Lewis’s original sketch for the octet rule, he initially placed the electrons at the corners of a cube rather than placing them as we do now.
To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. Fluorine, for example, with the electron configuration [He]2s22p5, has seven valence electrons, so its Lewis dot symbol is constructed as follows:
The number of dots in the Lewis dot symbol is the same as the number of valence electrons, which is the same as the last digit of the element’s group number in the periodic table. Lewis dot symbols for the elements in period 2 are given in .
Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in . The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds.
Figure 8.7 Lewis Dot Symbols for the Elements in Period 2
Lewis’s major contribution to bonding theory was to recognize that atoms tend to lose, gain, or share electrons to reach a total of eight valence electrons, called an octet. This so-called octet ruleThe tendency for atoms to lose, gain, or share electrons to reach a total of eight valence electrons. explains the stoichiometry of most compounds in the s and p blocks of the periodic table. We now know from quantum mechanics that the number eight corresponds to one ns and three np valence orbitals, which together can accommodate a total of eight electrons. Remarkably, though, Lewis’s insight was made nearly a decade before Rutherford proposed the nuclear model of the atom. An exception to the octet rule is helium, whose 1s2 electron configuration gives it a full n = 1 shell, and hydrogen, which tends to gain or share its one electron to achieve the electron configuration of helium.
Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows:
No dots are shown on Cs+ in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs+ ion, which has the valence electron configuration of Xe, and the F− ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement in that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements.
As you might expect for such a qualitative approach to bonding, there are exceptions to the octet rule, which we describe in . These include molecules in which one or more atoms contain fewer or more than eight electrons. In , however, we explain how to form molecular compounds by completing octets.
One convenient way to predict the number and basic arrangement of bonds in compounds is by using Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called octet rule. Hydrogen, with only two valence electrons, does not obey the octet rule.
The Lewis electron system is a simplified approach for understanding bonding in covalent and ionic compounds. Why do chemists still find it useful?
Is a Lewis dot symbol an exact representation of the valence electrons in an atom or ion? Explain your answer.
How can the Lewis electron dot system help to predict the stoichiometry of a compound and its chemical and physical properties?
Lewis dot symbols allow us to predict the number of bonds atoms will form, and therefore the stoichiometry of a compound. The Lewis structure of a compound also indicates the presence or absence of lone pairs of electrons, which provides information on the compound’s chemical reactivity and physical properties.
We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H2 molecule, which contains a purely covalent bond. Each hydrogen atom in H2 contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered ():
Figure 8.8 Attractive and Repulsive Interactions between Electrons and Nuclei in the Hydrogen Molecule
Electron–electron and proton–proton interactions are repulsive; electron–proton interactions are attractive. At the observed bond distance, the repulsive and attractive interactions are balanced.
A plot of the potential energy of the system as a function of the internuclear distance () shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from r = ∞, until the energy reaches a minimum at r = r0 (the observed internuclear distance in H2 is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between and , which described a system containing two oppositely charged ions. The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities.
Figure 8.9 A Plot of Potential Energy versus Internuclear Distance for the Interaction between Two Gaseous Hydrogen Atoms
At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms.
The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:
Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pairA pair of electrons in a Lewis structure that is shared by two atoms, thus forming a covalent bond.; the other three pairs of electrons on each chlorine atom are called lone pairsA pair of electrons in a Lewis structure that is not involved in covalent bonding.. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bondA covalent bond in which both electrons come from the same atom.. Examples of this type of bonding are presented in when we discuss atoms with less than an octet of electrons.
We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:
The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32−, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
2. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall from that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32−, for example, we add two electrons to the total because of the −2 charge.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom. We explain in that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.
4. Each H atom has a full valence shell of 2 electrons.
5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.
1. With only two atoms in the molecule, there is no central atom.
2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.
3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.
4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:
Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl− is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.
1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:
2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
Six electrons are used, and 6 are left over.
4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. There are no electrons left to place on the central atom.
6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.
Write the Lewis electron structure for each species.
Given: chemical species
Asked for: Lewis electron structures
Strategy:
Use the six-step procedure to write the Lewis electron structure for each species.
Solution:
Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:
Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.
In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:
Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:
Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:
Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:
All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.
Exercise
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Answer:
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of group 17 (the halogens), this number is one; for the elements of group 16 (the chalcogens), it is two; for group 15, three; and for group 14, four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group:
Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule.
Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule. Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom.
Lewis structures explain why the elements of groups 14–17 form neutral compounds with four, three, two, and one bonded atom(s), respectively.
Figure 8.10 The Three Allotropes of Phosphorus: White, Red, and Black
All three forms contain only phosphorus atoms, but they differ in the arrangement and connectivity of their atoms. White phosphorus contains P4 tetrahedra, red phosphorus is a network of linked P8 and P9 units, and black phosphorus forms sheets of six-membered rings. As a result, their physical and chemical properties differ dramatically.
It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal chargeThe difference between the number of valence electrons in a free atom and the number of electrons assigned to it in a particular Lewis electron structure. on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.
To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
For each atom, we then compute a formal charge:
Equation 8.11
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH3) whose Lewis electron structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e− + (6 bonding e−/2)]. Substituting into , we obtain
Equation 8.12
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e− + (2 bonding e−/2)]. Using to calculate the formal charge on hydrogen, we obtain
Equation 8.13
The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH3 molecule.
Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.
Calculate the formal charges on each atom in the NH4+ ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the NH4+ ion. Use the Lewis electron structure of NH4+ to identify the number of bonding and nonbonding electrons associated with each atom and then use to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the NH4+ ion is as follows:
The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using , the formal charge on the nitrogen atom is therefore
Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore
The formal charges on the atoms in the NH4+ ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.
Exercise
Write the formal charges on all atoms in BH4−.
Answer:
If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero.
An atom, molecule, or ion has a formal charge of zero if it has the number of bonds that is typical for that species.
As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO2. Both structures conform to the rules for Lewis electron structures.
1. C is less electronegative than O, so it is the central atom.
2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.
3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.
4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:
This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.
5. No electrons are left for the central atom.
6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:
Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO2 are as follows:
Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.
The Lewis structure with the set of formal charges closest to zero is usually the most stable.
The thiocyanate ion (SCN−), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.
Given: chemical species
Asked for: Lewis electron structures, formal charges, and preferred arrangement
Strategy:
A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN−.
B Calculate the formal charge on each atom using .
C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.
Solution:
A Possible Lewis structures for the SCN− ion are as follows:
B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore In (b), the sulfur atom has two bonding pairs and two lone pairs, giving it a formal charge of zero. In (c), sulfur has a formal charge of +1. Completing our calculations with nitrogen, in (a) the nitrogen atom has three bonding pairs, giving it a formal charge of zero. In (b), the nitrogen atom has two lone pairs and shares two bonding pairs, giving it a formal charge of In (c), nitrogen has a formal charge of −2.
C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2.
Exercise
Salts containing the fulminate ion (CNO−) are used in explosive detonators. Draw three Lewis electron structures for CNO− and use formal charges to predict which is more stable. (Note: N is the central atom.)
Answer:
The second structure is predicted to be more stable.
Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Such is the case for ozone (O3), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn in , if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structuresA Lewis electron structure that has different arrangements of electrons around atoms whose positions do not change.. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (C6H6) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
A Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
B Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
C Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Exercise
The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO2−).
Answer:
Resonance structures are particularly common in oxoanions of the p-block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene.
A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families. Neutral compounds of group 14 elements typically contain four bonds around each atom (a double bond counts as two, a triple bond as three), whereas neutral compounds of group 15 elements typically contain three bonds. In cases where it is possible to write more than one Lewis electron structure with octets around all the nonhydrogen atoms of a compound, the formal charge on each atom in alternative structures must be considered to decide which of the valid structures can be excluded and which is the most reasonable. The formal charge is the difference between the number of valence electrons of the free atom and the number of electrons assigned to it in the compound, where bonding electrons are divided equally between the bonded atoms. The Lewis structure with the lowest formal charges on the atoms is almost always the most stable one. Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures.
Compare and contrast covalent and ionic compounds with regard to
What are the similarities between plots of the overall energy versus internuclear distance for an ionic compound and a covalent compound? Why are the plots so similar?
Which atom do you expect to be the central atom in each of the following species?
Which atom is the central atom in each of the following species?
What is the relationship between the number of bonds typically formed by the period 2 elements in groups 14, 15, and 16 and their Lewis electron structures?
Although formal charges do not represent actual charges on atoms in molecules or ions, they are still useful. Why?
Why are resonance structures important?
In what types of compounds are resonance structures particularly common?
Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate?
Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate?
Based on Lewis dot symbols, predict the preferred oxidation state of Be, F, B, and Cs.
Based on Lewis dot symbols, predict the preferred oxidation state of Br, Rb, O, Si, and Sr.
Based on Lewis dot symbols, predict how many bonds gallium, silicon, and selenium will form in their neutral compounds.
Determine the total number of valence electrons in the following.
Determine the total number of valence electrons in the following.
Draw Lewis electron structures for the following.
Draw Lewis electron structures for the following.
Draw Lewis electron structures for CO2, NO2−, SO2, and NO2+. From your diagram, predict which pair(s) of compounds have similar electronic structures.
Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed.
Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed.
Use Lewis dot symbols to predict whether ICl and NO4− are chemically reasonable formulas.
Draw a plausible Lewis electron structure for a compound with the molecular formula Cl3PO.
Draw a plausible Lewis electron structure for a compound with the molecular formula CH4O.
While reviewing her notes, a student noticed that she had drawn the following structure in her notebook for acetic acid:
Why is this structure not feasible? Draw an acceptable Lewis structure for acetic acid. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures.
A student proposed the following Lewis structure shown for acetaldehyde.
Why is this structure not feasible? Draw an acceptable Lewis structure for acetaldehyde. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures.
Draw the most likely structure for HCN based on formal charges, showing the formal charge on each atom in your structure. Does this compound have any plausible resonance structures? If so, draw one.
Draw the most plausible Lewis structure for NO3−. Does this ion have any other resonance structures? Draw at least one other Lewis structure for the nitrate ion that is not plausible based on formal charges.
At least two Lewis structures can be drawn for BCl3. Using arguments based on formal charges, explain why the most likely structure is the one with three B–Cl single bonds.
Using arguments based on formal charges, explain why the most feasible Lewis structure for SO42− has two sulfur–oxygen double bonds.
At least two distinct Lewis structures can be drawn for N3−. Use arguments based on formal charges to explain why the most likely structure contains a nitrogen–nitrogen double bond.
Is H–O–N=O a reasonable structure for the compound HNO2? Justify your answer using Lewis electron dot structures.
Is H–O=C–H a reasonable structure for a compound with the formula CH2O? Use Lewis electron dot structures to justify your answer.
Explain why the following Lewis structure for SO32− is or is not reasonable.
Draw all the resonance structures for each ion.
[Ar]4s23d104p4
Selenium can accommodate two more electrons, giving the Se2− ion.
[Ar]4s23d104p6
Krypton has a closed shell electron configuration, so it cannot accommodate any additional electrons.
1s22s1
Lithium can accommodate one additional electron in its 2s orbital, giving the Li− ion.
[Kr]5s2
Strontium has a filled 5s subshell, and additional electrons would have to be placed in an orbital with a higher energy. Thus strontium has no tendency to accept an additional electron.
1s1
Hydrogen can accommodate one additional electron in its 1s orbital, giving the H− ion.
Be2+, F−, B3+, Cs+
K is the reductant; S is the oxidant. The final stoichiometry is K2S.
Sr is the reductant; Br is the oxidant. The final stoichiometry is SrBr2.
Al is the reductant; O is the oxidant. The final stoichiometry is Al2O3.
Mg is the reductant; Cl is the oxidant. The final stoichiometry is MgCl2.
The only structure that gives both oxygen and carbon an octet of electrons is the following:
The student’s proposed structure has two flaws: the hydrogen atom with the double bond has four valence electrons (H can only accommodate two electrons), and the carbon bound to oxygen only has six valence electrons (it should have an octet). An acceptable Lewis structure is
The formal charges on the correct and incorrect structures are as follows:
The most plausible Lewis structure for NO3− is:
There are three equivalent resonance structures for nitrate (only one is shown), in which nitrogen is doubly bonded to one of the three oxygens. In each resonance structure, the formal charge of N is +1; for each singly bonded O, it is −1; and for the doubly bonded oxygen, it is 0.
The following is an example of a Lewis structure that is not plausible:
This structure nitrogen has six bonds (nitrogen can form only four bonds) and a formal charge of –1.
With four S–O single bonds, each oxygen in SO42− has a formal charge of −1, and the central sulfur has a formal charge of +2. With two S=O double bonds, only two oxygens have a formal charge of –1, and sulfur has a formal charge of zero. Lewis structures that minimize formal charges tend to be lowest in energy, making the Lewis structure with two S=O double bonds the most probable.
Yes. This is a reasonable Lewis structure, because the formal charge on all atoms is zero, and each atom (except H) has an octet of electrons.
Lewis dot structures provide a simple model for rationalizing the bonding in most known compounds. However, there are three general exceptions to the octet rule: (1) molecules, such as NO, with an odd number of electrons; (2) molecules in which one or more atoms possess more than eight electrons, such as SF6; and (3) molecules such as BCl3, in which one or more atoms possess less than eight electrons.
Because most molecules or ions that consist of s- and p-block elements contain even numbers of electrons, their bonding can be described using a model that assigns every electron to either a bonding pair or a lone pair.Molecules or ions containing d-block elements frequently contain an odd number of electrons, and their bonding cannot adequately be described using the simple approach we have developed so far. Bonding in these compounds will be discussed in Chapter 23 "The ". There are, however, a few molecules containing only p-block elements that have an odd number of electrons. Some important examples are nitric oxide (NO), whose biochemical importance was described in earlier chapters; nitrogen dioxide (NO2), an oxidizing agent in rocket propulsion; and chlorine dioxide (ClO2), which is used in water purification plants. Consider NO, for example. With 5 + 6 = 11 valence electrons, there is no way to draw a Lewis structure that gives each atom an octet of electrons. Molecules such as NO, NO2, and ClO2 require a more sophisticated treatment of bonding, which will be developed in Chapter 9 "Molecular Geometry and Covalent Bonding Models".
The most common exception to the octet rule is a molecule or an ion with at least one atom that possesses more than an octet of electrons. Such compounds are found for elements of period 3 and beyond. Examples from the p-block elements include SF6, a substance used by the electric power industry to insulate high-voltage lines, and the SO42− and PO43− ions.
Let’s look at sulfur hexafluoride (SF6), whose Lewis structure must accommodate a total of 48 valence electrons [6 + (6 × 7) = 48]. If we arrange the atoms and electrons symmetrically, we obtain a structure with six bonds to sulfur; that is, it is six-coordinate. Each fluorine atom has an octet, but the sulfur atom has 12 electrons surrounding it rather than 8.The third step in our procedure for writing Lewis electron structures, in which we place an electron pair between each pair of bonded atoms, requires that an atom have more than 8 electrons whenever it is bonded to more than 4 other atoms.
The octet rule is based on the fact that each valence orbital (typically, one ns and three np orbitals) can accommodate only two electrons. To accommodate more than eight electrons, sulfur must be using not only the ns and np valence orbitals but additional orbitals as well. Sulfur has an [Ne]3s23p43d0 electron configuration, so in principle it could accommodate more than eight valence electrons by using one or more d orbitals. Thus species such as SF6 are often called expanded-valence moleculesA compound with more than an octet of electrons around an atom.. Whether or not such compounds really do use d orbitals in bonding is controversial, but this model explains why compounds exist with more than an octet of electrons around an atom.
There is no correlation between the stability of a molecule or an ion and whether or not it has an expanded valence shell. Some species with expanded valences, such as PF5, are highly reactive, whereas others, such as SF6, are very unreactive. In fact, SF6 is so inert that it has many commercial applications. In addition to its use as an electrical insulator, it is used as the coolant in some nuclear power plants, and it is the pressurizing gas in “unpressurized” tennis balls.
An expanded valence shell is often written for oxoanions of the heavier p-block elements, such as sulfate (SO42−) and phosphate (PO43−). Sulfate, for example, has a total of 32 valence electrons [6 + (4 × 6) + 2]. If we use a single pair of electrons to connect the sulfur and each oxygen, we obtain the four-coordinate Lewis structure (a). We know that sulfur can accommodate more than eight electrons by using its empty valence d orbitals, just as in SF6. An alternative structure (b) can be written with S=O double bonds, making the sulfur again six-coordinate. We can draw five other resonance structures equivalent to (b) that vary only in the arrangement of the single and double bonds. In fact, experimental data show that the S-to-O bonds in the SO42− ion are intermediate in length between single and double bonds, as expected for a system whose resonance structures all contain two S–O single bonds and two S=O double bonds. When calculating the formal charges on structures (a) and (b), we see that the S atom in (a) has a formal charge of +2, whereas the S atom in (b) has a formal charge of 0. Thus by using an expanded octet, a +2 formal charge on S can be eliminated.
In oxoanions of the heavier p-block elements, the central atom often has an expanded valence shell.
Molecules with atoms that possess less than an octet of electrons generally contain the lighter s- and p-block elements, especially beryllium, typically with just four electrons around the central atom, and boron, typically with six. One example, boron trichloride (BCl3) is used to produce fibers for reinforcing high-tech tennis rackets and golf clubs. The compound has 24 valence electrons and the following Lewis structure:
The boron atom has only six valence electrons, while each chlorine atom has eight. A reasonable solution might be to use a lone pair from one of the chlorine atoms to form a B-to-Cl double bond:
This resonance structure, however, results in a formal charge of +1 on the doubly bonded Cl atom and −1 on the B atom. The high electronegativity of Cl makes this separation of charge unlikely and suggests that this is not the most important resonance structure for BCl3. This conclusion is shown to be valid based on the three equivalent B–Cl bond lengths of 173 pm that have no double bond character. Electron-deficient compounds such as BCl3 have a strong tendency to gain an additional pair of electrons by reacting with species with a lone pair of electrons.
Molecules with atoms that have fewer than an octet of electrons generally contain the lighter s- and p-block elements.
Electron-deficient compounds have a strong tendency to gain electrons in their reactions.
Draw Lewis dot structures for each compound.
Include resonance structures where appropriate.
Given: two compounds
Asked for: Lewis electron structures
Strategy:
A Use the procedure given earlier to write a Lewis electron structure for each compound. If necessary, place any remaining valence electrons on the element most likely to be able to accommodate more than an octet.
B After all the valence electrons have been placed, decide whether you have drawn an acceptable Lewis structure.
Solution:
A Because it is the least electronegative element, Be is the central atom. The molecule has 16 valence electrons (2 from Be and 7 from each Cl). Drawing two Be–Cl bonds and placing three lone pairs on each Cl gives the following structure:
B Although this arrangement gives beryllium only 4 electrons, it is an acceptable Lewis structure for BeCl2. Beryllium is known to form compounds in which it is surrounded by less than an octet of electrons.
A Sulfur is the central atom because it is less electronegative than fluorine. The molecule has 34 valence electrons (6 from S and 7 from each F). The S–F bonds use 8 electrons, and another 24 are placed around the F atoms:
The only place to put the remaining 2 electrons is on the sulfur, giving sulfur 10 valence electrons:
B Sulfur can accommodate more than an octet, so this is an acceptable Lewis structure.
Exercise
Draw Lewis dot structures for XeF4.
Answer:
Molecules with an odd number of electrons are relatively rare in the s and p blocks but rather common among the d- and f-block elements. Compounds with more than an octet of electrons around an atom are called expanded-valence molecules. One model to explain their existence uses one or more d orbitals in bonding in addition to the valence ns and np orbitals. Such species are known for only atoms in period 3 or below, which contain nd subshells in their valence shell.
What regions of the periodic table contain elements that frequently form molecules with an odd number of electrons? Explain your answer.
How can atoms expand their valence shell? What is the relationship between an expanded valence shell and the stability of an ion or a molecule?
What elements are known to form compounds with less than an octet of electrons? Why do electron-deficient compounds form?
List three elements that form compounds that do not obey the octet rule. Describe the factors that are responsible for the stability of these compounds.
What is the major weakness of the Lewis system in predicting the electron structures of PCl6− and other species containing atoms from period 3 and beyond?
The compound aluminum trichloride consists of Al2Cl6 molecules with the following structure (lone pairs of electrons removed for clarity):
Does this structure satisfy the octet rule? What is the formal charge on each atom? Given the chemical similarity between aluminum and boron, what is a plausible explanation for the fact that aluminum trichloride forms a dimeric structure rather than the monomeric trigonal planar structure of BCl3?
Draw Lewis electron structures for ClO4−, IF5, SeCl4, and SbF5.
Draw Lewis electron structures for ICl3, Cl3PO, Cl2SO, and AsF6−.
Draw plausible Lewis structures for the phosphate ion, including resonance structures. What is the formal charge on each atom in your structures?
Draw an acceptable Lewis structure for PCl5, a compound used in manufacturing a form of cellulose. What is the formal charge of the central atom? What is the oxidation number of the central atom?
Using Lewis structures, draw all of the resonance structures for the BrO3− ion.
Draw an acceptable Lewis structure for xenon trioxide (XeO3), including all resonance structures.
ClO4− (one of four equivalent resonance structures)
The formal charge on phosphorus is 0, while three oxygen atoms have a formal charge of −1 and one has a formal charge of zero.
As you learned in , the Brønsted–Lowry concept of acids and bases defines a base as any species that can accept a proton, and an acid as any substance that can donate a proton. Lewis proposed an alternative definition that focuses on pairs of electrons instead.
A Lewis baseAny species that can donate a pair of electrons. is defined as any species that can donate a pair of electrons, and a Lewis acidAny species that can accept a pair of electrons. is any species that can accept a pair of electrons. All Brønsted–Lowry bases (proton acceptors), such as OH−, H2O, and NH3, are also electron-pair donors. Thus the Lewis definition of acids and bases does not contradict the Brønsted–Lowry definition. Rather, it expands the definition of acids to include substances other than the H+ ion.
Electron-deficient moleculesA compound that has less than an octet of electrons around one atom., such as BCl3, contain less than an octet of electrons around one atom and have a strong tendency to gain an additional pair of electrons by reacting with substances that possess a lone pair of electrons. Lewis’s definition, which is less restrictive than either the Brønsted–Lowry or the Arrhenius definition, grew out of his observation of this tendency.
Electron-deficient molecules (those with less than an octet of electrons) are Lewis acids.
A general Brønsted–Lowry acid–base reaction can be depicted in Lewis electron symbols as follows:
The proton (H+), which has no valence electrons, is a Lewis acid because it accepts a lone pair of electrons on the base to form a bond. The proton, however, is just one of many electron-deficient species that are known to react with bases. For example, neutral compounds of boron, aluminum, and the other group 13 elements, which possess only six valence electrons, have a very strong tendency to gain an additional electron pair. Such compounds are therefore potent Lewis acids that react with an electron-pair donor such as ammonia to form an acid–base adductThe product of a reaction between a Lewis acid and a Lewis base with a coordinate covalent bond., a new covalent bond, as shown here for boron trifluoride (BF3):
The bond formed between a Lewis acid and a Lewis base is a coordinate covalent bond because both electrons are provided by only one of the atoms (N, in the case of F3B:NH3). After it is formed, however, a coordinate covalent bond behaves like any other covalent single bond.
Species that are very weak Brønsted–Lowry bases can be relatively strong Lewis bases. For example, many of the group 13 trihalides are highly soluble in ethers (R–O–R′) because the oxygen atom in the ether contains two lone pairs of electrons, just as in H2O. Hence the predominant species in solutions of electron-deficient trihalides in ether solvents is a Lewis acid–base adduct. A reaction of this type is shown in for boron trichloride and diethyl ether:
Many molecules with multiple bonds can act as Lewis acids. In these cases, the Lewis base typically donates a pair of electrons to form a bond to the central atom of the molecule, while a pair of electrons displaced from the multiple bond becomes a lone pair on a terminal atom. A typical example is the reaction of the hydroxide ion with carbon dioxide to give the bicarbonate ion, as shown in Equation 8.21. The highly electronegative oxygen atoms pull electron density away from carbon, so the carbon atom acts as a Lewis acid. Arrows indicate the direction of electron flow.
Identify the acid and the base in each Lewis acid–base reaction.
Given: reactants and products
Asked for: identity of Lewis acid and Lewis base
Strategy:
In each equation, identify the reactant that is electron deficient and the reactant that is an electron-pair donor. The electron-deficient compound is the Lewis acid, whereas the other is the Lewis base.
Solution:
Exercise
Identify the acid and the base in each Lewis acid–base reaction.
Answer:
A Lewis acid is a compound with a strong tendency to accept an additional pair of electrons from a Lewis base, which can donate a pair of electrons. Such an acid–base reaction forms an adduct, which is a compound with a coordinate covalent bond in which both electrons are provided by only one of the atoms. Electron-deficient molecules, which have less than an octet of electrons around one atom, are relatively common. They tend to acquire an octet electron configuration by reacting with an atom having a lone pair of electrons.
Construct a table comparing how OH−, NH3, H2O, and BCl3 are classified according to the Arrhenius, the Brønsted–Lowry, and the Lewis definitions of acids and bases.
Describe how the proton (H+) can simultaneously behave as an Arrhenius acid, a Brønsted–Lowry acid, and a Lewis acid.
Would you expect aluminum to form compounds with covalent bonds or coordinate covalent bonds? Explain your answer.
Classify each compound as a Lewis acid or a Lewis base and justify your choice.
Explain how a carboxylate ion (RCO2−) can act as both a Brønsted–Lowry base and a Lewis base.
In each reaction, identify the Lewis acid and the Lewis base and complete the reaction by writing the products(s).
Use Lewis dot symbols to depict the reaction of BCl3 with dimethyl ether [(CH3)2O]. How is this reaction similar to that in which a proton is added to ammonia?
In proposing his theory that octets can be completed by two atoms sharing electron pairs, Lewis provided scientists with the first description of covalent bonding. In this section, we expand on this and describe some of the properties of covalent bonds.
When we draw Lewis structures, we place one, two, or three pairs of electrons between adjacent atoms. In the Lewis bonding model, the number of electron pairs that hold two atoms together is called the bond orderThe number of electron pairs that hold two atoms together.. For a single bond, such as the C–C bond in H3C–CH3, the bond order is one. For a double bond (such as H2C=CH2), the bond order is two. For a triple bond, such as HC≡CH, the bond order is three.
When analogous bonds in similar compounds are compared, bond length decreases as bond order increases. The bond length data in Table 8.5 "Bond Lengths and Bond Dissociation Energies for Bonds with Different Bond Orders in Selected Gas-Phase Molecules at 298 K", for example, show that the C–C distance in H3C–CH3 (153.5 pm) is longer than the distance in H2C=CH2 (133.9 pm), which in turn is longer than that in HC≡CH (120.3 pm). Additionally, as noted in Section 8.5 "Lewis Structures and Covalent Bonding", molecules or ions whose bonding must be described using resonance structures usually have bond distances that are intermediate between those of single and double bonds, as we demonstrated with the C–C distances in benzene. The relationship between bond length and bond order is not linear, however. A double bond is not half as long as a single bond, and the length of a C=C bond is not the average of the lengths of C≡C and C–C bonds. Nevertheless, as bond orders increase, bond lengths generally decrease.
Table 8.5 Bond Lengths and Bond Dissociation Energies for Bonds with Different Bond Orders in Selected Gas-Phase Molecules at 298 K
Compound | Bond Order | Bond Length (pm) | Bond Dissociation Energy (kJ/mol) | Compound | Bond Order | Bond Length (pm) | Bond Dissociation Energy (kJ/mol) |
---|---|---|---|---|---|---|---|
H3C–CH3 | 1 | 153.5 | 376 | H3C–NH2 | 1 | 147.1 | 331 |
H2C=CH2 | 2 | 133.9 | 728 | H2C=NH | 2 | 127.3 | 644 |
HC≡CH | 3 | 120.3 | 965 | HC≡N | 3 | 115.3 | 937 |
H2N–NH2 | 1 | 144.9 | 275.3 | H3C–OH | 1 | 142.5 | 377 |
HN=NH | 2 | 125.2 | 456 | H2C=O | 2 | 120.8 | 732 |
N≡N | 3 | 109.8 | 945.3 | O=C=O | 2 | 116.0 | 799 |
HO–OH | 1 | 147.5 | 213 | C≡O | 3 | 112.8 | 1076.5 |
O=O | 2 | 120.7 | 498.4 |
Sources: Data from CRC Handbook of Chemistry and Physics (2004); Lange’s Handbook of Chemistry (2005); http://cccbdb.nist.gov.
As shown in Table 8.5 "Bond Lengths and Bond Dissociation Energies for Bonds with Different Bond Orders in Selected Gas-Phase Molecules at 298 K", triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however. Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K" lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:
Table 8.6 Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K
Single Bonds | Multiple Bonds | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|
H–H | 432 | C–C | 346 | N–N | ≈167 | O–O | ≈142 | F–F | 155 | C=C | 602 |
H–C | 411 | C–Si | 318 | N–O | 201 | O–F | 190 | F–Cl | 249 | C≡C | 835 |
H–Si | 318 | C–N | 305 | N–F | 283 | O–Cl | 218 | F–Br | 249 | C=N | 615 |
H–N | 386 | C–O | 358 | N–Cl | 313 | O–Br | 201 | F–I | 278 | C≡N | 887 |
H–P | ≈322 | C–S | 272 | N–Br | 243 | O–I | 201 | Cl–Cl | 240 | C=O | 749 |
H–O | 459 | C–F | 485 | P–P | 201 | S–S | 226 | Cl–Br | 216 | C≡O | 1072 |
H–S | 363 | C–Cl | 327 | S–F | 284 | Cl–I | 208 | N=N | 418 | ||
H–F | 565 | C–Br | 285 | S–Cl | 255 | Br–Br | 190 | N≡N | 942 | ||
H–Cl | 428 | C–I | 213 | S–Br | 218 | Br–I | 175 | N=O | 607 | ||
H–Br | 362 | Si–Si | 222 | I–I | 149 | O=O | 494 | ||||
H–I | 295 | Si–O | 452 | S=O | 532 |
Source: Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry, 4th ed. (1993).
1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.11 "The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms").
2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.11 "The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms". Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore
Cl–Cl > Br–Br > F–F > I–ISimilar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column.
3. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.
Figure 8.11 The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms
The relative sizes of the region of space in which electrons are shared between (a) a hydrogen atom and lighter (smaller) vs. heavier (larger) atoms in the same periodic group; and (b) two lighter versus two heavier atoms in the same group. Although the absolute amount of shared space increases in both cases on going from a light to a heavy atom, the amount of space relative to the size of the bonded atom decreases; that is, the percentage of total orbital volume decreases with increasing size. Hence the strength of the bond decreases.
4. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, as we discussed for the SO42− ion in Section 8.6 "Exceptions to the Octet Rule". In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO2 has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO2 very different from those of CO2.
Bond strengths increase as bond order increases, while bond distances decrease.
Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is not the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table 8.7 "Energies for the Dissociation of Successive C–H Bonds in Methane"); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K" are average values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%.
Table 8.7 Energies for the Dissociation of Successive C–H Bonds in Methane
Reaction | D (kJ/mol) |
---|---|
CH4(g) → CH3(g) + H(g) | 439 |
CH3(g) → CH2(g) + H(g) | 462 |
CH2(g) → CH(g) + H(g) | 424 |
CH(g) → C(g) + H(g) | 338 |
Source: Data from CRC Handbook of Chemistry and Physics (2004).
We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction (ΔHrxn < 0):
Equation 8.14
The ≈ sign is used because we are adding together average bond energies; hence this approach does not give exact values for ΔHrxn.
Let’s consider the reaction of 1 mol of n-heptane (C7H16) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline:
Equation 8.15
CH3(CH2)5CH3(l) + 11 O2(g) → 7 CO2(g) + 8 H2O(g)In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of n-heptane, while 14 C=O bonds (two for each CO2) and 16 O–H bonds (two for each H2O) are formed. The energy changes can be tabulated as follows:
Bonds Broken (kJ/mol) | Bonds Formed (kJ/mol) | ||
---|---|---|---|
6 C–C | 346 × 6 = 2076 | 14 C=O | 799 × 14 = 11,186 |
16 C–H | 411 × 16 = 6576 | 16 O–H | 459 × 16 = 7344 |
11 O=O | 494 × 11 = 5434 | Total = 18,530 | |
Total = 14,086 |
The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that ΔHrxn is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction).
If we compare this approximation with the value obtained from measured values (ΔHrxn = −4817 kJ/mol), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual values are unknown. These approximations can be important for predicting whether a reaction is exothermic or endothermic—and to what degree.
The compound RDX (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K" to estimate the ΔHrxn per mole of RDX.
Given: chemical reaction, structure of reactant, and Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K"
Asked for: ΔHrxn per mole
Strategy:
A List the types of bonds broken in RDX, along with the bond energy required to break each type. Multiply the number of each type by the energy required to break one bond of that type and then add together the energies. Repeat this procedure for the bonds formed in the reaction.
B Use Equation 8.14 to calculate the amount of energy consumed or released in the reaction (ΔHrxn).
Solution:
We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO2) can be viewed as having one N–O single bond and one N=O double bond, as follows:
In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures.
A We can organize our data by constructing a table:
Bonds Broken (kJ/mol) | Bonds Formed (kJ/mol) | ||
---|---|---|---|
6 C–N | 305 × 6 = 1830 | 3 N≡N | 942 × 3 = 2826 |
6 C–H | 411 × 6 = 2466 | 6 C=O | 799 × 6 = 4794 |
3 N–N | 167 × 3 = 501 | 6 O–H | 459 × 6 = 2754 |
3 N–O | 201 × 3 = 603 | Total = 10,374 | |
3 N=O | 607 × 3 = 1821 | ||
1.5 O=O | 494 × 1.5 = 741 | ||
Total = 7962 |
B From Equation 8.14, we have
Thus this reaction is also highly exothermic.
Exercise
The molecule HCFC-142b, a hydrochlorofluorocarbon used in place of chlorofluorocarbons (CFCs) such as the Freons, can be prepared by adding HCl to 1,1-difluoroethylene:
Use tabulated bond energies to calculate ΔHrxn.
Answer: −54 kJ/mol
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms. Tabulated values of average bond energies can be used to calculate the enthalpy change of many chemical reactions. If the bonds in the products are stronger than those in the reactants, the reaction is exothermic and vice versa.
Which would you expect to be stronger—an S–S bond or an Se–Se bond? Why?
Which element—nitrogen, phosphorus, or arsenic—will form the strongest multiple bond with oxygen? Why?
Why do multiple bonds between oxygen and period 3 elements tend to be unusually strong?
What can bond energies tell you about reactivity?
Bond energies are typically reported as average values for a range of bonds in a molecule rather than as specific values for a single bond? Why?
If the bonds in the products are weaker than those in the reactants, is a reaction exothermic or endothermic? Explain your answer.
A student presumed that because heat was required to initiate a particular reaction, the reaction product would be stable. Instead, the product exploded. What information might have allowed the student to predict this outcome?
What is the bond order about the central atom(s) of hydrazine (N2H4), nitrogen, and diimide (N2H2)? Draw Lewis electron structures for each compound and then arrange these compounds in order of increasing N–N bond distance. Which of these compounds would you expect to have the largest N–N bond energy? Explain your answer.
What is the carbon–carbon bond order in ethylene (C2H4), BrH2CCH2Br, and FCCH? Arrange the compounds in order of increasing C–C bond distance. Which would you expect to have the largest C–C bond energy? Why?
From each pair of elements, select the one with the greater bond strength? Explain your choice in each case.
From each pair of elements, select the one with the greater bond strength? Explain your choice in each case.
Approximately how much energy per mole is required to completely dissociate acetone [(CH3)2CO] and urea [(NH2)2CO] into their constituent atoms?
Approximately how much energy per mole is required to completely dissociate ethanol, formaldehyde, and hydrazine into their constituent atoms?
Is the reaction of diimine (N2H2) with oxygen to produce nitrogen and water exothermic or endothermic? Quantify your answer.
N2H4, bond order 1; N2H2, bond order 2; N2, bond order 3; N–N bond distance: N2 < N2H2 < N2H4; Largest bond energy: N2; Highest bond order correlates with strongest and shortest bond.
In Chapter 2 "Molecules, Ions, and Chemical Formulas" and Section 8.1 "An Overview of Chemical Bonding", we described the two idealized extremes of chemical bonding: (1) ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and (2) covalent bonding, in which electrons are shared equally between two atoms. Most compounds, however, have polar covalent bondsA covalent bond in which the electrons are shared unequally between the bonded atoms., which means that electrons are shared unequally between the bonded atoms. Figure 8.12 "The Electron Distribution in a Nonpolar Covalent Bond, a Polar Covalent Bond, and an Ionic Bond Using Lewis Electron Structures" compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall from Chapter 4 "Reactions in Aqueous Solution", Section 4.1 "Aqueous Solutions" that a lowercase Greek delta () is used to indicate that a bonded atom possesses a partial positive charge, indicated by or a partial negative charge, indicated by and a bond between two atoms that possess partial charges is a polar bond.
Figure 8.12 The Electron Distribution in a Nonpolar Covalent Bond, a Polar Covalent Bond, and an Ionic Bond Using Lewis Electron Structures
In a purely covalent bond (a), the bonding electrons are shared equally between the atoms. In a purely ionic bond (c), an electron has been transferred completely from one atom to the other. A polar covalent bond (b) is intermediate between the two extremes: the bonding electrons are shared unequally between the two atoms, and the electron distribution is asymmetrical with the electron density being greater around the more electronegative atom. Electron-rich (negatively charged) regions are shown in blue; electron-poor (positively charged) regions are shown in red.
The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. In Chapter 7 "The Periodic Table and Periodic Trends", electronegativity (χ) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is nonpolar if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is polarized toward the more electronegative atom. A bond in which the electronegativity of B (χB) is greater than the electronegativity of A (χA), for example, is indicated with the partial negative charge on the more electronegative atom:
One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms: Δχ = χB − χA.
To predict the polarity of the bonds in Cl2, HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms: χCl = 3.16, χH = 2.20, and χNa = 0.93 (see Figure 7.14 "A Plot of Periodic Variation of Electronegativity with Atomic Number for the First Six Rows of the Periodic Table"). Cl2 must be nonpolar because the electronegativity difference (Δχ) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, Δχ is 2.23. This high value is typical of an ionic compound (Δχ ≥ ≈1.5) and means that the valence electron of sodium has been completely transferred to chlorine to form Na+ and Cl− ions. In HCl, however, Δχ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is
Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated.
Bond polarity and ionic character increase with an increasing difference in electronegativity.
As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in NaCl, Cl2, ClF5, and HClO4 would be exactly the same.
The asymmetrical charge distribution in a polar substance such as HCl produces a dipole momentThe product of the partial charge on the bonded atoms and the distance between the partial charges: , where is measured in coulombs (C) and in meters (m)., abbreviated by the Greek letter mu (µ). The dipole moment is defined as the product of the partial charge Q on the bonded atoms and the distance r between the partial charges:
Equation 8.16
µ = Qrwhere Q is measured in coulombs (C) and r in meters. The unit for dipole moments is the debye (D):
Equation 8.17
1 D = 3.3356 × 10−30 C·mWhen a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure 8.13 "Molecules That Possess a Dipole Moment Partially Align Themselves with an Applied Electric Field").
Figure 8.13 Molecules That Possess a Dipole Moment Partially Align Themselves with an Applied Electric Field
In the absence of a field (a), the HCl molecules are randomly oriented. When an electric field is applied (b), the molecules tend to align themselves with the field, such that the positive end of the molecular dipole points toward the negative terminal and vice versa.
We can measure the partial charges on the atoms in a molecule such as HCl using Equation 8.16. If the bonding in HCl were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of HCl is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is
Equation 8.18
By dividing this calculated value by the charge on a single electron (1.6022 × 10−19 C), we find that the charge on the Cl atom of an HCl molecule is about −0.18, corresponding to about 0.18 e−:
Equation 8.19
To form a neutral compound, the charge on the H atom must be equal but opposite. Thus the measured dipole moment of HCl indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing HCl as we can therefore indicate the charge separation quantitatively as
Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine χH = 2.20; χCl = 3.16, χCl − χH = 0.96), a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule.Mathematically, dipole moments are vectors, and they possess both a magnitude and a direction. The dipole moment of a molecule is the vector sum of the dipoles of the individual bonds. In HCl, for example, the dipole moment is indicated as follows:
The arrow shows the direction of electron flow by pointing toward the more electronegative atom.
The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure 8.14 "A Plot of the Percent Ionic Character of a Bond as Determined from Measured Dipole Moments versus the Difference in Electronegativity of the Bonded Atoms" shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as NaCl(g) and CsF(g) is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure 8.14 "A Plot of the Percent Ionic Character of a Bond as Determined from Measured Dipole Moments versus the Difference in Electronegativity of the Bonded Atoms" show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example 11.
Figure 8.14 A Plot of the Percent Ionic Character of a Bond as Determined from Measured Dipole Moments versus the Difference in Electronegativity of the Bonded Atoms
In the gas phase, even CsF, which has the largest possible difference in electronegativity between atoms, is not 100% ionic. Solid CsF, however, is best viewed as 100% ionic because of the additional electrostatic interactions in the lattice.
In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl.
Given: chemical species, dipole moment, and internuclear distance
Asked for: percent ionic character
Strategy:
A Compute the charge on each atom using the information given and Equation 8.16.
B Find the percent ionic character from the ratio of the actual charge to the charge of a single electron.
Solution:
A The charge on each atom is given by
Thus NaCl behaves as if it had charges of 1.272 × 10−19 C on each atom separated by 236.1 pm.
B The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron):
Exercise
In the gas phase, silver chloride (AgCl) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride?
Answer: 55.5%
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them.
Why do ionic compounds such as KI exhibit substantially less than 100% ionic character in the gas phase?
Of the compounds LiI and LiF, which would you expect to behave more like a classical ionic compound? Which would have the greater dipole moment in the gas phase? Explain your answers.
Predict whether each compound is purely covalent, purely ionic, or polar covalent.
Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond.
Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond.
Classify each species as having 0%–40% ionic character, 40%–60% ionic character, or 60%–100% ionic character based on the type of bonding you would expect. Justify your reasoning.
If the bond distance in HCl (dipole moment = 1.109 D) were double the actual value of 127.46 pm, what would be the effect on the charge localized on each atom? What would be the percent negative charge on Cl? At the actual bond distance, how would doubling the charge on each atom affect the dipole moment? Would this represent more ionic or covalent character?
Calculate the percent ionic character of HF (dipole moment = 1.826 D) if the H–F bond distance is 92 pm.
Calculate the percent ionic character of CO (dipole moment = 0.110 D) if the C–O distance is 113 pm.
Calculate the percent ionic character of PbS and PbO in the gas phase, given the following information: for PbS, r = 228.69 pm and µ = 3.59 D; for PbO, r = 192.18 pm and µ = 4.64 D. Would you classify these compounds as having covalent or polar covalent bonds in the solid state?
Problems marked with a ♦ involve multiple concepts.
Until recently, benzidine was used in forensic medicine to detect the presence of human blood: when mixed with human blood, benzidine turns a characteristic blue color. Because benzidine has recently been identified as a carcinogen, other indicators have replaced it. Draw the complete Lewis dot structure for benzidine. Would you expect this compound to behave as a Lewis acid or a Lewis base?
There are three possible ways to connect carbon, nitrogen, and oxygen to form a monoanion: CNO−, CON−, and OCN−. One is the cyanate ion, a common and stable species; one is the fulminate ion, salts of which are used as explosive detonators; and one is so unstable that it has never been isolated. Use Lewis electron structures and the concept of formal charge to determine which isomer is cyanate, which is the fulminate, and which is the least stable.
The colorless gas N2O4 is a deadly poison that has been used as an oxidizing agent in rocket fuel. The compound has a single N–N bond, with a formal charge of +1 on each nitrogen atom. Draw resonance structures for this molecule.
Naphthalene is an organic compound that is commonly used in veterinary medicine as the active ingredient in dusting powders; it is also used internally as an intestinal antiseptic. From its chemical structure and the ΔHf of CO2 and H2O, estimate the molar enthalpy of combustion and the enthalpy of formation of naphthalene.
♦ Compare the combustion of hydrazine (N2H4), which produces nitrogen and water, with the combustion of methanol. Use the chemical structures to estimate which has a higher heat of combustion. Given equal volumes of hydrazine (d = 1.004 g/mL) and methanol (d = 0.791 g/mL), which is the better fuel (i.e., which provides more energy per unit volume during combustion)? Can you think of a reason why hydrazine is not used in internal combustion engines?
♦ Race car drivers frequently prefer methanol to isooctane as a fuel. Is this justified based on enthalpies of combustion? If you had a choice between 10 gal of methanol (d = 0.791 g/mL) and the same volume of isooctane (d = 0.688 g/mL), which fuel would you prefer?
♦ An atmospheric reservoir species is a molecule that is rather unreactive, but it contains elements that can be converted to reactive forms. For example, chlorine nitrate (ClONO2) is a reservoir species for both chlorine and nitrogen dioxide. In fact, most of the chlorine in the atmosphere is usually bound up in chlorine nitrate as a result of the reaction of ClO with NO2.
Draw Lewis electron structures for each species in this reaction. What difficulty is associated with the structure of the reactants? How does this affect the reactivity of the compounds?
Chlorine nitrate can react in a surface reaction with water to form HClO and nitric acid.
♦ Aniline is an oily liquid used to prepare organic dyes, varnishes, and black shoe polishes.
The –NH2 bound to the ring contains a lone pair of electrons that can participate in resonance in the following way:
Draw a second Lewis structure for aniline that takes this interaction into account.
This molecule is likely to serve as a Lewis base because of the lone pair of electrons on each nitrogen atom.
The balanced chemical reaction is:
N2H4 + O2 → N2 + 2 H2OHydrazine:
ΔHcomb = −573 kJ/mol or −17.9 kJ/mlMethanol (CH3OH):
ΔHcomb = −1286 kJ/mol or –15.9 kJ/mlHydrazine is both extremely toxic and potentially explosive.
Both reactants have one unpaired electron, which makes them more reactive than might otherwise be expected.
ClONO2 + H2O → HClO + HONO2