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We continue our discussion of the chemistry of the main group elements with the p block of the periodic table. We will use the systematic approach developed in Chapter 21 "Periodic Trends and the ", which is based on valence electron configurations and periodic trends in atomic properties, while applying the unifying principles of chemical bonding, thermodynamics, and kinetics. The line that divides metals from nonmetals in the periodic table crosses the p block diagonally. As a result, the differences between metallic and nonmetallic properties are evident within each group, even though all members of each group have the same valence electron configuration. The p block is the only portion of the periodic table where we encounter the inert-pair effect. Moreover, as with the s-block elements, the chemistry of the lightest member of each group in the p block differs sharply from that of its heavier congeners but is similar to that of the element immediately below and to the right of it in the next group. Thus diagonal similarities in chemistry are seen across the p block.
An artist’s sketch of tetragonal boron, showing the linked B12icosahedra. The artist, Roger Hayward, served as Linus Pauling’s illustrator for several decades, most notably in The Architecture of Molecules, which was published by W. H. Freeman in 1964.
As you study the periodic trends in properties and the reactivity of the elements in groups 13–18, you will learn how “cobalt blue” glass, rubies, and sapphires are made and why the US military became interested in using boron hydrides as rocket fuels but then abandoned its effort. You will also discover the source of diamonds on Earth, why silicon-based life-forms are likely to exist only in science fiction, and why most compounds with N–N bonds are potentially explosive. You will also learn why phosphorus can cause a painful and lethal condition known as “phossy jaw” and why selenium is used in photocopiers.
Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B2O3 and AlCl3, respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13.
Group 13 elements are never found in nature in their free state.
As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores.
Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na2B4O5(OH)4·8H2O] are found in ancient lake beds (Figure 22.1 "Borax Deposits") and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH)3], which is then dehydrated to the oxide (B2O3). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure:
Equation 22.1
Equation 22.2
Figure 22.1 Borax Deposits
(a) Concentrated deposits of crystalline borax [Na2B4O5(OH)4·8H2O] are found in ancient lake beds, such as the Mojave Desert and Death Valley in the western United States. (b) Borax is used in various cleaning products, including 20 Mule Team Borax, a laundry detergent named for the teams of 20 mules that hauled wagons full of borax from desert deposits to railroad terminals in the 1880s.
Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl3 with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B2H6):
Equation 22.3
Equation 22.4
B2H6(g) → 2B(s) + 3H2(g)The reaction shown in Equation 22.3 is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware.
In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al2O3, are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH)4]−. Neutralization of the resulting solution with gaseous CO2 results in the precipitation of Al(OH)3:
Equation 22.5
2[Al(OH)4]−(aq) + CO2(g) → 2Al(OH)3(s) + CO32−(aq) + H2O(l)Thermal dehydration of Al(OH)3 produces Al2O3, and metallic aluminum is obtained by the electrolytic reduction of Al2O3 using the Hall–Heroult process described in Chapter 19 "Electrochemistry". Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass.
Source: Thomas D. Kelly and Grecia R. Matos, “Historical Statistics for Mineral and Material Commodities in the United States,” US Geological Survey Data Series 140, 2010, accessed July 20, 2011, http://pubs.usgs.gov/ds/2005/140/.
The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO2. Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As).
Table 22.1 "Selected Properties of the Group 13 Elements" summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns2np1 valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids.
Table 22.1 Selected Properties of the Group 13 Elements
Property | Boron | Aluminum* | Gallium | Indium | Thallium |
---|---|---|---|---|---|
atomic symbol | B | Al | Ga | In | Tl |
atomic number | 5 | 13 | 31 | 49 | 81 |
atomic mass (amu) | 10.81 | 26.98 | 69.72 | 114.82 | 204.38 |
valence electron configuration† | 2s22p1 | 3s23p1 | 4s24p1 | 5s25p1 | 6s26p1 |
melting point/boiling point (°C) | 2075/4000 | 660/2519 | 29.7/2204 | 156.6/2072 | 304/1473 |
density (g/cm3) at 25°C | 2.34 | 2.70 | 5.91 | 7.31 | 11.8 |
atomic radius (pm) | 87 | 118 | 136 | 156 | 156 |
first ionization energy (kJ/mol) | 801 | 578 | 579 | 558 | 589 |
most common oxidation state | +3 | +3 | +3 | +3 | +1 |
ionic radius (pm)‡ | −25 | 54 | 62 | 80 | 162 |
electron affinity (kJ/mol) | −27 | −42 | −40 | −39 | −37 |
electronegativity | 2.0 | 1.6 | 1.8 | 1.8 | 1.8 |
standard reduction potential (E°, V) | −0.87 | −1.66 | −0.55 | −0.34 | +0.741 of M3+(aq) |
product of reaction with O2 | B2O3 | Al2O3 | Ga2O3 | In2O3 | Tl2O |
type of oxide | acidic | amphoteric | amphoteric | amphoteric | basic |
product of reaction with N2 | BN | AlN | GaN | InN | none |
product of reaction with X2§ | BX3 | Al2X6 | Ga2X6 | In2X6 | TlX |
*This is the name used in the United States; the rest of the world inserts an extra i and calls it aluminium. | |||||
†The configuration shown does not include filled d and f subshells. | |||||
‡The values cited are for six-coordinate ions in the most common oxidation state, except for Al3+, for which the value for the four-coordinate ion is given. The B3+ ion is not a known species; the radius cited is an estimated four-coordinate value. | |||||
§X is Cl, Br, or I. Reaction with F2 gives the trifluorides (MF3) for all group 13 elements. |
Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids.
In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.
Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding.
Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group.
All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms.
Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding.
The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B12icosahedron. (For more information on B12, see Chapter 7 "The Periodic Table and Periodic Trends", Section 7.4 "The Chemical Families".) Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure 22.2 "Solid Boron Contains B"). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B4C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation 22.6); boron oxide (B2O3), which is formed when boron is heated with excess oxygen (Equation 22.7); and the boron trihalides (BX3), which are formed by heating boron with excess halogen (Equation 22.8).
Equation 22.6
Equation 22.7
Equation 22.8
Figure 22.2 Solid Boron Contains B12 Icosahedra
Unlike metallic solids, elemental boron consists of a regular array of B12 icosahedra rather than individual boron atoms. Note that each boron atom in the B12 icosahedron is connected to five other boron atoms within the B12 unit. (a) The allotrope of boron with the simplest structure is α-rhombohedral boron, which consists of B12 octahedra in an almost cubic close-packed lattice. (b) A side view of the structure shows that icosahedra do not pack as efficiently as spheres, making the density of solid boron less than expected.
As is typical of elements lying near the dividing line between metals and nonmetals, many compounds of boron are amphoteric, dissolving in either acid or base.
Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B3N3 rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B2O3) contains layers of trigonal planar BO3 groups (analogous to BX3) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO2, to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass.
At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB12 and CaB6—are shown in Figure 22.3 "The Structures of ScB". Because metal-rich borides such as ZrB2 and TiB2 are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles.
Figure 22.3 The Structures of ScB12 and CaB6, Two Boron-Rich Metal Borides
(a) The structure of ScB12 consists of B12 clusters and Sc atoms arranged in a faced-centered cubic lattice similar to that of NaCl, with B12 units occupying the anion positions and scandium atoms the cation positions. The B12 units here are not icosahedra but cubooctahedra, with alternating square and triangular faces. (b) The structure of CaB6 consists of octahedral B6 clusters and calcium atoms arranged in a body-centered cubic lattice similar to that of CsCl, with B6 units occupying the anion positions and calcium atoms the cation positions.
Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds, as discussed in Chapter 21 "Periodic Trends and the " (Figure 21.5 "A Three-Center Bond Uses Two Electrons to Link Three Atoms"). The simplest example is diborane (B2H6), which contains two bridging hydrogen atoms (part (a) in Figure 22.4 "The Structures of Diborane (B"). An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B12H122− ion, which has a polyhedral structure similar to the icosahedral B12 unit of elemental boron, with a single hydrogen atom bonded to each boron atom.
Figure 22.4 The Structures of Diborane (B2H6) and Aluminum Chloride (Al2Cl6)
(a) The hydrogen-bridged dimer B2H6 contains two three-center, two-electron bonds as described for the B2H7− ion in Figure 21.5 "A Three-Center Bond Uses Two Electrons to Link Three Atoms". (b) In contrast, the bonding in the halogen-bridged dimer Al2Cl6 can be described in terms of electron-pair bonds, in which a chlorine atom bonded to one aluminum atom acts as a Lewis base by donating a lone pair of electrons to another aluminum atom, which acts as a Lewis acid.
A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy.
The enthalpy of combustion of diborane (B2H6) is −2165 kJ/mol, one of the highest values known:
Equation 22.9
Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B2O3 proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity.
Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH4− ion are reductants, whereas electron-deficient species such as B2H6 act as oxidants.
For each reaction, explain why the given products form.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution:
Exercise
Predict the products of the reactions and write a balanced chemical equation for each reaction.
Answer:
All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry:
Equation 22.10
2M(s) + 3X2(s,l,g) → 2MX3(s) or M2X6The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI3, it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl+ salt of the triiodide ion (I3−). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state.
Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl3 and TlBr3, are more covalent in character and form halogen-bridged dimers (part (b) in Figure 22.4 "The Structures of Diborane (B"). Although the structure of these dimers is similar to that of diborane (B2H6), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct:
Equation 22.11
Al2Cl6(soln) + 2(CH3)3N(soln) → 2(CH3)3N:AlCl3(soln)In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide [M(OH)3]:
Equation 22.12
MX3(s) + 3H2O(l) → M(OH)3(s) + 3HX(aq)In a related reaction, Al2(SO4)3 is used to clarify drinking water by the precipitation of hydrated Al(OH)3, which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H2O)6]3+.
Of the group 13 halides, only the fluorides behave as typical ionic compounds.
Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct.
Like boron (Equation 22.7), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M2O3), although Tl2O3 is unstable:
Equation 22.13
Aluminum oxide (Al2O3), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al3+ ions in crystalline alumina with Cr3+ ions forms the gemstone ruby, whereas replacing Al3+ with a mixture of Fe2+, Fe3+, and Ti4+ produces blue sapphires. The gallium oxide compound MgGa2O4 gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH)4− ions.
Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M2Y3. However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S2−, Se2−, and Te2−, thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl2Y. Only aluminum, like boron, reacts directly with N2 (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. (For more information on band gaps, see Chapter 12 "Solids", Section 12.6 "Bonding in Metals and Semiconductors".) For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs1−x−yPxNy) is used in the displays of calculators and digital watches.
All group 13 oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric.
Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH3 is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH3 is unstable at room temperature.
Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H2O)63+], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H2O)5(OH)2+ ion:
Equation 22.14
[M(H2O)6]3+(aq) → M(H2O)5(OH)2+(aq) + H+(aq)Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases.
For each reaction, explain why the given products form.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution:
Exercise
Predict the products of the reactions and write a balanced chemical equation for each reaction.
Answer:
Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Zeff that arises from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state.
None of the group 13 elements was isolated until the early 19th century, even though one of these elements is the most abundant metal on Earth. Explain why the discovery of these elements came so late and describe how they were finally isolated.
Boron and aluminum exhibit very different chemistry. Which element forms complexes with the most ionic character? Which element is a metal? a semimetal? What single property best explains the difference in their reactivity?
The usual oxidation state of boron and aluminum is +3, whereas the heavier elements in group 13 have an increasing tendency to form compounds in the +1 oxidation state. Given that all group 13 elements have an ns2np1 electron configuration, how do you explain this difference between the lighter and heavier group 13 elements?
Do you expect the group 13 elements to be highly reactive in air? Why or why not?
Which of the group 13 elements has the least metallic character? Explain why.
Boron forms multicenter bonds rather than metallic lattices with delocalized valence electrons. Why does it prefer this type of bonding? Does this explain why boron behaves like a semiconductor rather than a metal? Explain your answer.
Because the B–N unit is isoelectronic with the C–C unit, compounds that contain these units tend to have similar chemistry, although they exhibit some important differences. Describe the differences in physical properties, conductivity, and reactivity of these two types of compounds.
Boron has a strong tendency to form clusters. Does aluminum have this same tendency? Why or why not?
Explain why a B–O bond is much stronger than a B–C bond.
The electron affinities of boron and aluminum are −27 and −42 kJ/mol, respectively. Contrary to the usual periodic trends, the electron affinities of the remaining elements in group 13 are between those of B and Al. How do you explain this apparent anomaly?
The reduction potentials of B and Al in the +3 oxidation state are −0.87 V and −1.66 V, respectively. Do you expect the reduction potentials of the remaining elements of group 13 to be greater than or less than these values? How do you explain the differences between the expected values and those given in Table 22.1 "Selected Properties of the Group 13 Elements"?
The compound Al2Br6 is a halide-bridged dimer in the vapor phase, similar to diborane (B2H6). Draw the structure of Al2Br6 and then compare the bonding in this compound with that found in diborane. Explain the differences.
The compound AlH3 is an insoluble, polymeric solid that reacts with strong Lewis bases, such as Me3N, to form adducts with 10 valence electrons around aluminum. What hybrid orbital set is formed to allow this to occur?
The high stability of compounds of the group 13 elements with oxygen required powerful reductants such as metallic potassium to be isolated. Al and B were initially prepared by reducing molten AlCl3 and B2O3, respectively, with potassium.
Due to its low electronegativity and small size, boron is an unreactive semimetal rather than a metal.
The B–N bond is significantly more polar than the C–C bond, which makes B–N compounds more reactive and generally less stable than the corresponding carbon compounds. Increased polarity results in less delocalization and makes the planar form of BN less conductive than graphite.
Partial pi bonding between O and B increases the B–O bond strength.
Periodic trends predict that the cations of the heavier elements should be easier to reduce, so the elements should have less negative reduction potentials. In fact, the reverse is observed because the heavier elements have anomalously high Zeff values due to poor shielding by filled (n − 1)d and (n − 2)f subshells.
dsp 3
Is B(OH)3 a strong or a weak acid? Using bonding arguments, explain why.
Using bonding arguments, explain why organoaluminum compounds are expected to be potent Lewis acids that react rapidly with weak Lewis bases.
Imagine that you are studying chemistry prior to the discovery of gallium, element 31. Considering its position in the periodic table, predict the following properties of gallium:
The halides of Al, Ga, In, and Tl dissolve in water to form acidic solutions containing the hydrated metal ions, but only the halides of aluminum and gallium dissolve in aqueous base to form soluble metal-hydroxide complexes. Show the formulas of the soluble metal–hydroxide complexes and of the hydrated metal ions. Explain the difference in their reactivities.
Complete and balance each chemical equation.
Complete and balance each chemical equation.
Write a balanced chemical equation for each reaction.
Write a balanced chemical equation for the reaction that occurs between Al and each species.
Write a balanced chemical equation that shows how you would prepare each compound from its respective elements or other common compounds.
Write a balanced chemical equation that shows how you would prepare each compound from its respective elements or other common compounds.
Diborane is a spontaneously flammable, toxic gas that is prepared by reacting NaBH4 with BF3. Write a balanced chemical equation for this reaction.
Draw the Lewis electron structure of each reactant and product in each chemical equation. Then describe the type of bonding found in each reactant and product.
Draw the Lewis electron structure of each reactant and product in each chemical equation. Then describe the type of bonding found in each reactant and product.
B12(s) + 18Cl2(g) → 12BCl3(l)
BCl3(l) + 3H2O(l) → B(OH)3(aq) + 3HCl(aq)
The elements of group 14 show a greater range of chemical behavior than any other family in the periodic table. Three of the five elements—carbon, tin, and lead—have been known since ancient times. For example, some of the oldest known writings are Egyptian hieroglyphics written on papyrus with ink made from lampblack, a finely divided carbon soot produced by the incomplete combustion of hydrocarbons (Figure 22.5 "Very Small Particles of Noncrystalline Carbon Are Used to Make Black Ink"). Activated carbon is an even more finely divided form of carbon that is produced from the thermal decomposition of organic materials, such as sawdust. Because it adsorbs many organic and sulfur-containing compounds, activated carbon is used to decolorize foods, such as sugar, and to purify gases and wastewater.
Figure 22.5 Very Small Particles of Noncrystalline Carbon Are Used to Make Black Ink
(a) Since ancient times, ink sticks have been the major source of black ink in Asia. Plant oils or resinous woods such as pine are burned, and the resulting soot (lampblack) is collected, mixed with binders such as animal glues and minerals, compressed into a solid stick, and allowed to dry. Liquid ink is made by rubbing the ink stick against the surface of a special stone ink dish with small amounts of water. (b) A 19th-century Japanese painting illustrates how ink is made from an ink stick.
Tin and lead oxides and sulfides are easily reduced to the metal by heating with charcoal, a discovery that must have occurred by accident when prehistoric humans used rocks containing their ores for a cooking fire. However, because tin and copper ores are often found together in nature, their alloy—bronze—was probably discovered before either element, a discovery that led to the Bronze Age. The heaviest element in group 14, lead, is such a soft and malleable metal that the ancient Romans used thin lead foils as writing tablets, as well as lead cookware and lead pipes for plumbing. (Recall that the atomic symbols for tin and lead come from their Latin names: Sn for stannum and Pb for plumbum.)
Although the first glasses were prepared from silica (silicon oxide, SiO2) around 1500 BC, elemental silicon was not prepared until 1824 because of its high affinity for oxygen. Jöns Jakob Berzelius was finally able to obtain amorphous silicon by reducing Na2SiF6 with molten potassium. The crystalline element, which has a shiny blue-gray luster, was not isolated until 30 yr later. The last member of the group 14 elements to be discovered was germanium, which was found in 1886 in a newly discovered silver-colored ore by the German chemist Clemens Winkler, who named the element in honor of his native country.
The natural abundance of the group 14 elements varies tremendously. Elemental carbon, for example, ranks only 17th on the list of constituents of Earth’s crust. Pure graphite is obtained by reacting coke, an amorphous form of carbon used as a reductant in the production of steel, with silica to give silicon carbide (SiC). This is then thermally decomposed at very high temperatures (2700°C) to give graphite:
Equation 22.15
Equation 22.16
One allotrope of carbon, diamond, is metastable under normal conditions, with a of 2.9 kJ/mol versus graphite. At pressures greater than 50,000 atm, however, the diamond structure is favored and is the most stable form of carbon. Because the structure of diamond is more compact than that of graphite, its density is significantly higher (3.51 g/cm3 versus 2.2 g/cm3). Because of its high thermal conductivity, diamond powder is used to transfer heat in electronic devices.
The most common sources of diamonds on Earth are ancient volcanic pipes that contain a rock called kimberlite, a lava that solidified rapidly from deep inside the Earth. Most kimberlite formations, however, are much newer than the diamonds they contain. In fact, the relative amounts of different carbon isotopes in diamond show that diamond is a chemical and geological “fossil” older than our solar system, which means that diamonds on Earth predate the existence of our sun. Thus diamonds were most likely created deep inside Earth from primordial grains of graphite present when Earth was formed (part (a) in Figure 22.6 "Crystalline Samples of Carbon and Silicon, the Lightest Group 14 Elements"). Gem-quality diamonds can now be produced synthetically and have chemical, optical, and physical characteristics identical to those of the highest-grade natural diamonds.
Figure 22.6 Crystalline Samples of Carbon and Silicon, the Lightest Group 14 Elements
(a) The 78.86-carat Ahmadabad diamond, a historic Indian gem purchased in Gujarat in the 17th century by the French explorer Jean-Baptiste Tavernier and sold in 1995 for $4.3 million, is a rare example of a large single crystal of diamond, the less-stable allotrope of carbon. (b) Large single crystals of highly purified silicon are the basis of the modern electronics industry. They are sliced into very thin wafers that are highly polished and then cut into smaller pieces for use as chips.
Although oxygen is the most abundant element on Earth, the next most abundant is silicon, the next member of group 14. Pure silicon is obtained by reacting impure silicon with Cl2 to give SiCl4, followed by the fractional distillation of the impure SiCl4 and reduction with H2:
Equation 22.17
Several million tons of silicon are annually produced with this method. Amorphous silicon containing residual amounts of hydrogen is used in photovoltaic devices that convert light to electricity, and silicon-based solar cells are used to power pocket calculators, boats, and highway signs, where access to electricity by conventional methods is difficult or expensive. Ultrapure silicon and germanium form the basis of the modern electronics industry (part (b) in Figure 22.6 "Crystalline Samples of Carbon and Silicon, the Lightest Group 14 Elements").
In contrast to silicon, the concentrations of germanium and tin in Earth’s crust are only 1–2 ppm. The concentration of lead, which is the end product of the nuclear decay of many radionuclides, is 13 ppm, making lead by far the most abundant of the heavy group 14 elements. (For more information on radionuclides, see Chapter 20 "Nuclear Chemistry".) No concentrated ores of germanium are known; like indium, germanium is generally recovered from flue dusts obtained by processing the ores of metals such as zinc. Because germanium is essentially transparent to infrared radiation, it is used in optical devices.
Tin and lead are soft metals that are too weak for structural applications, but because tin is flexible, corrosion resistant, and nontoxic, it is used as a coating in food packaging. A “tin can,” for example, is actually a steel can whose interior is coated with a thin layer (1–2 µm) of metallic tin. Tin is also used in superconducting magnets and low-melting-point alloys such as solder and pewter. Pure lead is obtained by heating galena (PbS) in air and reducing the oxide (PbO) to the metal with carbon, followed by electrolytic deposition to increase the purity:
Equation 22.18
Equation 22.19
or
Equation 22.20
By far the single largest use of lead is in lead storage batteries. (For more information on batteries, see Chapter 19 "Electrochemistry".)
As you learned in Chapter 7 "The Periodic Table and Periodic Trends", the group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively. Because covalent bonds decrease in strength with increasing atomic size and the ionization energies for the heavier elements of the group are higher than expected due to a higher Zeff, the relative stability of the +2 oxidation state increases smoothly from carbon to lead.
The relative stability of the +2 oxidation state increases, and the tendency to form catenated compounds decreases, from carbon to lead in group 14.
Recall that many carbon compounds contain multiple bonds formed by π overlap of singly occupied 2p orbitals on adjacent atoms. (For more information on atomic orbitals, see Chapter 9 "Molecular Geometry and Covalent Bonding Models".) Compounds of silicon, germanium, tin, and lead with the same stoichiometry as those of carbon, however, tend to have different structures and properties. For example, CO2 is a gas that contains discrete O=C=O molecules, whereas the most common form of SiO2 is the high-melting solid known as quartz, the major component of sand. Instead of discrete SiO2 molecules, quartz contains a three-dimensional network of silicon atoms that is similar to the structure of diamond but with an oxygen atom inserted between each pair of silicon atoms. Thus each silicon atom is linked to four other silicon atoms by bridging oxygen atoms. (For more information on the properties of solids, see Chapter 12 "Solids", Section 12.1 "Crystalline and Amorphous Solids".) The tendency to catenate—to form chains of like atoms—decreases rapidly as we go down group 14 because bond energies for both the E–E and E–H bonds decrease with increasing atomic number (where E is any group 14 element). Consequently, inserting a CH2 group into a linear hydrocarbon such as n-hexane is exergonic (ΔG° = −45 kJ/mol), whereas inserting an SiH2 group into the silicon analogue of n-hexane (Si6H14) actually costs energy (ΔG° ≈ +25 kJ/mol). As a result of this trend, the thermal stability of catenated compounds decreases rapidly from carbon to lead.
In Table 22.2 "Selected Properties of the Group 14 Elements" we see, once again, that there is a large difference between the lightest element (C) and the others in size, ionization energy, and electronegativity. As in group 13, the second and third elements (Si and Ge) are similar, and there is a reversal in the trends for some properties, such as ionization energy, between the fourth and fifth elements (Sn and Pb). As for group 13, these effects can be explained by the presence of filled (n − 1)d and (n − 2)f subshells, whose electrons are relatively poor at screening the outermost electrons from the higher nuclear charge.
Table 22.2 Selected Properties of the Group 14 Elements
Property | Carbon | Silicon | Germanium | Tin | Lead |
---|---|---|---|---|---|
atomic symbol | C | Si | Ge | Sn | Pb |
atomic number | 6 | 14 | 32 | 50 | 82 |
atomic mass (amu) | 12.01 | 28.09 | 72.64 | 118.71 | 207.2 |
valence electron configuration* | 2s22p2 | 3s23p2 | 4s24p2 | 5s25p2 | 6s26p2 |
melting point/boiling point (°C) | 4489 (at 10.3 MPa)/3825 | 1414/3265 | 939/2833 | 232/2602 | 327/1749 |
density (g/cm3) at 25°C | 2.2 (graphite), 3.51 (diamond) | 2.33 | 5.32 | 7.27(white) | 11.30 |
atomic radius (pm) | 77 (diamond) | 111 | 125 | 145 | 154 |
first ionization energy (kJ/mol) | 1087 | 787 | 762 | 709 | 716 |
most common oxidation state | +4 | +4 | +4 | +4 | +4 |
ionic radius (pm)† | ≈29 | ≈40 | 53 | 69 | 77.5 |
electron affinity (kJ/mol) | −122 | −134 | −119 | −107 | −35 |
electronegativity | 2.6 | 1.9 | 2.0 | 2.0 | 1.8 |
standard reduction potential (E°, V) (for EO2 → E in acidic solution) | 0.21 | −0.86 | −0.18 | −0.12 | 0.79 |
product of reaction with O2 | CO2, CO | SiO2 | GeO2 | SnO2 | PbO |
type of oxide | acidic (CO2) | acidic neutral (CO) | amphoteric | amphoteric | amphoteric |
product of reaction with N2 | none | Si3N4 | none | Sn3N4 | none |
product of reaction with X2‡ | CX4 | SiX4 | GeX4 | SnX4 | PbX2 |
product of reaction with H2 | CH4 | none | none | none | none |
*The configuration shown does not include filled d and f subshells. | |||||
†The values cited are for six-coordinate +4 ions in the most common oxidation state, except for C4+ and Si4+, for which values for the four-coordinate ion are estimated. | |||||
‡X is Cl, Br, or I. Reaction with F2 gives the tetrafluorides (EF4) for all group 14 elements, where E represents any group 14 element. |
The group 14 elements follow the same pattern as the group 13 elements in their periodic properties.
Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion.
The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds (Figure 7.18 "Four Allotropes of Carbon"). Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–)n. Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers.
Although all the carbon tetrahalides (CX4) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X2) but by indirect methods such as the following reaction, where X is Cl or Br:
Equation 22.21
CH4(g) + 4X2(g) → CX4(l,s) + 4HX(g)The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CCl4 and CI4. Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F2C=CF2), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex).
The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding.
Carbon reacts with oxygen to form either CO or CO2, depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26 "Binding of O" for myoglobin). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid:
Equation 22.22
Carbon monoxide also reacts with the halogens to form the oxohalides (COX2). Probably the best known of these is phosgene (Cl2C=O), which is highly poisonous and was used as a chemical weapon during World War I:
Equation 22.23
Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes.
Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO2 reacts with water to form acidic solutions containing carbonic acid (H2CO3). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS2):
Equation 22.24
The selenium analogue CSe2 is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe2, highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap.
Pi bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap.
Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be2C and Al4C3, which formally contain the C4− ion derived from methane (CH4) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na2C2 and CaC2. Because these carbides contain the C4− ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. As discussed in Chapter 21 "Periodic Trends and the ", Section 21.4 "The Alkaline Earth Metals (Group 2)", reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH4 or C2H2. For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern.
19th-century miner’s lamp. The lamp uses burning acetylene, produced by the slow reaction of calcium carbide with water, to provide light.
The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe3C), which is a major component of steel.
Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation 22.15) and boron carbide (B4C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors.
Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert.
For each reaction, explain why the given product forms.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution:
Exercise
Predict the products of the reactions and write a balanced chemical equation for each reaction.
Answer:
Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions.
Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF62− ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO3 and H2SO4, but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb2+ salt, such as PbSO4.
All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl2 is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl2 can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl2 is relatively stable at temperatures below 20°C. In contrast, SnCl2 is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl2 is an insoluble crystalline solid with a structure similar to that of SnCl2.
The stability of the group 14 dichlorides increases dramatically from carbon to lead.
Although the first four elements of group 14 form tetrahalides (MX4) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF4. The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge):
Equation 22.25
MX4(s,l) + 2H2O(l) → MO2(s) + 4HX(aq)In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions.
Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O2 or S8 gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group.
The dioxides of the group 14 elements become increasingly basic down the group.
Because the Si–O bond is even stronger than the C–O bond (∼452 kJ/mol versus ∼358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or spn hybrid orbitals of O results in a partial π bond (Figure 22.7 "Pi Bonding between Silicon and Oxygen"). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Figure 22.7 Pi Bonding between Silicon and Oxygen
Silicon has relatively low-energy, empty 3d orbitals that can interact with filled 2p hybrid orbitals on oxygen. This interaction results in a partial π bond in which both electrons are supplied by oxygen, giving the Si–O bond partial double bond character and making it significantly stronger (and shorter) than expected for a single bond.
Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO44− unit:
The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively.
Opal gemstones.
The silicates include many glasses as well as the gemstone known as opal, which typically contains 10%–15% water. In a large and important class of materials called aluminosilicates, some of the Si atoms are replaced by Al atoms to give aluminosilicates such as zeolites, whose three-dimensional framework structures have large cavities connected by smaller tunnels (Figure 22.8 "Zeolites Are Aluminosilicates with Large Cavities Connected by Channels"). Because the cations in zeolites are readily exchanged, zeolites are used in laundry detergents as water-softening agents: the more loosely bound Na+ ions inside the zeolite cavities are displaced by the more highly charged Mg2+ and Ca2+ ions present in hard water, which bind more tightly. Zeolites are also used as catalysts and for water purification.
Figure 22.8 Zeolites Are Aluminosilicates with Large Cavities Connected by Channels
The cavities normally contain hydrated cations that are loosely bound to the oxygen atoms of the negatively charged framework by electrostatic interactions. The sizes and arrangements of the channels and cavities differ in different types of zeolites. For example, in zeolite A the aluminosilicate cages are arranged in a cubic fashion, and the channels connecting the cavities intersect at right angles. In contrast, the cavities in faujasite are much larger, and the channels intersect at 120° angles. In these idealized models, the oxygen atoms that connect each pair of silicon atoms have been omitted.
Silicon and germanium react with nitrogen at high temperature to form nitrides (M3N4):
Equation 22.26
3Si(l) + 2N2(g) → Si3N4(s)Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C.
Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg2Si, are ionic compounds that contain the Si4− ion. They react with aqueous acid to form silicon hydrides such as SiH4:
Equation 22.27
Mg2Si(s) + 4H+(aq) → 2Mg2+(aq) + SiH4(g)Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH4, and it slowly decomposes to elemental Sn and H2 at room temperature. The simplest lead hydride (PbH4) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction.
The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker.
The only important organic derivatives of lead are compounds such as tetraethyllead [(CH3CH2)4Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. (Example 6 in Section 22.3 "The Elements of Group 15 (The Pnicogens)" examines this problem.) The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world.
Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–)n backbone with organic groups attached to Si (Figure 22.9 "Silicones Are Polymers with Long Chains of Alternating Silicon and Oxygen Atoms"). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty.
Figure 22.9 Silicones Are Polymers with Long Chains of Alternating Silicon and Oxygen Atoms
The structure of a linear silicone polymer is similar to that of quartz, but two of the oxygen atoms attached to each silicon atom are replaced by the carbon atoms of organic groups, such as the methyl groups (–CH3) shown here. The terminal silicon atoms are bonded to three methyl groups. Silicones can be oily, waxy, flexible, or elastic, depending on the chain length, the extent of cross-linking between the chains, and the type of organic group.
A child playing with Silly Putty, a silicone polymer with unusual mechanical properties. Gentle pressure causes Silly Putty to flow or stretch, but it cannot be flattened when hit with a hammer.
For each reaction, explain why the given products form.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution:
Exercise
Predict the products of the reactions and write a balanced chemical equation for each reaction.
Answer:
The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22− (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions.
Why is the preferred oxidation state of lead +2 rather than +4? What do you expect the preferred oxidation state of silicon to be based on its position in the periodic table?
Carbon uses pπ–pπ overlap to form compounds with multiple bonds, but silicon does not. Why? How does this same phenomenon explain why the heavier elements in group 14 do not form catenated compounds?
Diamond is both an electrical insulator and an excellent thermal conductor. Explain this property in terms of its bonding.
The lighter chalcogens (group 16) form π bonds with carbon. Does the strength of these π bonds increase or decrease with increasing atomic number of the chalcogen? Why?
The heavier group 14 elements can form complexes that contain expanded coordination spheres. How does this affect their reactivity compared with the reactivity of carbon? Is this a thermodynamic effect or a kinetic effect? Explain your answer.
Refer to Table 22.2 "Selected Properties of the Group 14 Elements" for the values of the electron affinities of the group 14 elements. Explain any discrepancies between these actual values and the expected values based on usual periodic trends.
Except for carbon, the elements of group 14 can form five or six electron-pair bonds. What hybrid orbitals are used to allow this expanded coordination? Why does carbon not form more than four electron-pair bonds?
Which of the group 14 elements is least stable in the +4 oxidation state? Why?
Predict the products of each reaction and balance each chemical equation.
Write a balanced chemical equation to indicate how you would prepare each compound.
Write a balanced chemical equation to indicate how you would prepare each compound.
Like the group 14 elements, the lightest member of group 15, nitrogen, is found in nature as the free element, and the heaviest elements have been known for centuries because they are easily isolated from their ores.
Antimony (Sb) was probably the first of the pnicogens to be obtained in elemental form and recognized as an element. Its atomic symbol comes from its Roman name: stibium. It is found in stibnite (Sb2S3), a black mineral that has been used as a cosmetic (an early form of mascara) since biblical times, and it is easily reduced to the metal in a charcoal fire (). The Egyptians used antimony to coat copper objects as early as the third millennium BC, and antimony is still used in alloys to improve the tonal quality of bells.
Figure 22.10 The Ancient Egyptians Used Finely Ground Antimony Sulfide for Eye Makeup
(a) Crystals of the soft black mineral stibnite (Sb2S3) on a white mineral matrix. (b) A fragment of an Egyptian painting on limestone from the 16th–13th centuries BC shows the use of ground stibnite (“kohl”) as black eye shadow. Small vases of ground stibnite have been found among the funeral goods buried with Egyptian pharaohs.
In the form of its yellow sulfide ore, orpiment (As2S3), arsenic (As) has been known to physicians and professional assassins since ancient Greece, although elemental arsenic was not isolated until centuries later. The history of bismuth (Bi), in contrast, is more difficult to follow because early alchemists often confused it with other metals, such as lead, tin, antimony, and even silver (due to its slightly pinkish-white luster). Its name comes from the old German wismut, meaning “white metal.” Bismuth was finally isolated in the 15th century, and it was used to make movable type for printing shortly after the invention of the Gutenberg printing process in 1440. Bismuth is used in printing because it is one of the few substances known whose solid state is less dense than the liquid. Consequently, its alloys expand as they cool, filling a mold completely and producing crisp, clear letters for typesetting.
Phosphorus was discovered in 1669 by the German alchemist Hennig Brandt, who was looking for the “philosophers’ stone,” a mythical substance capable of converting base metals to silver or gold. Believing that human urine was the source of the key ingredient, Brandt obtained several dozen buckets of urine, which he allowed to putrefy. The urine was distilled to dryness at high temperature and then condensed; the last fumes were collected under water, giving a waxy white solid that had unusual properties. For example, it glowed in the dark and burst into flames when removed from the water. (Unfortunately for Brandt, however, it did not turn lead into gold.) The element was given its current name (from the Greek phos, meaning “light,” and phoros, meaning “bringing”) in the 17th century. For more than a century, the only way to obtain phosphorus was the distillation of urine, but in 1769 it was discovered that phosphorus could be obtained more easily from bones. During the 19th century, the demand for phosphorus for matches was so great that battlefields and paupers’ graveyards were systematically scavenged for bones. Early matches were pieces of wood coated with elemental phosphorus that were stored in an evacuated glass tube and ignited when the tube was broken (which could cause unfortunate accidents if the matches were kept in a pocket!).
Unfortunately, elemental phosphorus is volatile and highly toxic. It is absorbed by the teeth and destroys bone in the jaw, leading to a painful and fatal condition called “phossy jaw,” which for many years was accepted as an occupational hazard of working in the match industry.
Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens to be obtained in pure form. In 1772, Daniel Rutherford, working with Joseph Black (who discovered CO2), noticed that a gas remained when CO2 was removed from a combustion reaction. Antoine Lavoisier called the gas azote, meaning “no life,” because it did not support life. When it was discovered that the same element was also present in nitric acid and nitrate salts such as KNO3 (nitre), it was named nitrogen. About 90% of the nitrogen produced today is used to provide an inert atmosphere for processes or reactions that are oxygen sensitive, such as the production of steel, petroleum refining, and the packaging of foods and pharmaceuticals.
Because the atmosphere contains several trillion tons of elemental nitrogen with a purity of about 80%, it is a huge source of nitrogen gas. Distillation of liquefied air yields nitrogen gas that is more than 99.99% pure, but small amounts of very pure nitrogen gas can be obtained from the thermal decomposition of sodium azide:
Equation 22.28
In contrast, Earth’s crust is relatively poor in nitrogen. The only important nitrogen ores are large deposits of KNO3 and NaNO3 in the deserts of Chile and Russia, which were apparently formed when ancient alkaline lakes evaporated. Consequently, virtually all nitrogen compounds produced on an industrial scale use atmospheric nitrogen as the starting material. Phosphorus, which constitutes only about 0.1% of Earth’s crust, is much more abundant in ores than nitrogen. Like aluminum and silicon, phosphorus is always found in combination with oxygen, and large inputs of energy are required to isolate it.
The other three pnicogens are much less abundant: arsenic is found in Earth’s crust at a concentration of about 2 ppm, antimony is an order of magnitude less abundant, and bismuth is almost as rare as gold. All three elements have a high affinity for the chalcogens and are usually found as the sulfide ores (M2S3), often in combination with sulfides of other heavy elements, such as copper, silver, and lead. Hence a major source of antimony and bismuth is flue dust obtained by smelting the sulfide ores of the more abundant metals.
In group 15, as elsewhere in the p block, we see large differences between the lightest element (N) and its congeners in size, ionization energy, electron affinity, and electronegativity (). The chemical behavior of the elements can be summarized rather simply: nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. With their ns2np3 valence electron configurations, all form compounds by losing either the three np valence electrons to form the +3 oxidation state or the three np and the two ns valence electrons to give the +5 oxidation state, whose stability decreases smoothly from phosphorus to bismuth. In addition, the relatively large magnitude of the electron affinity of the lighter pnicogens enables them to form compounds in the −3 oxidation state (such as NH3 and PH3), in which three electrons are formally added to the neutral atom to give a filled np subshell. Nitrogen has the unusual ability to form compounds in nine different oxidation states, including −3, +3, and +5. Because neutral covalent compounds of the trivalent pnicogens contain a lone pair of electrons on the central atom, they tend to behave as Lewis bases.
Table 22.3 Selected Properties of the Group 15 Elements
Property | Nitrogen | Phosphorus | Arsenic | Antimony | Bismuth |
---|---|---|---|---|---|
atomic symbol | N | P | As | Sb | Bi |
atomic number | 7 | 15 | 33 | 51 | 83 |
atomic mass (amu) | 14.01 | 30.97 | 74.92 | 121.76 | 209.98 |
valence electron configuration* | 2s22p3 | 3s23p3 | 4s24p3 | 5s25p3 | 6s26p3 |
melting point/boiling point (°C) | −210/−196 | 44.15/281c | 817 (at 3.70 MPa)/603 (sublimes)† | 631/1587 | 271/1564 |
density (g/cm3) at 25°C | 1.15 | 1.82† | 5.75‡ | 6.68 | 9.79 |
atomic radius (pm) | 56 | 98 | 114 | 133 | 143 |
first ionization energy (kJ/mol) | 1402 | 1012 | 945 | 831 | 703 |
common oxidation state(s) | −3 to +5 | +5, +3, −3 | +5, +3 | +5, +3 | +3 |
ionic radius (pm)§ | 146 (−3), 16 (+3) | 212 (−3), 44 (+3) | 58 (+3) | 76 (+3), 60 (+5) | 103 (+3) |
electron affinity (kJ/mol) | 0 | −72 | −78 | −101 | −91 |
electronegativity | 3.0 | 2.2 | 2.2 | 2.1 | 1.9 |
standard reduction potential (E°, V) (EV → EIII in acidic solution)|| | +0.93 | −0.28 | +0.56 | +0.65 | — |
product of reaction with O2 | NO2, NO | P4O6, P4O10 | As4O6 | Sb2O5 | Bi2O3 |
type of oxide | acidic (NO2), neutral (NO, N2O) | acidic | acidic | amphoteric | basic |
product of reaction with N2 | — | none | none | none | none |
product of reaction with X2 | none | PX3, PX5 | AsF5, AsX3 | SbF5, SbCl5, SbBr3, SbI3 | BiF5, BiX3 |
product of reaction with H2 | none | none | none | none | none |
*The configuration shown does not include filled d and f subshells. | |||||
†For white phosphorus. | |||||
‡For gray arsenic. | |||||
§The values cited are for six-coordinate ions in the indicated oxidation states. The N5+, P5+, and As5+ ions are not known species. | |||||
||The chemical form of the elements in these oxidation states varies considerably. For N, the reaction is NO3− + 3H+ + 2e− → HNO2 + H2O; for P and As, it is H3EO4 + 2H+ + 2e− → H3EO3 + H2O; and for Sb it is Sb2O5 + 4e− + 10H+ → 2Sb3+ + 5H2O. |
In group 15, the stability of the +5 oxidation state decreases from P to Bi.
Because neutral covalent compounds of the trivalent group 15 elements have a lone pair of electrons on the central atom, they tend to be Lewis bases.
Like carbon, nitrogen has four valence orbitals (one 2s and three 2p), so it can participate in at most four electron-pair bonds by using sp3 hybrid orbitals. Unlike carbon, however, nitrogen does not form long chains because of repulsive interactions between lone pairs of electrons on adjacent atoms. These interactions become important at the shorter internuclear distances encountered with the smaller, second-period elements of groups 15, 16, and 17. (For more information on internuclear distance, see , and , .) Stable compounds with N–N bonds are limited to chains of no more than three N atoms, such as the azide ion (N3−).
Nitrogen is the only pnicogen that normally forms multiple bonds with itself and other second-period elements, using π overlap of adjacent np orbitals. Thus the stable form of elemental nitrogen is N2, whose N≡N bond is so strong (DN≡N = 942 kJ/mol) compared with the N–N and N=N bonds (DN–N = 167 kJ/mol; DN=N = 418 kJ/mol) that all compounds containing N–N and N=N bonds are thermodynamically unstable with respect to the formation of N2. In fact, the formation of the N≡N bond is so thermodynamically favored that virtually all compounds containing N–N bonds are potentially explosive.
Again in contrast to carbon, nitrogen undergoes only two important chemical reactions at room temperature: it reacts with metallic lithium to form lithium nitride, and it is reduced to ammonia by certain microorganisms. (For more information lithium, see .) At higher temperatures, however, N2 reacts with more electropositive elements, such as those in group 13, to give binary nitrides, which range from covalent to ionic in character. Like the corresponding compounds of carbon, binary compounds of nitrogen with oxygen, hydrogen, or other nonmetals are usually covalent molecular substances.
Few binary molecular compounds of nitrogen are formed by direct reaction of the elements. At elevated temperatures, N2 reacts with H2 to form ammonia, with O2 to form a mixture of NO and NO2, and with carbon to form cyanogen (N≡C–C≡N); elemental nitrogen does not react with the halogens or the other chalcogens. Nonetheless, all the binary nitrogen halides (NX3) are known. Except for NF3, all are toxic, thermodynamically unstable, and potentially explosive, and all are prepared by reacting the halogen with NH3 rather than N2. Both nitrogen monoxide (NO) and nitrogen dioxide (NO2) are thermodynamically unstable, with positive free energies of formation. Unlike NO, NO2 reacts readily with excess water, forming a 1:1 mixture of nitrous acid (HNO2) and nitric acid (HNO3):
Equation 22.29
2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq)Nitrogen also forms N2O (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with CO2 and can be represented as −N=N+=O. Like the other two oxides of nitrogen, nitrous oxide is thermodynamically unstable. The structures of the three common oxides of nitrogen are as follows:
Few binary molecular compounds of nitrogen are formed by the direct reaction of the elements.
At elevated temperatures, nitrogen reacts with highly electropositive metals to form ionic nitrides, such as Li3N and Ca3N2. These compounds consist of ionic lattices formed by Mn+ and N3− ions. Just as boron forms interstitial borides and carbon forms interstitial carbides, with less electropositive metals nitrogen forms a range of interstitial nitrides, in which nitrogen occupies holes in a close-packed metallic structure. Like the interstitial carbides and borides, these substances are typically very hard, high-melting materials that have metallic luster and conductivity.
Nitrogen also reacts with semimetals at very high temperatures to produce covalent nitrides, such as Si3N4 and BN, which are solids with extended covalent network structures similar to those of graphite or diamond. Consequently, they are usually high melting and chemically inert materials.
Ammonia (NH3) is one of the few thermodynamically stable binary compounds of nitrogen with a nonmetal. It is not flammable in air, but it burns in an O2 atmosphere:
Equation 22.30
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)About 10% of the ammonia produced annually is used to make fibers and plastics that contain amide bonds, such as nylons and polyurethanes, while 5% is used in explosives, such as ammonium nitrate, TNT (trinitrotoluene), and nitroglycerine. Large amounts of anhydrous liquid ammonia are used as fertilizer.
Nitrogen forms two other important binary compounds with hydrogen. Hydrazoic acid (HN3), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine (N2H4) is also potentially explosive; it is used as a rocket propellant and to inhibit corrosion in boilers.
B, C, and N all react with transition metals to form interstitial compounds that are hard, high-melting materials.
For each reaction, explain why the given products form when the reactants are heated.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution:
Exercise
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
Answer:
Like the heavier elements of group 14, the heavier pnicogens form catenated compounds that contain only single bonds, whose stability decreases rapidly as we go down the group. For example, phosphorus exists as multiple allotropes, the most common of which is white phosphorus, which consists of P4 tetrahedra and behaves like a typical nonmetal. As is typical of a molecular solid, white phosphorus is volatile, has a low melting point (44.1°C), and is soluble in nonpolar solvents. It is highly strained, with bond angles of only 60°, which partially explains why it is so reactive and so easily converted to more stable allotropes. Heating white phosphorus for several days converts it to red phosphorus, a polymer that is air stable, virtually insoluble, denser than white phosphorus, and higher melting, properties that make it much safer to handle. A third allotrope of phosphorus, black phosphorus, is prepared by heating the other allotropes under high pressure; it is even less reactive, denser, and higher melting than red phosphorus. As expected from their structures, white phosphorus is an electrical insulator, and red and black phosphorus are semiconductors. The three heaviest pnicogens—arsenic, antimony, and bismuth—all have a metallic luster, but they are brittle (not ductile) and relatively poor electrical conductors.
As in group 14, the heavier group 15 elements form catenated compounds that contain only single bonds, whose stability decreases as we go down the group.
The reactivity of the heavier pnicogens decreases as we go down the column. Phosphorus is by far the most reactive of the pnicogens, forming binary compounds with every element in the periodic table except antimony, bismuth, and the noble gases. Phosphorus reacts rapidly with O2, whereas arsenic burns in pure O2 if ignited, and antimony and bismuth react with O2 only when heated. None of the pnicogens reacts with nonoxidizing acids such as aqueous HCl, but all dissolve in oxidizing acids such as HNO3. Only bismuth behaves like a metal, dissolving in HNO3 to give the hydrated Bi3+ cation.
The reactivity of the heavier group 15 elements decreases as we go down the column.
The heavier pnicogens can use energetically accessible 3d, 4d, or 5d orbitals to form dsp3 or d2sp3 hybrid orbitals for bonding. Consequently, these elements often have coordination numbers of 5 or higher. Phosphorus and arsenic form halides (e.g., AsCl5) that are generally covalent molecular species and behave like typical nonmetal halides, reacting with water to form the corresponding oxoacids (in this case, H3AsO4). All the pentahalides are potent Lewis acids that can expand their coordination to accommodate the lone pair of a Lewis base:
Equation 22.31
AsF5(soln) + F−(soln) → AsF6−(soln)In contrast, bismuth halides have extended lattice structures and dissolve in water to produce hydrated ions, consistent with the stronger metallic character of bismuth.
Except for BiF3, which is essentially an ionic compound, the trihalides are volatile covalent molecules with a lone pair of electrons on the central atom. Like the pentahalides, the trihalides react rapidly with water. In the cases of phosphorus and arsenic, the products are the corresponding acids, H3PO3 and H3AsO3, where E is P or As:
Equation 22.32
EX3(l) + 3H2O(l) → H3EO3(aq) + 3HX(aq)Phosphorus halides are also used to produce insecticides, flame retardants, and plasticizers.
With energetically accessible d orbitals, phosphorus and, to a lesser extent, arsenic are able to form π bonds with second-period atoms such as N and O. This effect is even more important for phosphorus than for silicon, resulting in very strong P–O bonds and even stronger P=O bonds. The first four elements in group 15 also react with oxygen to produce the corresponding oxide in the +3 oxidation state. Of these oxides, P4O6 and As4O6 have cage structures formed by inserting an oxygen atom into each edge of the P4 or As4 tetrahedron (part (a) in ), and they behave like typical nonmetal oxides. For example, P4O6 reacts with water to form phosphorous acid (H3PO3). Consistent with its position between the nonmetal and metallic oxides, Sb4O6 is amphoteric, dissolving in either acid or base. In contrast, Bi2O3 behaves like a basic metallic oxide, dissolving in acid to give solutions that contain the hydrated Bi3+ ion. The two least metallic elements of the heavier pnicogens, phosphorus and arsenic, form very stable oxides with the formula E4O10 in the +5 oxidation state (part (b) in ). In contrast, Bi2O5 is so unstable that there is no absolute proof it exists.
Figure 22.11 The Structures of Some Cage Compounds of Phosphorus
(a, b) The structures of P4O6 and P4O10 are both derived from the structure of white phosphorus (P4) by inserting an oxygen atom into each of the six edges of the P4 tetrahedron; P4O10 contains an additional terminal oxygen atom bonded to each phosphorus atom. (c) The structure of P4S3 is also derived from the structure of P4 by inserting three sulfur atoms into three adjacent edges of the tetrahedron.
The heavier pnicogens form sulfides that range from molecular species with three-dimensional cage structures, such as P4S3 (part (c) in ), to layered or ribbon structures, such as Sb2S3 and Bi2S3, which are semiconductors. Reacting the heavier pnicogens with metals produces substances whose properties vary with the metal content. Metal-rich phosphides (such as M4P) are hard, high-melting, electrically conductive solids with a metallic luster, whereas phosphorus-rich phosphides (such as MP15) are lower melting and less thermally stable because they contain catenated Pn units. Many organic or organometallic compounds of the heavier pnicogens containing one to five alkyl or aryl groups are also known. Because of the decreasing strength of the pnicogen–carbon bond, their thermal stability decreases from phosphorus to bismuth.
Phosphorus has the greatest ability to form π bonds with elements such as O, N, and C.
The thermal stability of organic or organometallic compounds of group 15 decreases down the group due to the decreasing strength of the pnicogen–carbon bond.
For each reaction, explain why the given products form.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution:
Exercise
Predict the products of each reaction and write a balanced chemical equation for each reaction.
Answer:
In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. Nitrogen forms compounds in nine different oxidation states. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state. Because of the presence of a lone pair of electrons on the pnicogen, neutral covalent compounds of the trivalent pnicogens are Lewis bases. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character. Reaction with electropositive metals produces ionic nitrides, reaction with less electropositive metals produces interstitial nitrides, and reaction with semimetals produces covalent nitrides. The reactivity of the pnicogens decreases with increasing atomic number. Compounds of the heavier pnicogens often have coordination numbers of 5 or higher and use dsp3 or d2sp3 hybrid orbitals for bonding. Because phosphorus and arsenic have energetically accessible d orbitals, these elements form π bonds with second-period atoms such as O and N. Phosphorus reacts with metals to produce phosphides. Metal-rich phosphides are hard, high-melting, electrically conductive solids with metallic luster, whereas phosphorus-rich phosphides, which contain catenated phosphorus units, are lower melting and less thermally stable.
Nitrogen is the first diatomic molecule in the second period of elements. Why is N2 the most stable form of nitrogen? Draw its Lewis electron structure. What hybrid orbitals are used to describe the bonding in this molecule? Is the molecule polar?
The polymer (SN)n has metallic luster and conductivity. Are the constituent elements in this polymer metals or nonmetals? Why does the polymer have metallic properties?
Except for NF3, all the halides of nitrogen are unstable. Explain why NF3 is stable.
Which of the group 15 elements forms the most stable compounds in the +3 oxidation state? Explain why.
Phosphorus and arsenic react with the alkali metals to produce salts with the composition M3Z11. Compare these products with those produced by reaction of P and As with the alkaline earth metals. What conclusions can you draw about the types of structures favored by the heavier elements in this part of the periodic table?
PF3 reacts with F2 to produce PF5, which in turn reacts with F− to give salts that contain the PF6− ion. In contrast, NF3 does not react with F2, even under extreme conditions; NF5 and the NF6− ion do not exist. Why?
Red phosphorus is safer to handle than white phosphorus, reflecting their dissimilar properties. Given their structures, how do you expect them to compare with regard to reactivity, solubility, density, and melting point?
Bismuth oxalate [Bi2(C2O4)3] is a poison. Draw its structure and then predict its solubility in H2O, dilute HCl, and dilute HNO3. Predict its combustion products. Suggest a method to prepare bismuth oxalate from bismuth.
Small quantities of NO can be obtained in the laboratory by reaction of the iodide ion with acidic solutions of nitrite. Write a balanced chemical equation that represents this reaction.
Although pure nitrous acid is unstable, dilute solutions in water are prepared by adding nitrite salts to aqueous acid. Write a balanced chemical equation that represents this type of reaction.
Metallic versus nonmetallic behavior becomes apparent in reactions of the elements with an oxidizing acid, such as HNO3. Write balanced chemical equations for the reaction of each element of group 15 with nitric acid. Based on the products, predict which of these elements, if any, are metals and which, if any, are nonmetals.
Predict the product(s) of each reaction and then balance each chemical equation.
Write a balanced chemical equation to show how you would prepare each compound.
NaNO2(s) + HCl(aq) → HNO2(aq) + NaCl(aq)
The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element.
Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S8 in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO2 fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility.
Group 16 is the first group in the p block with no stable metallic elements.
Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO3, Ag2CO3, and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. (For more information on the Bessemer process, see .) Another important industrial use of oxygen is in the production of TiO2, which is commonly used as a white pigment in paints, paper, and plastics.
A crystalline sulfur deposit. This sulfur deposit is located around a volcanic vent in Kilauea Crater, Hawaii.
Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices.
Selenium (Se) was first isolated in 1817 by the Swedish chemist Jöns Jakob Berzelius, who also discovered silicon. He had invested money in a sulfuric acid plant and decided to investigate a foul-smelling contaminant that formed a red precipitate. Although he initially thought the contaminant was tellurium, further study showed that it was actually a new element similar to tellurium. To emphasize the similarities, Berzelius named the new element selenium (after the Greek selene, meaning “moon”). Selenium is used primarily as a minor ingredient to decolorize glass. Because it is photosensitive, selenium is also used to capture images in the photocopying process ().
Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium).
Figure 22.12 The Chemistry of Photocopying
Because amorphous selenium is a photosensitive semiconductor, exposing an electrostatically charged Se film to light causes the positive charge on the film to be discharged in all areas that are white in the original. Dark areas in the original block the light and generate an invisible, positively charged image. To produce an image on paper, negatively charged toner particles are attracted to the positive image, transferred to a negatively charged sheet of blank paper, and fused with the paper at high temperature to give a permanent image.
The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. (For more information on radioactivity and polonium, see , .) Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium.
Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO2:
Equation 22.33
(For more information on electrolysis, see . For more information on the alkali metals and the alkaline earth metals, see .)
Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (). Sulfur is also recovered from H2S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS2).
Pyrite (FeS2). Because of its lustrous golden yellow cubic crystals, FeS2 is sometimes mistaken for gold, giving rise to its common name “fool’s gold.” Real gold, however, is much denser than FeS2, and gold is soft and malleable rather than hard and brittle.
Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH3)2Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.”
With their ns2np4 electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in . As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po2+ or Po4+ ion in aqueous solution, depending on conditions.
Table 22.4 Selected Properties of the Group 16 Elements
Property | Oxygen | Sulfur | Selenium | Tellurium | Polonium |
---|---|---|---|---|---|
atomic symbol | O | S | Se | Te | Po |
atomic number | 8 | 16 | 34 | 52 | 84 |
atomic mass (amu) | 16.00 | 32.07 | 78.96 | 127.60 | 209 |
valence electron configuration* | 2s22p4 | 3s23p4 | 4s24p4 | 5s25p4 | 6s26p4 |
melting point/boiling point (°C) | −219/−183 | 115/445 | 221/685 | 450/988 | 254/962 |
density (g/cm3) at 25°C | 1.31 | 2.07 | 4.81 | 6.24 | 9.20 |
atomic radius (pm) | 48 | 88 | 103 | 123 | 135 |
first ionization energy (kJ/mol) | 1314 | 1000 | 941 | 869 | 812 |
normal oxidation state(s) | −2 | +6, +4, −2 | +6, +4, −2 | +6, +4, −2 | +2 (+4) |
ionic radius (pm)† | 140 (−2) | 184 (−2), 29 (+6) | 198 (−2), 42 (+6) | 221 (−2), 56 (+6) | 230 (−2), 97 (+4) |
electron affinity (kJ/mol) | −141 | −200 | −195 | −190 | −180 |
electronegativity | 3.4 | 2.6 | 2.6 | 2.1 | 2.0 |
standard reduction potential (E°, V) (E0 → H2E in acidic solution) | +1.23 | +0.14 | −0.40 | −0.79 | −1.00 |
product of reaction with O2 | — | SO2 | SeO2 | TeO2 | PoO2 |
type of oxide | — | acidic | acidic | amphoteric | basic |
product of reaction with N2 | NO, NO2 | none | none | none | none |
product of reaction with X2 | O2F2 | SF6, S2Cl2, S2Br2 | SeF6, SeX4 | TeF6, TeX4 | PoF4, PoCl2, PoBr2 |
product of reaction with H2 | H2O | H2S | H2Se | none | none |
*The configuration shown does not include filled d and f subshells. | |||||
†The values cited for the hexacations are for six-coordinate ions and are only estimated values. |
As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O2, all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O3), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H2O2) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure:
Equation 22.34
As in groups 14 and 15, the lightest element in group 16 has the greatest tendency to form multiple bonds.
Despite the strength of the O=O bond ( = 494 kJ/mol), O2 is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O2 and related species, such as the peroxide and superoxide ions, are in . With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO4 is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens.
Table 22.5 Some Properties of O2 and Related Diatomic Species
Species | Bond Order | Number of Unpaired e− | O–O Distance (pm)* |
---|---|---|---|
O2+ | 2.5 | 1 | 112 |
O2 | 2 | 2 | 121 |
O2− | 1.5 | 1 | 133 |
O22− | 1 | 0 | 149 |
*Source of data: Lauri Vaska, “Dioxygen-Metal Complexes: Toward a Unified View,” Accounts of Chemical Research 9 (1976): 175. |
Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO2, are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (Eδ+–Oδ−). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid:
Equation 22.35
H2O(l) + SO3(g) → H2SO4(aq)The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect:
Equation 22.36
Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l)Equation 22.37
Al2O3(s) + 2OH−(aq) + 3H2O(l) → 2Al(OH)4−(aq)Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric.
For each reaction, explain why the given products form.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution:
Exercise
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
Answer:
Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column.
Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S8 rings packed together in a complex “crankshaft” arrangement (), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (Sn2−) and polyselenides (Sen2−), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group.
As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO3 to form oxoacids such as H2SO4. In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po2+ ion.
Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group.
Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF6), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (). In contrast, only four fluorides of selenium (SeF6, SeF4, FSeSeF, and SeSeF2) and only three of tellurium (TeF4, TeF6, and Te2F10) are known.
Figure 22.13 The Structures of the Known Fluorides of Sulfur
Five stable sulfur fluorides are known, containing sulfur in oxidation states ranging from +1 to +6. All are volatile molecular compounds that vary tremendously in stability and toxicity. Although both SF6 and S2F10 are very stable, S2F10 is toxic and SF6 is not. The other three are highly reactive substances.
Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO2 is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO2 is a white solid with an infinite chain structure (each Se is three coordinate), TeO2 is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO2 is a yellow ionic solid in which each Po4+ ion is eight coordinate.
The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H2YO3—sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H2SeO4) are strong acids, but telluric acid [Te(OH)6] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH)6, with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid.
The stability of the highest oxidation state of the chalcogens decreases down the column.
Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS2 and CSe2 are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO2. Because these double bonds are significantly weaker than the C=O bond, however, CS2, CSe2, and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y2−), as in Na2S and FeS, or polychalcogenide ions (Yn2−), as in FeS2 and Na2S5.
The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group.
Ionic chalcogenides like Na2S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H2S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous.
For each reaction, explain why the given product forms or no reaction occurs.
Given: balanced chemical equations
Asked for: why the given products (or no products) form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs.
Solution:
One of the reactants (Cl2) is an oxidant. If the other reactant can be oxidized, then a redox reaction is likely. Sulfur dioxide contains sulfur in the +4 oxidation state, which is 2 less than its maximum oxidation state. Sulfur dioxide is also known to be a mild reducing agent in aqueous solution, producing sulfuric acid as the oxidation product. Hence a redox reaction is probable. The simplest reaction is the formation of SO2Cl2 (sulfuryl chloride), which is a tetrahedral species with two S–Cl and two S=O bonds.
Exercise
Predict the products of each reaction and write a balanced chemical equation for each reaction.
Answer:
Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group.
Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Why?
Classify each oxide as basic, acidic, amphoteric, or neutral.
Classify each oxide as basic, acidic, amphoteric, or neutral.
Polarization of an oxide affects its solubility in acids or bases. Based on this, do you expect RuO2 to be an acidic, a basic, or a neutral oxide? Is the compound covalent? Justify your answers.
Arrange CrO3, Al2O3, Sc2O3, and BaO in order of increasing basicity.
As the atomic number of the group 16 elements increases, the complexity of their allotropes decreases. What factors account for this trend? Which chalcogen do you expect to polymerize the most readily? Why?
Arrange H3BO3, HIO4, and HNO2 in order of increasing acid strength.
Of OF2, SO2, P4O6, SiO2, and Al2O3, which is most ionic?
Of CO2, NO2, O2, SO2, Cl2O, H2O, NH3, and CH4, which do you expect to have the
Of Na2O2, MgO, Al2O3, and SiO2, which is most acidic?
Give an example of
The Si–O bond is shorter and stronger than expected. What orbitals are used in this bond? Do you expect Si to interact with Br in the same way? Why or why not?
Oxygen has the second highest electronegativity of any element; consequently, it prefers to share or accept electrons from other elements. Only with fluorine does oxygen form compounds in positive oxidation states.
CrO3 < Al2O3 < Sc2O3 < BaO
H3BO3 < HNO2 < HIO4
Most polar: H2O; least polar: O2
Considering its position in the periodic table, predict the following properties of selenium:
Using arguments based on electronegativity, explain why ZnO is amphoteric. What product would you expect when ZnO reacts with an aqueous
Write a balanced chemical equation for the reaction of sulfur with
Because the halogens are highly reactive, none is found in nature as the free element. Hydrochloric acid, which is a component of aqua regia (a mixture of HCl and HNO3 that dissolves gold), and the mineral fluorspar (CaF2) were well known to alchemists, who used them in their quest for gold. Despite their presence in familiar substances, none of the halogens was even recognized as an element until the 19th century.
Because the halogens are highly reactive, none is found in nature as the free element.
Chlorine was the first halogen to be obtained in pure form. In 1774, Carl Wilhelm Scheele (the codiscoverer of oxygen) produced chlorine by reacting hydrochloric acid with manganese dioxide. Scheele was convinced, however, that the pale green gas he collected over water was a compound of oxygen and hydrochloric acid. In 1811, Scheele’s “compound” was identified as a new element, named from the Greek chloros, meaning “yellowish green” (the same stem as in chlorophyll, the green pigment in plants). That same year, a French industrial chemist, Bernard Courtois, accidentally added too much sulfuric acid to the residue obtained from burned seaweed. A deep purple vapor was released, which had a biting aroma similar to that of Scheele’s “compound.” The purple substance was identified as a new element, named iodine from the Greek iodes, meaning “violet.” Bromine was discovered soon after by a young French chemist, Antoine Jérôme Balard, who isolated a deep red liquid with a strong chlorine-like odor from brine from the salt marshes near Montpellier in southern France. Because many of its properties were intermediate between those of chlorine and iodine, Balard initially thought he had isolated a compound of the two (perhaps ICl). He soon realized, however, that he had discovered a new element, which he named bromine from the Greek bromos, meaning “stench.” Currently, organic chlorine compounds, such as PVC (polyvinylchloride), consume about 70% of the Cl2 produced annually; organobromine compounds are used in much smaller quantities, primarily as fire retardants.
Because of the unique properties of its compounds, fluorine was believed to exist long before it was actually isolated. The mineral fluorspar (now called fluorite [CaF2]) had been used since the 16th century as a “flux,” a low-melting-point substance that could dissolve other minerals and ores. In 1670, a German glass cutter discovered that heating fluorspar with strong acid produced a solution that could etch glass. The solution was later recognized to contain the acid of a new element, which was named fluorine in 1812. Elemental fluorine proved to be very difficult to isolate, however, because both HF and F2 are extraordinarily reactive and toxic. After being poisoned three times while trying to isolate the element, the French chemist Henri Moissan succeeded in 1886 in electrolyzing a sample of KF in anhydrous HF to produce a pale green gas (). For this achievement, among others, Moissan narrowly defeated Mendeleev for the Nobel Prize in Chemistry in 1906. Large amounts of fluorine are now consumed in the production of cryolite (Na3AlF6), a key intermediate in the production of aluminum metal. Fluorine is also found in teeth as fluoroapatite [Ca5(PO4)3F], which is formed by reacting hydroxyapatite [Ca5(PO4)3OH] in tooth enamel with fluoride ions in toothpastes, rinses, and drinking water.
A crystal of the mineral fluorite (CaF2). The purple color of some fluorite crystals is due to small inclusions of highly oxidizing impurities, which generate detectable amounts of ozone when the crystals are crushed.
Figure 22.14 Isolation of Elemental Fluorine
The French chemist Henri Moissan was the first person to isolate elemental fluorine. A reproduction of the U-shaped electrolysis cell with which Moissan first isolated elemental fluorine in 1866 is shown with samples of cryolite (left) and fluorspar (right). Fluorspar is the raw material from which anhydrous hydrofluoric acid (HF) is prepared. Cryolite is a rare mineral that contains the fluoride ion.
The heaviest halogen is astatine (At), which is continuously produced by natural radioactive decay. All its isotopes are highly radioactive, and the most stable has a half-life of only about 8 h. Consequently, astatine is the least abundant naturally occurring element on Earth, with less than 30 g estimated to be present in Earth’s crust at any one time.
All the halogens except iodine are found in nature as salts of the halide ions (X−), so the methods used for preparing F2, Cl2, and Br2 all involve oxidizing the halide. Reacting CaF2 with concentrated sulfuric acid produces gaseous hydrogen fluoride:
Equation 22.38
CaF2(s) + H2SO4(l) → CaSO4(s) + 2HF(g)Fluorine is produced by the electrolysis of a 1:1 mixture of HF and K+HF2− at 60–300°C in an apparatus made of Monel, a highly corrosion-resistant nickel–copper alloy:
Equation 22.39
Fluorine is one of the most powerful oxidants known, and both F2 and HF are highly corrosive. Consequently, the production, storage, shipping, and handling of these gases pose major technical challenges.
Figure 22.15 A Subterranean Salt Mine
Subterranean deposits of rock salt are located worldwide, such as this one at Petralia in Sicily.
Although chlorine is significantly less abundant than fluorine, elemental chlorine is produced on an enormous scale. Fortunately, large subterranean deposits of rock salt (NaCl) are found around the world (), and seawater consists of about 2% NaCl by mass, providing an almost inexhaustible reserve. Inland salt lakes such as the Dead Sea and the Great Salt Lake are even richer sources, containing about 23% and 8% NaCl by mass, respectively. Chlorine is prepared industrially by the chloralkali process, which uses the following reaction:
Equation 22.40
Bromine is much less abundant than fluorine or chlorine, but it is easily recovered from seawater, which contains about 65 mg of Br− per liter. Salt lakes and underground brines are even richer sources; for example, the Dead Sea contains 4 g of Br− per liter. Iodine is the least abundant of the nonradioactive halogens, and it is a relatively rare element. Because of its low electronegativity, iodine tends to occur in nature in an oxidized form. Hence most commercially important deposits of iodine, such as those in the Chilean desert, are iodate salts such as Ca(IO3)2. The production of iodine from such deposits therefore requires reduction rather than oxidation. The process is typically carried out in two steps: reduction of iodate to iodide with sodium hydrogen sulfite, followed by reaction of iodide with additional iodate:
Equation 22.41
2IO3−(aq) + 6HSO3−(aq) → 2I−(aq) + 6SO42−(aq) + 6H+(aq)Equation 22.42
5I−(aq) + IO3−(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)Because the halogens all have ns2np5 electron configurations, their chemistry is dominated by a tendency to accept an additional electron to form the closed-shell ion (X−). Only the electron affinity and the bond dissociation energy of fluorine differ significantly from the expected periodic trends shown in . Electron–electron repulsion is important in fluorine because of its small atomic volume, making the electron affinity of fluorine less than that of chlorine. Similarly, repulsions between electron pairs on adjacent atoms are responsible for the unexpectedly low F–F bond dissociation energy. (As discussed earlier, this effect is also responsible for the weakness of O–O, N–N, and N–O bonds.)
Oxidative strength decreases down group 17.
Electrostatic repulsions between lone pairs of electrons on adjacent atoms cause single bonds between N, O, and F to be weaker than expected.
Table 22.6 Selected Properties of the Group 17 Elements
Property | Fluorine | Chlorine | Bromine | Iodine | Astatine |
---|---|---|---|---|---|
atomic symbol | F | Cl | Br | I | At |
atomic number | 9 | 17 | 35 | 53 | 85 |
atomic mass (amu) | 19.00 | 35.45 | 79.90 | 126.90 | 210 |
valence electron configuration* | 2s22p5 | 3s23p5 | 4s24p5 | 5s25p5 | 6s26p5 |
melting point/boiling point (°C) | −220/−188 | −102/−34.0 | −7.2/58.8 | 114/184 | 302/— |
density (g/cm3) at 25°C | 1.55 | 2.90 | 3.10 | 4.93 | — |
atomic radius (pm) | 42 | 79 | 94 | 115 | 127 |
first ionization energy (kJ/mol) | 1681 | 1251 | 1140 | 1008 | 926 |
normal oxidation state(s) | −1 | −1 (+1, +3, +5, +7) | −1 (+1, +3, +5, +7) | −1 (+1, +3, +5, +7) | −1, +1 |
ionic radius (pm)† | 133 | 181 | 196 | 220 | — |
electron affinity (kJ/mol) | −328 | −349 | −325 | −295 | −270 |
electronegativity | 4.0 | 3.2 | 3.0 | 2.7 | 2.2 |
standard reduction potential (E°, V) (X2 → X− in basic solution) | +2.87 | +1.36 | +1.07 | +0.54 | +0.30 |
dissociation energy of X2(g) (kJ/mol) | 158.8 | 243.6 | 192.8 | 151.1 | ~80 |
product of reaction with O2 | O2F2 | none | none | none | none |
type of oxide | acidic | acidic | acidic | acidic | acidic |
product of reaction with N2 | none | none | none | none | none |
product of reaction with H2 | HF | HCl | HBr | HI | HAt |
*The configuration shown does not include filled d and f subshells. | |||||
†The values cited are for the six-coordinate anion (X−). |
Because it is the most electronegative element in the periodic table, fluorine forms compounds in only the −1 oxidation state. Notice, however, that all the halogens except astatine have electronegativities greater than 2.5, making their chemistry exclusively that of nonmetals. The halogens all have relatively high ionization energies, but the energy required to remove electrons decreases substantially as we go down the column. Hence the heavier halogens also form compounds in positive oxidation states (+1, +3, +5, and +7), derived by the formal loss of ns and np electrons.
Because ionization energies decrease down the group, the heavier halogens form compounds in positive oxidation states (+1, +3, +5, and +7).
Fluorine is the most reactive element in the periodic table, forming compounds with every other element except helium, neon, and argon. The reactions of fluorine with most other elements range from vigorous to explosive; only O2, N2, and Kr react slowly. There are three reasons for the high reactivity of fluorine:
With highly electropositive elements, fluorine forms ionic compounds that contain the closed-shell F− ion. In contrast, with less electropositive elements (or with metals in very high oxidation states), fluorine forms covalent compounds that contain terminal F atoms, such as SF6. Because of its high electronegativity and 2s22p5 valence electron configuration, fluorine normally participates in only one electron-pair bond. Only a very strong Lewis acid, such as AlF3, can share a lone pair of electrons with a fluoride ion, forming AlF63−.
The halogens (X2) react with metals (M) according to the general equation
Equation 22.43
M(s,l) + nX2(s,l,g) → MXn(s,l)For elements that exhibit multiple oxidation states fluorine tends to produce the highest possible oxidation state and iodine the lowest. For example, vanadium reacts with the halogens to give VF5, VCl4, VBr4, and VI3.
Metal halides in the +1 or +2 oxidation state, such as CaF2, are typically ionic halides, which have high melting points and are often soluble in water. As the oxidation state of the metal increases, so does the covalent character of the halide due to polarization of the M–X bond. With its high electronegativity, fluoride is the least polarizable, and iodide, with the lowest electronegativity, is the most polarizable of the halogens. Halides of small trivalent metal ions such as Al3+ tend to be relatively covalent. For example, AlBr3 is a volatile solid that contains bromide-bridged Al2Br6 molecules. In contrast, the halides of larger trivalent metals, such as the lanthanides, are essentially ionic. For example, indium tribromide (InBr3) and lanthanide tribromide (LnBr3) are all high-melting-point solids that are quite soluble in water.
As the oxidation state of the metal increases, the covalent character of the corresponding metal halides also increases due to polarization of the M–X bond.
All halogens react vigorously with hydrogen to give the hydrogen halides (HX). Because the H–F bond in HF is highly polarized (Hδ+–Fδ−), liquid HF has extensive hydrogen bonds, giving it an unusually high boiling point and a high dielectric constant. As a result, liquid HF is a polar solvent that is similar in some ways to water and liquid ammonia; after a reaction, the products can be recovered simply by evaporating the HF solvent. (Hydrogen fluoride must be handled with extreme caution, however, because contact of HF with skin causes extraordinarily painful burns that are slow to heal.) Because fluoride has a high affinity for silicon, aqueous hydrofluoric acid is used to etch glass, dissolving SiO2 to give solutions of the stable SiF62− ion.
Glass etched with hydrogen flouride.
© Thinkstock
Except for fluorine, all the halogens react with water in a disproportionation reaction, where X is Cl, Br, or I:
Equation 22.44
X2(g,l,s) + H2O(l) → H+(aq) + X−(aq) + HOX(aq)The most stable oxoacids are the perhalic acids, which contain the halogens in their highest oxidation state (+7). The acid strengths of the oxoacids of the halogens increase with increasing oxidation state, whereas their stability and acid strength decrease down the group. Thus perchloric acid (HOClO3, usually written as HClO4) is a more potent acid and stronger oxidant than perbromic acid. Although all the oxoacids are strong oxidants, some, such as HClO4, react rather slowly at low temperatures. Consequently, mixtures of the halogen oxoacids or oxoanions with organic compounds are potentially explosive if they are heated or even agitated mechanically to initiate the reaction. Because of the danger of explosions, oxoacids and oxoanions of the halogens should never be allowed to come into contact with organic compounds.
Both the acid strength and the oxidizing power of the halogen oxoacids decrease down the group.
The halogens react with one another to produce interhalogen compounds, such as ICl3, BrF5, and IF7. In all cases, the heavier halogen, which has the lower electronegativity, is the central atom. The maximum oxidation state and the number of terminal halogens increase smoothly as the ionization energy of the central halogen decreases and the electronegativity of the terminal halogen increases. Thus depending on conditions, iodine reacts with the other halogens to form IFn (n = 1–7), ICl or ICl3, or IBr, whereas bromine reacts with fluorine to form only BrF, BrF3, and BrF5 but not BrF7. The interhalogen compounds are among the most powerful Lewis acids known, with a strong tendency to react with halide ions to give complexes with higher coordination numbers, such as the IF8− ion:
Equation 22.45
IF7(l) + KF(s) → KIF8(s)All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7). The interhalogen compounds are also potent oxidants and strong fluorinating agents; contact with organic materials or water can result in an explosion.
All group 17 elements form compounds in odd oxidation states (−1, +1, +3, +5, +7), but the importance of the higher oxidation states generally decreases down the group.
For each reaction, explain why the given products form.
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution:
Exercise
Predict the products of each reaction and write a balanced chemical equation for each reaction.
Answer:
The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X− ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group. Fluorine, the most reactive element in the periodic table, has a low F–F bond dissociation energy due to repulsions between lone pairs of electrons on adjacent atoms. Fluorine forms ionic compounds with electropositive elements and covalent compounds with less electropositive elements and metals in high oxidation states. All the halogens react with hydrogen to produce hydrogen halides. Except for F2, all react with water to form oxoacids, including the perhalic acids, which contain the halogens in their highest oxidation state. Halogens also form interhalogen compounds; the heavier halogen, with the lower electronegativity, is the central atom.
The lightest elements of groups 15, 16, and 17 form unusually weak single bonds. Why are their bonds so weak?
Fluorine has an anomalously low F–F bond energy. Why? Why does fluorine form compounds only in the −1 oxidation state, whereas the other halogens exist in multiple oxidation states?
Compare AlI3, InCl3, GaF3, and LaBr3 with respect to the type of M–X bond formed, melting point, and solubility in nonpolar solvents.
What are the formulas of the interhalogen compounds that will most likely contain the following species in the indicated oxidation states: I (+3), Cl (+3), I (−1), Br (+5)?
Consider this series of bromides: AlBr3, SiBr4, and PBr5. Does the ionic character of the bond between the Br atoms and the central atom decrease or increase in this series?
Chromium forms compounds in the +6, +3, and +2 oxidation states. Which halogen would you use to produce each oxidation state? Justify your selections.
Of ClF7, BrF5, IF7, BrF3, ICl3, IF3, and IF5, which one is least likely to exist? Justify your selection.
Electrostatic repulsions between lone pairs on adjacent atoms decrease bond strength.
Ionic character decreases as Δχ decreases from Al to P.
ClF7
SiF4 reacts easily with NaF to form SiF62−. In contrast, CF4 is totally inert and shows no tendency to form CF62− under even extreme conditions. Explain this difference.
Predict the products of each reaction and then balance each chemical equation.
Write a balanced chemical equation for the reaction of aqueous HF with
Oxyhalides of sulfur, such as the thionyl halides (SOX2, where X is F, Cl, or Br), are well known. Because the thionyl halides react vigorously with trace amounts of water, they are used for dehydrating hydrated metal salts. Write a balanced chemical equation to show the products of reaction of SOCl2 with water.
Write a balanced chemical equation describing each reaction.
Write the complete Lewis electron structure, the type of hybrid used by the central atom, and the number of lone pair electrons present on the central atom for each compound.
Carbon has no low energy d orbitals that can be used to form a set of d2sp3 hybrid orbitals. It is also so small that it is impossible for six fluorine atoms to fit around it at a distance that would allow for formation of strong C–F bonds.
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. (For more information on spectroscopy, see .) Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O2, CO2, and water vapor from air and showed that this residual gas was slightly denser than pure N2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens.
Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University.
Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh.
In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N2, so scuba divers often use gas mixtures that contain these gases, rather than N2, to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N2(g) that can occur when a diver returns to the surface too rapidly.
Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source.
The elements of group 18 all have closed-shell valence electron configurations, either ns2np6 or 1s2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements.
Table 22.7 Selected Properties of the Group 18 Elements
Property | Helium | Neon | Argon | Krypton | Xenon | Radon |
---|---|---|---|---|---|---|
atomic symbol | He | Ne | Ar | Kr | Xe | Rn |
atomic number | 2 | 10 | 18 | 36 | 54 | 86 |
atomic mass (amu) | 4.00 | 20.18 | 39.95 | 83.80 | 131.29 | 222 |
valence electron configuration* | 1s2 | 2s22p6 | 3s23p6 | 4s24p6 | 5s25p6 | 6s26p6 |
triple point/boiling point (°C) | —/−269† | −249 (at 43 kPa)/−246 | −189 (at 69 kPa)/−189 | −157/−153 | −112 (at 81.6 kPa)/−108 | −71/−62 |
density (g/L) at 25°C | 0.16 | 0.83 | 1.63 | 3.43 | 5.37 | 9.07 |
atomic radius (pm) | 31 | 38 | 71 | 88 | 108 | 120 |
first ionization energy (kJ/mol) | 2372 | 2081 | 1521 | 1351 | 1170 | 1037 |
normal oxidation state(s) | 0 | 0 | 0 | 0 (+2) | 0 (+2, +4, +6, +8) | 0 (+2) |
electron affinity (kJ/mol) | > 0 | > 0 | > 0 | > 0 | > 0 | > 0 |
electronegativity | — | — | — | — | 2.6 | — |
product of reaction with O2 | none | none | none | none | none | none |
type of oxide | — | — | — | — | acidic | — |
product of reaction with N2 | none | none | none | none | none | none |
product of reaction with X2 | none | none | none | KrF2 | XeF2, XeF4, XeF6 | RnF2 |
product of reaction with H2 | none | none | none | none | none | none |
*The configuration shown does not include filled d and f subshells. | ||||||
†This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given. |
For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (). Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H2O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved. In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century.
Figure 22.16 The Structure of Xenon Hydrate, a Clathrate
Small gaseous atoms or molecules such as Xe or CH4 can occupy cavities in a lattice of hydrogen-bonded water molecules to produce a stable structure with a fixed stoichiometry (in this case, Xe·5.75H2O). (The hydrogen atoms of the water molecules have been omitted for clarity.) Warming the solid hydrate or decreasing the pressure of the gas causes it to collapse, with the evolution of gas and the formation of liquid water.
“Burning snowballs.” Like xenon, methane (CH4) forms a crystalline clathrate with water: methane hydrate. When the solid is warmed, methane is released and can be ignited to give what appears to be burning snow.
The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF6, a compound used in the Manhattan Project, could oxidize O2. Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O2, Bartlett recognized that PtF6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF6 vapor, yellow-orange crystals immediately formed (). Although Bartlett initially postulated that they were Xe+PtF6−, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the XeF+ ion:
Equation 22.46
Xe(g) + PtF6(g) → [XeF+][PtF5−](s)Figure 22.17 The Synthesis of the First Chemical Compound of Xenon
(a) An apparatus containing platinum hexafluoride, the red vapor at the bottom left, and xenon, the colorless gas in the small tube at the upper right. (b) When the glass seal separating the two gases is broken and the gases are allowed to mix, a bright yellow solid is formed, which is best described as XeF+PtF5−.
Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF2, XeF4, or XeF6, depending on conditions; one such reaction is as follows:
Equation 22.47
Xe(g) + 2F2(g) → XeF4(s)The ionization energies of helium, neon, and argon are so high () that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F2 and Cl2. Although XeCl2 and KrF2 can be prepared directly from the elements, they are substantially less stable than the xenon fluorides.
The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known.
Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF6 produces CsXeF7, which gives Cs2XeF8 when heated:
Equation 22.48
XeF6(s) + CsF(s) → CsXeF7(s)Equation 22.49
The XeF82− ion contains eight-coordinate xenon and has the square antiprismatic structure shown here, which is essentially identical to that of the IF8− ion. Cs2XeF8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs2XeF8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF6.)
For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between O and Xe. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either XeF4 or XeF6 produces XeO3, an explosive white solid:
Equation 22.50
XeF6(aq) + 3H2O(l) → XeO3(aq) + 6HF(aq)Treating a solution of XeO3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO4, a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO64−), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO3 is isoelectronic with the iodate ion (IO3−), and XeF82− is isoelectronic with the IF8− ion.
Xenon has a high affinity for both fluorine and oxygen.
Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored.
On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon?
Given: nonradioactive isotope of radon
Asked for: summary of its chemistry
Strategy:
Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon.
Solution:
We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF2, RnF4, RnF6, and possibly RnF8 (due to its large radius). The ion RnF82− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO3 and possibly RnO4. The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl2 and possibly even RnCl4.
Exercise
Predict the stoichiometry of the product formed by reacting XeF6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion.
Answer: KXeF7; the xenon atom in XeF7− has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s2 electrons behave like an inert pair that does not participate in bonding.
The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
The chemistry of the noble gases is largely dictated by a balance between two competing properties. What are these properties? How do they affect the reactivity of these elements?
Of the group 18 elements, only krypton, xenon, and radon form stable compounds with other atoms and then only with very electronegative elements. Why?
Give the type of hybrid orbitals used by xenon in each species.
Which element is the least metallic—B, Ga, Tl, Pb, Ne, or Ge?
Of Br, N, Ar, Bi, Se, He, and S, which would you expect to form positive ions most easily? negative ions most easily?
Of BCl3, BCl4−, CH4, H3N·BF3, PCl3, PCl5, XeO3, H2O, and F−, which species do you expect to be
Of HCl, HClO4, HBr, H2S, HF, KrF2, and PH3, which is the strongest acid?
Of CF4, NH3, NF3, H2O, OF2, SiF4, H2S, XeF4, and SiH4, which is the strongest base?
Write a balanced chemical equation showing how you would prepare each compound from its elements and other common compounds.
Write a balanced chemical equation showing how you would make each compound.
In an effort to synthesize XeF6, a chemist passed fluorine gas through a glass tube containing xenon gas. However, the product was not the one expected. What was the actual product?
Write a balanced chemical equation to describe the reaction of each species with water.
Using heavy water (D2O) as the source of deuterium, how could you prepare each compound?
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
SiF4; SiO2(s) + 2F2(g) → SiF4(l) + 2O2(g)
2Na(s) + 2D2O(l) → D2(g) + 2NaOD(aq)
2Li(s) + D2(g) → 2LiD(s)
4LiD(s) + AlCl3(soln) → LiAlD4(s) + 3LiCl(soln)
Borax (Na2B4O5(OH)4·8H2O) is used as a flux during welding operations. As brass is heated during welding, for example, borax cleans the surface of Cu2O and prevents further oxidation of the fused metal. Explain why borax is effective at cleaning the surface and preventing surface oxidation.
Extensive research is being conducted into using GaAs as a material for computer memory chips. It has been found, for example, that chips made from GaAs are up to 10 times faster than those made from silicon. Propose an explanation for this increase in speed.
Cement that has a high content of alumina (Al2O3) is particularly resistant to corrosion, so it is used for structures that must be resistant to seawater and acidic conditions. Why is this material so effective under these service conditions? Failure occurs under prolonged exposure to a hot, wet environment. Why?
Aluminum is light and ductile. If you were considering using aluminum rather than steel as a structural material for building a high-speed ferry, what disadvantages would you need to consider in using aluminum for these service conditions?
Life on Earth is based on carbon. A possible explanation is that no other element in the periodic table forms compounds that are so diverse in their chemistry and physical properties. Discuss the chemistry of carbon with regard to
Then compare B, Al, Si, N, and P with C in terms of these properties.
After a traffic accident in which a tanker truck carrying liquid nitrogen overturned, a reporter at the scene warned of a danger to residents in the vicinity of the accident because nitrogen would react with hydrocarbons in the asphalt to produce ammonia gas. Comment on the credibility of this statement.
Nitrogen forms a hydride called hydrazoic acid (HN3), which is a colorless, highly toxic, explosive substance that boils at 37°C. The thermal decomposition of one of its salts—NaN3—is used to inflate automotive air bags. The N3− ion is isoelectronic with CO2.
Hydrazine (N2H4), a rocket fuel, is a colorless, oily liquid with a melting point of 1.4°C, and it is a powerful reducing agent. The physical properties of hydrazine presumably reflect the presence of multiple hydrogen-bond acceptors and donors within a single molecule. Explain the basis for this statement.
Because the N–C bond is almost as strong as the N–H bond, organic analogues of ammonia, hydrazine, and hydroxylamine are stable and numerous. Conceptually at least, they are formed by the successive replacement of H atoms by alkyl or aryl groups. Methylhydrazine and dimethylhydrazine, for example, were used as fuels in the US Apollo space program. They react spontaneously and vigorously with liquid N2O4, thus eliminating the need for an ignition source. Write balanced chemical equations for these reactions and calculate ΔG° for each reaction.
In an effort to remove a troublesome stain from a sink, a member of the cleaning staff of a commercial building first used bleach on the stain and then decided to neutralize the bleach with ammonia. What happened? Why?
A slow reaction that occurs on the ocean floor is the conversion of carbonate to bicarbonate, which absorbs CO2. Write a balanced chemical equation for this reaction. Silicate sediments play an important role in controlling the pH of seawater. Given the reaction, propose a chemical explanation for this.
Marketing surveys have shown that customers prefer to buy a bright red steak rather than a dull gray one. It is known that NO combines with myoglobin to form a bright red NO complex. What would you add to beef during processing to ensure that this reaction occurs and yields the desired appearance?
Covalent azides are used as detonators and explosives. Ionic azides, in contrast, are usually much more stable and are used in dyestuffs. Why is there such a difference between these two types of compounds? The N3− ion is considered a pseudohalide. Why?
The heads of modern “strike anywhere” matches contain a mixture of a nonvolatile phosphorus sulfide (P4S3) and an oxidizing agent (KClO3), which is ignited by friction when the match is struck against a rough object. Safety matches separate the oxidant and the reductant by putting KClO3 in the head and a paste containing nonvolatile red phosphorus on the match box or cover. Write a balanced chemical equation for the reaction that occurs when a match is rubbed against the abrasive end of a matchbox.
Paris green was a common pigment in paints and wallpaper of the Napoleonic era. It is a mixed acetate/arsenite salt of copper with the formula Cu2(OAc)2(AsO3). In damp conditions, certain fungi are able to convert arsenite salts to volatile, toxic organoarsenic compounds. Shortly after his exile in 1815 to the remote island of St. Helena in the southern Atlantic Ocean, Napoleon died. As a forensic scientist investigating the cause of Napoleon’s mysterious death, you notice that the walls of his enclosed bedchamber are covered in green wallpaper. What chemical clues would you look for to determine the cause of his death?
Selenium, an element essential to humans, appears to function biologically in an enzyme that destroys peroxides. Why is selenium especially suited for this purpose? Would sulfur or tellurium be as effective? Why or why not?
One way to distinguish between fool’s gold (FeS2, or iron pyrite) and real gold is to heat the sample over a fire. If your sample of “gold” were actually fool’s gold, what would happen?
Calcium hypochlorite is sold as swimming pool bleach. It is formed by the hydrolysis of Cl2O, which gives only one product, followed by neutralization with lime [Ca(OH)2]. Write balanced chemical equations for these reactions.
There is much interest in the superheavy elements beyond Z = 111 because of their potentially unique properties. Predict the valence electron configurations, preferred oxidation states, and products of the reaction with aqueous acid for elements 113 and 115.
Zeolites have become increasingly important in chemical engineering. They can be used as desiccants because the dehydrated zeolite absorbs small molecules, such as water. To be retained by the zeolite frame, a molecule must satisfy two conditions. What are they? Why can linear CO2 and tetrahedral CH4 not be held by a typical zeolite, even though they can penetrate it easily?
Three resonance structures for the azide ion may be reasonably drawn:
Examine samples of Napoleon’s hair and/or fingernails from museums or collections to determine arsenic concentrations.
Upon heating, pyrite will react with oxygen to form SO2(g), which has a pungent smell.
Element 113: 5f146d107s27p1, +1, E+(aq); element 115: 5f146d107s27p3, +3, E3+(aq)
Chapter 21 "Periodic Trends and the " and Chapter 22 "The " described the chemistry of the s-block and p-block elements. In this chapter, we survey the chemistry of the d-block elements, which are also called the transition metals. We again use valence electron configurations and periodic trends, as well as the principles of chemical bonding, thermodynamics, and kinetics, as tools to describe the properties and reactivity of these elements. Because all the d-block elements are metals, they do not have the extreme variability in chemistry that we saw among the elements of the p block. Instead, these elements exhibit significant horizontal and vertical similarities in chemistry, and all have a common set of characteristic properties due to partially filled d subshells.
Titanium metal is light and corrosion resistant. The Guggenheim Museum in Bilbao, Spain, is the largest titanium-clad building in the world. The exterior is covered with 344,000 ft2 of 0.016-in.-thick titanium pieces, each with a unique shape.
Alloys and compounds of the d-block elements are important components of the materials the modern world depends on for its continuing technological development, while most of the first-row transition metals are essential for life. This chapter introduces some of the key industrial and biological roles of these elements. You will learn, for example, why copper, silver, and gold have been used for coins and jewelry since ancient times, how Cr3+ impurities can be responsible for the characteristic colors of both rubies and emeralds, why an iron oxide was used in primitive compasses, why insects have greenish-blue blood, and why cobalt is an essential component of vitamin B12.
The transition metals, groups 3–12 in the periodic table, are generally characterized by partially filled d subshells in the free elements or their cations. (Although the metals of group 12 do not have partially filled d shells, their chemistry is similar in many ways to that of the preceding groups, and we therefore include them in our discussion.) Unlike the s-block and p-block elements, the transition metals exhibit significant horizontal similarities in chemistry in addition to their vertical similarities.
The valence electron configurations of the first-row transition metals are given in . As we go across the row from left to right, electrons are added to the 3d subshell to neutralize the increase in the positive charge of the nucleus as the atomic number increases. With two important exceptions, the 3d subshell is filled as expected based on the aufbau principle and Hund’s rule. Unexpectedly, however, chromium has a 4s13d5 electron configuration rather than the 4s23d4 configuration predicted by the aufbau principle, and copper is 4s13d10 rather than 4s23d9. In , we attributed these anomalies to the extra stability associated with half-filled subshells. Because the ns and (n − 1)d subshells in these elements are similar in energy, even relatively small effects are enough to produce apparently anomalous electron configurations.
Table 23.1 Valence Electron Configurations of the First-Row Transition Metals
Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
---|---|---|---|---|---|---|---|---|---|
4s23d1 | 4s23d2 | 4s23d3 | 4s13d5 | 4s23d5 | 4s23d6 | 4s23d7 | 4s23d8 | 4s13d10 | 4s23d10 |
In the second-row transition metals, electron–electron repulsions within the 4d subshell cause additional irregularities in electron configurations that are not easily predicted. For example, Nb and Tc, with atomic numbers 41 and 43, both have a half-filled 5s subshell, with 5s14d4 and 5s14d6 valence electron configurations, respectively. Further complications occur among the third-row transition metals, in which the 4f, 5d, and 6s orbitals are extremely close in energy. Although La has a 6s25d1 valence electron configuration, the valence electron configuration of the next element—Ce—is 6s25d04f2. From this point through element 71, added electrons enter the 4f subshell, giving rise to the 14 elements known as the lanthanides. After the 4f subshell is filled, the 5d subshell is populated, producing the third row of the transition metals. Next comes the seventh period, where the actinides have three subshells (7s, 6d, and 5f) that are so similar in energy that their electron configurations are even more unpredictable.
As we saw in the s-block and p-block elements, the size of neutral atoms of the d-block elements gradually decreases from left to right across a row, due to an increase in the effective nuclear charge (Zeff) with increasing atomic number. In addition, the atomic radius increases down a group, just as it does in the s and p blocks. Because of the lanthanide contraction, however, the increase in size between the 3d and 4d metals is much greater than between the 4d and 5d metals (). (For more information on the lanthanides, see , .) The effects of the lanthanide contraction are also observed in ionic radii, which explains why, for example, there is only a slight increase in radius from Mo3+ to W3+.
Figure 23.1 The Metallic Radii of the First-, Second-, and Third-Row Transition Metals
Because of the lanthanide contraction, the second- and third-row transition metals are very similar in size.
As you learned in , electrons in (n − 1)d and (n − 2)f subshells are only moderately effective at shielding the nuclear charge; as a result, the effective nuclear charge experienced by valence electrons in the d-block and f-block elements does not change greatly as the nuclear charge increases across a row. Consequently, the ionization energies of these elements increase very slowly across a given row (). In addition, as we go from the top left to the bottom right corner of the d block, electronegativities generally increase, densities and electrical and thermal conductivities increase, and enthalpies of hydration of the metal cations decrease in magnitude, as summarized in . Consistent with this trend, the transition metals become steadily less reactive and more “noble” in character from left to right across a row. The relatively high ionization energies and electronegativities and relatively low enthalpies of hydration are all major factors in the noble character of metals such as Pt and Au.
Figure 23.2 Some Trends in Properties of the Transition Metals
The electronegativity of the elements increases, and the hydration energies of the metal cations decrease in magnitude from left to right and from top to bottom of the d block. As a result, the metals in the lower right corner of the d block are so unreactive that they are often called the “noble metals.”
The similarity in ionization energies and the relatively small increase in successive ionization energies lead to the formation of metal ions with the same charge for many of the transition metals. This in turn results in extensive horizontal similarities in chemistry, which are most noticeable for the first-row transition metals and for the lanthanides and actinides. Thus all the first-row transition metals except Sc form stable compounds that contain the 2+ ion, and, due to the small difference between the second and third ionization energies for these elements, all except Zn also form stable compounds that contain the 3+ ion. The relatively small increase in successive ionization energies causes most of the transition metals to exhibit multiple oxidation states separated by a single electron. Manganese, for example, forms compounds in every oxidation state between −3 and +7. Because of the slow but steady increase in ionization potentials across a row, high oxidation states become progressively less stable for the elements on the right side of the d block. The occurrence of multiple oxidation states separated by a single electron causes many, if not most, compounds of the transition metals to be paramagnetic, with one to five unpaired electrons. This behavior is in sharp contrast to that of the p-block elements, where the occurrence of two oxidation states separated by two electrons is common, which makes virtually all compounds of the p-block elements diamagnetic.
Due to a small increase in successive ionization energies, most of the transition metals have multiple oxidation states separated by a single electron.
Most compounds of transition metals are paramagnetic, whereas virtually all compounds of the p-block elements are diamagnetic.
The electronegativities of the first-row transition metals increase smoothly from Sc (χ = 1.4) to Cu (χ = 1.9). Thus Sc is a rather active metal, whereas Cu is much less reactive. The steady increase in electronegativity is also reflected in the standard reduction potentials: thus E° for the reaction M2+(aq) + 2e− → M0(s) becomes progressively less negative from Ti (E° = −1.63 V) to Cu (E° = +0.34 V). Exceptions to the overall trends are rather common, however, and in many cases, they are attributable to the stability associated with filled and half-filled subshells. For example, the 4s23d10 electron configuration of zinc results in its strong tendency to form the stable Zn2+ ion, with a 3d10 electron configuration, whereas Cu+, which also has a 3d10 electron configuration, is the only stable monocation formed by a first-row transition metal. Similarly, with a half-filled subshell, Mn2+ (3d5) is much more difficult to oxidize than Fe2+ (3d6). The chemistry of manganese is therefore primarily that of the Mn2+ ion, whereas both the Fe2+ and Fe3+ ions are important in the chemistry of iron.
The transition metals form cations by the initial loss of the ns electrons of the metal, even though the ns orbital is lower in energy than the (n − 1)d subshell in the neutral atoms. This apparent contradiction is due to the small difference in energy between the ns and (n − 1)d orbitals, together with screening effects. The loss of one or more electrons reverses the relative energies of the ns and (n − 1)d subshells, making the latter lower in energy. Consequently, all transition-metal cations possess dnvalence electron configurations, as shown in for the 2+ ions of the first-row transition metals.
All transition-metal cations have dn electron configurations; the ns electrons are always lost before the (n − 1)d electrons.
Table 23.2 d-Electron Configurations of the Dications of the First-Row Transition Metals
Ti2+ | V2+ | Cr2+ | Mn2+ | Fe2+ | Co2+ | Ni2+ | Cu2+ | Zn2+ |
---|---|---|---|---|---|---|---|---|
d 2 | d 3 | d 4 | d 5 | d 6 | d 7 | d 8 | d 9 | d 10 |
The most common oxidation states of the first-row transition metals are shown in . The second- and third-row transition metals behave similarly but with three important differences:
The highest possible oxidation state, corresponding to the formal loss of all valence electrons, becomes increasingly less stable as we go from group 3 to group 8, and it is never observed in later groups.
In the transition metals, the stability of higher oxidation states increases down a column.
Table 23.3 Common Oxidation States of the First-Row Transition Metals*
Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | |
---|---|---|---|---|---|---|---|---|---|---|
electronic structure | s 2d 1 | s 2d 2 | s 2d 3 | s 1d 5 | s 2d 5 | s 2d 6 | s 2d 7 | s 2d 8 | s 1d 10 | s 2d 10 |
oxidation states | I | I | ||||||||
II | II | II | II | II | II | II | II | II | ||
III | III | III | III | III | III | III | III | III | ||
IV | IV | IV | IV | IV | IV | IV | ||||
V | V | V | V | V | ||||||
VI | VI | VI | ||||||||
VII | ||||||||||
*The convention of using roman numerals to indicate the oxidation states of a metal is used here. |
Binary transition-metal compounds, such as the oxides and sulfides, are usually written with idealized stoichiometries, such as FeO or FeS, but these compounds are usually cation deficient and almost never contain a 1:1 cation:anion ratio. Thus a substance such as ferrous oxide is actually a nonstoichiometric compound with a range of compositions.
The acid–base character of transition-metal oxides depends strongly on the oxidation state of the metal and its ionic radius. Oxides of metals in lower oxidation states (less than or equal to +3) have significant ionic character and tend to be basic. Conversely, oxides of metals in higher oxidation states are more covalent and tend to be acidic, often dissolving in strong base to form oxoanions.
Two of the group 8 metals (Fe, Ru, and Os) form stable oxides in the +8 oxidation state. Identify these metals; predict the stoichiometry of the oxides; describe the general physical and chemical properties, type of bonding, and physical state of the oxides; and decide whether they are acidic or basic oxides.
Given: group 8 metals
Asked for: identity of metals and expected properties of oxides in +8 oxidation state
Strategy:
Refer to the trends outlined in , , , , and to identify the metals. Decide whether their oxides are covalent or ionic in character, and, based on this, predict the general physical and chemical properties of the oxides.
Solution:
The +8 oxidation state corresponds to a stoichiometry of MO4. Because the heavier transition metals tend to be stable in higher oxidation states, we expect Ru and Os to form the most stable tetroxides. Because oxides of metals in high oxidation states are generally covalent compounds, RuO4 and OsO4 should be volatile solids or liquids that consist of discrete MO4 molecules, which the valence-shell electron-pair repulsion (VSEPR) model predicts to be tetrahedral. Finally, because oxides of transition metals in high oxidation states are usually acidic, RuO4 and OsO4 should dissolve in strong aqueous base to form oxoanions.
Exercise
Predict the identity and stoichiometry of the stable group 9 bromide in which the metal has the lowest oxidation state and describe its chemical and physical properties.
Answer: Because the lightest element in the group is most likely to form stable compounds in lower oxidation states, the bromide will be CoBr2. We predict that CoBr2 will be an ionic solid with a relatively high melting point and that it will dissolve in water to give the Co2+(aq) ion.
The transition metals are characterized by partially filled d subshells in the free elements and cations. The ns and (n − 1)d subshells have similar energies, so small influences can produce electron configurations that do not conform to the general order in which the subshells are filled. In the second- and third-row transition metals, such irregularities can be difficult to predict, particularly for the third row, which has 4f, 5d, and 6s orbitals that are very close in energy. The increase in atomic radius is greater between the 3d and 4d metals than between the 4d and 5d metals because of the lanthanide contraction. Ionization energies and electronegativities increase slowly across a row, as do densities and electrical and thermal conductivities, whereas enthalpies of hydration decrease. Anomalies can be explained by the increased stabilization of half-filled and filled subshells. Transition-metal cations are formed by the initial loss of ns electrons, and many metals can form cations in several oxidation states. Higher oxidation states become progressively less stable across a row and more stable down a column. Oxides of small, highly charged metal ions tend to be acidic, whereas oxides of metals with a low charge-to-radius ratio are basic.
The transition metals show significant horizontal similarities in chemistry in addition to their vertical similarities, whereas the same cannot be said of the s-block and p-block elements. Explain why this is so.
The energy of the d subshell does not change appreciably in a given period. Why? What effect does this have on the ionization potentials of the transition metals? on their electronegativities?
Standard reduction potentials vary across the first-row transition metals. What effect does this have on the chemical reactivity of the first-row transition metals? Which two elements in this period are more active than would be expected? Why?
Many transition metals are paramagnetic (have unpaired electrons). How does this affect electrical and thermal conductivities across the rows?
What is the lanthanide contraction? What effect does it have on the radii of the transition metals of a given group? What effect does it have on the chemistry of the elements in a group?
Why are the atomic volumes of the transition elements low compared with the elements of groups 1 and 2? Ir has the highest density of any element in the periodic table (22.65 g/cm3). Why?
Of the elements Ti, Ni, Cu, and Cd, which do you predict has the highest electrical conductivity? Why?
The chemistry of As is most similar to the chemistry of which transition metal? Where in the periodic table do you find elements with chemistry similar to that of Ge? Explain your answers.
The coinage metals (group 11) have significant noble character. In fact, they are less reactive than the elements of group 12. Explain why this is so, referring specifically to their reactivity with mineral acids, electronegativity, and ionization energies. Why are the group 12 elements more reactive?
Give the valence electron configurations of the 2+ ion for each first-row transition element. Which two ions do you expect to have the most negative E° value? Why?
Arrange Ru3+, Cu2+, Zn, Ti4+, Cr3+, and Ni2+ in order of increasing radius.
Arrange Pt4+, Hg2+, Fe2+, Zr4+, and Fe3+ in order of decreasing radius.
Of Ti2+, V2+, Mn2+, Fe2+, Co2+, Ni2+, and Zn2+, which divalent ion has the smallest ionic radius? Explain your reasoning.
Ti2+, 3d2; V2+, 3d3; Cr2+, 3d4; Mn2+, 3d5; Fe2+, 3d6; Co2+, 3d7; Ni2+, 3d8; Cu2+, 3d9; Zn2+, 3d10. Because Zeff increases from left to right, Ti2+ and V2+ will have the most negative reduction potentials (be most difficult to reduce).
Hg2+ > Fe2+ > Zr4+ > Fe3+ > Pt4+
We turn now to a brief survey of the chemistry of the transition metals, beginning with group 3. As we shall see, the two heaviest members of each group usually exhibit substantial similarities in chemical behavior and are quite different from the lightest member.
As shown in , the observed trends in the properties of the group 3 elements are similar to those of groups 1 and 2. Due to their ns2(n − 1)d1 valence electron configurations, the chemistry of all four elements is dominated by the +3 oxidation state formed by losing all three valence electrons. As expected based on periodic trends, these elements are highly electropositive metals and powerful reductants, with La (and Ac) being the most reactive. In keeping with their highly electropositive character, the group 3 metals react with water to produce the metal hydroxide and hydrogen gas:
Equation 23.1
2M(s) + 6H2O(l) → 2M(OH)3(s) + 3H2(g)The chemistry of the group 3 metals is almost exclusively that of the M3+ ion; the elements are powerful reductants.
Moreover, all dissolve readily in aqueous acid to produce hydrogen gas and a solution of the hydrated metal ion: M3+(aq).
Table 23.4 Some Properties of the Elements of Groups 3, 4, and 5
Group | Element | Z | Valence Electron Configuration | Electronegativity | Metallic Radius (pm) | Melting Point (°C) | Density |
---|---|---|---|---|---|---|---|
3 | Sc | 21 | 4s23d1 | 1.36 | 162 | 1541 | 2.99 |
Y | 39 | 5s24d1 | 1.22 | 180 | 1522 | 4.47 | |
La | 57 | 6s25d1 | 1.10 | 183 | 918 | 6.15 | |
Ac | 89 | 7s26d1 | 1.10 | 188 | 1051 | 10.07 | |
4 | Ti | 22 | 4s23d2 | 1.54 | 147 | 1668 | 4.51 |
Zr | 40 | 5s24d2 | 1.33 | 160 | 1855 | 6.52 | |
Hf | 72 | 6s25d24f14 | 1.30 | 159 | 2233 | 13.31 | |
5 | V | 23 | 4s23d3 | 1.63 | 134 | 1910 | 6.00 |
Nb | 41 | 5s24d3 | 1.60 | 146 | 2477 | 8.57 | |
Ta | 73 | 6s25d34f14 | 1.50 | 146 | 3017 | 16.65 |
The group 3 metals react with nonmetals to form compounds that are primarily ionic in character. For example, reacting group 3 metals with the halogens produces the corresponding trihalides: MX3. The trifluorides are insoluble in water because of their high lattice energies, but the other trihalides are very soluble in water and behave like typical ionic metal halides. All group 3 elements react with air to form an oxide coating, and all burn in oxygen to form the so-called sesquioxides (M2O3), which react with H2O or CO2 to form the corresponding hydroxides or carbonates, respectively. Commercial uses of the group 3 metals are limited, but “mischmetal,” a mixture of lanthanides containing about 40% La, is used as an additive to improve the properties of steel and make flints for cigarette lighters.
Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns2(n − 1)d2 valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti:
Equation 23.2
2FeTiO3(s) + 6C(s) + 7Cl2(g) → 2TiCl4(g) + 2FeCl3(g) + 6CO(g)followed by reduction of the tetrachlorides with an active metal such as Mg.
The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states.
In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO2-containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines.
Consistent with the periodic trends shown in , the group 4 metals become denser, higher melting, and more electropositive down the column (). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns2(n − 1)d2 valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti2+ are usually powerful reductants. In fact, the Ti2+(aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas.
Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX4), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl2 is an ionic salt, whereas TiCl4 is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO2) and dichalcogenides (MY2). Industrially, TiO2, which is used as a white pigment in paints, is prepared by reacting TiCl4 with oxygen at high temperatures:
Equation 23.3
TiCl4(g) + O2(g) → TiO2(s) + 2Cl2(g)The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH2), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB2), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity.
Figure 23.3 The Layered Structure of TiS2
Each titanium atom is surrounded by an octahedral arrangement of six sulfur atoms that are shared to form extended layers of atoms. Because the layers are held together by only van der Waals forces between adjacent sulfur atoms, rather than covalent bonds, the layers slide past one another relatively easily when a mechanical stress is applied.
Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores.
Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V4C3, which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V2O5, an important catalyst for the industrial conversion of SO2 to SO3 in the contact process for the production of sulfuric acid. (For more information on sulfuric acid production, see , .) In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb3Zr and Nb3Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones.
The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4.
As indicated in , the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M2O5), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H2O)6]2+; the blue-green [V(H2O)6]3+ ion is acidic, dissociating to form small amounts of the [V(H2O)5(OH)]2+ ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H2O)4VO]2+, which contains a formal V=O bond (). Consistent with its covalent character, V2O5 is acidic, dissolving in base to give the vanadate ion ([VO4]3−), whereas both Nb2O5 and Ta2O5 are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric.
Figure 23.4 Aqueous Solutions of Vanadium Ions in Oxidation States of +2 to +5
Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO2]+ ion; V(IV) is the blue vanadyl ion [VO]2+; and V(III) and V(II) exist as the hydrated V3+ (blue-green) and V2+ (violet) ions, respectively.
Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY2), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools.
As an illustration of the trend toward increasing polarizability as we go from left to right across the d block, in group 6 we first encounter a metal (Mo) that occurs naturally as a sulfide ore rather than as an oxide. Molybdenite (MoS2) is a soft black mineral that can be used for writing, like PbS and graphite. Because of this similarity, people long assumed that these substances were all the same. In fact, the name molybdenum is derived from the Greek molybdos, meaning “lead.” More than 90% of the molybdenum produced annually is used to make steels for cutting tools, which retain their sharp edge even when red hot. In addition, molybdenum is the only second- or third-row transition element that is essential for humans. The major chromium ore is chromite (FeCr2O4), which is oxidized to the soluble [CrO4]2− ion under basic conditions and reduced successively to Cr2O3 and Cr with carbon and aluminum, respectively. Pure chromium can be obtained by dissolving Cr2O3 in sulfuric acid followed by electrolytic reduction; a similar process is used for electroplating metal objects to give them a bright, shiny, protective surface layer. Pure tungsten is obtained by first converting tungsten ores to WO3, which is then reduced with hydrogen to give the metal.
Consistent with periodic trends, the group 6 metals are slightly less electropositive than those of the three preceding groups, and the two heaviest metals are essentially the same size because of the lanthanide contraction (). All three elements have a total of six valence electrons, resulting in a maximum oxidation state of +6. Due to extensive polarization of the anions, compounds in the +6 oxidation state are highly covalent. As in groups 4 and 5, the lightest element exhibits variable oxidation states, ranging from Cr2+, which is a powerful reductant, to CrO3, a red solid that is a powerful oxidant. For Mo and W, the highest oxidation state (+6) is by far the most important, although compounds in the +4 and +5 oxidation states are known.
The metals become increasing polarizable across the d block.
Table 23.5 Some Properties of the Elements of Groups 6 and 7
Group | Element | Z | Valence Electron Configuration | Electronegativity | Metallic Radius (pm) | Melting Point (°C) | Density |
---|---|---|---|---|---|---|---|
6 | Cr | 24 | 4s13d5 | 1.66 | 128 | 1907 | 7.15 |
Mo | 42 | 5s14d5 | 2.16 | 139 | 2623 | 10.20 | |
W | 74 | 6s25d44f14 | 1.70 | 139 | 3422 | 19.30 | |
7 | Mn | 25 | 4s23d5 | 1.55 | 127 | 1246 | 7.30 |
Tc | 43 | 5s24d5 | 2.10 | 136 | 2157 | 11.50 | |
Re | 75 | 6s25d54f14 | 1.90 | 137 | 3186 | 20.80 |
The chemistry of the two heaviest group 6 metals (Mo and W) is dominated by the +6 oxidation state. The chemistry of the lightest element (Cr) is dominated by lower oxidation states.
As observed in previous groups, the group 6 halides become more covalent as the oxidation state of the metal increases: their volatility increases, and their melting points decrease. Recall that as the electronegativity of the halogens decreases from F to I, they are less able to stabilize high oxidation states; consequently, the maximum oxidation state of the corresponding metal halides decreases. Thus all three metals form hexafluorides, but CrF6 is unstable at temperatures above −100°C, whereas MoF6 and WF6 are stable. Consistent with the trend toward increased stability of the highest oxidation state for the second- and third-row elements, the other halogens can oxidize chromium to only the trihalides, CrX3 (X is Cl, Br, or I), while molybdenum forms MoCl5, MoBr4, and MoI3, and tungsten gives WCl6, WBr5, and WI4.
Both Mo and W react with oxygen to form the covalent trioxides (MoO3 and WO3), but Cr reacts to form only the so-called sesquioxide (Cr2O3). Chromium will form CrO3, which is a highly toxic compound that can react explosively with organic materials. All the trioxides are acidic, dissolving in base to form the corresponding oxoanions ([MO4]2−). Consistent with periodic trends, the sesquioxide of the lightest element in the group (Cr2O3) is amphoteric. The aqueous chemistry of molybdate and tungstate is complex, and at low pH they form a series of polymeric anions called isopolymetallates, such as the [Mo8O26]4− ion, whose structure is as follows:
An isopolymolybdate cluster. The [Mo8O26]4− ion, shown here in both side and top views, is typical of the oxygen-bridged clusters formed by Mo(VI) and W(VI) in aqueous solution.
Reacting molybdenum or tungsten with heavier chalcogens gives binary chalcogenide phases, most of which are nonstoichiometric and electrically conducting. One of the most stable is MoS2; it has a layered structure similar to that of TiS2 (), in which the layers are held together by only weak van der Waals forces, which allows them to slide past one another rather easily. Consequently, both MoS2 and WS2 are used as lubricants in a variety of applications, including automobile engines. Because tungsten itself has an extraordinarily high melting point (3380°C), lubricants described as containing “liquid tungsten” actually contain a suspension of very small WS2 particles.
As in groups 4 and 5, the elements of group 6 form binary nitrides, carbides, and borides whose stoichiometries and properties are similar to those of the preceding groups. Tungsten carbide (WC), one of the hardest compounds known, is used to make the tips of drill bits.
Continuing across the periodic table, we encounter the group 7 elements (). One group 7 metal (Mn) is usually combined with iron in an alloy called ferromanganese, which has been used since 1856 to improve the mechanical properties of steel by scavenging sulfur and oxygen impurities to form MnS and MnO. Technetium is named after the Greek technikos, meaning “artificial,” because all its isotopes are radioactive. One isotope, 99mTc (m for metastable), has become an important biomedical tool for imaging internal organs. (For more information on biomedical imaging, see , .) Because of its scarcity, Re is one of the most expensive elements, and its applications are limited. It is, however, used in a bimetallic Pt/Re catalyst for refining high-octane gasoline.
All three group 7 elements have seven valence electrons and can form compounds in the +7 oxidation state. Once again, the lightest element exhibits multiple oxidation states. Compounds of Mn in oxidation states ranging from −3 to +7 are known, with the most common being +2 and +4 (). In contrast, compounds of Tc and Re in the +2 oxidation state are quite rare. Because the electronegativity of Mn is anomalously low, elemental manganese is unusually reactive. In contrast, the chemistry of Tc is similar to that of Re because of their similar size and electronegativity, again a result of the lanthanide contraction. Due to the stability of the half-filled 3d5 electron configuration, the aqueous Mn3+ ion, with a 3d4 valence electron configuration, is a potent oxidant that is able to oxidize water. It is difficult to generalize about other oxidation states for Tc and Re because their stability depends dramatically on the nature of the compound.
Figure 23.5 Compounds of Manganese in Oxidation States +2 to +7
Like vanadium, compounds of manganese in different oxidation states have different numbers of d electrons, which leads to compounds with different colors: the Mn2+(aq) ion is pale pink; Mn(OH)3, which contains Mn(III), is a dark brown solid; MnO2, which contains Mn(IV), is a black solid; and aqueous solutions of Mn(VI) and Mn(VII) contain the green manganate ion [MnO4]2− and the purple permanganate ion [MnO4]−, respectively.
Consistent with higher oxidation states being more stable for the heavier transition metals, reacting Mn with F2 gives only MnF3, a high-melting, red-purple solid, whereas Re reacts with F2 to give ReF7, a volatile, low-melting, yellow solid. Again, reaction with the less oxidizing, heavier halogens produces halides in lower oxidation states. Thus reaction with Cl2, a weaker oxidant than F2, gives MnCl2 and ReCl6. Reaction of Mn with oxygen forms only Mn3O4, a mixed-valent compound that contains two Mn(II) and one Mn(III) per formula unit and is similar in both stoichiometry and structure to magnetite (Fe3O4). In contrast, Tc and Re form high-valent oxides, the so-called heptoxides (M2O7), consistent with the increased stability of higher oxidation states for the second and third rows of transition metals. Under forced conditions, manganese will form Mn2O7, an unstable, explosive, green liquid. Also consistent with this trend, the permanganate ion [MnO4]2− is a potent oxidant, whereas [TcO4]− and [ReO4]− are much more stable. Both Tc and Re form disulfides and diselenides with layered structures analogous to that of MoS2, as well as more complex heptasulfides (M2S7). As is typical of the transition metals, the group 7 metals form binary nitrides, carbides, and borides that are generally stable at high temperatures and exhibit metallic properties.
The chemistry of the group 7 metals (Mn, Tc, and Re) is dominated by lower oxidation states. Compounds in the maximum possible oxidation state (+7) are readily reduced.
In many older versions of the periodic table, groups 8, 9, and 10 were combined in a single group (group VIII) because the elements of these three groups exhibit many horizontal similarities in their chemistry, in addition to the similarities within each column. In part, these horizontal similarities are due to the fact that the ionization potentials of the elements, which increase slowly but steadily across the d block, have now become so large that the oxidation state corresponding to the formal loss of all valence electrons is encountered only rarely (group 8) or not at all (groups 9 and 10). As a result, the chemistry of all three groups is dominated by intermediate oxidation states, especially +2 and +3 for the first-row metals (Fe, Co, and Ni). The heavier elements of these three groups are called precious metals because they are rather rare in nature and mostly chemically inert.
The chemistry of groups 8, 9, and 10 is dominated by intermediate oxidation states such as +2 and +3.
The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age.
Cobalt is one of the least abundant of the first-row transition metals. Its oxide ores, however, have been used in glass and pottery for thousands of years to produce the brilliant color known as “cobalt blue,” and its compounds are consumed in large quantities in the paint and ceramics industries. The heavier elements of group 9 are also rare, with terrestrial abundances of less than 1 ppb; they are generally found in combination with the heavier elements of groups 8 and 10 in Ni–Cu–S ores.
Nickel silicates are easily processed; consequently, nickel has been known and used since antiquity. In fact, a 75:25 Cu:Ni alloy was used for more than 2000 years to mint “silver” coins, and the modern US nickel uses the same alloy. In contrast to nickel, palladium and platinum are rare (their terrestrial abundance is about 10–15 ppb), but they are at least an order of magnitude more abundant than the heavier elements of groups 8 and 9. Platinum and palladium are used in jewelry, the former as the pure element and the latter as the Pd/Au alloy known as white gold.
Over 2000 years ago, the Bactrian civilization in Western Asia used a 75:25 alloy of copper and nickel for its coins. A modern US nickel has the same composition, but a modern Canadian nickel is nickel-plated steel and contains only 2.5% nickel by mass.
Some properties of the elements in groups 8–10 are summarized in . As in earlier groups, similarities in size and electronegativity between the two heaviest members of each group result in similarities in chemistry. We are now at the point in the d block where there is no longer a clear correlation between the valence electron configuration and the preferred oxidation state. For example, all the elements of group 8 have eight valence electrons, but only Ru and Os have any tendency to form compounds in the +8 oxidation state, and those compounds are powerful oxidants. The predominant oxidation states for all three group 8 metals are +2 and +3. Although the elements of group 9 possess a total of nine valence electrons, the +9 oxidation state is unknown for these elements, and the most common oxidation states in the group are +3 and +1. Finally, the elements of group 10 all have 10 valence electrons, but all three elements are normally found in the +2 oxidation state formed by losing the ns2 valence electrons. In addition, Pd and Pt form numerous compounds and complexes in the +4 oxidation state.
Table 23.6 Some Properties of the Elements of Groups 8, 9, and 10
Group | Element | Z | Valence Electron Configuration | Electronegativity | Metallic Radius (pm) | Melting Point (°C) | Density |
---|---|---|---|---|---|---|---|
8 | Fe | 26 | 4s23d6 | 1.83 | 126 | 1538 | 7.87 |
Ru | 44 | 5s14d7 | 2.20 | 134 | 2334 | 12.10 | |
Os | 76 | 6s25d64f14 | 2.20 | 135 | 3033 | 22.59 | |
9 | Co | 27 | 4s23d7 | 1.88 | 125 | 1495 | 8.86 |
Rh | 45 | 5s14d8 | 2.28 | 134 | 1964 | 12.40 | |
Ir | 77 | 6s25d74f14 | 2.20 | 136 | 2446 | 22.50 | |
10 | Ni | 28 | 4s23d8 | 1.91 | 124 | 1455 | 8.90 |
Pd | 46 | 4d10 | 2.20 | 137 | 1555 | 12.00 | |
Pt | 78 | 6s25d84f14 | 2.20 | 139 | 1768 | 21.50 |
We stated that higher oxidation states become less stable as we go across the d-block elements and more stable as we go down a group. Thus Fe and Co form trifluorides, but Ni forms only the difluoride NiF2. In contrast to Fe, Ru and Os form a series of fluorides up to RuF6 and OsF7. The hexafluorides of Rh and Ir are extraordinarily powerful oxidants, and Pt is the only element in group 10 that forms a hexafluoride. Similar trends are observed among the oxides. For example, Fe forms only FeO, Fe2O3, and the mixed-valent Fe3O4 (magnetite), all of which are nonstoichiometric. In contrast, Ru and Os form the dioxides (MO2) and the highly toxic, volatile, yellow tetroxides, which contain formal M=O bonds. As expected for compounds of metals in such high oxidation states, the latter are potent oxidants. The tendency of the metals to form the higher oxides decreases rapidly as we go farther across the d block.
Higher oxidation states become less stable across the d-block, but more stable down a group.
Reactivity with the heavier chalcogens is rather complex. Thus the oxidation state of Fe, Ru, Os, Co, and Ni in their disulfides is +2 because of the presence of the disulfide ion (S22−), but the disulfides of Rh, Ir, Pd, and Pt contain the metal in the +4 oxidation state together with sulfide ions (S2−). This combination of highly charged cations and easily polarized anions results in substances that are not simple ionic compounds and have significant covalent character.
The groups 8–10 metals form a range of binary nitrides, carbides, and borides. By far the most important of these is cementite (Fe3C), which is used to strengthen steel. At high temperatures, Fe3C is soluble in iron, but slow cooling causes the phases to separate and form particles of cementite, which gives a metal that retains much of its strength but is significantly less brittle than pure iron. Palladium is unusual in that it forms a binary hydride with the approximate composition PdH0.5. Because the H atoms in the metal lattice are highly mobile, thin sheets of Pd are highly permeable to H2 but essentially impermeable to all other gases, including He. Consequently, diffusion of H2 through Pd is an effective method for separating hydrogen from other gases.
The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times.
This 1 kg gold nugget was found in Australia; in 2005, it was for sale in Hong Kong at an asking price of more than US$64,000.
Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. (For more information on button batteries, see , .) Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe2). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer.
The Chuquicamata copper mine in northern Chile, the world’s largest open-pit copper mine, is 4.3 km long, 3 km wide, and 825 m deep. Each gigantic truck in the foreground (and barely visible in the lower right center) can hold 330 metric tn (330,000 kg) of copper ore.
Some properties of the coinage metals are listed in . The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns1(n − 1)d10 valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au− with a 6s25d10 valence electron configuration.
Table 23.7 Some Properties of the Elements of Groups 11 and 12
Group | Element | Z | Valence Electron Configuration | Electronegativity | Metallic Radius (pm) | Melting Point (°C) | Density |
---|---|---|---|---|---|---|---|
11 | Cu | 29 | 4s13d10 | 1.90 | 128 | 1085 | 8.96 |
Ag | 47 | 5s14d10 | 1.93 | 144 | 962 | 10.50 | |
Au | 79 | 6s15d104f14 | 2.40 | 144 | 1064 | 19.30 | |
12 | Zn | 30 | 4s23d10 | 1.65 | 134 | 420 | 7.13 |
Cd | 48 | 5s24d10 | 1.69 | 149 | 321 | 8.69 | |
Hg | 80 | 6s25d104f14 | 1.90 | 151 | −38.8 | 13.53 |
All group 11 elements are relatively unreactive, and their reactivity decreases from Cu to Au. Hence they are noble metals that are particularly well suited for use in coins and jewelry. Copper reacts with O2 at high temperatures to produce Cu2O and with sulfur to form Cu2S. Neither silver nor gold reacts directly with oxygen, although oxides of these elements can be prepared by other routes. Silver reacts with sulfur compounds to form the black Ag2S coating known as tarnish. Gold is the only metal that does not react with sulfur; it also does not react with nitrogen, carbon, or boron. All the coinage metals do, however, react with oxidizing acids. Thus both Cu and Ag dissolve in HNO3 and in hot concentrated H2SO4, while Au dissolves in the 3:1 HCl:HNO3 mixture known as aqua regia. Furthermore, all three metals dissolve in basic cyanide solutions in the presence of oxygen to form very stable [M(CN)2]− ions, a reaction that is used to separate gold from its ores. (For more information about gold processing, see , .)
Although the most important oxidation state for group 11 is +1, the elements are relatively unreactive, with reactivity decreasing from Cu to Au.
All the monohalides except CuF and AuF are known (including AgF). Once again, iodine is unable to stabilize the higher oxidation states (Au3+ and Cu2+). Thus all the copper(II) halides except the iodide are known, but the only dihalide of silver is AgF2. In contrast, all the gold trihalides (AuX3) are known, again except the triiodide. No binary nitrides, borides, or carbides are known for the group 11 elements.
We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. (For more information on corrosion, see , .) Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries. (For more information on the uses of mercury, see , .)
As shown in , the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns2(n − 1)d10 valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg22+.
The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble.
All the possible group 12 dihalides (MX2) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl2). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d10 subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO3 and H2SO4. All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur.
For each reaction, explain why the indicated products form.
Given: balanced chemical equation
Asked for: why the indicated products form
Strategy:
Refer to the periodic trends in this section, , , , , , , , , , and to explain why these products form.
Solution:
Exercise
Predict the products of each reactions and then balance each chemical equation.
Answer:
The group 3 transition metals are highly electropositive metals and powerful reductants. They react with nonmetals to form compounds that are largely ionic and with oxygen to form sesquioxides (M2O3). The group 4 metals also have a high affinity for oxygen. In their reactions with halogens, the covalent character of the halides increases as the oxidation state of the metal increases because the high charge-to-radius ratio causes extensive polarization of the anions. The dichalcogenides have layered structures similar to graphite, and the hydrides, nitrides, carbides, and borides are all hard, high-melting-point solids with metallic conductivity. The group 5 metals also have a high affinity for oxygen. Consistent with periodic trends, only the lightest (vanadium) has any tendency to form compounds in oxidation states lower than +5. The oxides are sufficiently polarized to make them covalent in character. These elements also form layered chalcogenides, as well as nitrides, carbides, borides, and hydrides that are similar to those of the group 4 elements. As the metals become more polarizable across the row, their affinity for oxygen decreases. The group 6 metals are less electropositive and have a maximum oxidation state of +6, making their compounds in high oxidation states largely covalent in character. As the oxidizing strength of the halogen decreases, the maximum oxidation state of the metal also decreases. All three trioxides are acidic, but Cr2O3 is amphoteric. The chalcogenides of the group 6 metals are generally nonstoichiometric and electrically conducting, and these elements also form nitrides, carbides, and borides that are similar to those in the preceding groups. The metals of group 7 have a maximum oxidation state of +7, but the lightest element, manganese, exhibits an extensive chemistry in lower oxidation states. As with the group 6 metals, reaction with less oxidizing halogens produces metals in lower oxidation states, and disulfides and diselenides of Tc and Re have layered structures. The group 7 metals also form nitrides, carbides, and borides that are stable at high temperatures and have metallic properties. In groups 8, 9, and 10, the ionization potentials of the elements are so high that the oxidation state corresponding to the formal loss of all valence electrons is encountered rarely (group 8) or not at all (groups 9 and 10). Compounds of group 8 metals in their highest oxidation state are powerful oxidants. The reaction of metals in groups 8, 9, and 10 with the chalcogens is complex, and these elements form a range of binary nitrides, carbides, and borides. The coinage metals (group 11) have the highest electrical and thermal conductivities and are the most ductile and malleable of the metals. Although they are relatively unreactive, they form halides but not nitrides, borides, or carbides. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands.
The valence electron configuration of Sc is 4s23d1, yet it does not lose the d1 electron to form 1+ ion. Why?
Give the ground-state electron configuration for Mn, Mn2+, Au, Au3+, Mo, and Mo5+.
A great deal of research is being conducted on the use of titanium alloys as materials for transportation applications (airplanes, ships, automobiles, etc.). Why is Ti particularly suited to this purpose? What is the primary disadvantage that needs to be overcome?
Both Ti and Ta are used for bioimplants because they are highly resistant to corrosion. Their uses also extend to other applications where corrosion must be avoided. Why are these metals so corrosion resistant?
Give two reasons why Zr is used to make the casing for UO2 fuel in water-cooled nuclear reactors.
Why is chromium added to steel to form stainless steel? What other elements might also be effective additives for this purpose? Why did you select these elements?
Tungsten is commonly used as the filament in electric light bulbs. Why is tungsten particularly suited to this purpose?
Palladium metal is used to purify H2 by removing other gases. Why is Pd so permeable to H2?
Give the valence electron configuration for Sc, Fe, Re, Ag, Zr, Co, V, Pr, Hg, Cr, Ni, Ce, Cu, and Tb.
The Hg–Hg bond is much stronger than the Cd–Cd bond, reversing the trend found among the other transition-metal groups. Explain this anomaly.
Which of the transition metals are most likely to form compounds in the +6 oxidation state? Why?
Do you expect TiCl4, TiCl3, TiCl2, and Ti to be oxidized, reduced, or hydrolyzed by water? Explain your reasoning.
The atomic radii of vanadium, niobium, and tantalum are 134 pm, 146 pm, and 146 pm, respectively. Why does the radius increase from vanadium to niobium but not from niobium to tantalum?
The most stable oxidation state for the metals of groups 3, 4, and 5 is the highest oxidation state possible. In contrast, for nearly all the metals of groups 8, 9, and 10, intermediate oxidation states are most stable. Why?
Most of the transition metals can form compounds in multiple oxidation states. Ru, for example, can form compounds in the +8, +6, +4, +3, +2, and −2 oxidation states. Give the valence electron configuration of Ru in each oxidation state. Why does Ru exhibit so many oxidation states? Which ones are the most stable? Why?
Predict the maximum oxidation states of Cu, Cr, Mo, Rh, Zr, Y, Ir, Hg, and Fe.
In the +4 oxidation state all three group 7 metals form the dioxides (MO2). Which of the three metals do you expect to form the most stable dioxide? Why?
Of [Fe(H2O)6]+, OsBr7, CoF4, PtF6, FeI3, [Ni(H2O)6]2+, OsO4, IrO4, NiO, RhS2, and PtH, which do not exist? Why?
The chemistry of gold is somewhat anomalous for a metal. With which elements does it form the Au− ion? Does it form a stable sulfide?
Of Os4+, Pt10+, Cr6+, Ir9+, Ru8+, Re7+, and Ni10+, which are not likely to exist? Why?
Of Ag2S, Cu2S, AuI3, CuF, AuF, AgN, and AuO, which are not likely to exist?
There is evidence that the Au− ion exists. What would be its electron configuration? The compound CsAu has been isolated; it does not exhibit a metallic luster and does not conduct electricity. Is this compound an alloy? What type of bonding is involved? Explain your answers.
Of Hg2Cl2, ZnO, HgF2, Cs2[ZnCl5], and HgNa, which are not likely to exist?
Mercurous oxide (Hg2O) and mercurous hydroxide [Hg2(OH)2] have never been prepared. Why not? What products are formed if a solution of aqueous sodium hydroxide is added to an aqueous solution of mercurous nitrate [Hg2(NO3)2]?
Arrange Fe2O3, TiO2, V2O5, MoO3, Mn2O7, and OsO4 in order of increasing basicity.
Mercurous sulfide has never been prepared. What products are formed when H2S gas is bubbled through an aqueous solution of mercurous nitrate?
Arrange Sc2O3, VO, V2O5, Cr2O3, Fe2O3, Fe3O4, and ZnO in order of increasing acidity.
Arrange Sc2O3, V2O5, CrO3, Mn2O7, MnO2, and VO2 in order of increasing basicity.
Predict the products of each reaction and then balance each chemical equation.
Predict the products of each reaction and then balance each chemical equation.
What do you predict to be the coordination number of Pt2+, Au+, Fe3+, and Os2+?
Of La, Sc, Cr, and Hf, which is most likely to form stable compounds in the +4 oxidation state? Why?
Give the most common oxidation state for Y, W, Ru, Ag, Hg, Zn, Cr, Nb, and Ti.
Give the most common oxidation state for Os, Cd, Hf, V, Ac, Ni, Mn, Pt, and Fe.
Give the highest oxidation state observed for Zr, Fe, Re, Hg, Ni, La, and Mo.
Give the highest oxidation state observed for Ag, Co, Os, Au, W, and Mn.
Arrange La, Cs, Y, Pt, Cd, Mo, Fe, Co, and Ir in order of increasing first ionization energy.
Briefly explain the following trends within the transition metals.
Propose a method to prepare each of the following compounds: TiCl4[(CH3)2O]2, Na2TiO3, V2O5, and Na2Cr2O7.
Of the group 5 elements, which
Pt10+, Ir9+, and Ni10+. Because ionization energies increase from left to right across the d block, by the time you reach group 9, it is impossible to form compounds in the oxidation state that corresponds to loss of all the valence electrons.
Hg22+(aq) + H2S(g) → Hg(l) + HgS(s) + 2H+(aq)
Mn2O7 < CrO3 < V2O5 < MnO2 ≈ VO2 < Sc2O3
Os, +4; Cd, +2; Hf, +4; V, +5; Ac, +3; Ni, +2; Mn, +2; Pt, +2 & +4; Fe, +2 & +3
Ag, +3; Co, +4; Os, +8; Au, +5; W, +6; Mn, +7
Very few of the transition metals are found in nature as free metals. Consequently, almost all metallic elements must be isolated from metal oxide or metal sulfide ores. MetallurgyA set of processes by which metals are extracted from their ores and converted to more useful forms. is the set of processes by which metals are extracted from their ores and converted to more useful forms.
Metallurgy consists of three general steps: (1) mining the ore, (2) separating and concentrating the metal or the metal-containing compound, and (3) reducing the ore to the metal. Additional processes are sometimes required to improve the mechanical properties of the metal or increase its purity. Many ores contain relatively low concentrations of the desired metal; for example, copper ores that contain even 1% Cu by mass are considered commercially useful.
After an ore has been mined, the first step in processing is usually to crush it because the rate of chemical reactions increases dramatically with increased surface area. Next, one of three general strategies is used to separate and concentrate the compound(s) of interest: settling and flotation, which are based on differences in density between the desired compound and impurities; pyrometallurgy, which uses chemical reduction at high temperatures; and hydrometallurgy, which employs chemical or electrochemical reduction of an aqueous solution of the metal. Other methods that take advantage of unusual physical or chemical properties of a particular compound may also be used. For example, crystals of magnetite (Fe3O4) are tiny but rather powerful magnets; in fact, magnetite (also known as lodestone) was used to make the first compasses in China during the first century BC. If a crushed ore that contains magnetite is passed through a powerful magnet, the Fe3O4 particles are attracted to the poles of the magnet, allowing them to be easily separated from other minerals.
Metallurgy depends on the separation of a metal compound from its ore and reduction to the metal at high temperature (pyrometallurgy) or in aqueous solution (hydrometallurgy).
Settling and flotation have been used for thousands of years to separate particles of dense metals such as gold, using the technique known as panning, in which a sample of gravel or sand is swirled in water in a shallow metal pan. Because the density of gold (19.3 g/cm3) is so much greater than that of most silicate minerals (about 2.5 g/cm3), silicate particles settle more slowly and can be poured off with the water, leaving dense gold particles on the bottom of the pan. Conversely, in flotation, the compound of interest is made to float on top of a solution. Blowing air through a suspension of the crude ore in a mixture of water and an organic liquid, such as pine tar, produces a “froth” that contains tiny particles of hydrophobic solids, such as metal sulfides, while more hydrophilic oxide minerals remain suspended in the aqueous phase (Figure 23.6 "Froth Flotation"). To make the separation more efficient, small amounts of an anionic sulfur-containing compound, such as Na+C2H5OCS2−, are added; the additive binds to the sulfur-rich surface of the metal sulfide particles and makes the metal sulfide particles even more hydrophobic. The resulting froth is highly enriched in the desired metal sulfide(s), which can be removed simply by skimming. This method works even for compounds as dense as PbS (7.5 g/cm3).
Figure 23.6 Froth Flotation
(a) When air is blown through a mixture of a finely ground metal sulfide ore and water, the more hydrophobic metal sulfides form a froth that can be easily removed, allowing them to be separated from more hydrophilic metal oxides and silicates. (b) A froth containing precious metal sulfides is formed as a by-product during the production of metallic nickel. (c) An anionic sulfur additive with hydrophobic “tails” can be used to enhance the hydrophobic character of metal sulfide particles, which causes them to be attracted to the air/water interface in the foam.
In pyrometallurgy, an ore is heated with a reductant to obtain the metal. Theoretically, it should be possible to obtain virtually any metal from its ore by using coke, an inexpensive form of crude carbon, as the reductant. An example of such a reaction is as follows:
Equation 23.4
Unfortunately, many of the early transition metals, such as Ti, react with carbon to form stable binary carbides. Consequently, more expensive reductants, such as hydrogen, aluminum, magnesium, or calcium, must be used to obtain these metals. Many metals that occur naturally as sulfides can be obtained by heating the sulfide in air, as shown for lead in the following equation:
Equation 23.5
The reaction is driven to completion by the formation of SO2, a stable gas.
Pyrometallurgy is also used in the iron and steel industries. The overall reaction for the production of iron in a blast furnace is as follows:
Equation 23.6
The actual reductant is CO, which reduces Fe2O3 to give Fe(l) and CO2(g); the CO2 is then reduced back to CO by reaction with excess carbon. As the ore, lime, and coke drop into the furnace (Figure 23.7 "A Blast Furnace for Converting Iron Oxides to Iron Metal"), any silicate minerals in the ore react with the lime to produce a low-melting mixture of calcium silicates called slag, which floats on top of the molten iron. Molten iron is then allowed to run out the bottom of the furnace, leaving the slag behind. Originally, the iron was collected in pools called pigs, which is the origin of the name pig iron.
Figure 23.7 A Blast Furnace for Converting Iron Oxides to Iron Metal
(a) The furnace is charged with alternating layers of iron ore (largely Fe2O3) and a mixture of coke (C) and limestone (CaCO3). Blasting hot air into the mixture from the bottom causes it to ignite, producing CO and raising the temperature of the lower part of the blast furnace to about 2000°C. As the CO that is formed initially rises, it reduces Fe2O3 to form CO2 and elemental iron, which absorbs heat and melts as it falls into the hottest part of the furnace. Decomposition of CaCO3 at high temperatures produces CaO (lime) and additional CO2, which reacts with excess coke to form more CO. (b) This blast furnace in Magnitogorsk, Russia, was the largest in the world when it was built in 1931.
Iron that is obtained directly from a blast furnace has an undesirably low melting point (about 1100°C instead of 1539°C) because it contains a large amount of dissolved carbon. It contains other impurities (such as Si, S, P, and Mn from contaminants in the iron ore that were also reduced during processing) that must be removed because they make iron brittle and unsuitable for most structural applications. In the Bessemer process, oxygen is blown through the molten pig iron to remove the impurities by selective oxidation because these impurities are more readily oxidized than iron (Figure 23.8 "A Basic Oxygen Furnace for Converting Crude Iron to Steel"). In the final stage of this process, small amounts of other metals are added at specific temperatures to produce steel with the desired combination of properties.
Figure 23.8 A Basic Oxygen Furnace for Converting Crude Iron to Steel
(a) A blast of oxygen is used to agitate the molten iron and oxidize impurities to products that dissolve in the less dense slag layer. The slag and the molten steel are removed by tilting the entire furnace and pouring the liquids out through the taphole. (b) A basic oxygen furnace is being filled with molten iron from a blast furnace.
The most selective methods for separating metals from their ores are based on the formation of metal complexes. For example, gold is often found as tiny flakes of the metal, usually in association with quartz or pyrite deposits. In those circumstances, gold is typically extracted by using cyanide leaching, which forms a stable gold–cyanide complex—[Au(CN)2]−:
Equation 23.7
4Au(s) + 8NaCN(aq) + O2(g) + 2H2O(l) → 4Na[Au(CN)2](aq) + 4NaOH(aq)Virtually pure gold can be obtained by adding powdered zinc to the solution:
Equation 23.8
Zn(s) + 2[Au(CN)2]−(aq) → [Zn(CN)4]2−(aq) + 2Au(s)A related method, which is used to separate Co3+, Ni2+, and Cu+ from Fe, Mn, and Ti, is based on the formation of stable, soluble ammonia complexes of ions of the late transition metals.
Suppose you are working in the chemistry laboratory of a mining company that has discovered a new source of tungsten ore containing about 5% WS2 in a granite matrix (granite is a complex aluminosilicate mineral). You have been asked to outline an economical procedure for isolating WS2 from the ore and then converting it to elemental tungsten in as few steps as possible. What would you recommend?
Given: composition of ore
Asked for: procedure to isolate metal sulfide
Strategy:
Determine which method would be most effective for separating the metal sulfide from the ore. Then determine the best method for reducing the metal to the pure element.
Solution:
You need to separate and concentrate the WS2, convert it to a suitable form so it can be reduced to the metal (if necessary), and then carry out the reduction. Because the new ore is a binary metal sulfide, you could take advantage of the hydrophilic nature of most metal sulfides to separate WS2 by froth flotation. Then, because most metal sulfides cannot be reduced directly to the metal using carbon, you will probably need to convert WS2 to an oxide for subsequent reduction. One point to consider is whether the oxide can be reduced using carbon because many transition metals react with carbon to form stable carbides. Here is one possible procedure for producing tungsten from this new ore:
Reduce the oxide with hydrogen gas at high temperature to avoid carbide formation:
Exercise
Propose an economical procedure for converting a silicate mineral deposit containing BaCO3 to the pure Ba metal.
Answer:
The conversion of metals from their ores to more useful forms is called metallurgy, which consists of three general steps: mining, separation and concentration, and reduction. Settling and flotation are separation methods based on differences in density, whereas pyrometallurgy is based on a chemical reduction at elevated temperatures, and hydrometallurgy uses chemical or electrochemical reduction of an aqueous solution. In pyrometallurgy, a reductant must be used that does not form stable compounds with the metal of interest. In hydrometallurgy, metals are separated via the formation of metal complexes.
Coke is a plentiful and inexpensive reductant that is used to isolate metals from their ores. Of Cr, Co, W, Cu, Ni, Os, Fe, Mn, La, and Hf, which cannot be isolated using this reductant? Why?
Hydrometallurgy is the preferred method for separating late transition metals from their ores. What types of ligands are most effective in this process?
Coke cannot be used as a reductant for metals that form stable carbides, such as the early transition metals (La, Hf, and W).
Tantalum and niobium are frequently found together in ores. These elements can be separated from other metals present by treatment with a solution of HF. Explain why this is an effective separation technique.
A commercially important ore of chromium is chromite (FeCr2O4), which is an analogue of magnetite (Fe3O4). Based on what you know about the oxidation states of iron in magnetite, predict the oxidation states of the metal ions in chromite.
Pure vanadium is obtained by reducing VCl4 with H2 or Mg or by reducing V2O5 with Ca. Write a balanced chemical equation for each reaction. Why is carbon not used for the reduction?
Manganese is an important additive in steel because of its reactivity with oxygen and sulfur, both of which contribute to brittleness. Predict the products of reacting Mn with these species.
The diagram of a blast furnace in Figure 23.7 "A Blast Furnace for Converting Iron Oxides to Iron Metal" illustrates several important features of the reduction of Fe2O3 to iron. Write a balanced chemical equation for each step of the process described in the figure and give the overall equation for the conversion. Oxygen is blown through the final product to remove impurities. Why does this step not simply reverse the process and produce iron oxides?
Metallic Zr is produced by the Kroll method, which uses Na as the reductant. Write a balanced chemical equation for each reaction involved in this process. The product is frequently contaminated with Hf. Propose a feasible method for separating the two elements.
The compound Cr2O3 is important commercially; among other things, it is used as a pigment in paint and as a catalyst for the manufacture of butadiene. Write a balanced chemical equation to show how you would produce this compound from
Carbon cannot be used as a reductant because vanadium forms stable carbides, such as VC and VC2.
One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complexA chemical compound composed of a central metal atom or ion bonded to one or more ligands. consists of a central metal atom or ion that is bonded to one or more ligandsAn ion or a molecule that contains one or more pairs of electrons that can be shared with the central metal in a metal complex. (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH3)3Cl3; positively charged, such as [Nd(H2O)9]3+; or negatively charged, such as [UF8]4−. Electrically charged metal complexes are sometimes called complex ionsAn ionic species formed between a central metal ion and one or more surrounding ligands because of a Lewis acid–base interaction.. A coordination compoundA chemical compound with one or more metal complexes. contains one or more metal complexes.
Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions. Metal complexes are so important in biology that we consider the topic separately in .
Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: KFe2(CN)6. The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as AlF3·3KF, Fe(CN)2·4KCN, and ZnCl2·2CsCl, which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should AlF3·3KF exist but not AlF3·4KF or AlF3·2KF? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, CoCl3·6NH3, which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, CoCl3·6NH3, CoCl3·5NH3, CoCl3·4NH3, and CoCl3·3NH3 were all known and had very different properties, but despite all attempts, chemists could not prepare CoCl3·2NH3 or CoCl3·NH3.
The Great Wave Off Kanagawa. The Japanese artist Katsushika Hokusai used Prussian blue to create this famous woodcut.
Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl3·4NH3 shown here.
The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl4 were known at the time: PtCl4·nNH3 (n = 2–6). Some of Werner’s original data on these compounds are shown in . The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag+(aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl4·6NH3 in showed that all the chloride ions were present as free chloride. In contrast, PtCl4·2NH3 was a neutral molecule that contained no free chloride ions.
Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry.
Table 23.8 Werner’s Data on Complexes of Ammonia with PtCl4
Complex | Conductivity (ohm−1) | Number of Ions per Formula Unit | Number of Cl− Ions Precipitated by Ag+ |
---|---|---|---|
PtCl4·6NH3 | 523 | 5 | 4 |
PtCl4·5NH3 | 404 | 4 | 3 |
PtCl4·4NH3 | 299 | 3 | 2 |
PtCl4·3NH3 | 97 | 2 | 1 |
PtCl4·2NH3 | 0 | 0 | 0 |
These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence (oxidation state) that corresponds to the positive charge on the metal ion and (2) a secondary valence (coordination number) that is the total number of ligands bound to the metal ion. If Pt had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the PtCl4·NH3 adducts by the following reactions, where the metal complex is enclosed in square brackets:
Equation 23.9
Further work showed that the two missing members of the series—[Pt(NH3)Cl5]− and [PtCl6]2−—could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co3+ and Cr3+ and 4 for Pt2+ and Pd2+.
Werner’s studies on the analogous Co3+ complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH3)6]Cl3 (yellow) and [Co(NH3)5Cl]Cl2 (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH3)4Cl2]Cl compounds were known: one was red, and the other was green (part (a) in ). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH3)4Cl2]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (part (b) in ). His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH3)3Cl3.
Figure 23.9 Complexes with Different Arrangements of the Same Ligands Have Different Properties
The [Co(NH3)4Cl2]+ ion can have two different arrangements of the ligands, which results in different colors: if the two Cl− ligands are next to each other, the complex is red (a), but if they are opposite each other, the complex is green (b).
In Werner’s time, many complexes of the general formula MA4B2 were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA4B2 complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA4B2 complex were known tell you about the three-dimensional structures of these complexes?
Given: three possible structures and the number of different forms known for MA4B2 complexes
Asked for: number of different arrangements of ligands for MA4B2 complex for each structure
Strategy:
Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand.
Solution:
The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA6 complex and simply replace two of the A ligands in each structure to give an MA4B2 complex:
For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand.
Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements.
In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other.
The number of possible MA4B2 arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA4B2 complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA3B3), however, the data were best explained by an octahedral structure for six-coordinate metal complexes.
Exercise
Determine the maximum number of structures that are possible for a four-coordinate MA2B2 complex with either a square planar or a tetrahedral symmetrical structure.
Answer: square planar, 2; tetrahedral, 1
The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion (VSEPR) model in because they correspond to the lowest-energy arrangements of n electron pairs around a central atom.
Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands.
Although it is rare for most metals, this coordination number is surprisingly common for d10 metal ions, especially Cu+, Ag+, Au+, and Hg2+. An example is the [Au(CN)2]− ion, which is used to extract gold from its ores, as described in . As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here.
Although it is also rare, this coordination number is encountered with d10 metal ions such as Cu+ and Hg2+. Among the few known examples is the HgI3− ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model.
Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF4]2−, and d10 ions, such as [ZnCl4]2−. It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl4]− and [FeCl4]2−). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d8 electron configurations, such as Rh+ and Pd2+, and they are also encountered in some complexes of Ni2+ and Cu2+.
This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes.
This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes.
This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid.
This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube.
This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H2O)9]3+.
The thermodynamic stability of a metal complex depends greatly on the properties of the ligand and the metal ion and on the type of bonding. Recall that the metal–ligand interaction is an example of a Lewis acid–base interaction. Lewis bases can be divided into two categories: hard basesA type of Lewis base with small, relatively nonpolarizable donor atoms., which contain small, relatively nonpolarizable donor atoms (such as N, O, and F), and soft basesA type of Lewis base with large, relatively polarizable donor atoms., which contain larger, relatively polarizable donor atoms (such as P, S, and Cl). Metal ions with the highest affinities for hard bases are hard acidsAn acid with the highest affinity for hard bases. It is relatively nonpolarizable and has a relatively high charge-to-radius ratio., whereas metal ions with the highest affinity for soft bases are soft acidsAn acid with the highest affinity for soft bases. It tends to be a cation of a less electropositive metal.. Some examples of hard and soft acids and bases are given in . Notice that hard acids are usually cations of electropositive metals; consequently, they are relatively nonpolarizable and have higher charge-to-radius ratios. Conversely, soft acids tend to be cations of less electropositive metals; consequently, they have lower charge-to-radius ratios and are more polarizable. Chemists can predict the relative stabilities of complexes formed by the d-block metals with a remarkable degree of accuracy by using a simple rule: hard acids prefer to bind to hard bases, and soft acids prefer to bind to soft bases.
Table 23.9 Examples of Hard and Soft Acids and Bases
Acids | Bases | |
---|---|---|
hard | H+ | NH3, RNH2, N2H4 |
Li+, Na+, K+ | H2O, ROH, R2O | |
Be2+, Mg2+, Ca2+, VO2+ | OH−, F−, Cl−, CH3CO2− | |
Al3+, Sc3+, Cr3+ | CO32− | |
Ti4+ | PO43− | |
soft | BF3, Al2Cl6, CO2, SO3 | |
Cu+, Ag+, Au+, Tl+, Hg22+ | H− | |
Pd2+, Pt2+, Hg2+ | CN−, SCN−, I−, RS− | |
GaCl3, GaBr3, GaI3 | CO, R2S |
Because the interaction between hard acids and hard bases is primarily electrostatic in nature, the stability of complexes involving hard acids and hard bases increases as the positive charge on the metal ion increases and as its radius decreases. For example, the complex of Al3+ (r = 53.5 pm) with four fluoride ligands (AlF4−) is about 108 times more stable than InF4−, the corresponding fluoride complex of In3+ (r = 80 pm). In general, the stability of complexes of divalent first-row transition metals with a given ligand varies inversely with the radius of the metal ion, as shown in the following series:The inversion in the order at copper is due to the anomalous structure of copper(II) complexes, which will be discussed shortly.
Because a hard metal interacts with a base in much the same way as a proton, by binding to a lone pair of electrons on the base, the stability of complexes of hard acids with hard bases increases as the ligand becomes more basic. For example, because ammonia is a stronger base than water, metal ions bind preferentially to ammonia. Consequently, adding ammonia to aqueous solutions of many of the first-row transition-metal cations results in the formation of the corresponding ammonia complexes.
In contrast, the interaction between soft metals (such as the second- and third-row transition metals and Cu+) and soft bases is largely covalent in nature. Most soft-metal ions have a filled or nearly filled d subshell, which suggests that metal-to-ligand π bonding is important. Complexes of soft metals with soft bases are therefore much more stable than would be predicted based on electrostatic arguments.
Hard acids prefer to bind to hard bases, and soft acids prefer to bind to soft bases.
The hard acid–hard base/soft acid–soft base concept also allows us to understand why metals are found in nature in different kinds of ores. Recall from that most of the first-row transition metals are isolated from oxide ores but that copper and zinc tend to occur naturally in sulfide ores. This is consistent with the increase in the soft character of the metals across the first row of the transition metals from left to right. Recall also that most of the second- and third-row transition metals occur in nature as sulfide ores, consistent with their greater soft character.
Ligands like chloride, water, and ammonia are said to be monodentate (one-toothed, from the Greek mono, meaning “one,” and the Latin dent-, meaning “tooth”): they are attached to the metal via only a single atom. Ligands can, however, be bidentate (two-toothed, from the Greek di, meaning “two”), tridentate (three-toothed, from the Greek tri, meaning “three”), or, in general, polydentate (many-toothed, from the Greek poly, meaning “many”), indicating that they are attached to the metal at two, three, or several sites, respectively. Ethylenediamine (H2NCH2CH2NH2, often abbreviated as en) and diethylenetriamine (H2NCH2CH2NHCH2CH2NH2, often abbreviated as dien) are examples of a bidentate and a tridentate ligand, respectively, because each nitrogen atom has a lone pair that can be shared with a metal ion. When a bidentate ligand such as ethylenediamine binds to a metal such as Ni2+, a five-membered ring is formed. A metal-containing ring like that shown is called a chelate ring (from the Greek chele, meaning “claw”). Correspondingly, a polydentate ligand is a chelating agent, and complexes that contain polydentate ligands are called chelate complexes.
Experimentally, it is observed that metal complexes of polydentate ligands are significantly more stable than the corresponding complexes of chemically similar monodentate ligands; this increase in stability is called the chelate effect. For example, the complex of Ni2+ with three ethylenediamine ligands, [Ni(en)3]2+, should be chemically similar to the Ni2+ complex with six ammonia ligands, [Ni(NH3)6]2+. In fact, the equilibrium constant for the formation of [Ni(en)3]2+ is almost 10 orders of magnitude larger than the equilibrium constant for the formation of [Ni(NH3)6]2+:
Equation 23.10
Chelate complexes are more stable than the analogous complexes with monodentate ligands.
The stability of a chelate complex depends on the size of the chelate rings. For ligands with a flexible organic backbone like ethylenediamine, complexes that contain five-membered chelate rings, which have almost no strain, are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings. For example, the complex of copper(II) with two ethylenediamine ligands is about 1000 times more stable than the corresponding complex with triethylenediamine (H2NCH2CH2CH2NH2, abbreviated as trien):
Equation 23.11
Arrange [Cr(en)3]3+, [CrCl6]3−, [CrF6]3−, and [Cr(NH3)6]3+ in order of increasing stability.
Given: four Cr(III) complexes
Asked for: relative stabilities
Strategy:
A Determine the relative basicity of the ligands to identify the most stable complexes.
B Decide whether any complexes are further stabilized by a chelate effect and arrange the complexes in order of increasing stability.
Solution:
A The metal ion is the same in each case: Cr3+. Consequently, we must focus on the properties of the ligands to determine the stabilities of the complexes. Because the stability of a metal complex increases as the basicity of the ligands increases, we need to determine the relative basicity of the four ligands. Our earlier discussion of acid–base properties suggests that ammonia and ethylenediamine, with nitrogen donor atoms, are the most basic ligands. The fluoride ion is a stronger base (it has a higher charge-to-radius ratio) than chloride, so the order of stability expected due to ligand basicity is [CrCl6]3− < [CrF6]3− < [Cr(NH3)6]3+ ≈ [Cr(en)3]3+.
B Because of the chelate effect, we expect ethylenediamine to form a stronger complex with Cr3+ than ammonia. Consequently, the likely order of increasing stability is [CrCl6]3− < [CrF6]3− < [Cr(NH3)6]3+ < [Cr(en)3]3+.
Exercise
Arrange [Co(NH3)6]3+, [CoF6]3−, and [Co(en)3]3+ in order of decreasing stability.
Answer: [Co(en)3]3+ > [Co(NH3)6]3+ > [CoF6]3−
As we discussed earlier in this section, the existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomersTwo or more compounds with the same molecular formula but different arrangements of their atoms.. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall from that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique. (For more information on isomers in organic compounds, see , .)
Isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another are called structural isomersTwo or more compounds that have the same molecular formula but differ in which atoms are bonded to one another.. Isobutane and n-butane are examples of structural isomers. One kind of isomerism consists of two compounds that have the same empirical formula but differ in the number of formula units present in the molecular formula. An example in coordination compounds is two compounds with the empirical formula Pt(NH3)2Cl2. One is a simple square planar platinum(II) complex, Pt(NH3)2Cl2, and the other is an ionic compound that contains the [Pt(NH3)4]2+ cation and the [PtCl4]2− anion, [Pt(NH3)4][PtCl4]. As you might expect, these compounds have very different physical and chemical properties. One arrangement of the Cl− and NH3 ligands around the platinum ion in the former gives the anticancer drug cisplatin, whereas the other arrangement has no known biomedical applications.
Metal complexes that differ only in which ligands are adjacent to one another (cisA type of geometrical isomer in which the ligands or the substituents are adjacent to one another in a rigid molecule or a metal complex.) or directly across from one another (transA type of geometrical isomer in which the ligands or the substituents are directly across from each other in a rigid molecule or a metal complex.) in the coordination sphere of the metal are called geometrical isomersComplexes that differ only in which ligands are adjacent to one another or directly across from one another in the coordination sphere of a metal.. They are most important for square planar and octahedral complexes.
Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA3B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB3 case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space:
For an MA2B2 complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent:
Because there is no way to convert the cis structure to the trans by rotating or flipping the molecule in space, they are fundamentally different arrangements of atoms in space. Probably the best-known examples of cis and trans isomers of an MA2B2 square planar complex are cis-Pt(NH3)2Cl2, also known as cisplatin, and trans-Pt(NH3)2Cl2, which is actually toxic rather than therapeutic.
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs (the testes in males and the ovaries in females), which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin and went on to win an unprecedented seven Tour de France bicycle races.
Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en)2]2+, have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands:
Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA5B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA5B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA5B structure are as follows:
If two ligands in an octahedral complex are different from the other four, giving an MA4B2 complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH3)4Cl2]Cl are examples of this type of system:
Replacing another A ligand by B gives an MA3B3 complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the facAn isomer in which three ligands occupy opposite triangular faces of an octahedron. isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the merAn isomer in which three ligands lie on a spherical meridian. isomer (for meridional):
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl]−, where ox is −O2CCO2−, which stands for oxalate.
Given: formula of complex
Asked for: structures of geometrical isomers
Solution:
This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H2O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here:
In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens:
This complex can therefore exist as four different geometrical isomers.
Exercise
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer:
Two geometrical isomers are possible: trans and cis.
Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d10 metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. The stability of metal complexes with first-row transition metals in a +2 oxidation state varies inversely with their ionic radius. Lewis bases can be hard bases, which have small, relatively nonpolarizable donor atoms, or soft bases, with larger, relatively polarizable donor atoms. Hard acids have the highest affinity for hard bases, and soft acids have the highest affinity for soft bases. Soft metals and soft bases form complexes that are more stable than would be predicted based on electrostatic arguments, which suggests that metal-to-ligand π bonding is important. Ligands that are strong bases form the most stable complexes with metal ions that are hard acids. Exceptionally stable complexes are formed by chelates, which are polyatomic ligands with two or more donor atoms; this enhanced stability is known as the chelate effect. Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans.
Give two reasons a metal can bind to only a finite number of ligands. Based on this reasoning, what do you predict is the maximum coordination number of Ti? of Ac?
Can a tetrahedral MA2B2 complex form cis and trans isomers? Explain your answer.
The group 12 elements are never found in their native (free) form but always in combination with one other element. What element is this? Why? Which of the group 12 elements has the highest affinity for the element you selected?
The group 12 metals are rather soft and prefer to bind to a soft anion such as sulfide rather than to a hard anion like oxide; hence they are usually found in nature as sulfide ores. Because it is the softest of these metals, mercury has the highest affinity for sulfide.
Complexes of metals in the +6 oxidation state usually contain bonds to which two Lewis bases? Why are these bonds best described as covalent rather than ionic? Do Ca, Sr, and Ba also form covalent bonds with these two Lewis bases, or is their bonding best described as ionic?
Cr, Mn, Fe, Co, and Ni form stable CO complexes. In contrast, the earlier transition metals do not form similar stable complexes. Why?
The transition metals Cr through Ni form very stable cyanide complexes. Why are these complexes so much more stable than similar compounds formed from the early transition metals?
Of Co(en)33+, CoF63−, Co(NH3)63+, and Co(dien)23+, which species do you expect to be the most stable? Why?
Of Ca2+, Ti2+, V2+, Mn2+, Fe2+, Co2+, Ni2+, and Zn2+, which divalent metal ions forms the most stable complexes with ligands such as NH3? Why?
Match each Lewis base with the metal ions with which it is most likely to form a stable complex:
Lewis bases: NH3, F−, RS−, OH−, and Cl−
Metals: Sc3+, Cu+, W6+, Mg2+, V3+, Fe3+, Zr4+, Co2+, Ti4+, Au+, Al3+, and Mn7+
Of ReF2, ReCl5, MnF6, Mn2O7, and ReO, which are not likely to exist?
Of WF2, CrF6, MoBr6, WI6, CrO3, MoS2, W2S3, and MoH, which are not likely to exist?
Metals in the +6 oxidation state are stabilized by oxide (O2−) and fluoride (F−). The M−F and M−O bonds are polar covalent due to extreme polarization of the anions by the highly charged metal. Ca, Sr, and Ba can be oxidized only to the dications (M2+), which form ionic oxides and fluorides.
Cyanide is a relatively soft base, and the early transition-metal cations are harder acids than the later transition metals.
The formation of complexes between NH3 and a divalent cation is largely due to electrostatic interactions between the negative end of the ammonia dipole moment and the positively charged cation. Thus the smallest divalent cations (Ni2+, Zn2+, and Cu2+) will form the most stable complexes with ammonia.
Re2+ is a very soft cation, and F− and O2− are very hard bases, so ReO and ReF2 are unlikely to exist. MnF6 is also unlikely to exist: although fluoride should stabilize high oxidation states, in this case Mn6+ is probably too small to accommodate six F− ions.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit ( and ). In this section, we describe crystal field theory (CFT)A bonding model based on the assumption that metal–ligand interactions are purely electrostatic in nature, which explains many important properties of transition-metal complexes., a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy.
CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model in , the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands.
We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall from that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (part (a) in ). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in part (b) in , the and orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
Figure 23.10 An Octahedral Arrangement of Six Negative Charges around a Metal Ion Causes the Five d Orbitals to Split into Two Sets with Different Energies
(a) Distributing a charge of −6 uniformly over a spherical surface surrounding a metal ion causes the energy of all five d orbitals to increase due to electrostatic repulsions, but the five d orbitals remain degenerate. Placing a charge of −1 at each vertex of an octahedron causes the d orbitals to split into two groups with different energies: the and orbitals increase in energy, while the, dxy, dxz, and dyz orbitals decrease in energy. The average energy of the five d orbitals is the same as for a spherical distribution of a −6 charge, however. Attractive electrostatic interactions between the negatively charged ligands and the positively charged metal ion (far right) cause all five d orbitals to decrease in energy but does not affect the splittings of the orbitals. (b) The two eg orbitals (left) point directly at the six negatively charged ligands, which increases their energy compared with a spherical distribution of negative charge. In contrast, the three t2g orbitals (right) point between the negatively charged ligands, which decreases their energy compared with a spherical distribution of charge.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energyThe difference in energy between the set of orbitals and and the set of orbitals , , that results when the five orbitals are placed in an octahedral crystal field. (Δo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is 2(0.6Δo) + 3(−0.4Δo) = 0.
Crystal field splitting does not change the total energy of the d orbitals.
Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in part (a) in ).
We can use the d-orbital energy-level diagram in to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in , for d1–d3 systems—such as [Ti(H2O)6]3+, [V(H2O)6]3+, and [Cr(H2O)6]3+, respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H2O)6]3+, for example, by saying that the chromium ion has a d3 electron configuration or, more succinctly, Cr3+ is a d3 ion.
Figure 23.11 The Possible Electron Configurations for Octahedral dn Transition-Metal Complexes (n = 1–10)
Two different configurations are possible for octahedral complexes of metals with d4, d5, d6, and d7 configurations; the magnitude of Δo determines which configuration is observed.
When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall from that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P)The energy that must be overcome to place an electron in an orbital that already has one electron.. If Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H2O)6]2+ ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN)6]3− ion, is called a low-spin complex. Similarly, metal ions with the d5, d6, or d7 electron configurations can be either high spin or low spin, depending on the magnitude of Δo.
In contrast, only one arrangement of d electrons is possible for metal ions with d8–d10 electron configurations. For example, the [Ni(H2O)6]2+ ion is d8 with two unpaired electrons, the [Cu(H2O)6]2+ ion is d9 with one unpaired electron, and the [Zn(H2O)6]2+ ion is d10 with no unpaired electrons.
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in .
Table 23.10 Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes
Octahedral Complexes | Δo (cm−1) | Octahedral Complexes | Δo (cm−1) | Tetrahedral Complexes | Δt (cm−1) |
---|---|---|---|---|---|
[Ti(H2O)6]3+ | 20,300 | [Fe(CN)6]4− | 32,800 | VCl4 | 9010 |
[V(H2O)6]2+ | 12,600 | [Fe(CN)6]3− | 35,000 | [CoCl4]2− | 3300 |
[V(H2O)6]3+ | 18,900 | [CoF6]3− | 13,000 | [CoBr4]2− | 2900 |
[CrCl6]3− | 13,000 | [Co(H2O)6]2+ | 9300 | [CoI4]2− | 2700 |
[Cr(H2O)6]2+ | 13,900 | [Co(H2O)6]3+ | 27,000 | ||
[Cr(H2O)6]3+ | 17,400 | [Co(NH3)6]3+ | 22,900 | ||
[Cr(NH3)6]3+ | 21,500 | [Co(CN)6]3− | 34,800 | ||
[Cr(CN)6]3− | 26,600 | [Ni(H2O)6]2+ | 8500 | ||
Cr(CO)6 | 34,150 | [Ni(NH3)6]2+ | 10,800 | ||
[MnCl6]4− | 7500 | [RhCl6]3− | 20,400 | ||
[Mn(H2O)6]2+ | 8500 | [Rh(H2O)6]3+ | 27,000 | ||
[MnCl6]3− | 20,000 | [Rh(NH3)6]3+ | 34,000 | ||
[Mn(H2O)6]3+ | 21,000 | [Rh(CN)6]3− | 45,500 | ||
[Fe(H2O)6]2+ | 10,400 | [IrCl6]3− | 25,000 | ||
[Fe(H2O)6]3+ | 14,300 | [Ir(NH3)6]3+ | 41,000 | ||
*Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. |
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H2O)6]2+, Δo = 11,800 cm−1; for [V(H2O)6]3+, Δo = 17,850 cm−1.
For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent group 9 metals illustrate this point:
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom increases. For example, Δo values for halide complexes generally decrease in the order F− > Cl− > Br− > I− because smaller, more localized charges, such as we see for F−, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F−. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical seriesAn ordering of ligands by their crystal field splitting energies., shown here in order of decreasing Δo:
The values of Δo listed in illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
The largest Δos are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition (). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δo.
Figure 23.12 A d–d Transition
In a d–d transition, an electron in one of the t2g orbitals of an octahedral complex such as the [Cr(H2O)6]3+ ion absorbs a photon of light with energy equal to Δo, which causes the electron to move to an empty or singly occupied eg orbital.
Recall from that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel in End-of-Chapter Application Problem 6 in . Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δo, which depends on the structure of the complex. For example, the complex [Cr(NH3)6]3+ has strong-field ligands and a relatively large Δo. Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H2O)6]3+ ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color.
We can now understand why emeralds and rubies have such different colors, even though both contain Cr3+ in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al2O3 in rubies and Be3Al2Si6O18 in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δo relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si6O18]12− silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δo. Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion.
Gem-quality crystals of ruby and emerald. The colors of both minerals are due to the presence of small amounts of Cr3+ impurities in octahedral sites in an otherwise colorless metal oxide lattice.
Recall from that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 complex such as [Ti(H2O)6]3+ is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE)The additional stabilization of a metal complex by selective population of the lower-energy orbitals (the t2g orbitals).. The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration.
Table 23.11 CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin | CFSE (Δo) | Low Spin | CFSE (Δo) | |||
---|---|---|---|---|---|---|
d 0 | 0 | |||||
d 1 | ↿ | 0.4 | ||||
d 2 | ↿ ↿ | 0.8 | ||||
d 3 | ↿ ↿ ↿ | 1.2 | ||||
d 4 | ↿ ↿ ↿ | ↿ | 0.6 | ↿⇂ ↿ ↿ | 1.6 | |
d 5 | ↿ ↿ ↿ | ↿ ↿ | 0.0 | ↿⇂ ↿⇂ ↿ | 2.0 | |
d 6 | ↿⇂ ↿ ↿ | ↿ ↿ | 0.4 | ↿⇂ ↿⇂ ↿⇂ | 2.4 | |
d 7 | ↿⇂ ↿⇂ ↿ | ↿ ↿ | 0.8 | ↿⇂ ↿⇂ ↿⇂ | ↿ | 1.8 |
d 8 | ↿⇂ ↿⇂ ↿⇂ | ↿ ↿ | 1.2 | |||
d 9 | ↿⇂ ↿⇂ ↿⇂ | ↿⇂ ↿ | 0.6 | |||
d 10 | ↿⇂ ↿⇂ ↿⇂ | ↿⇂ ↿⇂ | 0.0 |
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences.
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
If two trans ligands in an octahedral complex are either chemically different from the other four, as in the trans-[Co(NH3)4Cl2]+ ion, or at a different distance from the metal than the other four, the result is a tetragonally distorted octahedral complex. The electronic structures of such complexes are best viewed as the result of distorting an octahedral complex. Consider, for example, an octahedral complex such as [Co(NH3)6]3+ and then slowly remove two trans NH3 molecules by moving them away from the metal along the ±z axes, as shown in the top half of . As the two axial Co–N distances increase simultaneously, the d orbitals that interact most strongly with the two axial ligands will decrease in energy due to a decrease in electrostatic repulsions between electrons in these orbitals and the negative ends of the ligand dipoles. The affected d orbitals are those with a component along the ±z axes—namely, , dxz, and dyz. They will not be affected equally, however. Because the orbital points directly at the two ligands being removed, its energy will decrease much more rapidly than the energy of the other two, as shown in the bottom half of . In addition, the positive charge on the metal will increase somewhat as the axial ligands are removed, causing the four remaining in-plane ligands to be more strongly attracted to the metal. This will increase their interactions with the other two d orbitals and increase their energy. Again, the two d orbitals will not be affected equally. Because the orbital points directly at the four in-plane ligands, its energy will increase more rapidly than the energy of the dxy orbital, which points between the in-plane ligands. If we remove the two axial ligands to an infinite distance, we obtain a square planar complex. The energies of the and dxy orbitals actually cross as the axial ligands are removed, and the largest orbital spliting in a square planar complex is identical in magnitude to Δo.
Figure 23.13 d-Orbital Splittings for Tetragonal and Square Planar Complexes
Moving the two axial ligands away from the metal ion along the z axis initially gives an elongated octahedral complex (center) and eventually produces a square planar complex (right). As shown below the structures, an axial elongation causes the dxz and dyz orbitals to decrease in energy and the and dxy orbitals to increase in energy. As explained in the text, the change in energy is not the same for all five d orbitals. Removing the two axial ligands completely causes the energy of the orbital to decrease so much that the order of the and dxy orbitals is reversed.
In a tetrahedral arrangement of four ligands around a metal ion, none of the ligands lies on any of the three coordinate axes (part (a) in ); consequently, none of the five d orbitals points directly at the ligands. Nonetheless, the dxy, dxz, and dyz orbitals interact more strongly with the ligands than do and again resulting in a splitting of the five d orbitals into two sets. The splitting of the energies of the orbitals in a tetrahedral complex (Δt) is much smaller than that for Δo, however, for two reasons. First, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement. Second, there are only four negative charges rather than six, which decreases the electrostatic interactions by one-third if all other factors are equal. It can be shown that, for complexes of the same metal ion with the same charge, the same ligands, and the same M–L distance, The relationship between the splitting of the five d orbitals in octahedral and tetrahedral crystal fields imposed by the same ligands is shown schematically in part (b) in .
Figure 23.14 d-Orbital Splittings for a Tetrahedral Complex
(a) In a tetrahedral complex, none of the five d orbitals points directly at or between the ligands. (b) Because the dxy, dxz, and dyz orbitals (the t2g orbitals) interact more strongly with the ligands than do the and orbitals (the eg orbitals), the order of orbital energies in a tetrahedral complex is the opposite of the order in an octahedral complex.
Δt < Δo because of weaker d-orbital–ligand interactions and decreased electrostatic interactions.
Because Δo is so large for the second- and third-row transition metals, all four-coordinate complexes of these metals are square planar due to the much higher CFSE for square planar versus tetrahedral structures. The only exception is for d10 metal ions such as Cd2+, which have zero CFSE and are therefore tetrahedral as predicted by the VSEPR model. Four-coordinate complexes of the first-row transition metals can be either square planar or tetrahedral. The former is favored by strong-field ligands, whereas the latter is favored by weak-field ligands. For example, the [Ni(CN)4]2− ion is square planar, while the [NiCl4]2− ion is tetrahedral.
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
A From the number of ligands, determine the coordination number of the compound.
B Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
C Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
D Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons.
Solution:
A With six ligands, we expect this complex to be octahedral.
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
A This complex has four ligands, so it is either square planar or tetrahedral.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dyz, and dxz (these are degenerate); dxy; and
D The eight electrons occupy the first four of these orbitals, leaving the orbital empty. Thus there are no unpaired electrons.
Exercise
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
Answer:
The splitting of the d orbitals because of their interaction with the ligands in a complex has important consequences for the chemistry of transition-metal complexes; they can be divided into structural effects and thermodynamic effects. Although the two kinds of effects are interrelated, we will consider them separately.
There are two major kinds of structural effects: effects on the ionic radius of metal ions with regular octahedral or tetrahedral geometries, and structural distortions that are observed for specific electron configurations.
is a plot of the ionic radii of the divalent fourth-period metal ions versus atomic number. Only Ca2+(d0), Mn2+ (high-spin d5), and Zn2+ (d10) fall on the smooth curve calculated based on the effective nuclear charge (Zeff), which assumes that the distribution of d electrons is spherically symmetrical. All the other divalent ions fall below this curve because they have asymmetrical distributions of d electrons. (The points shown for Cr2+ and Cu2+ are only estimated values; as you will learn shortly, these two ions do not form any truly octahedral complexes.) To see why an asymmetrical distribution of d electrons makes a metal ion smaller than expected, consider the Ti2+ ion, which has a d2 configuration with both electrons in the t2g orbitals. Because the t2g orbitals are directed between the ligands, the two d electrons are unable to shield the ligands from the nuclear charge. Consequently, the ligands experience a higher effective nuclear charge than expected, the metal–ligand distance is unusually short, and the ionic radius is smaller than expected. If instead the two electrons were distributed uniformly over all five d orbitals, they would be much more effective at screening the ligands from the nuclear charge, making the metal–ligand distances longer and giving a larger ionic radius.
Figure 23.15 The Effect of d-Orbital Splittings on the Radii of the Divalent Ions of the Fourth-Period Metals
Because these radii are based on the structures of octahedral complexes and Cr2+ and Cu2+ do not form truly octahedral complexes, the points for these ions are shown as open circles. The dashed line represents the behavior predicted based on the effects of screening and variation in effective nuclear charge (Zeff), assuming a spherical distribution of the 3d electrons.
A similar effect is observed for the V2+ ion, which has a d3 configuration. Because the three electrons in the t2g orbitals provide essentially no shielding of the ligands from the metal, the ligands experience the full increase of +1 in nuclear charge that occurs in going from Ti2+ to V2+. Consequently, the observed ionic radius of the V2+ ion is significantly smaller than that of the Ti2+ ion.
Skipping the Cr2+ ion for the moment, we next consider the d5 Mn2+ ion. Because the nuclear charge increases by +2 from V2+ to Mn2+, we might expect Mn2+ to be smaller than V2+. The two electrons that are also added from V2+ to Mn2+ occupy the eg orbitals, however, which point directly at the six ligands. Because these electrons are localized directly between the metal ion and the ligands, they are effective at screening the ligands from the increased nuclear charge. As a result, the ionic radius actually increases significantly as we go from V2+ to Mn2+, despite the higher nuclear charge of the latter.
Exactly the same effects are seen in the second half of the first-row transition metals. In the Fe2+, Co2+, and Ni2+ ions, the extra electrons are added successively to the t2g orbitals, resulting in poor shielding of the ligands from the nuclear charge and abnormally small ionic radii. Skipping over Cu2+, we again see that adding the last two electrons causes a significant increase in the ionic radius of Zn2+, despite its higher nuclear charge.
Because simple octahedral complexes are not known for the Cr2+ and Cu2+ ions, only estimated values for their radii are shown in . We see in that both the Cr2+ and Cu2+ ions have electron configurations with an odd number of electrons in the eg orbitals. Because the single electron (in the case of Cr2+) or the third electron (in the case of Cu2+) can occupy either one of two degenerate eg orbitals, they have what is called a degenerate ground state. The Jahn–Teller theoremA theory that states that a non-linear molecule with a spatially degenerate electronic ground state will undergo a geometrical distortion to remove the degeneracy and lower the overall energy of the system. states that such non-linear systems are not stable; they will undergo a distortion that makes the complex less symmetrical and splits the degenerate states, which decreases the energy of the system. The distortion and resulting decrease in energy are collectively referred to as the Jahn–Teller effect. Neither the nature of the distortion nor its magnitude is specified, and in fact, they are difficult to predict. In principle, Jahn–Teller distortions are possible for many transition-metal ions; in practice, however, they are observed only for systems with an odd number of electrons in the eg orbitals, such as the Cr2+ and Cu2+ ions.
To see how a geometrical distortion can decrease the energy of such a system, consider an octahedral Cu2+ complex, the [Cu(H2O)6]2+ ion, which has been elongated along the z axis. As indicated in , this kind of distortion splits both the eg and t2g sets of orbitals. Because the axial ligands interact most strongly with the orbital, the splitting of the eg set (δ1) is significantly larger than the splitting of the t2g set (δ2), but both δ1 and δ2 are much, much smaller than the Δo. This splitting does not change the center of gravity of the energy within each set, so a Jahn–Teller distortion results in no net change in energy for a filled or half-filled set of orbitals. If, however, the eg set contains one (as in the d4 ions, Cr2+ and Mn3+) or three (as in the d9 ion, Cu2+) electrons, the distortion decreases the energy of the system. For Cu2+, for example, the change in energy after distortion is 2(−δ1/2) + 1(δ1/2) = −δ1/2. For Cu2+ complexes, the observed distortion is always an elongation along the z axis by as much as 50 pm; in fact, many Cu2+ complexes are so distorted that they are effectively square planar. In contrast, the distortion observed for most Cr2+ complexes is a compression along the z axis. In both cases, however, the net effect is the same: the distorted system is more stable than the undistorted system.
Jahn–Teller distortions are most important for d9 and high-spin d4 complexes; the distorted system is more stable than the undistorted one.
Figure 23.16 The Jahn–Teller Effect
Increasing the axial metal–ligand distances in an octahedral d9 complex is an example of a Jahn–Teller distortion, which causes the degenerate pair of eg orbitals to split in energy by an amount δ1; δ1 and δ2 are much smaller than Δo. As a result, the distorted system is more stable (lower in energy) than the undistorted complex by δ1/2.
As we previously noted, CFSEs can be as large as several hundred kilojoules per mole, which is the same magnitude as the strength of many chemical bonds or the energy change in most chemical reactions. Consequently, CFSEs are important factors in determining the magnitude of hydration energies, lattice energies, and other thermodynamic properties of the transition metals.
The hydration energy of a metal ion is defined as the change in enthalpy for the following reaction:
Equation 23.12
M2+(g) + H2O(l) → M2+(aq)Although hydration energies cannot be measured directly, they can be calculated from experimentally measured quantities using thermochemical cycles. As shown in part (a) in , a plot of the hydration energies of the fourth-period metal dications versus atomic number gives a curve with two valleys. Note the relationship between the plot in part (a) in and the plot of ionic radii in : the overall shapes are essentially identical, and only the three cations with spherically symmetrical distributions of d electrons (Ca2+, Mn2+, and Zn2+) lie on the dashed lines. In part (a) in , the dashed line corresponds to hydration energies calculated based solely on electrostatic interactions. Subtracting the CFSE values for the [M(H2O)6]2+ ions from the experimentally determined hydration energies gives the points shown as open circles, which lie very near the calculated curve. Thus CFSEs are primarily responsible for the differences between the measured and calculated values of hydration energies.
Figure 23.17 Thermochemical Effects of d-Orbital Splittings
(a) A plot of the hydration energies of the divalent fourth-period metal ions versus atomic number (solid circles) shows large deviations from the smooth curve calculated, assuming a spherical distribution of d electrons (dashed line). Correcting for CFSE gives the points shown as open circles, which, except for Ti2+ and Cr2+, are close to the calculated values. The apparent deviations for these ions are caused by the fact that solutions of the Ti2+ ion in water are not stable, and Cr2+ does not form truly octahedral complexes. (b) A plot of the lattice energies for the fourth-period metal dichlorides versus atomic number shows similar deviations from the smooth curve calculated, assuming a spherical distribution of d electrons (dashed lines), again illustrating the importance of CFSEs.
Values of the lattice energies for the fourth-period metal dichlorides are plotted versus atomic number in part (b) in . Recall that the lattice energy is defined as the negative of the enthalpy change for the following reaction. Like hydration energies, lattice energies are determined indirectly by using a thermochemical cycle:
Equation 23.13
M2+(g) + 2Cl−(g) → MCl2(s)The shape of the lattice-energy curve is essentially the mirror image of the hydration-energy curve in part (a) in , with only Ca2+, Mn2+, and Zn2+ lying on the smooth curve. It is not surprising that the explanation for the deviations from the curve is exactly the same as for the hydration energy data: all the transition-metal dichlorides, except MnCl2 and ZnCl2, are more stable than expected due to CFSE.
Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT. Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δt. The crystal field stabilization energy (CFSE) is the additional stabilization of a complex due to placing electrons in the lower-energy set of d orbitals. CFSE explains the unusual curves seen in plots of ionic radii, hydration energies, and lattice energies versus atomic number. The Jahn–Teller theorem states that a non-linear molecule with a spatially degenerate electronic ground state will undergo a geometrical distortion to remove the degeneracy and lower the overall energy of the system.
Describe crystal field theory in terms of its
In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?
Will the value of Δo increase or decrease if I− ligands are replaced by NO2− ligands? Why?
For an octahedral complex of a metal ion with a d6 configuration, what factors favor a high-spin configuration versus a low-spin configuration?
How can CFT explain the color of a transition-metal complex?
Do strong-field ligands favor a tetrahedral or a square planar structure? Why?
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
The ionic radii of V2+, Fe2+, and Zn2+ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.
In , you learned that 19 of the elements in the periodic table are essential elements that are necessary for most organisms, including humans, and in we discussed some of the biological functions of these elements. In this section, we describe several systems that illustrate the roles transition metals play in biological systems. Our goal is for you to understand why the chemical properties of these elements make them essential for life. We begin with a discussion of the strategies organisms use to extract transition metals from their environment. The section continues with a brief discussion of the use of transition metals in reactions that involve the transfer of electrons, reactions of small molecules such as O2, Lewis-acid catalysis, and the generation of reactive organic radicals.
In , we described the three possible dietary levels for any essential element: deficient, optimal, and toxic, in order of increasing concentration in the diet (). If the concentration of an essential element in the diet is too low, an organism must be able to extract the element from the environment and concentrate it. If the concentration of an essential element in the diet is too high, an organism must be able to limit its intake to avoid toxic effects. Moreover, organisms must be able to switch off the uptake process rapidly if dietary levels rise suddenly, and they must be able to store essential elements for future use.
Three distinct steps are involved in transition metal uptake. First, the metal must be “mobilized” from the environment and brought into contact with a cell in a form that can be absorbed. Second, the metal must be transported across the cell membrane into the cell. Third, the element must be transported to its point of utilization within a cell or to other cells within the organism. In our discussion, we focus on the uptake, transport, and storage of iron, which illustrates the most important points. Because iron deficiency (anemia) is the most widespread nutritional deficiency known in humans, the uptake of iron is especially well understood.
Iron complexes in biological systems. Iron(III) forms very stable octahedral complexes with hydroxamate and catecholate ligands.
In , you learned that the solubility of metal ions such as Fe3+, which form highly insoluble hydroxides, depends on the pH and the presence of complexing agents. In an oxygen-containing atmosphere, iron exists as Fe(III) because of the positive reduction potential of Fe3+ (Fe3+ + e− → Fe2+; E° = +0.77 V). Because ferric hydroxide [Fe(OH)3] is highly insoluble (Ksp ≈ 1 × 10−39), the equilibrium concentration of Fe3+(aq) at pH 7.0 is very low, about 10−18 M. You would have to drink 2 × 1013 L of iron-saturated water per day (roughly 5 mi3) to consume the recommended daily intake of Fe for humans, which is about 1 mg/day. Animals such as humans can overcome this problem by consuming concentrated sources of iron, such as red meat, but microorganisms cannot.
Consequently, most microorganisms synthesize and secrete organic molecules called siderophoresAn organic ligand that has a high affinity for Fe(III) and is secreted into the surrounding medium to increase the total concentration of dissolved iron. to increase the total concentration of available iron in the surrounding medium. Siderophores are generally cyclic compounds that use bidentate ligands, such as the hydroxamate and catecholate groups shown here, to bind Fe3+ in an octahedral arrangement. Typical siderophores are ferrichrome, a cyclic peptide produced by fungi, and enterobactin, a cyclic ester produced by bacteria (). Attaching the three iron ligands to a cyclic framework greatly increases the stability of the resulting Fe3+ complex due to the chelate effect described in . The formation constants for the Fe3+ complexes of ferrichrome and enterobactin are about 1032 and 1040, respectively, which are high enough to allow them to dissolve almost any Fe(III) compound.
Figure 23.18 Siderophores
Ferrichrome (a) and enterobactin (b) are siderophores that use hydroxamate and catecholate ligands, respectively, to bind Fe3+. The “wrapped” drawing of enterobactin (c) shows how the cyclic ester framework places the three catecholate ligands in the correct orientation to bind to a single Fe3+ ion, which is an application of the chelate effect. The actual structure of ferrichrome is similar to that of enterobactin, with the three hydroxamate ligands adjacent to one another for optimal binding of Fe3+. Note: For clarity, most or all hydrogen atoms have been omitted in this and the following structures.
Siderophores increase the [Fe3+] in solution, providing the bacterium that synthesized them (as well as any competitors) with a supply of iron. In addition, siderophores neutralize the positive charge on the metal ion and provide a hydrophobic “wrapping” that enables the Fe3+–siderophore complex to be recognized by a specific protein that transports it into the interior of a cell. Once it is inside a cell, the iron is reduced to Fe2+, which has a much lower affinity for the siderophore and spontaneously dissociates.
In contrast, multicellular organisms can increase the concentration of iron in their diet by lowering the pH in the gastrointestinal tract. At pH 1.0 (the approximate pH of the stomach), most Fe(III) salts dissolve to form Fe3+(aq), which is absorbed by specific proteins in the intestinal wall. A protein called transferrin forms a complex with iron(III), allowing it to be transported to other cells. Proteins that bind tightly to Fe(III) can also be used as antibacterial agents because iron is absolutely essential for bacterial growth. For example, milk, tears, and egg white all contain proteins similar to transferrin, and their high affinity for Fe3+ allows them to sequester iron, thereby preventing bacteria from growing in these nutrient-rich media.
Iron is released from transferrin by reduction to Fe2+, and then it is either used immediately (e.g., for the synthesis of hemoglobin) or stored in a very large protein called ferritin for future use (). Ferritin uses oxygen to oxidize Fe2+ to Fe3+, which at neutral pH precipitates in the central cavity of the protein as a polymeric mixture of Fe(OH)3 and FePO4. Because a fully loaded ferritin molecule can contain as many as 4500 Fe atoms, which corresponds to about 25% Fe by mass, ferritin is an effective way to store iron in a highly concentrated form. When iron is needed by a cell, the Fe3+ is reduced to the much more soluble Fe2+ by a reductant such as ascorbic acid (vitamin C). The structure of ferritin contains channels at the junctions of the subunits, which provide pathways for iron to enter and leave the interior of a molecule.
Figure 23.19 Ferritin, an Iron Storage Protein
A schematic drawing of the structure of iron-loaded ferritin, showing the almost spherical protein shell inside which the iron hydroxide/phosphate core is formed.
A protein that contains one or more metal ions tightly bound to amino acid side chains is called a metalloproteinA protein that contains one or more tightly bound metal ions.; some of the most common ligands provided by amino acids are shown here. A metalloprotein that catalyzes a chemical reaction is a metalloenzymeA protein that contains one or more tightly bound metal ions and catalyzes a biochemical reaction.. Thus all metalloenzymes are metalloproteins, but the converse is not true. Recent estimates suggest that more than 40% of all known enzymes require at least one metal ion for activity, including almost all the enzymes responsible for the synthesis, duplication, and repair of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
Ligands used in biological systems. These metal ligands are commonly found in metalloproteins.
Proteins whose function is to transfer electrons from one place to another are called electron-transfer proteins. Because they do not catalyze a chemical reaction, electron-transfer proteins are not enzymes; they are biochemical reductants or oxidants consumed in an enzymatic reaction. The general reaction for an electron-transfer protein is as follows:
Equation 23.14
Because many transition metals can exist in more than one oxidation state, electron-transfer proteins usually contain one or more metal ions that can undergo a redox reaction. Incorporating a metal ion into a protein has three important biological consequences:
Three important classes of metalloproteins transfer electrons: blue copper proteins, cytochromes, and iron–sulfur proteins, which generally transfer electrons at high (> 0.20 V), intermediate (±0 V), and low (−0.20 to −0.50 V) potentials, respectively (). Although these electron-transfer proteins contain different metals with different structures, they are all designed to ensure rapid electron transfer to and from the metal. Thus when the protein collides with its physiological oxidant or reductant, electron transfer can occur before the two proteins diffuse apart. For electron transfer to be rapid, the metal sites in the oxidized and reduced forms of the protein must have similar structures.
Table 23.12 Some Properties of the Most Common Electron-Transfer Proteins
Figure 23.20 A Blue Copper Protein
In both the oxidized and reduced forms of a blue copper protein, the copper is coordinated by four ligands (two histidine imidazole nitrogen atoms, a cysteine thiolate sulfur, and a thioether sulfur of a methionine) in a roughly tetrahedral arrangement.
Blue copper proteins were first isolated from bacteria in the 1950s and from plant tissues in the early 1960s. The intense blue color of these proteins is due to a strong absorption band at a wavelength of about 600 nm. Although simple Cu2+ complexes, such as [Cu(H2O)6]2+ and [Cu(NH3)4]2+, are also blue due to an absorption band at 600 nm, the intensity of the absorption band is about 100 times less than that of a blue copper protein. Moreover, the reduction potential for the Cu2+/Cu+ couple in a blue copper protein is usually +0.3 to +0.5 V, considerably more positive than that of the aqueous Cu2+/Cu+ couple (+0.15 V).
The copper center in blue copper proteins has a distorted tetrahedral structure, in which the copper is bound to four amino acid side chains (). Although the most common structures for four-coordinate Cu2+ and Cu+ complexes are square planar and tetrahedral, respectively, the structures of the oxidized (Cu2+) and reduced (Cu+) forms of the protein are essentially identical. Thus the protein forces the Cu2+ ion to adopt a higher-energy structure that is more suitable for Cu+, which makes the Cu2+ form easier to reduce and raises its reduction potential.
Moreover, by forcing the oxidized and reduced forms of the metal complex to have essentially the same structure, the protein ensures that electron transfer to and from the copper site is rapid because only minimal structural reorganization of the metal center is required. Kinetics studies on simple metal complexes have shown that electron-transfer reactions tend to be slow when the structures of the oxidized and reduced forms of a metal complex are very different, and fast when they are similar. You will see that other metal centers used for biological electron-transfer reactions are also set up for minimal structural reorganization after electron transfer, which ensures the rapid transfer of electrons.
The cytochromes (from the Greek cytos, meaning “cell”, and chroma, meaning “color”) were first identified in the 1920s by spectroscopic studies of cell extracts. Based on the wavelength of the maximum absorption in the visible spectrum, they were classified as cytochromes a (with the longest wavelength), cytochromes b (intermediate wavelength), and cytochromes c (shortest wavelength). It quickly became apparent that there was a correlation between their spectroscopic properties and other physical properties. For examples, cytochromes c are generally small, soluble proteins with a reduction potential of about +0.25 V, whereas cytochromes b are larger, less-soluble proteins with reduction potentials of about 0 V.
All cytochromes contain iron, and the iron atom in all cytochromes is coordinated by a planar array of four nitrogen atoms provided by a cyclic tetradentate ligand called a porphyrin. The iron–porphyrin unit is called a heme group. The structures of a typical porphyrin (protoporphyrin IX) and its iron complex (protoheme) are shown here. In addition to the four nitrogen atoms of the porphyrin, the iron in a cytochrome is usually bonded to two additional ligands provided by the protein, as shown in .
A cytochrome. Shown here is protoporphyrin IX and its iron complex, protoheme.
Figure 23.21 A Cytochrome c
In a cytochrome c, the heme iron is coordinated to the nitrogen atom of a histidine imidazole and the sulfur atom of a methionine thioether, in addition to the four nitrogen atoms provided by the porphyrin.
In contrast to the blue copper proteins, two electron configurations are possible for both the oxidized and reduced forms of a cytochrome, and this has significant structural consequences. Thus Fe2+ is d6 and can be either high spin (with four unpaired electrons) or low spin (with no unpaired electrons; ). Similarly, Fe3+ is d5 and can also be high spin (with five unpaired electrons) or low spin (with one unpaired electron). In low-spin heme complexes, both the Fe2+ and the Fe3+ ions are small enough to fit into the “hole” in the center of the porphyrin; hence the iron atom lies almost exactly in the plane of the four porphyrin nitrogen atoms in both cases. Because cytochromes b and c are low spin in both their oxidized and reduced forms, the structures of the oxidized and reduced cytochromes are essentially identical. Hence minimal structural changes occur after oxidation or reduction, which makes electron transfer to or from the heme very rapid.
Electron transfer reactions occur most rapidly when minimal structural changes occur during oxidation or reduction.
Although all known bacteria, plants, and animals use iron–sulfur proteins to transfer electrons, the existence of these proteins was not recognized until the late 1950s. Iron–sulfur proteins transfer electrons over a wide range of reduction potentials, and their iron content can range from 1 to more than 12 Fe atoms per protein molecule. In addition, most iron–sulfur proteins contain stoichiometric amounts of sulfide (S2−).
These properties are due to the presence of four different kinds of iron–sulfur units, which contain one, two, three, or four iron atoms per Fe–S complex (). In all cases, the Fe2+ and Fe3+ ions are coordinated to four sulfur ligands in a tetrahedral environment. Due to tetrahedral coordination by weak-field sulfur ligands, the iron is high spin in both the Fe3+ and Fe2+ oxidation states, which results in similar structures for the oxidized and reduced forms of the Fe–S complexes. Consequently, only small structural changes occur after oxidation or reduction of the Fe–S center, which results in rapid electron transfer.
Figure 23.22 Fe–S Centers in Proteins
Four kinds of iron–sulfur centers, containing one, two, three, and four iron atoms, respectively, are known in electron-transfer proteins. Although they differ in the number of sulfur atoms provided by cysteine thiolates versus sulfide, in all cases the iron is coordinated to four sulfur ligands in a roughly tetrahedral environment.
Although small molecules, such as O2, N2, and H2, do not react with organic compounds under ambient conditions, they do react with many transition-metal complexes. Consequently, virtually all organisms use metalloproteins to bind, transport, and catalyze the reactions of these molecules. Probably the best-known example is hemoglobin, which is used to transport O2 in many multicellular organisms.
Under ambient conditions, small molecules, such as O2, N2, and H2, react with transition-metal complexes but not with organic compounds.
Many microorganisms and most animals obtain energy by respiration, the oxidation of organic or inorganic molecules by O2. At 25°C, however, the concentration of dissolved oxygen in water in contact with air is only about 0.25 mM. Because of their high surface area-to-volume ratio, aerobic microorganisms can obtain enough oxygen for respiration by passive diffusion of O2 through the cell membrane. As the size of an organism increases, however, its volume increases much more rapidly than its surface area, and the need for oxygen depends on its volume. Consequently, as a multicellular organism grows larger, its need for O2 rapidly outstrips the supply available through diffusion. Unless a transport system is available to provide an adequate supply of oxygen for the interior cells, organisms that contain more than a few cells cannot exist. In addition, O2 is such a powerful oxidant that the oxidation reactions used to obtain metabolic energy must be carefully controlled to avoid releasing so much heat that the water in the cell boils. Consequently, in higher-level organisms, the respiratory apparatus is located in internal compartments called mitochondria, which are the power plants of a cell. Oxygen must therefore be transported not only to a cell but also to the proper compartment within a cell.
Three different chemical solutions to the problem of oxygen transport have developed independently in the course of evolution, as indicated in . Mammals, birds, reptiles, fish, and some insects use a heme protein called hemoglobin to transport oxygen from the lungs to the cells, and they use a related protein called myoglobin to temporarily store oxygen in the tissues. Several classes of invertebrates, including marine worms, use an iron-containing protein called hemerythrin to transport oxygen, whereas other classes of invertebrates (arthropods and mollusks) use a copper-containing protein called hemocyanin. Despite the presence of the hem- prefix, hemerythrin and hemocyanin do not contain a metal–porphyrin complex.
Table 23.13 Some Properties of the Three Classes of Oxygen-Transport Proteins
Protein | Source | M per Subunit | M per O2 Bound | Color (deoxy form) | Color (oxy form) |
---|---|---|---|---|---|
hemoglobin | mammals, birds, fish, reptiles, some insects | 1 Fe | 1 Fe | red-purple | red |
hemerythrin | marine worms | 2 Fe | 2 Fe | colorless | red |
hemocyanin | mollusks, crustaceans, spiders | 2 Cu | 2 Cu | colorless | blue |
Myoglobin is a relatively small protein that contains 150 amino acids. The functional unit of myoglobin is an iron–porphyrin complex that is embedded in the protein (). In myoglobin, the heme iron is five-coordinate, with only a single histidine imidazole ligand from the protein (called the proximal histidine because it is near the iron) in addition to the four nitrogen atoms of the porphyrin. A second histidine imidazole (the distal histidine because it is more distant from the iron) is located on the other side of the heme group, too far from the iron to be bonded to it. Consequently, the iron atom has a vacant coordination site, which is where O2 binds. In the ferrous form (deoxymyoglobin), the iron is five-coordinate and high spin. Because high-spin Fe2+ is too large to fit into the “hole” in the center of the porphyrin, it is about 60 pm above the plane of the porphyrin. When O2 binds to deoxymyoglobin to form oxymyoglobin, the iron is converted from five-coordinate (high spin) to six-coordinate (low spin; ). Because low-spin Fe2+ and Fe3+ are smaller than high-spin Fe2+, the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. The O2 pressure at which half of the molecules in a solution of myoglobin are bound to O2 (P1/2) is about 1 mm Hg (1.3 × 10−3 atm).
A vacant coordination site at a metal center in a protein usually indicates that a small molecule will bind to the metal ion, whereas a coordinatively saturated metal center is usually involved in electron transfer.
Figure 23.23 The Structure of Deoxymyoglobin, Showing the Heme Group
The iron in deoxymyoglobin is five-coordinate, with one histidine imidazole ligand from the protein. Oxygen binds at the vacant site on iron.
Figure 23.24 Oxygen Binding to Myoglobin and Hemoglobin
(a) The Fe2+ ion in deoxymyoglobin is high spin, which makes it too large to fit into the “hole” in the center of the porphyrin. (b) When O2 binds to deoxymyoglobin, the iron is converted to low-spin Fe3+, which is smaller, allowing the iron to move into the plane of the four nitrogen atoms of the porphyrin to form an octahedral complex.
Hemoglobin consists of two subunits of 141 amino acids and two subunits of 146 amino acids, both similar to myoglobin; it is called a tetramer because of its four subunits. Because hemoglobin has very different O2-binding properties, however, it is not simply a “super myoglobin” that can carry four O2 molecules simultaneously (one per heme group). The shape of the O2-binding curve of myoglobin (Mb; ) can be described mathematically by the following equilibrium:
Equation 23.15
In contrast, the O2-binding curve of hemoglobin is S shaped (). As shown in the curves, at low oxygen pressures, the affinity of deoxyhemoglobin for O2 is substantially lower than that of myoglobin, whereas at high O2 pressures the two proteins have comparable O2 affinities. The physiological consequences of the unusual S-shaped O2-binding curve of hemoglobin are enormous. In the lungs, where O2 pressure is highest, the high oxygen affinity of deoxyhemoglobin allows it to be completely loaded with O2, giving four O2 molecules per hemoglobin. In the tissues, however, where the oxygen pressure is much lower, the decreased oxygen affinity of hemoglobin allows it to release O2, resulting in a net transfer of oxygen to myoglobin.
Figure 23.25 The O2-Binding Curves of Myoglobin and Hemoglobin
The curve for myoglobin can be described by a simple equilibrium between deoxy- and oxymyoglobin, but the S-shaped curve for hemoglobin can be described only in terms of a cooperative interaction between the four hemes.
The S-shaped O2-binding curve of hemoglobin is due to a phenomenon called cooperativity, in which the affinity of one heme for O2 depends on whether the other hemes are already bound to O2. Cooperativity in hemoglobin requires an interaction between the four heme groups in the hemoglobin tetramer, even though they are more than 3000 pm apart, and depends on the change in structure of the heme group that occurs with oxygen binding. The structures of deoxyhemoglobin and oxyhemoglobin are slightly different, and as a result, deoxyhemoglobin has a much lower O2 affinity than myoglobin, whereas the O2 affinity of oxyhemoglobin is essentially identical to that of oxymyoglobin. Binding of the first two O2 molecules to deoxyhemoglobin causes the overall structure of the protein to change to that of oxyhemoglobin; consequently, the last two heme groups have a much higher affinity for O2 than the first two.
Oxygen is not unique in its ability to bind to a ferrous heme complex; small molecules such as CO and NO bind to deoxymyoglobin even more tightly than does O2. The interaction of the heme iron with oxygen and other diatomic molecules involves the transfer of electron density from the filled t2g orbitals of the low-spin d6 Fe2+ ion to the empty π* orbitals of the ligand. In the case of the Fe2+–O2 interaction, the transfer of electron density is so great that the Fe–O2 unit can be described as containing low-spin Fe3+ (d5) and O2−. We can therefore represent the binding of O2 to deoxyhemoglobin and its release as a reversible redox reaction:
Equation 23.16
Fe2+ + O2 ⇌ Fe3+–O2−As shown in , the Fe–O2 unit is bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in CO are empty and those in NO are singly occupied, these ligands interact more strongly with Fe2+ than does O2, in which the π* orbitals of the neutral ligand are doubly occupied.
Figure 23.26 Binding of O2 and CO to the Iron of Myoglobin
Because the Fe–O–O unit is bent, while the Fe–C–O unit is linear, the imidazole group of the distal histidine in hemoglobin interferes with CO binding and decreases the affinity of hemoglobin for CO.
Although CO has a much greater affinity for a ferrous heme than does O2 (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O2, which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O2 and CO provides a plausible explanation for the difference in affinities. As shown in , the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O2 unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O2 affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO.
Hemerythrin is used to transport O2 in a variety of marine invertebrates. It is an octamer (eight subunits), with each subunit containing two iron atoms and binding one molecule of O2. Deoxyhemerythrin contains two Fe2+ ions per subunit and is colorless, whereas oxyhemerythrin contains two Fe3+ ions and is bright reddish violet. These invertebrates also contain a monomeric form of hemerythrin that is located in the tissues, analogous to myoglobin. The binding of oxygen to hemerythrin and its release can be described by the following reaction, where the HO2− ligand is the hydroperoxide anion derived by the deprotonation of hydrogen peroxide (H2O2):
Equation 23.17
2Fe2+ + O2 + H+ ⇌ 2Fe3+–O2HThus O2 binding is accompanied by the transfer of two electrons (one from each Fe2+) and a proton to O2.
Hemocyanin is used for oxygen transport in many arthropods (spiders, crabs, lobsters, and centipedes) and in mollusks (shellfish, octopi, and squid); it is responsible for the bluish-green color of their blood. The protein is a polymer of subunits that each contain two copper atoms (rather than iron), with an aggregate molecular mass of greater than 1,000,000 amu. Deoxyhemocyanin contains two Cu+ ions per subunit and is colorless, whereas oxyhemocyanin contains two Cu2+ ions and is bright blue. As with hemerythrin, the binding and release of O2 correspond to a two-electron reaction:
Equation 23.18
2Cu+ + O2 ⇌ Cu2+–O22−–Cu2+Although hemocyanin and hemerythrin perform the same basic function as hemoglobin, these proteins are not interchangeable. In fact, hemocyanin is so foreign to humans that it is one of the major factors responsible for the common allergies to shellfish.
Myoglobin, hemoglobin, hemerythrin, and hemocyanin all use a transition-metal complex to transport oxygen.
Many of the enzymes involved in the biological reactions of oxygen contain metal centers with structures that are similar to those used for O2 transport. Many of these enzymes also contain metal centers that are used for electron transfer, which have structures similar to those of the electron-transfer proteins discussed previously. In this section, we briefly describe two of the most important examples: dioxygenases and methane monooxygenase.
Dioxygenases are enzymes that insert both atoms of O2 into an organic molecule. In humans, dioxygenases are responsible for cross-linking collagen in connective tissue and for synthesizing complex organic molecules called prostaglandins, which trigger inflammation and immune reactions. Iron is by far the most common metal in dioxygenases; and the target of the most commonly used drug in the world, aspirin, is an iron enzyme that synthesizes a specific prostaglandin. Aspirin inhibits this enzyme by binding to the iron atom at the active site, which prevents oxygen from binding.
Methane monooxygenase catalyzes the conversion of methane to methanol. The enzyme is a monooxygenase because only one atom of O2 is inserted into an organic molecule, while the other is reduced to water:
Equation 23.19
CH4 + O2 + 2e− + 2H+ → CH3OH + H2OBecause methane is the major component of natural gas, there is enormous interest in using this reaction to convert methane to a liquid fuel (methanol) that is much more convenient to ship and store. Because the C–H bond in methane is one of the strongest C–H bonds known, however, an extraordinarily powerful oxidant is needed for this reaction. The active site of methane monooxygenase contains two Fe atoms that bind O2, but the details of how the bound O2 is converted to such a potent oxidant remain unclear.
Reactions catalyzed by metal ions that do not change their oxidation states during the reaction are usually group transfer reactionsReactions involving the transfer of a group, catalyzed by metal ions that do not change their oxidation states during the reaction., in which a group such as the phosphoryl group (−PO32−) is transferred. These enzymes usually use metal ions such as Zn2+, Mg2+, and Mn2+, and they range from true metalloenzymes, in which the metal ion is tightly bound, to metal-activated enzymes, which require the addition of metal ions for activity. Because tight binding is usually the result of specific metal–ligand interactions, metalloenzymes tend to be rather specific for a particular metal ion. In contrast, the binding of metal ions to metal-activated enzymes is largely electrostatic in nature; consequently, several different metal ions with similar charges and sizes can often be used to give an active enzyme.
Metalloenzymes generally contain a specific metal ion, whereas metal-activated enzymes can use any of several metal ions of similar size and charge.
A metal ion that acts as a Lewis acid can catalyze a group transfer reaction in many different ways, but we will focus on only one of these, using a zinc enzyme as an example. Carbonic anhydrase is found in red blood cells and catalyzes the reaction of CO2 with water to give carbonic acid.
Equation 23.20
Although this reaction occurs spontaneously in the absence of a catalyst, it is too slow to absorb all the CO2 generated during respiration. Without a catalyst, tissues would explode due to the buildup of excess CO2 pressure. Carbonic anhydrase contains a single Zn2+ ion per molecule, which is coordinated by three histidine imidazole ligands and a molecule of water. Because Zn2+ is a Lewis acid, the pKa of the Zn2+–OH2 unit is about 8 versus 14 for pure water. Thus at pH 7–8, a significant fraction of the enzyme molecules contain the Zn2+–OH− group, which is much more reactive than bulk water. When carbon dioxide binds in a nonpolar site next to the Zn2+–OH− unit, it reacts rapidly to give a coordinated bicarbonate ion that dissociates from the enzyme:
Equation 23.21
Zn2+–OH− + CO2 ⇌ Zn2+–OCO2H− ⇌ Zn2+ + HCO3−The active site of carbonic anhydrase.
Thus the function of zinc in carbonic anhydrase is to generate the hydroxide ion at pH 7.0, far less than the pH required in the absence of the metal ion.
An organic radical is an organic species that contains one or more unpaired electrons. Chemists often consider organic radicals to be highly reactive species that produce undesirable reactions. For example, they have been implicated in some of the irreversible chemical changes that accompany aging. It is surprising, however, that organic radicals are also essential components of many important enzymes, almost all of which use a metal ion to generate the organic radical within the enzyme. These enzymes are involved in the synthesis of hemoglobin and DNA, among other important biological molecules, and they are the targets of pharmaceuticals for the treatment of diseases such as anemia, sickle-cell anemia, and cancer. In this section, we discuss one class of radical enzymes that use vitamin B12.
Vitamin B12 was discovered in the 1940s as the active agent in the cure of pernicious anemia, which does not respond to increased iron in the diet. Humans need only tiny amounts of vitamin B12, and the average blood concentration in a healthy adult is only about 3.5 × 10−8 M. The structure of vitamin B12, shown in , is similar to that of a heme, but it contains cobalt instead of iron, and its structure is much more complex. In fact, vitamin B12 has been called the most complex nonpolymeric biological molecule known and was the first naturally occurring organometallic compound to be isolated. When vitamin B12 (the form present in vitamin tablets) is ingested, the axial cyanide ligand is replaced by a complex organic group.
Figure 23.27 Vitamin B12
In the body, the axial cyanide ligand found in the vitamin is replaced by a complex organic unit. Heterolytic cleavage of the Co–C bond in the resulting organometallic complex generates an organic radical for the catalysis of rearrangement reactions.
The cobalt–carbon bond in the enzyme-bound form of vitamin B12 and related compounds is unusually weak, and it is particularly susceptible to homolytic cleavage:
Equation 23.22
Homolytic cleavage of the Co3+–CH2R bond produces two species, each of which has an unpaired electron: a d7 Co2+ derivative and an organic radical, ·CH2R, which is used by vitamin B12-dependent enzymes to catalyze a wide variety of reactions. Virtually all vitamin B12-catalyzed reactions are rearrangements in which an H atom and an adjacent substituent exchange positions:
In the conversion of ethylene glycol to acetaldehyde, the initial product is the hydrated form of acetaldehyde, which rapidly loses water:
The enzyme uses the ·CH2R radical to temporarily remove a hydrogen atom from the organic substrate, which then rearranges to give a new radical. Transferring the hydrogen atom back to the rearranged radical gives the product and regenerates the ·CH2R radical.
The metal is not involved in the actual catalytic reaction; it provides the enzyme with a convenient mechanism for generating an organic radical, which does the actual work. Many examples of similar reactions are now known that use metals other than cobalt to generate an enzyme-bound organic radical.
Nearly all vitamin B12-catalyzed reactions are rearrangements that occur via a radical reaction.
Three separate steps are required for organisms to obtain essential transition metals from their environment: mobilization of the metal, transport of the metal into the cell, and transfer of the metal to where it is needed within a cell or an organism. The process of iron uptake is best understood. To overcome the insolubility of Fe(OH)3, many bacteria use organic ligands called siderophores, which have high affinity for Fe(III) and are secreted into the surrounding medium to increase the total concentration of dissolved iron. The iron–siderophore complex is absorbed by a cell, and the iron is released by reduction to Fe(II). Mammals use the low pH of the stomach to increase the concentration of dissolved iron. Iron is absorbed in the intestine, where it forms an Fe(III) complex with a protein called transferrin that is transferred to other cells for immediate use or storage in the form of ferritin.
Proteins that contain one or more tightly bound metal ions are called metalloproteins, and metalloproteins that catalyze biochemical reactions are called metalloenzymes. Proteins that transfer electrons from one place to another are called electron-transfer proteins. Most electron-transfer proteins are metalloproteins, such as iron–sulfur proteins, cytochromes, and blue copper proteins that accept and donate electrons. The oxidized and reduced centers in all electron-transfer proteins have similar structures to ensure that electron transfer to and from the metal occurs rapidly. Metalloproteins also use the ability of transition metals to bind small molecules, such as O2, N2, and H2, to transport or catalyze the reactions of these small molecules. For example, hemoglobin, hemerythrin, and hemocyanin, which contain heme iron, nonheme iron, and copper, respectively, are used by different kinds of organisms to bind and transfer O2. Other metalloenzymes use transition-metal ions as Lewis acids to catalyze group transfer reactions. Finally, some metalloenzymes use homolytic cleavage of the cobalt–carbon bond in derivatives of vitamin B12 to generate an organic radical that can abstract a hydrogen atom and thus cause molecular rearrangements to occur.
What are the advantages of having a metal ion at the active site of an enzyme?
Why does the structure of the metal center in a metalloprotein that transfers electrons show so little change after oxidation or reduction?
In enzymes, explain how metal ions are particularly suitable for generating organic radicals.
A common method for treating carbon-monoxide poisoning is to have the patient inhale pure oxygen. Explain why this treatment is effective.
Tungsten bronzes are semimetallic solids that are inert to strong acids and bases, lustrous, and good conductors of electricity; they are used in the production of bronze or metallic paints. These nonstoichiometric compounds have the general formula MxWO3, where M = Na, K, a group 2 metal, or a lanthanide and x < 1. The properties of tungsten bronzes suggest that at least some of the tungsten atoms are in the +6 oxidation state. Given the high oxidation state, why do these solids conduct electricity?
Gout is a painful disorder caused by the overproduction of uric acid, which is deposited as sodium urate crystals in joints. The enzyme that catalyzes the production of uric acid contains Fe3+ and Mo6+. Molecular oxygen is a substrate in this reaction. Based on the oxidation states of the metals, what do you expect one of the products of the reaction to be?
A laboratory technician added aqueous ammonia to an aqueous solution of Mn2+, which produced a pale pink precipitate. She left the solution exposed to air and went home. The next day she returned to the lab and found that her pink precipitate had turned brown-black. Write balanced chemical equations to show what had happened.
Plants use Mn(IV) during photosynthesis to produce dioxygen from water. Write a balanced chemical equation showing this reaction and suggest why Mn is well suited for this purpose.
Rust stains (Fe2O3) can be removed from fabrics by oxalic acid (HO2CCO2H). Write a balanced chemical equation showing the reaction that occurs and predict the solubility of the product in water.
It has been suggested that complexes that can coordinate N2 are used by bacteria to fix atmospheric nitrogen. One such complex is [Ru(NH3)5·N2]Cl2, which was first discovered in 1965. Sigma bonding with N2 in this complex would be weak because the N2 molecule is symmetrical and has zero dipole moment.
Monel metal, which contains 68% Ni, 32% Cu, and traces of Fe and Mn, is highly corrosion resistant. It is used, for example, to make items that will be used in marine environments and to hold corrosive fluorine compounds. Based on its composition, why is Monel particularly suitable for these purposes?
From 1845 to 1850, a fungus known as “potato blight” caused a potato famine in Ireland. Approximately 25% of the Irish population either died or emigrated as a direct result. A spray called Bordeaux mixture is now used to kill the fungus; it is made by the reaction of CuSO4 with Ca(OH)2. What is the formula of the Bordeaux mixture? Write a balanced chemical equation for the reaction.
Many transition metals and their compounds are used as catalysts. Given MnO2, FeCl3, Pt, and Ni, which would you select for each purpose and why?
The Fe2+ site in hemoglobin binds oxygen reversibly; consequently, it is suitable for transporting oxygen in blood. Various other small molecules can bind to the iron instead, thus preventing oxygen transport. Based on the type of bonding, why are CO and NO particularly toxic? Would they be as toxic if hemoglobin contained a V2+ center instead of Fe2+? Why?
In 1951, G. Wilkinson reported a surprising iron–hydrocarbon compound that is now called ferrocene. Ferrocene is orange and has a structure in which the metal is sandwiched between two planar cyclopentadienyl (C5H5−) rings. It can be viewed as a compound of Fe2+ with two C5H5− rings. Ferrocene does not contain Fe–C σ bonds but another type of bond formed by the lateral overlap of orbitals. Which metal orbitals are involved? A similar structure is obtained with Ru2+. One of these metals forms a sandwich complex that has a staggered conformation, and the other forms a complex that is eclipsed. Which metal produces which conformation? Why?
The Ziegler–Natta catalyst is used for the polymerization of ethylene to form high-density polyethylene, a widely used lightweight plastic. The active form of the catalyst is believed to be TiCl3CH2CH3, and the first step in the polymerization reaction is believed to be binding of the double bond in ethylene to Ti. If you were interested in developing a similar catalyst for this same purpose, would you choose chlorides of Zr, Hf, V, Nb, Ta, Cr, Mo, or W? Why?
Cobalt(II) chloride is used as a visual indicator of humidity because it exists as a blue complex when dry and a pink complex when exposed to moisture in the air. The bonding environment around the cobalt in one of these complexes is octahedral; in the other, it is tetrahedral. What reaction occurs to produce the color change? Write a balanced chemical equation for this reaction, indicating the species present.
A platinum complex that is widely available commercially is chloroplatinic acid {H2[PtCl6]}, which is used to make platinized asbestos, a catalyst. What is the structure of chloroplatinic acid? Is it distorted from an idealized geometry? Do you expect it to be colored? Justify your answers.
The tungsten bronzes can be viewed as the products of partial reduction of WO3 by an active metal to give a mixture of W(VI) and W(V). In the solid, the d orbitals of the metal overlap to form a partially filled band, which is responsible for the luster and metallic conductivity.
In , you were introduced to the major classes of organic compounds, covalent compounds composed primarily of carbon and hydrogen. Organic substances have been used throughout this text to illustrate the differences between ionic and covalent bonding and to demonstrate the intimate connection between the structures of compounds and their chemical reactivity. You learned, for example, that even though NaOH and alcohols (ROH) both have OH in their formula, NaOH is an ionic compound that dissociates completely in water to produce a basic solution containing Na+ and OH− ions, whereas alcohols are covalent compounds that do not dissociate in water and instead form neutral aqueous solutions. You also learned that an amine (RNH2), with its lone pairs of electrons, is a base, whereas a carboxylic acid (RCO2H), with its dissociable proton, is an acid. (For more information on acids and bases, see , .)
The structure of a solid with a hybrid metal-organic framework. Organic and inorganic groups of the proper structure can be used to synthesize solids with very large pores (central sphere) that can accommodate a variety of small molecules. The rigid benzene rings are used as “props” to hold the metal units (carboxylate-bridged copper dimers) apart. Such solids have potential applications in hydrogen storage for use in fuel cells or automobiles.
Carbon is unique among the elements in its ability to catenate, to form a wide variety of compounds that contain long chains and/or rings of carbon atoms. (For more information on carbon, see , , and , .) Some of the most complex chemical structures known are those of the organic molecules found in living organisms. (For more information on biopolymers, see , .) In spite of their size and complexity, these biological molecules obey the same chemical principles as simpler organic molecules. Thus we can use Lewis electron structures to understand the preferred mode of reactivity of a variety of organic compounds, relative electronegativities and bond polarities to predict how certain groups of atoms will react, and molecular orbital theory to explain why certain organic species that contain multiple bonds are especially stable or undergo particular reactions when they interact with light. (For more information on Lewis electron structures, see , . For more information on bonding, see , . For more information on light interactions, see , .) In this chapter, we continue our description of organic compounds by focusing on their molecular structures and reactivity; we will also introduce some of the fundamental types of reactions and reaction mechanisms you will encounter in organic and biological chemistry. We discuss why butter is a solid and oils are liquids despite the apparent similarities in their structures, why the widely used anti-inflammatory drug ibuprofen takes longer than half an hour to relieve pain, and the identity of the major carcinogen in grilled meats and cigarette smoke. The chapter concludes with a brief introduction to the molecules of life, which will explain how the consumption of lactose can result in mental retardation and cirrhosis of the liver in some individuals, how hibernating animals survive during the winter, and how certain groups of antibiotics kill bacteria that are harmful to humans.
In Chapter 2 "Molecules, Ions, and Chemical Formulas" and Chapter 5 "Energy Changes in Chemical Reactions", you were introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groupsThe structural units that chemists use to classify organic compounds and predict their reactivities under a given set of conditions., which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO2H). The major families of organic compounds are characterized by their functional groups. Figure 24.1 "Major Classes of Organic Compounds" summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group.
Figure 24.1 Major Classes of Organic Compounds
The first family listed in Figure 24.1 "Major Classes of Organic Compounds" is the hydrocarbons. These include alkanes, with the general molecular formula CnH2n+2 where n is an integer; alkenes, represented by CnH2n; alkynes, represented by CnH2n−2; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO2H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO2 group.
As you learned in Chapter 2 "Molecules, Ions, and Chemical Formulas", Section 2.4 "Naming Covalent Compounds", the systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH3)2C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde.
Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH3 group in dimethyl benzene is indicated with a 1, but the second CH3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure 24.2 "Common Nomenclature for Aromatic Ring Substitutions"). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene.
Figure 24.2 Common Nomenclature for Aromatic Ring Substitutions
We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached.
Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom.
Can two substances have the same systematic name and be different compounds?
Is a carbon–carbon multiple bond considered a functional group?
In earlier discussions of organic compounds, we focused on differences in how the functional groups were connected to the carbon framework. Differences in connectivity resulted in different chemical compounds with different names. You learned, for example, that although 1-propanol (n-propanol) and 2-propanol (isopropanol) have the same molecular formula (C3H8O), they have different physical and chemical properties. Just as with metal complexes, compounds that have the same molecular formula but different arrangements of atoms are called isomers. (For more information on metal complexes, see , .) In this section, we describe various types of isomers, beginning with those whose three-dimensional structures differ only as the result of rotation about a C–C bond.
The C–C single bonds in ethane, propane, and other alkanes are formed by the overlap of an sp3 hybrid orbital on one carbon atom with an sp3 hybrid orbital on another carbon atom, forming a σ bond (). Each sp3 hybrid orbital is cylindrically symmetrical (all cross-sections are circles), resulting in a carbon–carbon single bond that is also cylindrically symmetrical about the C–C axis. Because rotation about the carbon–carbon single bond can occur without changing the overlap of the sp3 hybrid orbitals, there is no significant electronic energy barrier to rotation. Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer)Isomers whose three-dimensional structures differ because of rotation about a σ bond..
Conformational isomers differ in their three-dimensional structure due to rotation about a σ bond.
Figure 24.3 Carbon–Carbon Bonding in Alkanes
Overlapping sp3 hybrid orbitals on adjacent carbon atoms form a cylindrically symmetrical σ bond. Because rotation about the bond does not affect the overlap of the bonding orbitals, there is no electronic energy barrier to rotation.
The simplest alkane to have a conformational isomer is ethane. Differences between the conformations of ethane are depicted especially clearly in drawings called Newman projections, such as those shown in part (a) in . In a Newman projection, the ethane molecule is viewed along the C–C axis, with the carbon that is in front shown as a vertex and the carbon that is in back shown as a circle. The three hydrogen atoms nearest the viewer are shown bonded to the front carbon, and the three hydrogen atoms farthest from the viewer are shown bonded to the circle. In one extreme, called the eclipsed conformation, the C–H bonds on adjacent carbon atoms lie in the same plane. In the other extreme, called the staggered conformation, the hydrogen atoms are positioned as far from one another as possible. Rotation about the C–C bond produces an infinite number of conformations between these two extremes, but the staggered conformation is the most stable because it minimizes electrostatic repulsion between the hydrogen atoms on adjacent carbons.
Figure 24.4 Eclipsed and Staggered Conformations of Ethane
(a) In a Newman projection, the molecule is viewed along a C–C axis. The carbon in front is represented as a vertex, whereas the carbon that is bonded to it is represented as a circle. In ethane, the C–H bonds to each carbon are positioned at 120° from each other. In the fully eclipsed conformation, the C–H bonds on adjacent carbon atoms are parallel and lie in the same plane. In the staggered conformation, the hydrogen atoms are positioned as far apart as possible. (b) The eclipsed conformation is 12.6 kJ/mol higher in energy than the staggered conformation because of electrostatic repulsion between the hydrogen atoms. An infinite number of conformations of intermediate energy exist between the two extremes.
In a Newman projection, the angles between adjacent C–H bonds on the same carbon are drawn at 120°, although H–C–H angles in alkanes are actually tetrahedral angles of 109.5°, which for chains of more than three carbon atoms results in a kinked structure. (For more information on bond angles and molecular modeling, see , .) Despite this three-dimensional inaccuracy, Newman projections are useful for predicting the relative stability of conformational isomers. As shown in part (b) in , the higher energy of the eclipsed conformation represents an energy barrier of 12.6 kJ/mol that must be overcome for rotation about the C–C bond to occur. This barrier is so low, however, that rotation about the C–C bond in ethane is very fast at room temperature and occurs several million times per second for each molecule.
Longer-chain alkanes can also be represented by Newman projections. In more complex alkanes and alkane derivatives, rotation can occur about each C–C bond in a molecule. Newman projections are therefore useful for revealing steric barriers to rotation at a particular C–C bond due to the presence of bulky substituents. shows a plot of potential energy versus the angle of rotation about the central C–C bond (between carbon atoms 2 and 3) of n-butane (C4H10). The structure that minimizes electrostatic repulsion is the one in which the methyl groups, corresponding to carbon atoms 1 and 4, are as far apart as possible; that is, the staggered conformation. Notice that because the substituents on C2 and C3 in n-butane are not all the same, energetically nonequivalent eclipsed and staggered conformations are possible; most molecules interconvert rapidly between these conformations by a series of simple rotations.
Figure 24.5 Potential Energy Plot and Newman Projections of Eclipsed and Staggered Conformations of n-Butane
In these projections, the molecule is viewed along the C2–C3 axis. The least stable structure is the eclipsed conformation in which the two methyl groups (C1 and C4) are adjacent to each other. The most stable structure is the staggered conformation in which the methyl groups are as far apart as possible. Because the substituents on each central carbon atom are not all the same, a 120° rotation about the C2–C3 bond generates energetically nonequivalent eclipsed and staggered conformations.
Draw Newman projections showing the staggered and eclipsed conformations of 1,1,1-trichloroethane (CCl3CH3).
Given: organic molecule
Asked for: staggered and eclipsed conformations
Strategy:
A Identify the C–C bond of interest. Then draw the Newman projection by representing one carbon as a vertex and the other as a circle.
B Draw bonds to each carbon at 120° angles from one another, with one arrangement representing the staggered conformation and the other the eclipsed conformation.
C Complete the Newman projections by attaching the appropriate atoms or substituent groups to the central C atoms in each conformation.
Solution:
A There is only one C–C bond: C1 is connected to three Cl atoms and C2 to three H atoms. We draw C1 as a point and C2 as a circle.
B Now we draw bonds on each carbon at 120° angles from one another to represent the staggered conformation and the eclipsed conformation.
C We then attach the H and Cl atoms to the carbon atoms in each conformation as shown.
Exercise
Draw Newman projections to illustrate the staggered and eclipsed conformations of propane (C3H8) as viewed along the C1–C2 axis.
Answer:
Unlike conformational isomers, which do not differ in connectivity, structural isomersIsomers that have the same molecular formula but differ in which atoms are bonded to one another. differ in connectivity, as illustrated here for 1-propanol and 2-propanol. (For more information on structural isomers, see , .) Although these two alcohols have the same molecular formula (C3H8O), the position of the –OH group differs, which leads to differences in their physical and chemical properties.
In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule. Consider, for example, the following five structures represented by the formula C5H12:
Of these structures, (a) and (d) represent the same compound, as do (b) and (c). No bonds have been broken and reformed; the molecules are simply rotated about a 180° vertical axis. Only three—n-pentane (a) and (d), 2-methylbutane (b) and (c), and 2,2-dimethylpropane (e)—are structural isomers. Because no bonds are broken in going from (a) to (d) or from (b) to (c), these alternative representations are not structural isomers. The three structural isomers—either (a) or (d), either (b) or (c), and (e)—have distinct physical and chemical properties.
Structural isomers differ in their connectivity.
Draw all the structural isomers of C6H14.
Given: organic molecule
Asked for: all structural isomers
Strategy:
A Draw the simplest structural isomer, which is often the straight-chain alkane.
B Obtain branched isomers by substituting one hydrogen along the chain with an appropriate group from the chain.
C If possible, substitute more than one hydrogen with appropriate groups to obtain isomers that are more highly branched.
Solution:
A The simplest structural isomer is the straight-chain alkane n-hexane (CH3CH2CH2CH2CH2CH3).
B Removing a methyl group from one end and reattaching it to adjacent carbons while substituting hydrogen in its place give two other structures:
C To obtain yet another structural isomer, move two methyl groups to create a molecule with two branches:
We create one more structural isomer by attaching two methyl groups to the same carbon atom:
Thus there are four structural isomers of C6H14.
Exercise
Draw all the structural isomers of C4H9Cl.
Answer:
Molecules with the same connectivity but different arrangements of the atoms in space are called stereoisomersMolecules that have the same connectivity but whose component atoms have different orientations in space.. There are two types of stereoisomers: geometric and optical. Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. (For more information on stereoisomers, see , .) Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement (part (a) in ). Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds.
Stereoisomers have the same connectivity but different arrangements of atoms in space.
Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands, feet, and ears. As shown in part (a) in , your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks in part (b) in .
Figure 24.6 Chiral and Achiral Objects
(a) Objects that are nonsuperimposable mirror images of each other are chiral, such as the left and the right hand. (b) The unmarked flask is achiral because it can be superimposed on its mirror image.
Most chiral organic molecules have at least one carbon atom that is bonded to four different groups, as occurs in the bromochlorofluoromethane molecule shown in part (a) in . This carbon, often designated by an asterisk in structural drawings, is called a chiral center or asymmetric carbon atom. If the bromine atom is replaced by another chlorine (part (b) in ), the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Asymmetric carbon atoms are found in many naturally occurring molecules, such as lactic acid, which is present in milk and muscles, and nicotine, a component of tobacco. A molecule and its nonsuperimposable mirror image are called enantiomers (from the Greek enantiou, meaning “opposite”).
Figure 24.7 Comparison of Chiral and Achiral Molecules
(a) Bromochlorofluoromethane is a chiral molecule whose stereocenter is designated with an asterisk. Rotation of its mirror image does not generate the original structure. To superimpose the mirror images, bonds must be broken and reformed. (b) In contrast, dichlorofluoromethane and its mirror image can be rotated so they are superimposable.
Draw the cis and trans isomers of each compound.
Given: organic compounds
Asked for: cis and trans isomers
Strategy:
Draw the unsubstituted compound corresponding to the systematic name given. Then place substituents on the same side to obtain the cis isomer and on opposite sides to obtain the trans isomer.
Solution:
The name tells us that this compound contains a five-carbon ring with two methyl groups attached. The 1,3 notation means that the methyl groups are not adjacent in the five-membered ring:
Placing the methyl substituents on the same side of the ring gives the cis isomer, whereas placing them on opposite sides of the ring gives the trans isomer:
The compound 3-hexene can exist as a cis or trans isomer:
Replacing the hydrogen atoms on the third and fourth carbons by chlorine does not change the overall structures of the isomers:
Exercise
Draw the cis and trans isomers of each compound.
Answer:
Which of these compounds exist as at least one pair of enantiomers?
Given: organic compounds
Asked for: existence of enantiomers
Strategy:
Determine whether the compound is chiral. In most cases, this means that at least one carbon is bonded to four different groups. If the compound is chiral, it exists as enantiomers.
Solution:
Exercise
Which of these compounds have at least one pair of enantiomers?
Answer:
(c)
Although enantiomers have identical densities, melting and boiling points, colors, and solubility in most solvents, they differ in their interaction with plane-polarized light, which consists of electromagnetic waves oscillating in a single plane. In contrast, normal (unpolarized) light consists of electromagnetic waves oscillating in all directions perpendicular to the axis of propagation. When normal light is passed through a substance called a polarizer, only light oscillating in one direction is transmitted. A polarizer selectively filters out light that oscillates in any but the desired plane ().
Figure 24.8 Detecting the Optical Activity of Chiral Substances
When polarized light is passed through a solution that contains an achiral compound, there is no net rotation of the plane of polarization of the light. In contrast, when polarized light is passed through a solution that contains one enantiomer of a chiral compound, as shown here, the light is rotated either clockwise [dextrorotatory, (+) enantiomer] or counterclockwise [levorotatory, (−) enantiomer] by an angle that depends on the molecular structure and concentration of the compound, the path length, and the wavelength of the light.
When plane-polarized light is passed through a solution, electromagnetic radiation interacts with the solute and solvent molecules. If the solution contains an achiral compound, the plane-polarized light enters and leaves the solution unchanged because achiral molecules cause it to rotate in random directions. The solute is therefore said to be optically inactive. If the solution contains a single enantiomer of a chiral compound, however, the plane-polarized light is rotated in only one direction, and the solute is said to be optically active. A clockwise rotation is called dextrorotatory (from the Latin dextro, meaning “to the right”) and is indicated in the name of the compound by (+), whereas a counterclockwise rotation is called levorotatory (from the Latin levo, meaning “to the left”) and is designated (−). As you will soon discover, this designation is important in understanding how chiral molecules interact with one another.
Chiral molecules are optically active; achiral molecules are not.
The magnitude of the rotation of plane-polarized light is directly proportional to the number of chiral molecules in a solution; it also depends on their molecular structure, the temperature, and the wavelength of the light. Because of these variables, every chiral compound has a specific rotationThe amount (in degrees) by which the plane of polarized light is rotated when the light is passed through a solution that contains 1.0 g of a solute per 1.0 mL of solvent in a tube 10.0 cm long., which is defined as the amount (in degrees) by which the plane of polarized light is rotated when the light is passed through a solution containing 1.0 g of solute per 1.0 mL of solvent in a tube 10.0 cm long. A chiral solution that contains equal concentrations of a pair of enantiomers is called a racemic mixture. In such a solution, the optical rotations exactly cancel one another, so there is no net rotation, and the solution is optically inactive. The categories of stereoisomers are summarized in .
Figure 24.9 Classification of Stereoisomers
In both types of stereoisomer—geometric and optical—isomeric molecules have identical connectivity, but the arrangement of atoms in space differs. Cis and trans isomers exhibit different physical and chemical properties, whereas enantiomers differ only in their interaction with plane-polarized light and reactions in asymmetric environments. Depending on the direction in which they rotate polarized light, enantiomers are identified as (+) or (−). The designations L- and D- represent an alternative labeling system.
In living organisms, virtually every molecule that contains a chiral center is found as a single enantiomer, not a racemic mixture. At the molecular level, our bodies are chiral and interact differently with the individual enantiomers of a particular compound. For example, the two enantiomers of carvone produce very different responses in humans: (−)-carvone is the substance responsible for the smell of spearment oil, and (+)-carvone—the major flavor component of caraway seeds—is responsible for the characteristic aroma of rye bread.
A pharmaceutical example of a chiral compound is ibuprofen, a common analgesic and anti-inflammatory agent that is the active ingredient in pain relievers such as Motrin and Advil (). The drug is sold as a racemic mixture that takes approximately 38 minutes to achieve its full effect in relieving pain and swelling in an adult human. Because only the (+) enantiomer is active in humans, however, the same mass of medication would relieve symptoms in only about 12 minutes if it consisted of only the (+) enantiomer. Unfortunately, isolating only the (+) enantiomer would substantially increase the cost of the drug. Conversion of the (−) to (+) enantiomer in the human body accounts for the delay in feeling the full effects of the drug. A racemic mixture of another drug, the sedative thalidomide, was sold in Europe from 1956 to the early 1960s. It was prescribed to treat nausea during pregnancy, but unfortunately only the (+) enantiomer was safe for that purpose. The (−) enantiomer was discovered to be a relatively potent teratogen, a substance that causes birth defects, which caused the children of many women who had taken thalidomide to be born with missing or undeveloped limbs. As a result, thalidomide was quickly banned for this use. It is currently used to treat leprosy, however, and it has also shown promise as a treatment for AIDS (acquired immunodeficiency syndrome).
These examples dramatically illustrate the point that the biological activities of enantiomers may be very different. But how can two molecules that differ only by being nonsuperimposable mirror images cause such different responses? The biological effects of many substances—including molecules such as carvone that have a scent and drugs such as ibuprofen and thalidomide—depend on their interaction with chiral sites on specific receptor proteins. As schematically illustrated in , only one enantiomer of a chiral substance interacts with a particular receptor, thereby initiating a response. The other enantiomer may not bind at all, or it may bind to another receptor, producing a different response.
Figure 24.10 The Interaction of Chiral Molecules with Biological Receptors
Only one enantiomer of a chiral molecule fits into a chiral receptor site, which typically is a small portion of a large protein. The binding of a molecule to its receptor elicits a characteristic response. The other enantiomer cannot fit into the same site and thus elicits no response. It may, however, produce a different response by binding to another site.
Isomers are different compounds that have the same molecular formula. For an organic compound, rotation about a σ bond can produce different three-dimensional structures called conformational isomers (or conformers). In a Newman projection, which represents the view along a C–C axis, the eclipsed conformation has the C–H bonds on adjacent carbon atoms parallel to each other and in the same plane, representing one conformational extreme. In the staggered conformation, the opposite extreme, the hydrogen atoms are as far from one another as possible. Electrostatic repulsions are minimized in the staggered conformation. Structural isomers differ in the connectivity of the atoms. Structures that have the same connectivity but whose components differ in their orientations in space are called stereoisomers. Stereoisomers can be geometric isomers, which differ in the placement of substituents in a rigid molecule, or optical isomers, nonsuperimposable mirror images. Molecules that are nonsuperimposable mirror images are chiral molecules. A molecule and its nonsuperimposable mirror image are called enantiomers. These differ in their interaction with plane-polarized light, light that oscillates in only one direction. A compound is optically active if its solution rotates plane-polarized light in only one direction and optically inactive if its rotations cancel to produce no net rotation. A clockwise rotation is called dextrorotatory and is indicated in the compound’s name by (+), whereas a counterclockwise rotation is called levorotatory, designated by (−). The specific rotation is the amount (in degrees) by which the plane of polarized light is rotated when light is passed through a solution containing 1.0 g of solute per 1.0 mL of solvent in a tube 10.0 cm long. A solution that contains equal concentrations of each enantiomer in a pair is a racemic mixture; such solutions are optically inactive.
What hybrid orbitals are used to form C–C bonds in saturated hydrocarbons? Describe the bond.
How are conformational isomers related? Sketch two conformational isomers of propane, looking along the C1–C2 axis.
Why do alkanes with more than two carbons have a kinked structure? Explain why a kinked structure is so stable.
Are n-pentane and 2-methylbutane conformational isomers or structural isomers? How would you separate these compounds from a mixture of the two?
How are structural isomers different from stereoisomers? Do stereoisomers have free rotation about all carbon–carbon bonds? Explain your answers.
Which of these objects is chiral?
Which of these objects is chiral?
Are all stereoisomers also enantiomers? Are all enantiomers stereoisomers? Explain your answers.
sp3; it is a σ bond that is cylindrically symmetrical (all cross sections perpendicular to the internuclear axis are circles).
The sp3 hybridized orbitals form bonds at tetrahedral angles (109.5°), which forces the carbon atoms to form a zigzag chain.
(a), (c), and (d)
Single bonds between carbon atoms are free to rotate 360°.
Draw Newman projections of the n-hexane conformations corresponding to the energy minima and maxima in the diagram, which shows potential energy versus degrees of rotation about the C3–C4 axis.
Sketch all the structural isomers of each compound.
Draw all the possible structural isomers of each compound.
Sketch all the isomers of each compound. Identify the cis- and trans-isomers.
Which molecules are chiral? On the structural formulas of the chiral molecules, identify any chiral centers with an asterisk.
Which molecules are chiral? On the structural formulas of the chiral molecules, identify any chiral centers with an asterisk.
Draw the structures of the enantiomers of each compound.
Draw the structures of the enantiomers of each compound.
Draw the structures of the enantiomers of each compound.
Draw the structures of the enantiomers of each compound.
(b) and (d);
Understanding why organic molecules react as they do requires knowing something about the structure and properties of the transient species that are generated during chemical reactions. Identifying transient intermediates enables chemists to elucidate reaction mechanisms, which often allows them to control the products of a reaction. In designing the synthesis of a molecule, such as a new drug, for example, chemists must be able to understand the mechanisms of intermediate reactions to maximize the yield of the desired product and minimize the occurrence of unwanted reactions. Moreover, by recognizing the common reaction mechanisms of simple organic molecules, we can understand how more complex systems react, including the much larger molecules encountered in biochemistry.
Nearly all chemical reactions, whether organic or inorganic, proceed because atoms or groups of atoms having a positive charge or a partial positive charge interact with atoms or groups of atoms having a negative charge or a partial negative charge. Thus when a bond in a hydrocarbon is cleaved during a reaction, identifying the transient species formed, some of which are charged, allows chemists to determine the mechanism and predict the products of a reaction.
Chemists often find that the reactivity of a molecule is affected by the degree of substitution of a carbon that is bonded to a functional group. These carbons are designated as primary, secondary, or tertiary. A primary carbon is bonded to only one other carbon and a functional group, a secondary carbon is bonded to two other carbons and a functional group, and a tertiary carbon is bonded to three other carbons and a functional group.
Cleaving a C–H bond can generate either –C+ and H−, −C· and H· or −C− and H+, all of which are unstable and therefore highly reactive. The most common species formed is –C+, which is called a carbocationA highly reactive species that can form when a C–H bond is cleaved, carbocations have only six valence electrons and are electrophiles. (part (a) in ). A carbocation has only six valence electrons and is therefore electron deficient. It is an electrophileAn electron-deficient species that needs electrons to complete its octet. (from “electron” and the Greek suffix phile, meaning “loving”), which is a species that needs electrons to complete its octet. (Recall that electron-deficient compounds, such as those of the group 13 elements, act as Lewis acids in inorganic reactions.) In general, when a highly electronegative atom, such as Cl, is bonded to a carbocation, it draws electrons away from the carbon and destabilizes the positive charge. In contrast, alkyl groups and other species stabilize the positive charge by increasing electron density at the carbocation. Thus a tertiary carbocation (R3C+) is more stable than a primary carbocation (RCH2+).
The reactivity of a molecule is often affected by the degree of substitution of the carbon bonded to a functional group.
Adding one electron to a carbocation produces a neutral species called a radicalHighly reactive species that have an unpaired valence electron.. (For more information on radicals, see , .) An example is the methyl radical (·CH3), shown in part (b) in . Because the carbon still has less than an octet of electrons, it is electron deficient and also behaves as an electrophile. Like carbocations, radicals can be stabilized by carbon substituents that can donate some electron density to the electron-deficient carbon center. Like carbocations, a tertiary radical (R3C·) is more stable than a primary radical (RCH2·).
Figure 24.11 Transient Intermediates in Organic Reactions
(a) The simplest carbocation is the methyl cation (CH3+), which has six valence electrons and is an electrophile. Its structure is trigonal planar, with an sp2 hybridized carbon and a vacant p orbital. (b) The methyl radical (·CH3) is a radical that, like the carbocation, is trigonal planar and an electrophile. It is also sp2 hybridized, but there is a single electron in the unhybridized p orbital. (c) The simplest organic carbanion is CH3−, which has a trigonal pyramidal structure with an sp3 hybridized carbon that has a lone pair of electrons. Because it has a strong tendency to share its lone pair with another atom or molecule, a carbanion is a nucleophile.
Adding an electron to a radical produces a carbanionA highly reactive species that can form when a C–H bond is cleaved, carbanions have eight valence electrons and are nucleophiles., which contains a negatively charged carbon with eight valence electrons (part (c) in ). The methyl anion (CH3−) has a structure that is similar to NH3 with its lone pair of electrons, but it has a much stronger tendency to share its lone pair with another atom or molecule. A carbanion is a nucleophileAn electron-rich species that has a pair of electrons available to be shared with another atom. (from “nucleus” and phile), an electron-rich species that has a pair of electrons available to share with another atom. Carbanions are destabilized by groups that donate electrons, so the relationship between their structure and reactivity is exactly the opposite of carbocations and radicals. That is, a tertiary carbanion (R3C−) is less stable than a primary carbanion (RCH2−). Carbanions are most commonly encountered in organometallic compounds such as methyllithium (CH3Li) or methylmagnesium chloride (CH3MgCl), where the more electropositive metal ion stabilizes the negative charge on the more electronegative carbon atom.
Electrophiles such as carbocations seek to gain electrons and thus have a strong tendency to react with nucleophiles, which are negatively charged species or substances with lone pairs of electrons. Reacting electrophiles with nucleophiles is a central theme in organic reactions.
Electrophiles react with nucleophiles.
Classify each species as an electrophile, a nucleophile, or neither.
Given: molecular formulas
Asked for: mode of reactivity
Strategy:
Determine whether the compound is electron deficient, in which case it is an electrophile; electron rich, in which case it is a nucleophile; or neither.
Solution:
Exercise
Classify each compound as an electrophile, a nucleophile, or neither.
Answer:
The reactivity of a molecule is often affected by the degree of substitution of the carbon bonded to a functional group; the carbon is designated as primary, secondary, or tertiary. Identifying the transient species formed in a chemical reaction, some of which are charged, enables chemists to predict the mechanism and products of the reaction. One common transient species is a carbocation, a carbon with six valence electrons that is an electrophile; that is, it needs electrons to complete its octet. A radical is a transient species that is neutral but electron deficient and thus acts as an electrophile. In contrast, a carbanion has eight valence electrons and is negatively charged. It is an electron-rich species that is a nucleophile because it can share a pair of electrons. In chemical reactions, electrophiles react with nucleophiles.
Arrange CH2F+, CHCl2+, CH3+, and CHF2+ in order of increasing stability. Explain your reasoning.
Arrange CH3CH2+, CHBr2+, CH3+, and CHBrCl+ in order of decreasing stability. Explain your reasoning.
Identify the electrophile and the nucleophile in each pair.
Identify the electrophile and the nucleophile in each pair.
CHF2+ < CHCl2+ < CH2F+ < CH3+; electronegative substituents destabilize the positive charge. The greater the number of electronegative substituents and the higher their electronegativity, the more unstable the carbocation.
Draw Lewis electron structures of the products of carbon–hydrogen cleavage reactions. What is the charge on each species?
Identify the electrophile and the nucleophile in each reaction; then complete each chemical equation.
Certain patterns are encountered repeatedly in organic reactions, many reflecting the interactions of nucleophiles and electrophiles. In this section, we discuss five common types of organic reactions: substitution reactions, elimination reactions, addition reactions, radical reactions, and oxidation–reduction reactions. You have encountered many of these types of reactions previously, such as the formation of peptides by the elimination of water, the oxidation–reduction reactions that generate voltage in batteries, and chain reactions that involve organic radicals. (For more information on peptide formation, see Chapter 12 "Solids", Section 12.8 "Polymeric Solids". For more information on batteries, see Chapter 19 "Electrochemistry". For more information on radicals, see Chapter 14 "Chemical Kinetics", Section 14.6 "Reaction Rates—A Microscopic View".) In this section, we expand our discussion to include some of the mechanisms behind these reactions.
In a substitution reactionA chemical reaction in which one atom or a group of atoms in a substance is replaced by another atom or a group of atoms from another substance., one atom or a group of atoms in a substance is replaced by another atom or group of atoms from another substance. A typical substitution reaction is reacting the hydroxide ion with methyl chloride:
Equation 24.1
CH3Cl + OH− → CH3OH + Cl−Methyl chloride has a polar C–Cl bond, with the carbon atom having a partial positive charge. In Equation 24.1, the electronegative Cl atom is replaced by another electronegative species that is a stronger nucleophile, in this case OH−. Reactions of this sort are called nucleophilic substitution reactions. For this type of reaction to occur, the nucleophilic reactant must possess a pair of electrons and have a greater affinity for the electropositive carbon atom than the original substituent.
One type of nucleophilic substitution reaction is shown in Equation 24.1. It proceeds by a mechanism in which the lone pair of electrons on the entering nucleophile (OH−) attacks the partially positively charged carbon atom of the polar C–Cl bond, causing the C–Cl bond to weaken and break:
Figure 24.12
In nucleophilic substitution reactions, the nucleophile must possess a pair of electrons and have a greater affinity for the electropositive species than the original substituent.
The convention for writing such a mechanism is to draw arrows showing the direction of electron flow—that is, from the electron-rich center (the nucleophile) to the electron-poor center (the electrophile). The intermediate species, enclosed by square brackets, represents a transient arrangement of atoms that is only postulated to exist. If the atom under attack (in this case, the partially positively charged carbon atom) had –CH3 groups bonded to it rather than H atoms, the bulky methyl groups would interfere with the attack by OH−, making the reaction sterically hindered. The reaction would then proceed in two discrete steps in a second type of substitution reaction: the C–Cl bond would break, forming the (CH3)3C+ carbocation (the electrophile), which would then react with hydroxide (the nucleophile) in a separate step to give the product, (CH3)3COH.
Mustard gas.
An example of a nucleophilic substitution reaction involves the chemical warfare agent known as mustard gas [(ClCH2CH2)2S], which caused about 400,000 casualties during World War I. Mustard gas is toxic because it contains a chloride that can be displaced by nucleophilic amino groups in proteins, thereby allowing the molecule to irreversibly bond to a protein. Because the other product of the reaction is HCl, mustard gas causes severe burns to mucous membranes in the respiratory tract. If mustard gas reacts with DNA (deoxyribonucleic acid), cross-linking of the DNA strands through sulfur occurs, which results in coding errors, the inhibition of replication, and disruption of other DNA functions. If mustard gas reacts with RNA (ribonucleic acid), protein synthesis is altered (see Section 24.6 "The Molecules of Life").
Some reactions involve the removal, or “elimination,” of adjacent atoms from a molecule. This results in the formation of a multiple bond and the release of a small molecule, so they are called elimination reactionsA chemical reaction in which adjacent atoms are removed, or “eliminated,” from a molecule, resulting in the formation of a multiple bond and a small molecule.. They have the general form
Figure 24.13
and are similar to cleavage reactions in inorganic compounds. (For more information on cleavage reactions, see Chapter 3 "Chemical Reactions", Section 3.5 "Classifying Chemical Reactions".) A typical example is the conversion of ethyl chloride to ethylene:
Equation 24.2
CH3CH2Cl → CH2=CH2 + HClElimination reactions are similar to cleavage reactions in inorganic compounds.
Much of the approximately 26 million tons of ethylene produced per year in the United States is used to synthesize plastics, such as polyethylene. In Equation 24.2, the A–B molecule eliminated is HCl, whose components are eliminated as H+ from the carbon atom on the left and Cl− from the carbon on the right. When an acid is produced, as occurs here, the reaction is generally carried out in the presence of a base (such as NaOH) to neutralize the acid.
A reaction in which the components of a species A–B are added to adjacent atoms across a carbon–carbon multiple bond is called an addition reactionA chemical reaction in which the components of a species A–B are added to adjacent atoms across a carbon-carbon multiple bond.. An example is the reverse of the reaction shown in Equation 24.2, reacting HCl with ethylene to give ethyl chloride:
Equation 24.3
HCl + CH2=CH2 → CH3CH2ClAn addition reaction is the reverse of an elimination reaction.
Although a multiple bond is stronger than a single bond, the π bonds of the multiple bond are weaker than the σ bond. The high electron density located between multiply bonded carbon atoms, however, causes alkenes and alkynes to behave like nucleophiles, where nucleophilic attack occurs from the more weakly bound π electrons. Hence alkenes and alkynes are regarded as functional groups. Nucleophilic attack occurs on the Hδ+ atom of the polar HCl bond, initially producing a species with a carbon that has only three bonds, a carbocation. In a second nucleophilic attack, Cl−, the electrophile in Equation 24.3, attacks the carbocation:
Alcohols, an important class of organic compounds, are often produced by addition reactions. Initial attack by the π bond of an alkene on a Hδ+ of H3O+ produces a carbocation. The carbocation then undergoes nucleophilic attack by a lone pair of electrons from H2O followed by elimination of H+ to form the alcohol.
Many important organic reactions involve radicals, such as the combustion of fuels. Probably the best known is reacting a saturated hydrocarbon, such as ethane, with a halogen, such as Br2. The overall reaction is as follows:
Equation 24.4
Radical chain reactions occur in three stages: initiation, propagation, and termination. (For more information on radicals, see Chapter 14 "Chemical Kinetics", Section 14.6 "Reaction Rates—A Microscopic View".) At high temperature or in the presence of light, the relatively weak Br–Br bond is broken in an initiation step that produces an appreciable number of Br atoms (Br·). During propagation, a bromine atom attacks ethane, producing a radical, which then reacts with another bromine molecule to produce ethyl bromide:
Equation 24.5
The sum of the two propagation steps corresponds to the balanced chemical equation for the overall reaction. There are three possible termination steps: the combination of (1) two bromine atoms, (2) two ethyl radicals, or (3) an ethyl and a bromine radical:
Equation 24.6
Because radicals are powerful nucleophiles and hence highly reactive, such reactions are not very selective. For example, the chlorination of n-butane gives a roughly 70:30 mixture of 2-chlorobutane, formed from the more stable radical by reacting a secondary carbon and 1-chlorobutane.
Because radicals are highly reactive, radical reactions are usually not very selective.
Oxidation–reduction reactions, which are common in organic chemistry, can often be identified by changes in the number of oxygen atoms at a particular position in the hydrocarbon skeleton or in the number of bonds between carbon and oxygen at that position. An increase in either corresponds to an oxidation, whereas a decrease corresponds to a reduction. Conversely, an increase in the number of hydrogen atoms in a hydrocarbon is often an indication of a reduction. We can illustrate these points by considering how the oxidation state of the carbon atom changes in the series of compounds, which is shown in part (a) in Figure 24.14 "The Oxidation State of Carbon in Oxygen- and Nitrogen-Containing Functional Groups". (For a review of oxidation states and formal changes, see Chapter 3 "Chemical Reactions", Section 3.5 "Classifying Chemical Reactions", and Chapter 8 "Ionic versus Covalent Bonding", Section 8.5 "Lewis Structures and Covalent Bonding"). The number of oxygen atoms or the number of bonds to oxygen changes throughout the series. Hence the conversion of methane to formic acid is an oxidation, whereas the conversion of carbon dioxide to methanol is a reduction. Also, the number of hydrogen atoms increases in going from the most oxidized to least oxidized compound. As expected, as the oxidation state of carbon increases, the carbon becomes a more potent electrophile. Thus the carbon of CO2 is a stronger electrophile (i.e., more susceptible to nucleophilic attack) than the carbon of an alkane such as methane.
Figure 24.14 The Oxidation State of Carbon in Oxygen- and Nitrogen-Containing Functional Groups
(a) In a hydrocarbon, oxidation is indicated by an increase in the number of oxygen atoms or carbon–oxygen bonds or a decrease in the number of hydrogen atoms. (b) In nitrogen-containing compounds, the number of carbon–nitrogen bonds changes with the oxidation state of carbon.
Similarly, in compounds with a carbon–nitrogen bond, the number of bonds between the C and N atoms increases as the oxidation state of the carbon increases (part (b) in Figure 24.14 "The Oxidation State of Carbon in Oxygen- and Nitrogen-Containing Functional Groups"). In a nitrile, which contains the –C≡N group, the carbon has the same oxidation state (+2) as in a carboxylic acid, characterized by the –CO2H group. We therefore expect the carbon of a nitrile to be a rather strong electrophile.
Write an equation to describe each reaction. Identify the electrophile and the nucleophile in each reaction.
Given: reactants, products, and reaction mechanism
Asked for: equation and identification of electrophile and nucleophile
Strategy:
Use the mechanisms described to show how the indicated products are formed from the reactants.
Solution:
The CN− ion of KCN is a potent nucleophile that can displace the chlorine atom of 1-chloropropane, releasing a chloride ion. Substitution results in the formation of a new C–C bond:
The carbon bonded to chlorine is an electrophile because of the highly polar C–Cl bond.
In the electrophilic addition of a hydrogen halide to an alkene, the reaction is as follows:
The first step is nucleophilic attack of the π electrons of the double bond on the electrophilic hydrogen of the polar H–Br bond to generate the transient carbocation, followed by nucleophilic attack by the halide to give the product. Thus the alkene is the nucleophile, and the proton of the acid is the electrophile.
Exercise
Write an equation to describe each reaction. In each reaction, identify the electrophile and nucleophile.
Answer:
Cyclopentene is the nucleophile, and H3O+ is the electrophile.
There are common patterns to how organic reactions occur. In a substitution reaction, one atom or a group of atoms in a substance is replaced by another atom or a group of atoms from another substance. Bulky groups that prevent attack cause the reaction to be sterically hindered. In an elimination reaction, adjacent atoms are removed with subsequent formation of a multiple bond and a small molecule. An addition reaction is the reverse of an elimination reaction. Radical reactions are not very selective and occur in three stages: initiation, propagation, and termination. Oxidation–reduction reactions in organic chemistry are identified by the change in the number of oxygens in the hydrocarbon skeleton or the number of bonds between carbon and oxygen or carbon and nitrogen.
Identify the nucleophile and the electrophile in the nucleophilic substitution reaction of 2-bromobutane with KCN.
Identify the nucleophile and the electrophile in the nucleophilic substitution reaction of 1-chloropentane with sodium methoxide.
Do you expect an elimination reaction to be favored by a strong or a weak base? Why?
Why do molecules with π bonds behave as nucleophiles when mixed with strong electrophiles?
CN− is the nucleophile, and C2H5Cδ+HBrCH3 is the electrophile.
Sketch the mechanism for the nucleophilic substitution reaction of potassium cyanide with iodoethane.
Sketch the mechanism for the nucleophilic substitution reaction of NaSH with 1-bromopropane.
Sketch the mechanism for the elimination reaction of cyclohexylchloride with potassium ethoxide. Identify the electrophile and the nucleophile in this reaction.
What is the product of the elimination reaction of 1-bromo-2-methylpropane with sodium ethoxide?
Write the structure of the product expected from the electrophilic addition of HBr to cis-3-hexene.
Write the structure of the product expected from the electrophilic addition of 1-methylcyclopentene to HBr. Identify the electrophile and the nucleophile, and then write a mechanism for this reaction.
Write a synthetic scheme for making propene from propane. After synthesizing propene, how would you make 2-bromopropane?
Write a synthetic scheme for making ethylene from ethane. After synthesizing ethylene, how would you make iodoethane?
From the high-temperature reaction of Br2 with 3-methylpentane, how many monobrominated isomers would you expect to be produced? Which isomer is produced from the most stable radical?
For the photochemical reaction of Cl2 with 2,4-dimethylpentane, how many different monochlorinated isomers would you expect to be produced? Which isomer is produced from the most stable precursor radical?
How many different radicals can be formed from the photochemical reaction of Cl2 with 3,3,4-trimethylhexane?
How many monobrominated isomers would you expect from the photochemical reaction of Br2 with
Arrange acetone, ethane, carbon dioxide, acetaldehyde, and ethanol in order of increasing oxidation state of carbon.
What product(s) do you expect from the reduction of a ketone? the oxidation of an aldehyde?
What product(s) do you expect from the reduction of formaldehyde? the oxidation of ethanol?
four; 3-bromo-3-methylpentane
seven
methanol; acetaldehyde, followed by acetic acid and finally CO2
The general properties and reactivity of each class of organic compounds () is largely determined by its functional groups. In this section, we describe the relationships between structure, physical properties, and reactivity for the major classes of organic compounds. We also show you how to apply these relationships to understand some common reactions that chemists use to synthesize organic compounds.
The boiling points of alkanes increase smoothly with increasing molecular mass. They are similar to those of the corresponding alkenes and alkynes because of similarities in molecular mass between analogous structures (). In contrast, the melting points of alkanes, alkenes, and alkynes with similar molecular masses show a much wider variation because the melting point strongly depends on how the molecules stack in the solid state. It is therefore sensitive to relatively small differences in structure, such as the location of a double bond and whether the molecule is cis or trans.
Table 24.1 Boiling Points (in °C) of Alkanes, Alkenes, and Alkynes of Comparable Molecular Mass
Length of Carbon Chain | |||
---|---|---|---|
Class | Two C Atoms | Three C Atoms | Four C Atoms |
alkane | −88.6 | −42.1 | −0.5 |
alkene | −103.8 | −47.7 | −6.3 |
alkyne | −84.7 | −23.2 | 8.1 |
Because alkanes contain only C–C and C–H bonds, which are strong and not very polar (the electronegativities of C and H are similar; ), they are not easily attacked by nucleophiles or electrophiles. Consequently, their reactivity is limited, and often their reactions occur only under extreme conditions. For example, catalytic cracking can be used to convert straight-chain alkanes to highly branched alkanes, which are better fuels for internal combustion engines. Catalytic cracking is one example of a pyrolysis reactionA high-temperature decomposition reaction that can be used to form fibers of synthetic polymers. (from the Greek pyros, meaning “fire,” and lysis, meaning “loosening”), in which alkanes are heated to a sufficiently high temperature to induce cleavage of the weakest bonds: the C–C single bonds. The result is a mixture of radicals derived from essentially random cleavage of the various C–C bonds in the chain. Pyrolysis of n-pentane, for example, is nonspecific and can produce these four radicals:
Equation 24.7
Recombination of these radicals (a termination step) can produce ethane, propane, butane, n-pentane, n-hexane, n-heptane, and n-octane. Radicals that are formed in the middle of a chain by cleaving a C–H bond tend to produce branched hydrocarbons. In catalytic cracking, lighter alkanes are removed from the mixture by distillation.
Radicals are also produced during the combustion of alkanes, with CO2 and H2O as the final products. As discussed in , radicals are stabilized by the presence of multiple carbon substituents that can donate electron density to the electron-deficient carbon. The chemical explanation of octane ratings, as described in , , rests partly on the stability of radicals produced from the different hydrocarbon fuels. Recall that n-heptane, which does not burn smoothly, has an octane rating of 0, and 2,2,4-trimethylpentane (“isooctane”), which burns quite smoothly, has a rating of 100 (). Isooctane has a branched structure and is capable of forming tertiary radicals that are comparatively stable.
In contrast, the radicals formed during the combustion of n-heptane, whether primary or secondary, are less stable and hence more reactive, which partly explains why burning n-heptane causes premature ignition and engine knocking.
In , we explained that rotation about the carbon–carbon multiple bonds of alkenes and alkynes cannot occur without breaking a π bond, which therefore constitutes a large energy barrier to rotation (). Consequently, the cis and trans isomers of alkenes generally behave as distinct compounds with different chemical and physical properties. A four-carbon alkene has four possible isomeric forms: three structural isomers, which differ in their connectivity, plus a pair of geometric isomers from one structural isomer (2-butene). These two geometric isomers are cis-2-butene and trans-2-butene. The four isomers have significantly different physical properties.
Figure 24.15 Carbon–Carbon Bonding in Alkenes and Interconversion of Cis and Trans Isomers
In butane, there is only a small energy barrier to rotation about the C2–C3 σ bond. In the formation of cis- or trans-2-butene from butane, the p orbitals on C2 and C3 overlap to form a π bond. To convert cis-2-butene to trans-2-butene or vice versa through rotation about the double bond, the π bond must be broken. Because this interconversion is energetically unfavorable, cis and trans isomers are distinct compounds that generally have different physical and chemical properties.
Alkynes in which the triple bond is located at one end of a carbon chain are called terminal alkynes and contain a hydrogen atom attached directly to a triply bonded carbon: R–C≡C–H. Terminal alkynes are unusual in that the hydrogen atom can be removed relatively easily as H+, forming an acetylide ion (R–C≡C−). Acetylide ions are potent nucleophiles that are especially useful reactants for making longer carbon chains by a nucleophilic substitution reaction. As in earlier examples of such reactions, the nucleophile attacks the partially positively charged atom in a polar bond, which in the following reaction is the carbon of the Br–C bond:
Alkenes and alkynes are most often prepared by elimination reactions (). A typical example is the preparation of 2-methyl-1-propene, whose derivative, 3-chloro-2-methyl-1-propene, is used as a fumigant and insecticide. The parent compound can be prepared from either 2-hydroxy-2-methylpropane or 2-bromo-2-methylpropane:
The reaction on the left proceeds by eliminating the elements of water (H+ plus OH−), so it is a dehydration reactionA reaction that proceeds by eliminating the elements of water . If an alkane contains two properly located functional groups, such as –OH or –X, both of them may be removed as H2O or HX with the formation of a carbon–carbon triple bond:
Alkenes and alkynes are most often prepared by elimination reactions.
Most arenes that contain a single six-membered ring are volatile liquids, such as benzene and the xylenes, although some arenes with substituents on the ring are solids at room temperature. In the gas phase, the dipole moment of benzene is zero, but the presence of electronegative or electropositive substituents can result in a net dipole moment that increases intermolecular attractive forces and raises the melting and boiling points. For example, 1,4-dichlorobenzene, a compound used as an alternative to naphthalene in the production of mothballs, has a melting point of 52.7°C, which is considerably greater than the melting point of benzene (5.5°C). (For more information on 1,4-dichlorobenzene, see , .)
Certain aromatic hydrocarbons, such as benzene and benz[a]pyrene, are potent liver toxins and carcinogens. In 1775, a British physician, Percival Pott, described the high incidence of cancer of the scrotum among small boys used as chimney sweeps and attributed it to their exposure to soot. His conclusions were correct: benz[a]pyrene, a component of chimney soot, charcoal-grilled meats, and cigarette smoke, was the first chemical carcinogen to be identified.
Although arenes are usually drawn with three C=C bonds, benzene is about 150 kJ/mol more stable than would be expected if it contained three double bonds. This increased stability is due to the delocalization of the π electron density over all the atoms of the ring. (For more information on delocalization, see , .) Compared with alkenes, arenes are poor nucleophiles. Consequently, they do not undergo addition reactions like alkenes; instead, they undergo a variety of electrophilic aromatic substitution reactionsA reaction in which a −H of an arene is replaced (substituted) by an electrophilic group in a two-step process. that involve the replacement of –H on the arene by a group –E, such as –NO2, –SO3H, a halogen, or an alkyl group, in a two-step process. The first step involves addition of the electrophile (E) to the π system of benzene, forming a carbocation. In the second step, a proton is lost from the adjacent carbon on the ring:
The carbocation formed in the first step is stabilized by resonance.
Arenes undergo substitution reactions rather than elimination because of increased stability arising from delocalization of their π electron density.
Many substituted arenes have potent biological activity. Some examples include common drugs and antibiotics such as aspirin and ibuprofen, illicit drugs such as amphetamines and peyote, the amino acid phenylalanine, and hormones such as adrenaline ().
Figure 24.16 Biologically Active Substituted Arenes
Aspirin (antifever activity), ibuprofen (antifever and anti-inflammatory activity), and amphetamine (stimulant) have pharmacological effects. Phenylalanine is an amino acid. Adrenaline is a hormone that elicits the “fight or flight” response to stress. Chiral centers are indicated with an asterisk.
Both alcohols and ethers can be thought of as derivatives of water in which at least one hydrogen atom has been replaced by an organic group, as shown here. Because of the electronegative oxygen atom, the individual O–H bond dipoles in alcohols cannot cancel one another, resulting in a substantial dipole moment that allows alcohols to form hydrogen bonds. Alcohols therefore have significantly higher boiling points than alkanes or alkenes of comparable molecular mass, whereas ethers, without a polar O–H bond, have intermediate boiling points due to the presence of a small dipole moment (). The larger the alkyl group in the molecule, however, the more “alkane-like” the alcohol is in its properties. Because of their polar nature, alcohols and ethers tend to be good solvents for a wide range of organic compounds.
Table 24.2 Boiling Points of Alkanes, Ethers, and Alcohols of Comparable Molecular Mass
Name | Formula | Molecular Mass (amu) | Boiling Point (°C) | |
---|---|---|---|---|
alkane | propane | C3H8 | 44 | −42.1 |
n-pentane | C5H12 | 72 | 36.1 | |
n-heptane | C7H16 | 100 | 98.4 | |
ether | dimethylether | (CH3)2O | 46 | −24.8 |
diethylether | (CH3CH2)2O | 74 | 34.5 | |
di-n-propylether | (CH3CH2CH2)2O | 102 | 90.1 | |
alcohol | ethanol | CH3CH2OH | 46 | 78.3 |
n-butanol | CH3(CH2)3OH | 74 | 117.7 | |
n-hexanol | CH3(CH2)5OH | 102 | 157.6 |
Alcohols are usually prepared by adding water across a carbon–carbon double bond or by a nucleophilic substitution reaction of an alkyl halide using hydroxide, a potent nucleophile (). As you will see in , alcohols can also be prepared by reducing compounds that contain the carbonyl functional group (C=O; part (a) in ). Alcohols are classified as primary, secondary, or tertiary, depending on whether the –OH group is bonded to a primary, secondary, or tertiary carbon. For example, the compound 5-methyl-3-hexanol is a secondary alcohol.
Ethers, especially those with two different alkyl groups (ROR′), can be prepared by a substitution reaction in which a nucleophilic alkoxide ion (RO−) attacks the partially positively charged carbon atom of the polar C–X bond of an alkyl halide (R′X):
Although both alcohols and phenols have an –OH functional group, phenols are 106–108 more acidic than alcohols. This is largely because simple alcohols have the –OH unit attached to an sp3 hybridized carbon, whereas phenols have an sp2 hybridized carbon atom bonded to the oxygen atom. The negative charge of the phenoxide ion can therefore interact with the π electrons in the ring, thereby delocalizing and stabilizing the negative charge through resonance. (For more information on resonance, see , .) In contrast, the negative charge on an alkoxide ion cannot be stabilized by these types of interactions.
Alcohols undergo two major types of reactions: those involving cleavage of the O–H bond and those involving cleavage of the C–O bond. Cleavage of an O–H bond is a reaction characteristic of an acid, but alcohols are even weaker acids than water. The acidic strength of phenols, however, is about a million times greater than that of ethanol, making the pKa of phenol comparable to that of the NH4+ ion (9.89 versus 9.25, respectively):
Equation 24.8
Alcohols undergo two major types of reactions: cleavage of the O–H bond and cleavage of the C–O bond.
Cleavage of the C–O bond in alcohols occurs under acidic conditions. The –OH is first protonated, and nucleophilic substitution follows:
In the absence of a nucleophile, however, elimination can occur, producing an alkene ().
Ethers lack the –OH unit that is central to the reactivity of alcohols, so they are comparatively unreactive. Their low reactivity makes them highly suitable as solvents for carrying out organic reactions.
Aromatic aldehydes, which have intense and characteristic flavors and aromas, are the major components of such well-known flavorings as vanilla and cinnamon (). Many ketones, such as camphor and jasmine, also have intense aromas. Ketones are found in many of the hormones responsible for sex differentiation in humans, such as progesterone and testosterone. (For more information on aldehydes and ketones, see , .)
Figure 24.17 Some Familiar Aldehydes and Their Uses
In compounds containing a carbonyl group, nucleophilic attack can occur at the carbon atom of the carbonyl, whereas electrophilic attack occurs at oxygen.
Aldehydes and ketones contain the carbonyl functional group, which has an appreciable dipole moment because of the polar C=O bond. The presence of the carbonyl group results in strong intermolecular interactions that cause aldehydes and ketones to have higher boiling points than alkanes or alkenes of comparable molecular mass (). As the mass of the molecule increases, the carbonyl group becomes less important to the overall properties of the compound, and the boiling points approach those of the corresponding alkanes.
Table 24.3 Boiling Points of Alkanes, Aldehydes, and Ketones of Comparable Molecular Mass
Name | Formula | Molecular Mass (amu) | Boiling Point (°C) | |
---|---|---|---|---|
alkane | n-butane | C4H10 | 58 | −0.5 |
n-pentane | C5H12 | 72 | 36.1 | |
aldehyde | propionaldehyde (propanal) | C3H6O | 58 | 48.0 |
butyraldehyde (butanal) | C4H8O | 72 | 74.8 | |
ketone | acetone (2-propanone) | C3H6O | 58 | 56.1 |
methyl ethyl ketone (2-butanone) | C4H8O | 72 | 79.6 |
Aldehydes and ketones are typically prepared by oxidizing alcohols (part (a) in ). In their reactions, the partially positively charged carbon atom of the carbonyl group is an electrophile that is subject to nucleophilic attack. Conversely, the lone pairs of electrons on the oxygen atom of the carbonyl group allow electrophilic attack to occur. Aldehydes and ketones can therefore undergo both nucleophilic attack (at the carbon atom) and electrophilic attack (at the oxygen atom).
Nucleophilic attack occurs at the partially positively charged carbon of a carbonyl functional group. Electrophilic attack occurs at the lone pairs of electrons on the oxygen atom.
Aldehydes and ketones react with many organometallic compounds that contain stabilized carbanions. One of the most important classes of such compounds are the Grignard reagentsAn organometallic compound that has stabilized carbanions, whose general formula is RMgX, where X is Cl, Br, or I., organomagnesium compounds with the formula RMgX (X is Cl, Br, or I) that are so strongly polarized that they can be viewed as containing R− and MgX+. These reagents are named for the French chemist Victor Grignard (1871–1935), who won a Nobel Prize in Chemistry in 1912 for their development. In a Grignard reaction, the carbonyl functional group is converted to an alcohol, and the carbon chain of the carbonyl compound is lengthened by the addition of the R group from the Grignard reagent. One example is reacting cyclohexylmagnesium chloride, a Grignard reagent, with formaldehyde:
The nucleophilic carbanion of the cyclohexyl ring attacks the electrophilic carbon atom of the carbonyl group. Acidifying the solution results in protonation of the intermediate to give the alcohol. Aldehydes can also be prepared by reducing a carboxylic acid group (–CO2H) (part (a) in ), and ketones can be prepared by reacting a carboxylic acid derivative with a Grignard reagent. The former reaction requires a powerful reducing agent, such as a metal hydride.
Explain how each reaction proceeds to form the indicated product.
Given: chemical reaction
Asked for: how products are formed
Strategy:
A Identify the functional group and classify the reaction.
B Use the mechanisms described to propose the initial steps in the reaction.
Solution:
A One reactant is an alcohol that undergoes a substitution reaction.
B In the product, a bromide group is substituted for a hydroxyl group. The first step in this reaction must therefore be protonation of the –OH group of the alcohol by H+ of HBr, followed by the elimination of water to give the carbocation:
The bromide ion is a good nucleophile that can react with the carbocation to give an alkyl bromide:
A One reactant is a Grignard reagent, and the other contains a carbonyl functional group. Carbonyl compounds act as electrophiles, undergoing nucleophilic attack at the carbonyl carbon.
B The nucleophile is the phenyl carbanion of the Grignard reagent:
The product is benzyl alcohol.
Exercise
Predict the product of each reaction.
Answer:
The pungent odors of many carboxylic acids are responsible for the smells we associate with sources as diverse as Swiss cheese, rancid butter, manure, goats, and sour milk. The boiling points of carboxylic acids tend to be somewhat higher than would be expected from their molecular masses because of strong hydrogen-bonding interactions between molecules. In fact, most simple carboxylic acids form dimers in the liquid and even in the vapor phase. (For more information on the vapor phase, see , .) The four lightest carboxylic acids are completely miscible with water, but as the alkyl chain lengthens, they become more “alkane-like,” so their solubility in water decreases.
Compounds that contain the carboxyl functional group are acidic because carboxylic acids can easily lose a proton: the negative charge in the carboxylate ion (RCO2−) is stabilized by delocalization of the π electrons:
As a result, carboxylic acids are about 1010 times more acidic than the corresponding simple alcohols whose anions (RO−) are not stabilized through resonance.
Carboxylic acids are typically prepared by oxidizing the corresponding alcohols and aldehydes (part (a) in ). They can also be prepared by reacting a Grignard reagent with CO2, followed by acidification:
Equation 24.9
The initial step in the reaction is nucleophilic attack by the R− group of the Grignard reagent on the electrophilic carbon of CO2:
Delocalization of π bonding over three atoms (O–C–O) makes carboxylic acids and their derivatives less susceptible to nucleophilic attack than aldehydes and ketones with their single π bond. The reactions of carboxylic acids are dominated by two factors: their polar –CO2H group and their acidity. Reaction with strong bases, for example, produce carboxylate salts, such as sodium stearate:
Equation 24.10
RCO2H + NaOH → RCO2−Na+ + H2Owhere R is CH3(CH2)16. As you learned in , , long-chain carboxylate salts are used as soaps.
Delocalization of π bonding over three atoms makes carboxylic acids and their derivatives less susceptible to nucleophilic attack as compared with aldehydes and ketones.
Replacing the –OH of a carboxylic acid with groups that have different tendencies to participate in resonance with the C=O functional group produces derivatives with rather different properties. Resonance structures have significant effects on the reactivity of carboxylic acid derivatives, but their influence varies substantially, being least important for halides and most important for the nitrogen of amides. In this section, we take a brief look at the chemistry of two of the most familiar and important carboxylic acid derivatives: esters and amides.
Esters have the general formula RCO2R′, where R and R′ can be virtually any alkyl or aryl group. Esters are often prepared by reacting an alcohol (R′OH) with a carboxylic acid (RCO2H) in the presence of a catalytic amount of strong acid. (For more information on esters and catalysts, see , .) The purpose of the acid (an electrophile) is to protonate the doubly bonded oxygen atom of the carboxylic acid (a nucleophile) to give a species that is more electrophilic than the parent carboxylic acid.
The nucleophilic oxygen atom of the alcohol attacks the electrophilic carbon atom of the protonated carboxylic acid to form a new C–O bond. The overall reaction can be written as follows:
Figure 24.18
Because water is eliminated, this is a dehydration reaction. If an aqueous solution of an ester and strong acid or base is heated, the reverse reaction will occur, producing the parent alcohol R′OH and either the carboxylic acid RCO2H (under strongly acidic conditions) or the carboxylate anion RCO2− (under basic conditions).
As stated earlier, esters are familiar to most of us as fragrances, such as banana and pineapple. Other esters with intense aromas function as sex attractants, or pheromones, such as the pheromone from the oriental fruit fly. Research on using synthetic insect pheromones as a safer alternative to insecticides for controlling insect populations, such as cockroaches, is a rapidly growing field in organic chemistry.
In the general structure of an amide,
the two substituents on the amide nitrogen can be hydrogen atoms, alkyl groups, aryl groups, or any combination of those species. Although amides appear to be derived from an acid and an amine, in practice they usually cannot be prepared by this synthetic route. In principle, nucleophilic attack by the lone electron pair of the amine on the carbon of the carboxylic acid could occur, but because carboxylic acids are weak acids and amines are weak bases, an acid–base reaction generally occurs instead:
Equation 24.11
RCO2H + R′NH2 → RCO2− + R′NH3+Amides are therefore usually prepared by the nucleophilic reaction of amines with more electrophilic carboxylic acid derivatives, such as esters.
The lone pair of electrons on the nitrogen atom of an amide can participate in π bonding with the carbonyl group, thus reducing the reactivity of the amide () and inhibiting free rotation about the C–N bond. Amides are therefore the least reactive of the carboxylic acid derivatives. The stability of the amide bond is crucially important in biology because amide bonds form the backbones of peptides and proteins. (For more information on peptides and proteins, see , .) The amide bond is also found in many other biologically active and commercially important molecules, including penicillin; urea, which is used as fertilizer; saccharin, a sugar substitute; and valium, a potent tranquilizer. (For more information on the structure of penicillin, see , .)
Amides are the least reactive of the carboxylic acid derivatives because amides participate in π bonding with the carbonyl group.
Figure 24.19 The Electronic Structure of an Amide
(a) An unhybridized 2pz orbital on nitrogen, containing a lone electron pair of electrons, can interact with the π orbital of the carbonyl group to give a three-center, four-electron bond. This interaction reduces the reactivity of the amide, making amides the least reactive of the carboxylic acid derivatives. (b) A comparison of the electrostatic potential maps of acetaldehyde and formamide shows that the negative charge (indicated in blue) is more localized on the oxygen atom of acetaldehyde than it is in formamide. Formamide is therefore less reactive.
Amines are derivatives of ammonia in which one or more hydrogen atoms have been replaced by alkyl or aryl groups. They are therefore analogous to alcohols and ethers. Like alcohols, amines are classified as primary, secondary, or tertiary, but in this case the designation refers to the number of alkyl groups bonded to the nitrogen atom, not to the number of adjacent carbon atoms. In primary amines, the nitrogen is bonded to two hydrogen atoms and one alkyl group; in secondary amines, the nitrogen is bonded to one hydrogen and two alkyl groups; and in tertiary amines, the nitrogen is bonded to three alkyl groups. With one lone pair of electrons and C–N bonds that are less polar than C–O bonds, ammonia and simple amines have much lower boiling points than water or alcohols with similar molecular masses. Primary amines tend to have boiling points intermediate between those of the corresponding alcohol and alkane. Moreover, secondary and tertiary amines have lower boiling points than primary amines of comparable molecular mass.
Tertiary amines form cations analogous to the ammonium ion (NH4+), in which all four H atoms are replaced by alkyl groups. Such substances, called quaternary ammonium saltsA salt that consist of an anion and a cation in which all four H atoms of the ammonium ion are replaced by alkyl groups., can be chiral if all four substituents are different. (Amines with three different substituents are also chiral because the lone pair of electrons represents a fourth substituent.)
Alkylamines can be prepared by nucleophilic substitution reactions of alkyl halides with ammonia or other amines:
Equation 24.12
RCl + NH3 → RNH2 + HClEquation 24.13
RCl + R′NH2 → RR′NH + HClEquation 24.14
RCl + R′R″NH → RR′R″N + HClThe primary amine formed in the first reaction () can react with more alkyl halide to generate a secondary amine (), which in turn can react to form a tertiary amine (). Consequently, the actual reaction mixture contains primary, secondary, and tertiary amines and even quaternary ammonium salts.
The reactions of amines are dominated by two properties: their ability to act as weak bases and their tendency to act as nucleophiles, both of which are due to the presence of the lone pair of electrons on the nitrogen atom. Amines typically behave as bases by accepting a proton from an acid to form an ammonium salt, as in the reaction of triethylamine (the ethyl group is represented as Et) with aqueous HCl (the lone pair of electrons on nitrogen is shown):
Equation 24.15
Et3N:(l) + HCl(aq) → Et3NH+Cl−(aq)which gives triethylammonium chloride. Amines can react with virtually any electrophile, including the carbonyl carbon of an aldehyde, a ketone, or an ester. Aryl amines such as aniline (C6H5NH2) are much weaker bases than alkylamines because the lone pair of electrons on nitrogen interacts with the π bonds of the aromatic ring, delocalizing the lone pair through resonance ().
The reactions of amines are dominated by their ability to act as weak bases and their tendency to act as nucleophiles.
Figure 24.20 Structures and Basicity of Aniline and Cyclohexylamine
Delocalization of the lone electron pair on N over the benzene ring reduces the basicity of aryl amines, such as aniline, compared with that of alkylamines, such as cyclohexylamine. These electrostatic potential maps show that the electron density on the N of cyclohexylamine is more localized than it is in aniline, which makes cyclohexylamine a stronger base.
Predict the products formed in each reaction and show the initial site of attack and, for part (b), the final products.
Given: reactants
Asked for: products and mechanism of reaction
Strategy:
Use the strategy outlined in Example 7.
Solution:
Exercise
Predict the products of each reaction. State the initial site of attack.
Answer:
Initial attack occurs with protonation of the oxygen of the carbonyl. The products are:
Initial attack occurs at the carbon of the carbonyl group. The products are:
Reactions like we have discussed in this section and are used to synthesize a wide range of organic compounds. When chemists plan the synthesis of an organic molecule, however, they must take into consideration various factors, such as the availability and cost of reactants, the need to minimize the formation of undesired products, and the proper sequencing of reactions to maximize the yield of the target molecule and minimize the formation of undesired products. Because the synthesis of many organic molecules requires multiple steps, in designing a synthetic scheme for such molecules, chemists must often work backward from the desired product in a process called retrosynthesis. Using this process, they can identify the reaction steps needed to synthesize the desired product from the available reactants.
There are strong connections among the structure, the physical properties, and the reactivity for compounds that contain the major functional groups. Hydrocarbons that are alkanes undergo catalytic cracking, which can convert straight-chain alkanes to highly branched alkanes. Catalytic cracking is one example of a pyrolysis reaction, in which the weakest bond is cleaved at high temperature, producing a mixture of radicals. The multiple bond of an alkene produces geometric isomers (cis and trans). Terminal alkynes contain a hydrogen atom directly attached to a triply bonded carbon. Removal of the hydrogen forms an acetylide ion, a potent nucleophile used to make longer carbon chains. Arenes undergo substitution rather than elimination because of enhanced stability from delocalization of their π electron density. An alcohol is often prepared by adding the elements of water across a double bond or by a substitution reaction. Alcohols undergo two major types of reactions: those involving cleavage of the O–H bond and those involving cleavage of the C–O bond. Phenols are acidic because of π interactions between the oxygen atom and the ring. Ethers are comparatively unreactive. Aldehydes and ketones are generally prepared by oxidizing alcohols. Their chemistry is characterized by nucleophilic attack at the carbon atom of the carbonyl functional group and electrophilic attack at the oxygen atom. Grignard reagents (RMgX, where X is Cl, Br, or I) convert the carbonyl functional group to an alcohol and lengthen the carbon chain. Compounds that contain the carboxyl functional group are weakly acidic because of delocalization of the π electrons, which causes them to easily lose a proton and form the carboxylate anion. Carboxylic acids are generally prepared by oxidizing alcohols and aldehydes or reacting a Grignard reagent with CO2. Carboxylic acid derivatives include esters, prepared by reacting a carboxylic acid and an alcohol, and amides, prepared by the nucleophilic reaction of amines with more electrophilic carboxylic acid derivatives, such as esters. Amides are relatively unreactive because of π bonding interactions between the lone pair on nitrogen and the carbonyl group. Amines can also be primary, secondary, or tertiary, depending on the number of alkyl groups bonded to the amine. Quaternary ammonium salts have four substituents attached to nitrogen and can be chiral. Amines are often prepared by a nucleophilic substitution reaction between a polar alkyl halide and ammonia or other amines. They are nucleophiles, but their base strength depends on their substituents.
Why do branched-chain alkanes have lower melting points than straight-chain alkanes of comparable molecular mass?
Describe alkanes in terms of their orbital hybridization, polarity, and reactivity. What is the geometry about each carbon of a straight-chain alkane?
Why do alkenes form cis and trans isomers, whereas alkanes do not? Do alkynes form cis and trans isomers? Why or why not?
Which compounds can exist as cis and trans isomers?
Which compounds can exist as cis and trans isomers?
Which compounds have a net dipole moment?
Why is the boiling point of an alcohol so much greater than that of an alkane of comparable molecular mass? Why are low-molecular-mass alcohols reasonably good solvents for some ionic compounds, whereas alkanes are not?
Is an alcohol a nucleophile or an electrophile? What determines the mode of reactivity of an alcohol? How does the reactivity of an alcohol differ from that of an ionic compound containing OH, such as KOH?
How does the reactivity of ethers compare with that of alcohols? Why? Ethers can be cleaved under strongly acidic conditions. Explain how this can occur.
What functional group is common to aldehydes, ketones, carboxylic acids, and esters? This functional group can react with both nucleophiles and electrophiles. Where does nucleophilic attack on this functional group occur? Where does electrophilic attack occur?
What key feature of a Grignard reagent allows it to engage in a nucleophilic attack on a carbonyl carbon?
Do you expect carboxylic acids to be more or less water soluble than ketones of comparable molecular mass? Why?
Because amides are formally derived from an acid plus an amine, why can they not be prepared by the reaction of an acid with an amine? How are they generally prepared?
Is an amide susceptible to nucleophilic attack, electrophilic attack, or both? Specify where the attack occurs.
What factors determine the reactivity of amines?
(c) and (d)
The presence of a nucleophilic Cδ− resulting from a highly polar interaction with an electropositive Mg
Their ability to act as weak bases and their tendency to act as nucleophiles
What is the product of the reaction of 2-butyne with excess HBr?
What is the product of the reaction of 3-hexyne with excess HCl?
What elements are eliminated during the dehydrohalogenation of an alkyl halide? What products do you expect from the dehydrohalogenation of 2-chloro-1-pentene?
What elements are eliminated during the dehydration of an alcohol? What products do you expect from the dehydration of ethanol?
Predict the products of each reaction.
Show the mechanism and predict the organic product of each reaction.
A Grignard reagent can be used to generate a carboxylic acid. Show the mechanism for the first step in this reaction using CH3CH2MgBr as the Grignard reagent. What is the geometry about the carbon of the –CH2 of the intermediate species formed in this first step?
Draw a molecular orbital picture showing the bonding in an amide. What orbital is used for the lone pair of electrons on nitrogen?
What is the product of the reaction of
Develop a synthetic scheme to generate
2,2-dibromobutane
All the functional groups described in this chapter are found in the organic molecules that are constantly synthesized and destroyed by every living organism on Earth. A detailed understanding of the reactions that occur in living organisms is the goal of biochemistry, which deals with a wide variety of organic structures and reactions. The most abundant substances found in living systems belong to four major classes: proteins, carbohydrates, lipids, and nucleic acids. Here we briefly describe the structure and some functions of these biological molecules.
In , , we described proteinsA biological polymer with more than 50 amino acid residues linked together by amide bonds. as biologically active polymers formed from amino acids linked together by amide bonds. In addition to an amine group and a carboxylic acid group, each amino acid contains a characteristic R group (). In the simplest amino acid, glycine, the R group is hydrogen (–H), but in other naturally occurring amino acids, the R group may be an alkyl group or a substituted alkyl group, a carboxylic group, or an aryl group. The nature of the R group determines the particular chemical properties of each amino acid. In , all the amino acids found in proteins except glycine are chiral compounds, which suggests that their interactions with other chiral compounds are selective. Some proteins, called enzymes, catalyze biological reactions, whereas many others have structural, contractile, or signaling functions. Because we have described proteins previously, we will not discuss them further.
The general structure of an amino acid. An amino acid is chiral except when R is an H atom.
Carbohydrates are the most abundant of the organic compounds found in nature. They constitute a substantial portion of the food we consume and provide us with the energy needed to support life. Table sugar, milk, honey, and fruits all contain low-molecular-mass carbohydrates that are easily assimilated by the human body. In contrast, the walls of plant cells and wood contain high-molecular-mass carbohydrates that we cannot digest.
Once thought to be hydrates of carbon with the general formula Cn(H2O)m, carbohydratesA polyhydroxy aldehyde or a polyhydroxy ketone with the general formula are actually polyhydroxy aldehydes or polyhydroxy ketones (i.e., aldehydes or ketones with several –OH groups attached to the parent hydrocarbon). The simplest carbohydrates consist of unbranched chains of three to eight carbon atoms: one carbon atom is part of a carbonyl group, and some or all of the others are bonded to hydroxyl groups. The structure of a carbohydrate can be drawn either as a hydrocarbon chain, using a Fischer projection, or as a ring, using a Haworth projection (). The Haworth projection is named after the British chemist Sir Walter Norman Haworth, who was awarded a Nobel Prize in Chemistry in 1937 for his discovery that sugars exist mainly in their cyclic forms, as well as for his collaboration on the synthesis of vitamin C. The cyclic form is the product of nucleophilic attack by the oxygen of a hydroxyl group on the electrophilic carbon of the carbonyl group within the same molecule, producing a stable ring structure composed of five or six carbons that minimizes bond strain (). The substituents on the right side of the carbon chain in a Fischer projection are in the “down” position in the corresponding Haworth projection. Attack by the hydroxyl group on either side of the carbonyl group leads to the formation of two cyclic forms, called anomers: an α form, with the –OH in the “down” position, and a β form, with the –OH in the “up” position.
At age 14, Walter Norman Haworth left school to join his father to learn linoleum design and manufacturing, but he became interested in chemistry through his use of dyes. Private tutoring enabled him to pass the entrance exam of the University of Manchester, where he received his doctorate in 1911. During World War I, Haworth organized the laboratories at St. Andrews for the production of chemicals and drugs, returning to the investigation of carbohydrates after the war.
Figure 24.21 Fischer Projection and Haworth Projection of Glucose
In solution, simple sugars exist predominantly in the ring form, the product of nucleophilic attack by the oxygen of a hydroxyl group on the electrophilic carbon of the carbonyl group. The α and β forms, called anomers, differ in the configuration at C1.
Carbohydrates are classified according to the number of single saccharide, or sugar, units they contain (from the Latin saccharum, meaning “sugar”). The simplest are monosaccharides; a disaccharide consists of two linked monosaccharide units; a trisaccharide has three linked monosaccharide units; and so forth. Glucose is a monosaccharide, and sucrose (common table sugar) is a disaccharide. The hydrolysis of sucrose produces glucose and another monosaccharide, fructose, in a reaction catalyzed by an enzyme or by acid:
Polysaccharides hydrolyze to produce more than 10 monosaccharide units.
The common monosaccharides contain several chiral carbons and exist in several isomeric forms. One isomer of glucose, for example, is galactose, which differs from glucose in the position of the –OH bond at carbon-4:
Because carbons-2, -3, -4, and -5 of glucose are chiral, changing the position of the –OH on carbon-4 does not produce an enantiomer of glucose but a different compound, galactose, with distinct physical and chemical properties. Galactose is a hydrolysis product of lactose, a disaccharide found in milk. People who suffer from galactosemia lack the enzyme needed to convert galactose to glucose, which is then metabolized to CO2 and H2O, releasing energy. Galactose accumulates in their blood and tissues, leading to mental retardation, cataracts, and cirrhosis of the liver.
Because carbohydrates have a carbonyl functional group and several hydroxyl groups, they can undergo a variety of biochemically important reactions. The carbonyl group, for example, can be oxidized to form a carboxylic acid or reduced to form an alcohol. The hydroxyl groups can undergo substitution reactions, resulting in derivatives of the original compound. One such derivative is Sucralose, an artificial sweetener that is six times sweeter than sucrose; it is made by replacing two of the hydroxyl groups on sucrose with chlorine. Carbohydrates can also eliminate hydroxyl groups, producing alkenes.
Because carbohydrates have a carbonyl functional group and several hydroxyl groups, they can undergo a variety of reactions.
Two familiar polysaccharides are starch and cellulose, which both hydrolyze to produce thousands of glucose units. They differ only in the connection between glucose units and the amount of branching in the molecule (). Starches can be coiled or branched and are hydrolyzed by the enzymes in our saliva and pancreatic juices. Animal starch, called glycogen, is stored in the liver and muscles. It consists of branched glucose units linked by bonds that produce a coiled structure. The glucose units in cellulose, in contrast, are linked to give long, unbranched chains. The chains in cellulose stack in parallel rows held together by hydrogen bonds between hydroxyl groups. This arrangement produces a rigid structure that is insoluble in water.
Figure 24.22 The Polysaccharides Starch and Cellulose
Starches (a) and cellulose (b) differ in the connection between glucose units and the amount of branching in the molecule. Starches can be coiled or branched, whereas cellulose, the primary structural material of plants, has long, unbranched chains held together by hydrogen bonds.
Cellulose is the primary structural material of plants and one of the most abundant organic substances on Earth. Because our enzymes are not able to hydrolyze the bonds between the glucose units in cellulose, we are unable to digest it. A recently marketed product containing a high percentage of cellulose was sold as a dietetic substance for rapid weight loss, but those who consumed it experienced severe intestinal discomfort because the cellulose could not be digested. The product was quickly removed from the market.
The Fischer projection of xylose, found in many varieties of apples, is shown. Draw the ring form (Haworth projection) of xylose.
Given: Fischer projection of a sugar
Asked for: cyclic structure
Strategy:
A Identify the nucleophile and the electrophile. Indicate the point of attack, remembering that cyclic structures are most stable when they contain at least five atoms in the ring to prevent bond strain from bond angles that are too small.
B Draw the cyclic form of the structure.
Solution:
A The carbonyl carbon (C1) is a good electrophile, and each oxygen is a good nucleophile. Nucleophilic attack occurs from the –OH group on C4, producing a stable five-membered ring.
B Because of rotation about the bond between C1 and C2, ring formation gives both a and b anomers, with the following structures (H atoms have been omitted for clarity):
Exercise
Draw the cyclic form(s) of galactose, whose Fischer projection is shown in the previous discussion.
Answer:
LipidsA family of compounds that includes fats, waxes, some vitamins, and steroids and characterized by their insolubility in water. (from the Greek lipos, meaning “fat” or “lard”) are characterized by their insolubility in water. They form a family of compounds that includes fats, waxes, some vitamins, and steroids. Fatty acids, the simplest lipids, have a long hydrocarbon chain that ends with a carboxylic acid functional group. In saturated fatty acids, the hydrocarbon chains contain only C–C bonds, so they can stack in a regular array (part (a) in ). In contrast, unsaturated fatty acids have a single double bond in the hydrocarbon chain (monounsaturated) or more than one double bond (polyunsaturated). These double bonds give fatty acid chains a kinked structure, which prevents the molecules from packing tightly (part (b) in ). As a result of reduced van der Waals interactions, the melting point of an unsaturated fatty acid is lower than that of a saturated fatty acid of comparable molecule mass, thus producing an oil rather than a solid. (For more information on van der Waals interactions, see , .) Fish oils and vegetable oils, for example, have a higher concentration of unsaturated fatty acids than does butter.
Figure 24.23 Fatty Acids, the Simplest Class of Lipids
Fatty acids are composed of a long chain that terminates in a carboxylic acid functional group. (a) Molecules of saturated fatty acids, which contain no carbon–carbon double bonds, can stack in a regular array. (b) Molecules of unsaturated fatty acids, which contain one or more cis carbon–carbon double bonds, have kinked structures that cannot pack closely together.
The double bonds of unsaturated fatty acids can by hydrogenated in an addition reaction that produces a saturated fatty acid:
Equation 24.16
They can also be oxidized to produce an aldehyde or carboxylic acid. (For more information on hydrogenation, see , .)
Unsaturated fatty acids are the starting compounds for the biosynthesis of prostaglandins. These hormone-like substances are involved in regulating blood pressure, tissue inflammation, and contracting and relaxing smooth muscles. Drugs such as aspirin and ibuprofen inhibit the production of prostaglandins, thereby reducing inflammation.
Waxes are esters produced by the nucleophilic attack of an alcohol on the carbonyl carbon of a long-chain carboxylic acid (). For example, the wax used in shoe polish and wax paper, which is derived from beeswax, is formed from a straight-chain alcohol with 15 carbon atoms and a fatty acid with 31 carbon atoms. Triacylglycerols are a particularly important type of ester in living systems; they are used by the body to store fats and oils. These compounds are formed from one molecule of glycerol (1,2,3-trihydroxypropane) and three fatty acid molecules. During warmer months of the year, animals that hibernate consume large quantities of plants, seeds, and nuts that have a high fat and oil content. They convert the fatty acids to triacylglycerols and store them. Hydrolysis of stored triacylglycerols during hibernation (the reverse of ) releases alcohols and carboxylic acids that the animal uses to generate energy for maintaining cellular activity, respiration, and heart rate. Derivatives of triacylglycerols with a phosphate group are major components of all cell membranes.
Steroids are lipids whose structure is composed of three cyclohexane rings and one cyclopentane ring fused together. The presence of various substituents, including double bonds, on the basic steroid ring structure produces a large family of steroid compounds with different biological activities. For example, cholesterol, a steroid found in cellular membranes, contains a double bond in one ring and four substituents: a hydroxyl group, two methyl groups, and a hydrocarbon chain.
Cholesterol is the starting point for the biosynthesis of steroid hormones, including testosterone, the primary male sex hormone, and progesterone, which helps maintain pregnancy. These cholesterol derivatives lack the long hydrocarbon side chain, and most contain one or more ketone groups.
Cholesterol is synthesized in the human body in a multistep pathway that begins with a derivative of acetic acid. We also consume cholesterol in our diets: eggs, meats, fish, and diary products all contain cholesterol, but vegetables and other plant-derived foods do not contain cholesterol. Excess cholesterol in the human body can cause gallstones, which are composed of nearly 100% cholesterol, or lipid deposits called plaque in arteries. A buildup of plaque can block a coronary artery and result in a heart attack ().
Figure 24.24 Plaque in an Artery
Plaque, a lipid deposit, forms from excess cholesterol in the body. This artery is nearly blocked by a thick deposit of plaque, which greatly increases the risk of a heart attack due to reduced blood flow to the heart.
Nucleic acidsA linear polymer of nucleotides that is the basic structural component of DNA and RNA. are the basic structural components of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), the biochemical substances found in the nuclei of all cells that transmit the information needed to direct cellular growth and reproduction. Their structures are derived from cyclic nitrogen-containing compounds called pyrimidines and purines, which can engage in hydrogen bonding through the lone electron pair on nitrogen (in pyrimidine and purine) or through the hydrogen of the amine (in purine):
The same cyclic structures are found in substances such as caffeine, a purine that is a stimulant, and the antifungal agent flucytosine, a pyrimidine. (For more information on the structure of caffeine, see , .)
When a pyrimidine or a purine is linked to a sugar by a bond called a glycosidic bond, a nucleoside is formed. Adding a phosphoric acid group to the sugar then produces a nucleotide (part (a) in ). The linkage of nucleotides forms a polymeric chain that consists of alternating sugar and phosphate groups, which is the backbone of DNA and RNA (part (b) in ).
While the function of DNA is to preserve genetic information, RNA translates the genetic information in DNA and carries that information to cellular sites where proteins are synthesized. Many antibiotics function by interfering with the synthesis of proteins in one or more kinds of bacteria. Chloramphenicol, for example, is used against infections of the eye or outer ear canal; it inhibits the formation of peptide bonds between amino acids in a protein chain. Puromycin, which is used against herpes simplex type I, interrupts extension of a peptide chain, causing the release of an incomplete protein and the subsequent death of the virus.
Figure 24.25 The Formation of Nucleic Acids
(a) When pyrimidine or purine and a sugar react to form a glycosidic bond, a nucleoside is produced. Adding a phosphoric acid group to the sugar of a nucleoside produces a nucleotide. (b) Nucleotides link together to form long polymeric chains. A DNA molecule consists of two such chains held together by hydrogen bonding between the purine and pyrimidine components on different chains.
Mutations in the DNA of an organism may lead to the synthesis of defective proteins. Phenylketonuria (PKU), for example, is a condition caused by a defective enzyme. Left untreated, it produces severe brain damage and mental retardation. Albinism is caused by a defective enzyme that is unable to produce melanin, the pigment responsible for the color of skin and hair. Cystic fibrosis, the most common inherited disease in the United States, blocks pancreatic function and causes thick mucus secretions that make breathing difficult. An area of intense research in combating cancer involves the synthesis of drugs that stop uncontrolled cell growth by interfering with DNA replication.
The four major classes of organic compounds found in biology are proteins, carbohydrates, lipids, and nucleic acids. Their structures and reactivity are determined by the functional groups present.
Proteins are biologically active polymers formed from amino acids linked together by amide bonds. All the amino acids in proteins are chiral compounds except glycine. The most common organic compounds found in nature are the carbohydrates, polyhydroxy aldehydes or polyhydroxy ketones in unbranched chains of three to eight carbons. They are classified according to the number of sugar, or saccharide, units, and they can be drawn as a chain in a Fischer projection or in a cyclic form called a Haworth projection. The two cyclic forms in a Haworth projection are called anomers. Many sugars contain at least one chiral center. With their carbonyl and hydroxyl functional groups, carbohydrates can undergo a variety of biochemically relevant reactions. Starch and cellulose differ only in the connectivity between glucose units. Starches can be branched or unbranched, but cellulose, the structural material of plants, is unbranched, and cannot be digested by humans. Lipids are insoluble in water. The simplest lipids, fatty acids, have a long hydrocarbon chain ending in a carboxylic acid functional group. Their physical properties depend on the number of double bonds in the chain. Prostaglandins, hormone-like substances, are formed from unsaturated fatty acids, and waxes are long-chain esters of saturated fatty acids. Triacylglycerols, which the body uses to store fats and oils, consist of glycerol esterified to three fatty acid molecules. Steroids, which include cholesterol and the steroid hormones, are characterized by three cyclohexane rings and one cyclopentane ring fused together. The basic structural units of DNA and RNA are the nucleic acids, whose structures are derived from nitrogen-containing cyclic compounds called pyrimidines and purines. These structures are linked to a sugar through a glycosidic bond, forming a nucleoside. Adding a phosphoric acid group produces a nucleotide. Nucleotides link to form a polymeric chain that is the backbone of DNA and RNA.
What are the strengths and limitations of using a Haworth projection? of using a Fischer projection?
Nutritionists will often state that a leafy salad contains no calories. Do you agree?
Would you expect margarine, a polyunsaturated fat, to have a higher or lower melting point than butter, a saturated fat?
Are all the naturally occurring amino acids chiral compounds? Do you expect proteins to contain both enantiomers of alanine and other amino acids? Explain your answer.
The structures of cholesterol and testosterone were shown in this section. Identify the functional groups in each.
The structures of glucose and purine were shown in this section. Identify the functional groups in each.
Use a condensation reaction:
After exercise, the concentration of lactic acid increases in both muscle tissue and blood. In fact, it is responsible for muscle cramps that may develop after vigorous exercise. Using the structure of lactic acid shown here, draw the conformational isomers of lactic acid as viewed along the C2–C3 axis, where C3 is the carbon of the methyl group. Which would you predict to be the most stable conformation? Why?
Cyclohexanecarboxylic acid is a liquid used in insecticide formulations. A derivative of this compound, 2-methylcyclohexanecarboxylic acid is shown here. Sketch the cis and trans isomers of this derivative. Arrange the conformations of its geometric isomers in order of increasing energy.
Coniine is a naturally occurring compound with insect-paralyzing properties. Ingestion of this compound causes weakness, vomiting, labored respiration, and eventual death. Does coniine have a chiral carbon? If so, indicate the carbon with an asterisk.
A compound that has been found to be an effective hypertensive is captopril, sold commercially as Hypertil and Tensoprel, among other names. Captopril has a chiral center at C2. Draw the enantiomers that result from this chiral center. Indicate any other chiral centers in captopril with an asterisk.
The structure of phenobarbital, a compound used to treat epilepsy, is shown here. Identify the functional groups.
The compound 2-amino-2-methyl-1-propanol is used not only in the synthesis of pharmaceuticals but also in cosmetic creams, polishes, and cleaning compounds. Is it a chiral compound?
Glycerol (1,2,3-propanetriol, or 1,2,3-trihydroxypropane) is produced from sugars by fermentation. It also is obtained from oils and fats as a by-product during the manufacture of soaps. Glycerol can be converted to glyceric acid by the following sequence:
Both glyceraldehyde and glyceric acid are derivatives of biochemical intermediates in sugar metabolism.
When exposed to light or heat, peroxides (ROOR) can undergo radical reactions. The first step is the dissociation of the peroxide into two alkoxy radicals. Each alkoxy radical then reacts with a species such as HBr to form an alcohol.
The structure of isophytol is shown here. This compound is used to prepare vitamins E and K. Does isophytol form cis and trans isomers? If so, draw these isomers.
Isopentyl acetate, also known as “oil of banana,” has the structure shown here. Show the first step in the mechanism for reaction of isopentyl acetate with C6H5MgBr, which gives the first ionic adduct.
An enkephalin is a pentapeptide that controls pain in humans. Enkephalins function by binding to the same specific sites in brain cells that are known to bind morphine and heroin. Methionine enkephalin has the structure Tyr-Gly-Gly-Phe-Met.
The first staggered conformation minimizes electrostatic repulsions between adjacent atoms:
yes
three amides and a phenyl group
Substance | ΔHf° (kJ/mol) | ΔGf° (kJ/mol) | S° (J/mol K) |
---|---|---|---|
Aluminum: | |||
Al(s) | 0.0 | 0.0 | 28.3 |
Al(g) | 330.0 | 289.4 | 164.6 |
AlCl3(s) | −704.2 | −628.8 | 109.3 |
Al2O3(s) | −1675.7 | −1582.3 | 50.9 |
Barium: | |||
Ba(s) | 0.0 | 0.0 | 62.5 |
Ba(g) | 180.0 | 146.0 | 170.2 |
BaO(s) | −548.0 | −520.3 | 72.1 |
BaCO3(s) | −1213.0 | −1134.4 | 112.1 |
BaSO4(s) | −1473.2 | −1362.2 | 132.2 |
Beryllium: | |||
Be(s) | 0.0 | 0.0 | 9.5 |
Be(g) | 324.0 | 286.6 | 136.3 |
Be(OH)2(s) | −902.5 | −815.0 | 45.5 |
BeO(s) | −609.4 | −580.1 | 13.8 |
Bismuth: | |||
Bi(s) | 0.0 | 0.0 | 56.7 |
Bi(g) | 207.1 | 168.2 | 187.0 |
Bromine: | |||
Br(g) | 111.9 | 82.4 | 175.0 |
Br2(l) | 0.0 | 0.0 | 152.2 |
Br−(aq) | −121.6 | −104.0 | 82.4 |
Br2(g) | 30.9 | 3.1 | 245.5 |
HBr(g) | −36.3 | −53.4 | 198.7 |
HBr(aq) | −121.6 | −104.0 | 82.4 |
Cadmium: | |||
Cd(s) | 0.0 | 0.0 | 51.8 |
Cd(g) | 111.8 | — | 167.7 |
CdCl2(s) | −391.5 | −343.9 | 115.3 |
CdS(s) | −161.9 | −156.5 | 64.9 |
Calcium: | |||
Ca(s) | 0.0 | 0.0 | 41.6 |
Ca(g) | 177.8 | 144.0 | 154.9 |
CaCl2(s) | −795.4 | −748.8 | 108.4 |
CaF2(s) | −1228.0 | −1175.6 | 68.5 |
Ca(OH)2(s) | −985.2 | −897.5 | 83.4 |
CaO(s) | −634.9 | −603.3 | 38.1 |
CaSO4(s) | −1434.5 | −1322.0 | 106.5 |
CaCO3(s, calcite) | −1207.6 | −1129.1 | 91.7 |
CaCO3(s, aragonite) | −1207.8 | −1128.2 | 88.0 |
Carbon: | |||
C(s, graphite) | 0.0 | 0.0 | 5.7 |
C(s, diamond) | 1.9 | 2.9 | 2.4 |
C(s, fullerene—C60) | 2327.0 | 2302.0 | 426.0 |
C(s, fullerene—C70) | 2555.0 | 2537.0 | 464.0 |
C(g) | 716.7 | 671.3 | 158.1 |
C(g, fullerene—C60) | 2502.0 | 2442.0 | 544.0 |
C(g, fullerene—C70) | 2755.0 | 2692.0 | 614.0 |
CBr4(s) | 29.4 | 47.7 | 212.5 |
CBr4(g) | 83.9 | 67.0 | 358.1 |
CCl2F2(g) | −477.4 | −439.4 | 300.8 |
CCl2O(g) | −219.1 | −204.9 | 283.5 |
CCl4(l) | −128.2 | −62.6 | 216.2 |
CCl4(g) | −95.7 | −53.6 | 309.9 |
CF4(g) | −933.6 | −888.3 | 261.6 |
CHCl3(l) | −134.1 | −73.7 | 201.7 |
CHCl3(g) | −102.7 | 6.0 | 295.7 |
CH2Cl2(l) | −124.2 | — | 177.8 |
CH2Cl2(g) | −95.4 | −68.9 | 270.2 |
CH3Cl(g) | −81.9 | −58.5 | 234.6 |
CH4(g) | −74.6 | −50.5 | 186.3 |
CH3COOH(l) | −484.3 | −389.9 | 159.8 |
CH3OH(l) | −239.2 | −166.6 | 126.8 |
CH3OH(g) | −201.0 | −162.3 | 239.9 |
CH3NH2(l) | −47.3 | 35.7 | 150.2 |
CH3NH2(g) | −22.5 | 32.7 | 242.9 |
CH3CN(l) | 40.6 | 86.5 | 149.6 |
CH3CN(g) | 74.0 | 91.9 | 243.4 |
CO(g) | −110.5 | −137.2 | 197.7 |
CO2(g) | −393.5 | −394.4 | 213.8 |
CS2(l) | 89.0 | 64.6 | 151.3 |
CS2(g) | 116.7 | 67.1 | 237.8 |
C2H2(g) | 227.4 | 209.9 | 200.9 |
C2H4(g) | 52.4 | 68.4 | 219.3 |
C2H6(g) | −84.0 | −32.0 | 229.2 |
C3H8(g) | −103.8 | −23.4 | 270.3 |
C3H6O3(s) (lactic acid) | −694.1 | −522.9 | 142.3 |
C6H6(l) | 49.1 | 124.5 | 173.4 |
C6H6(g) | 82.9 | 129.7 | 269.2 |
C6H12O6(s) (glucose) | −1273.3 | −910.4 | 212.1 |
C2H5OH(l) | −277.6 | −174.8 | 160.7 |
C2H5OH(g) | −234.8 | −167.9 | 281.6 |
(CH3)2O(l) | −203.3 | — | — |
(CH3)2O(g) | −184.1 | −112.6 | 266.4 |
CH3CO2−(aq) | −486.0 | −369.3 | 86.6 |
n-C12H26(l) (dodecane) | −350.9 | 28.1 | 490.6 |
Cesium: | |||
Cs(s) | 0.0 | 0.0 | 85.2 |
Cs(g) | 76.5 | 49.6 | 175.6 |
CsCl(s) | −443.0 | −414.5 | 101.2 |
Chlorine: | |||
Cl(g) | 121.3 | 105.3 | 165.2 |
Cl2(g) | 0.0 | 0.0 | 223.1 |
Cl−(aq) | −167.2 | −131.2 | 56.5 |
HCl(g) | −92.3 | −95.3 | 186.9 |
HCl(aq) | −167.2 | −131.2 | 56.5 |
ClF3(g) | −163.2 | −123.0 | 281.6 |
Chromium: | |||
Cr(s) | 0.0 | 0.0 | 23.8 |
Cr(g) | 396.6 | 351.8 | 174.5 |
CrCl3(s) | −556.5 | −486.1 | 123.0 |
CrO3(g) | −292.9 | — | 266.2 |
Cr2O3(s) | −1139.7 | −1058.1 | 81.2 |
Cobalt: | |||
Co(s) | 0.0 | 0.0 | 30.0 |
Co(g) | 424.7 | 380.3 | 179.5 |
CoCl2(s) | −312.5 | −269.8 | 109.2 |
Copper: | |||
Cu(s) | 0.0 | 0.0 | 33.2 |
Cu(g) | 337.4 | 297.7 | 166.4 |
CuCl(s) | −137.2 | −119.9 | 86.2 |
CuCl2(s) | −220.1 | −175.7 | 108.1 |
CuO(s) | −157.3 | −129.7 | 42.6 |
Cu2O(s) | −168.6 | −146.0 | 93.1 |
CuS(s) | −53.1 | −53.6 | 66.5 |
Cu2S(s) | −79.5 | −86.2 | 120.9 |
CuCN(s) | 96.2 | 111.3 | 84.5 |
Fluorine: | |||
F(g) | 79.4 | 62.3 | 158.8 |
F−(aq) | −332.6 | −278.8 | −13.8 |
F2(g) | 0.0 | 0.0 | 202.8 |
HF(g) | −273.3 | −275.4 | 173.8 |
HF(aq) | −332.6 | −278.8 | −13.8 |
Hydrogen: | |||
H(g) | 218.0 | 203.3 | 114.7 |
H2(g) | 0.0 | 0.0 | 130.7 |
H+(aq) | 0.0 | 0.0 | 0.0 |
Iodine: | |||
I(g) | 106.8 | 70.2 | 180.8 |
I−(aq) | −55.2 | −51.6 | 111.3 |
I2(s) | 0.0 | 0.0 | 116.1 |
I2(g) | 62.4 | 19.3 | 260.7 |
HI(g) | 26.5 | 1.7 | 206.6 |
HI(aq) | −55.2 | −51.6 | 111.3 |
Iron: | |||
Fe(s) | 0.0 | 0.0 | 27.3 |
Fe(g) | 416.3 | 370.7 | 180.5 |
Fe2+(aq) | −89.1 | −78.9 | −137.7 |
Fe3+(aq) | −48.5 | −4.7 | −315.9 |
FeCl2(s) | −341.8 | −302.3 | 118.0 |
FeCl3(s) | −399.5 | −334.0 | 142.3 |
FeO(s) | −272.0 | −251.4 | 60.7 |
Fe2O3(s) | −824.2 | −742.2 | 87.4 |
Fe3O4(s) | −1118.4 | −1015.4 | 146.4 |
FeS2(s) | −178.2 | −166.9 | 52.9 |
FeCO3(s) | −740.6 | −666.7 | 92.9 |
Lead: | |||
Pb(s) | 0.0 | 0.0 | 64.8 |
Pb(g) | 195.2 | 162.2 | 175.4 |
PbO(s, red or litharge) | −219.0 | −188.9 | 66.5 |
PbO(s, yellow or massicot) | −217.3 | −187.9 | 68.7 |
PbO2(s) | −277.4 | −217.3 | 68.6 |
PbCl2(s) | −359.4 | −314.1 | 136.0 |
PbS(s) | −100.4 | −98.7 | 91.2 |
PbSO4(s) | −920.0 | −813.0 | 148.5 |
PbCO3(s) | −699.1 | −625.5 | 131.0 |
Pb(NO3)2(s) | −451.9 | — | — |
Pb(NO3)2(aq) | −416.3 | −246.9 | 303.3 |
Lithium: | |||
Li(s) | 0.0 | 0.0 | 29.1 |
Li(g) | 159.3 | 126.6 | 138.8 |
Li+(aq) | −278.5 | −293.3 | 13.4 |
LiCl(s) | −408.6 | −384.4 | 59.3 |
Li2O(s) | −597.9 | −561.2 | 37.6 |
Magnesium: | |||
Mg(s) | 0.0 | 0.0 | 32.7 |
Mg(g) | 147.1 | 112.5 | 148.6 |
MgCl2(s) | −641.3 | −591.8 | 89.6 |
MgO(s) | −601.6 | −569.3 | 27.0 |
Mg(OH)2(s) | −924.5 | −833.5 | 63.2 |
MgSO4(s) | −1284.9 | −1170.6 | 91.6 |
MgS(s) | −346.0 | −341.8 | 50.3 |
Manganese: | |||
Mn(s) | 0.0 | 0.0 | 32.0 |
Mn(g) | 280.7 | 238.5 | 173.7 |
MnCl2(s) | −481.3 | −440.5 | 118.2 |
MnO(s) | −385.2 | −362.9 | 59.7 |
MnO2(s) | −520.0 | −465.1 | 53.1 |
KMnO4(s) | −837.2 | −737.6 | 171.7 |
MnO4−(aq) | −541.4 | −447.2 | 191.2 |
Mercury: | |||
Hg(l) | 0.0 | 0.0 | 75.9 |
Hg(g) | 61.4 | 31.8 | 175.0 |
HgCl2(s) | −224.3 | −178.6 | 146.0 |
Hg2Cl2(s) | −265.4 | −210.7 | 191.6 |
HgO(s) | −90.8 | −58.5 | 70.3 |
HgS(s, red) | −58.2 | −50.6 | 82.4 |
Hg2(g) | 108.8 | 68.2 | 288.1 |
Molybdenum: | |||
Mo(s) | 0.0 | 0.0 | 28.7 |
Mo(g) | 658.1 | 612.5 | 182.0 |
MoO2(s) | −588.9 | −533.0 | 46.3 |
MoO3(s) | −745.1 | −668.0 | 77.7 |
Nickel: | |||
Ni(s) | 0.0 | 0.0 | 29.9 |
Ni(g) | 429.7 | 384.5 | 182.2 |
NiCl2(s) | −305.3 | −259.0 | 97.7 |
Ni(OH)2(s) | −529.7 | −447.2 | 88.0 |
Nitrogen: | |||
N(g) | 472.7 | 455.5 | 153.3 |
N2(g) | 0.0 | 0.0 | 191.6 |
NH3(g) | −45.9 | −16.4 | 192.8 |
NH4+(aq) | −132.5 | −79.3 | 113.4 |
N2H4(l) | 50.6 | 149.3 | 121.2 |
N2H4(g) | 95.4 | 159.4 | 238.5 |
NH4Cl(s) | −314.4 | −202.9 | 94.6 |
NH4OH(l) | −361.2 | −254.0 | 165.6 |
NH4NO3(s) | −365.6 | −183.9 | 151.1 |
(NH4)2SO4(s) | −1180.9 | −901.7 | 220.1 |
NO(g) | 91.3 | 87.6 | 210.8 |
NO2(g) | 33.2 | 51.3 | 240.1 |
N2O(g) | 81.6 | 103.7 | 220.0 |
N2O4(l) | −19.5 | 97.5 | 209.2 |
N2O4(g) | 11.1 | 99.8 | 304.4 |
HNO2(g) | −79.5 | −46.0 | 254.1 |
HNO3(l) | −174.1 | −80.7 | 155.6 |
HNO3(g) | −133.9 | −73.5 | 266.9 |
HNO3(aq) | −207.4 | −111.3 | 146.4 |
NF3(g) | −132.1 | −90.6 | 260.8 |
HCN(l) | 108.9 | 125.0 | 112.8 |
HCN(g) | 135.1 | 124.7 | 201.8 |
Osmium: | |||
Os(s) | 0.0 | 0.0 | 32.6 |
Os(g) | 791.0 | 745.0 | 192.6 |
OsO4(s) | −394.1 | −304.9 | 143.9 |
OsO4(g) | −337.2 | −292.8 | 293.8 |
Oxygen: | |||
O(g) | 249.2 | 231.7 | 161.1 |
O2(g) | 0.0 | 0.0 | 205.2 |
O3(g) | 142.7 | 163.2 | 238.9 |
OH−(aq) | −230.0 | −157.2 | −10.8 |
H2O(l) | −285.8 | −237.1 | 70.0 |
H2O(g) | −241.8 | −228.6 | 188.8 |
H2O2(l) | −187.8 | −120.4 | 109.6 |
H2O2(g) | −136.3 | −105.6 | 232.7 |
Phosphorus: | |||
P(s, white) | 0.0 | 0.0 | 41.1 |
P(s, red) −17.6 | −17.6 | −12.5 | 22.8 |
P(s, black) | −39.3 | — | — |
P(g, white) | 316.5 | 280.1 | 163.2 |
P2(g) | 144.0 | 103.5 | 218.1 |
P4(g) | 58.9 | 24.4 | 280.0 |
PCl3(l) | −319.7 | −272.3 | 217.1 |
PCl3(g) | −287.0 | −267.8 | 311.8 |
POCl3(l) | −597.1 | −520.8 | 222.5 |
POCl3(g) | −558.5 | −512.9 | 325.5 |
PCl5(g) | −374.9 | −305.0 | 364.6 |
PH3(g) | 5.4 | 13.5 | 210.2 |
H3PO4(s) | −1284.4 | −1124.3 | 110.5 |
H3PO4(l) | −1271.7 | −1123.6 | 150.8 |
Potassium: | |||
K(s) | 0.0 | 0.0 | 64.7 |
K(g) | 89.0 | 60.5 | 160.3 |
KBr(s) | −393.8 | −380.7 | 95.9 |
KCl(s) | −436.5 | −408.5 | 82.6 |
KClO3(s) | −397.7 | −296.3 | 143.1 |
K2O(s) | −361.5 | −322.1 | 94.1 |
K2O2(s) | −494.1 | −425.1 | 102.1 |
KNO2(s) | −369.8 | −306.6 | 152.1 |
KNO3(s) | −494.6 | −394.9 | 133.1 |
KSCN(s) | −200.2 | −178.3 | 124.3 |
K2CO3(s) | −1151.0 | −1063.5 | 155.5 |
K2SO4(s) | −1437.8 | −1321.4 | 175.6 |
Rubidium: | |||
Rb(s) | 0.0 | 0.0 | 76.8 |
Rb(g) | 80.9 | 53.1 | 170.1 |
RbCl(s) | −435.4 | −407.8 | 95.9 |
Selenium: | |||
Se(s, gray) | 0.0 | 0.0 | 42.4 |
Se(g, gray) | 227.1 | 187.0 | 176.7 |
H2Se(g) | 29.7 | 15.9 | 219.0 |
Silicon: | |||
Si(s) | 0.0 | 0.0 | 18.8 |
Si(g) | 450.0 | 405.5 | 168.0 |
SiCl4(l) | −687.0 | −619.8 | 239.7 |
SiCl4(g) | −657.0 | −617.0 | 330.7 |
SiH4(g) | 34.3 | 56.9 | 204.6 |
SiC(s, cubic) | −65.3 | −62.8 | 16.6 |
SiC(s, hexagonal) | −62.8 | −60.2 | 16.5 |
Silver: | |||
Ag(s) | 0.0 | 0.0 | 42.6 |
Ag(g) | 284.9 | 246.0 | 173.0 |
Ag+(aq) | 105.6 | 77.1 | 72.7 |
AgBr(s) | −100.4 | −96.9 | 107.1 |
AgCl(s) | −127.0 | −109.8 | 96.3 |
AgNO3(s) | −124.4 | −33.4 | 140.9 |
Ag2O(s) | −31.1 | −11.2 | 121.3 |
Ag2S(s) | −32.6 | −40.7 | 144.0 |
Sodium: | |||
Na(s) | 0.0 | 0.0 | 51.3 |
Na(g) | 107.5 | 77.0 | 153.7 |
Na+(aq) | −240.1 | −261.9 | 59.0 |
NaF(s) | −576.6 | −546.3 | 51.1 |
NaF(aq) | −572.8 | −540.7 | 45.2 |
NaCl(s) | −411.2 | −384.1 | 72.1 |
NaCl(aq) | −407.3 | −393.1 | 115.5 |
NaBr(s) | −361.1 | −349.0 | 86.8 |
NaBr(g) | −143.1 | −177.1 | 241.2 |
NaBr(aq) | −361.7 | −365.8 | 141.4 |
NaO2(s) | −260.2 | −218.4 | 115.9 |
Na2O(s) | −414.2 | −375.5 | 75.1 |
Na2O2(s) | −510.9 | −447.7 | 95.0 |
NaCN(s) | −87.5 | −76.4 | 115.6 |
NaNO3(aq) | −447.5 | −373.2 | 205.4 |
NaNO3(s) | −467.9 | −367.0 | 116.5 |
NaN3(s) | 21.7 | 93.8 | 96.9 |
Na2CO3(s) | −1130.7 | −1044.4 | 135.0 |
Na2SO4(s) | −1387.1 | −1270.2 | 149.6 |
Sulfur: | |||
S(s, rhombic) | 0.0 | 0.0 | 32.1 |
S(g, rhombic) | 277.2 | 236.7 | 167.8 |
SO2(g) | −296.8 | −300.1 | 248.2 |
SO3(g) | −395.7 | −371.1 | 256.8 |
SO42−(aq) | −909.3 | −744.5 | 20.1 |
SOCl2(g) | −212.5 | −198.3 | 309.8 |
H2S(g) | −20.6 | −33.4 | 205.8 |
H2SO4(aq) | −909.3 | −744.5 | 20.1 |
Tin: | |||
Sn(s, white) | 0.0 | 0.0 | 51.2 |
Sn(s, gray) | −2.1 | 0.1 | 44.1 |
Sn(g, white) | 301.2 | 266.2 | 168.5 |
SnCl4(l) | −511.3 | −440.1 | 258.6 |
SnCl4(g) | −471.5 | −432.2 | 365.8 |
SnO2(s) | −557.6 | −515.8 | 49.0 |
Titanium: | |||
Ti(s) | 0.0 | 0.0 | 30.7 |
Ti(g) | 473.0 | 428.4 | 180.3 |
TiCl2(s) | −513.8 | −464.4 | 87.4 |
TiCl3(s) | −720.9 | −653.5 | 139.7 |
TiCl4(l) | −804.2 | −737.2 | 252.3 |
TiCl4(g) | −763.2 | −726.3 | 353.2 |
TiO2(s) | −944.0 | −888.8 | 50.6 |
Uranium: | |||
U(s) | 0.0 | 0.0 | 50.2 |
U(g) | 533.0 | 488.4 | 199.8 |
UO2(s) | −1085.0 | −1031.8 | 77.0 |
UO2(g) | −465.7 | −471.5 | 274.6 |
UF4(s) | −1914.2 | −1823.3 | 151.7 |
UF4(g) | −1598.7 | −1572.7 | 368.0 |
UF6(s) | −2197.0 | −2068.5 | 227.6 |
UF6(g) | −2147.4 | −2063.7 | 377.9 |
Vanadium: | |||
V(s) | 0.0 | 0.0 | 28.9 |
V(g) | 514.2 | 754.4 | 182.3 |
VCl3(s) | −580.7 | −511.2 | 131.0 |
VCl4(l) | −569.4 | −503.7 | 255.0 |
VCl4(g) | −525.5 | −492.0 | 362.4 |
V2O5(s) | −1550.6 | −1419.5 | 131.0 |
Zinc: | |||
Zn(s) | 0.0 | 0.0 | 41.6 |
Zn(g) | 130.4 | 94.8 | 161.0 |
ZnCl2(s) | −415.1 | −369.4 | 111.5 |
Zn(NO3)2(s) | −483.7 | — | — |
ZnS(s, sphalerite) | −206.0 | −201.3 | 57.7 |
ZnSO4(s) | −982.8 | −871.5 | 110.5 |
Zirconium: | |||
Zr(s) | 0.0 | 0.0 | 39.0 |
Zr(g) | 608.8 | 566.5 | 181.4 |
ZrCl2(s) | −502.0 | −386 | 110 |
ZrCl4(s) | −980.5 | −889.9 | 181.6 |
Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004).
Compound Name | Compound Formula | K sp |
---|---|---|
Aluminum phosphate | AlPO4 | 9.84 × 10−21 |
Barium bromate | Ba(BrO3)2 | 2.43 × 10−4 |
Barium carbonate | BaCO3 | 2.58 × 10−9 |
Barium chromate | BaCrO4 | 1.17 × 10−10 |
Barium fluoride | BaF2 | 1.84 × 10−7 |
Barium iodate | Ba(IO3)2 | 4.01 × 10−9 |
Barium nitrate | Ba(NO3)2 | 4.64 × 10−3 |
Barium sulfate | BaSO4 | 1.08 × 10−10 |
Barium sulfite | BaSO3 | 5.0 × 10−10 |
Beryllium hydroxide | Be(OH)2 | 6.92 × 10−22 |
Bismuth arsenate | BiAsO4 | 4.43 × 10−10 |
Bismuth iodide | BiI3 | 7.71 × 10−19 |
Cadmium carbonate | CdCO3 | 1.0 × 10−12 |
Cadmium fluoride | CdF2 | 6.44 × 10−3 |
Cadmium hydroxide | Cd(OH)2 | 7.2 × 10−15 |
Cadmium iodate | Cd(IO3)2 | 2.5 × 10−8 |
Cadmium phosphate | Cd3(PO4)2 | 2.53 × 10−33 |
Cadmium sulfide | CdS | 8.0 × 10−27 |
Calcium carbonate | CaCO3 | 3.36 × 10−9 |
Calcium fluoride | CaF2 | 3.45 × 10−11 |
Calcium hydroxide | Ca(OH)2 | 5.02 × 10−6 |
Calcium iodate | Ca(IO3)2 | 6.47 × 10−6 |
Calcium phosphate | Ca3(PO4)2 | 2.07 × 10−33 |
Calcium sulfate | CaSO4 | 4.93 × 10−5 |
Cesium perchlorate | CsClO4 | 3.95 × 10−3 |
Cesium periodate | CsIO4 | 5.16 × 10−6 |
Cobalt(II) arsenate | Co3(AsO4)2 | 6.80 × 10−29 |
Cobalt(II) hydroxide | Co(OH)2 | 5.92 × 10−15 |
Cobalt(II) phosphate | Co3(PO4)2 | 2.05 × 10−35 |
Copper(I) bromide | CuBr | 6.27 × 10−9 |
Copper(I) chloride | CuCl | 1.72 × 10−7 |
Copper(I) cyanide | CuCN | 3.47 × 10−20 |
Copper(I) iodide | CuI | 1.27 × 10−12 |
Copper(I) thiocyanate | CuSCN | 1.77 × 10−13 |
Copper(II) arsenate | Cu3(AsO4)2 | 7.95 × 10−36 |
Copper(II) oxalate | CuC2O4 | 4.43 × 10−10 |
Copper(II) phosphate | Cu3(PO4)2 | 1.40 × 10−37 |
Copper(II) sulfide | CuS | 6.3 × 10−36 |
Europium(III) hydroxide | Eu(OH)3 | 9.38 × 10−27 |
Gallium(III) hydroxide | Ga(OH)3 | 7.28 × 10−36 |
Iron(II) carbonate | FeCO3 | 3.13 × 10−11 |
Iron(II) fluoride | FeF2 | 2.36 × 10−6 |
Iron(II) hydroxide | Fe(OH)2 | 4.87 × 10−17 |
Iron(III) hydroxide | Fe(OH)3 | 2.79 × 10−39 |
Iron(III) sulfide | FeS | 6.3 × 10−18 |
Lanthanum iodate | La(IO3)3 | 7.50 × 10−12 |
Lead(II) bromide | PbBr2 | 6.60 × 10−6 |
Lead(II) carbonate | PbCO3 | 7.40 × 10−14 |
Lead(II) chloride | PbCl2 | 1.70 × 10−5 |
Lead(II) fluoride | PbF2 | 3.3 × 10−8 |
Lead(II) hydroxide | Pb(OH)2 | 1.43 × 10−20 |
Lead(II) iodate | Pb(IO3)2 | 3.69 × 10−13 |
Lead(II) iodide | PbI2 | 9.8 × 10−9 |
Lead(II)selenite | PbSeO4 | 1.37 × 10−7 |
Lead(II) sulfate | PbSO4 | 2.53 × 10−8 |
Lead(II) sulfide | PbS | 8.0 × 10−28 |
Lithium carbonate | Li2CO3 | 8.15 × 10−4 |
Lithium fluoride | LiF | 1.84 × 10−3 |
Lithium phosphate | Li3PO4 | 2.37 × 10−11 |
Magnesium carbonate | MgCO3 | 6.82 × 10−6 |
Magnesium fluoride | MgF2 | 5.16 × 10−11 |
Magnesium hydroxide | Mg(OH)2 | 5.61 × 10−12 |
Magnesium phosphate | Mg3(PO4)2 | 1.04 × 10−24 |
Manganese(II) carbonate | MnCO3 | 2.24 × 10−11 |
Manganese(II) iodate | Mn(IO3)2 | 4.37 × 10−7 |
Mercury(I) bromide | Hg2Br2 | 6.40 × 10−23 |
Mercury(I) carbonate | Hg2CO3 | 3.6 × 10−17 |
Mercury(I) chloride | Hg2Cl2 | 1.43 × 10−18 |
Mercury(I) fluoride | Hg2F2 | 3.10 × 10−6 |
Mercury(I) iodide | Hg2I2 | 5.2 × 10−29 |
Mercury(I) oxalate | Hg2C2O4 | 1.75 × 10−13 |
Mercury(I) sulfate | Hg2SO4 | 6.5 × 10−7 |
Mercury(I) thiocyanate | Hg2(SCN)2 | 3.2 × 10−20 |
Mercury(II) bromide | HgBr2 | 6.2 × 10−20 |
Mercury (II) iodide | HgI2 | 2.9 × 10−29 |
Mercury(II) sulfide (red) | HgS | 4 × 10−53 |
Mercury(II) sulfide (black) | HgS | 1.6 × 10−52 |
Neodymium carbonate | Nd2(CO3)3 | 1.08 × 10−33 |
Nickel(II) carbonate | NiCO3 | 1.42 × 10−7 |
Nickel(II) hydroxide | Ni(OH)2 | 5.48 × 10−16 |
Nickel(II) iodate | Ni(IO3)2 | 4.71 × 10−5 |
Nickel(II) phosphate | Ni3(PO4)2 | 4.74 × 10−32 |
Palladium(II) thiocyanate | Pd(SCN)2 | 4.39 × 10−23 |
Potassium hexachloroplatinate | K2PtCl6 | 7.48 × 10−6 |
Potassium perchlorate | KClO4 | 1.05 × 10−2 |
Potassium periodate | KIO4 | 3.71 × 10−4 |
Praseodymium hydroxide | Pr(OH)3 | 3.39 × 10−24 |
Rubidium perchlorate | RbClO4 | 3.00 × 10−3 |
Scandium fluoride | ScF3 | 5.81 × 10−24 |
Scandium hydroxide | Sc(OH)3 | 2.22 × 10−31 |
Silver(I) acetate | AgCH3CO2 | 1.94 × 10−3 |
Silver(I) arsenate | Ag3AsO4 | 1.03 × 10−22 |
Silver(I) bromate | AgBrO3 | 5.38 × 10−5 |
Silver(I) bromide | AgBr | 5.35 × 10−13 |
Silver(I) carbonate | Ag2CO3 | 8.46 × 10−12 |
Silver(I) chloride | AgCl | 1.77 × 10−10 |
Silver(I) chromate | Ag2CrO4 | 1.12 × 10−12 |
Silver(I) cyanide | AgCN | 5.97 × 10−17 |
Silver(I) iodate | AgIO3 | 3.17 × 10−8 |
Silver(I) iodide | AgI | 8.52 × 10−17 |
Silver(I) oxalate | Ag2C2O4 | 5.40 × 10−12 |
Silver(I) phosphate | Ag3PO4 | 8.89 × 10−17 |
Silver(I) sulfate | Ag2SO4 | 1.20 × 10−5 |
Silver(I) sulfide | Ag2S | 6.3 × 10−50 |
Silver(I) sulfite | Ag2SO3 | 1.50 × 10−14 |
Silver(I) thiocyanate | AgSCN | 1.03 × 10−12 |
Strontium arsenate | Sr3(AsO4)2 | 4.29 × 10−19 |
Strontium carbonate | SrCO3 | 5.60 × 10−10 |
Strontium fluoride | SrF2 | 4.33 × 10−9 |
Strontium iodate | Sr(IO3)2 | 1.14 × 10−7 |
Strontium sulfate | SrSO4 | 3.44 × 10−7 |
Thallium(I) bromate | TlBrO3 | 1.10 × 10−4 |
Thallium(I) bromide | TlBr | 3.71 × 10−6 |
Thallium(I) chloride | TlCl | 1.86 × 10−4 |
Thallium(I) chromate | Tl2CrO4 | 8.67 × 10−13 |
Thallium(I) iodate | TlIO3 | 3.12 × 10−6 |
Thallium(I) iodide | TlI | 5.54 × 10−8 |
Thallium(I) thiocyanate | TlSCN | 1.57 × 10−4 |
Thallium(III) hydroxide | Tl(OH)3 | 1.68 × 10−44 |
Tin(II) hydroxide | Sn(OH)2 | 5.45 × 10−27 |
Tin(II) sulfide | SnS | 1.0 × 10−25 |
Yttrium carbonate | Y2(CO3)3 | 1.03 × 10−31 |
Yttrium fluoride | YF3 | 8.62 × 10−21 |
Yttrium hydroxide | Y(OH)3 | 1.00 × 10−22 |
Yttrium iodate | Y(IO3)3 | 1.12 × 10−10 |
Zinc arsenate | Zn3(AsO4)2 | 2.8 × 10−28 |
Zinc carbonate | ZnCO3 | 1.46 × 10−10 |
Zinc fluoride | ZnF2 | 3.04 × 10−2 |
Zinc hydroxide | Zn(OH)2 | 3 × 10−17 |
Zinc selenide | ZnSe | 3.6 × 10−26 |
Zinc sulfide (wurtzite) | ZnS | 1.6 × 10−24 |
Zinc sulfide (sphalerite) | ZnS | 2.5 × 10−22 |
Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004); sulfide data from Lange’s Handbook of Chemistry, 15th Edition (1999).
Name | Formula | K a1 | pKa1 | K a2 | pKa2 | K a3 | pKa3 | K a4 | pKa4 |
---|---|---|---|---|---|---|---|---|---|
Acetic acid | CH3CO2H | 1.75 × 10−5 | 4.756 | ||||||
Arsenic acid | H3AsO4 | 5.5 × 10−3 | 2.26 | 1.7 × 10−7 | 6.76 | 5.1 × 10−12 | 11.29 | ||
Benzoic acid | C6H5CO2H | 6.25 × 10−5 | 4.204 | ||||||
Boric acid | H3BO3 | 5.4 × 10−10* | 9.27* | >1 × 10−14* | >14* | ||||
Bromoacetic acid | CH2BrCO2H | 1.3 × 10−3 | 2.90 | ||||||
Carbonic acid | H2CO3 | 4.5 × 10−7 | 6.35 | 4.7 × 10−11 | 10.33 | ||||
Chloroacetic acid | CH2ClCO2H | 1.3 × 10−3 | 2.87 | ||||||
Chlorous acid | HClO2 | 1.1 × 10−2 | 1.94 | ||||||
Chromic acid | H2CrO4 | 1.8 × 10−1 | 0.74 | 3.2 × 10−7 | 6.49 | ||||
Citric acid | C6H8O7 | 7.4 × 10−4 | 3.13 | 1.7 × 10−5 | 4.76 | 4.0 × 10−7 | 6.40 | ||
Cyanic acid | HCNO | 3.5 × 10−4 | 3.46 | ||||||
Dichloroacetic acid | CHCl2CO2H | 4.5 × 10−2 | 1.35 | ||||||
Fluoroacetic acid | CH2FCO2H | 2.6 × 10−3 | 2.59 | ||||||
Formic acid | CH2O2 | 1.8 × 10−4 | 3.75 | ||||||
Hydrazoic acid | HN3 | 2.5 × 10−5 | 4.6 | ||||||
Hydrocyanic acid | HCN | 6.2 × 10−10 | 9.21 | ||||||
Hydrofluoric acid | HF | 6.3 × 10−4 | 3.20 | ||||||
Hydrogen selenide | H2Se | 1.3 × 10−4 | 3.89 | 1.0× 10−11 | 11.0 | ||||
Hydrogen sulfide | H2S | 8.9 × 10−8 | 7.05 | 1 × 10−19 | 19 | ||||
Hydrogen telluride | H2Te | 2.5 × 10−3‡ | 2.6‡ | 1 × 10−11 | 11 | ||||
Hypobromous acid | HBrO | 2.8 × 10−9 | 8.55 | ||||||
Hypochlorous acid | HClO | 4.0 × 10−8 | 7.40 | ||||||
Hypoiodous acid | HIO | 3.2 × 10−11 | 10.5 | ||||||
Iodic acid | HIO3 | 1.7 × 10−1 | 0.78 | ||||||
Iodoacetic acid | CH2ICO2H | 6.6 × 10−4 | 3.18 | ||||||
Nitrous acid | HNO2 | 5.6 × 10−4 | 3.25 | ||||||
Oxalic acid | C2H2O4 | 5.6 × 10−2 | 1.25 | 1.5 × 10−4 | 3.81 | ||||
Periodic acid | HIO4 | 2.3 × 10−2 | 1.64 | ||||||
Phenol | C6H5OH | 1.0 × 10−10 | 9.99 | ||||||
Phosphoric acid | H3PO4 | 6.9 × 10−3 | 2.16 | 6.2 × 10−8 | 7.21 | 4.8 × 10−13 | 12.32 | ||
Phosphorous acid | H3PO3 | 5.0 × 10−2* | 1.3* | 2.0 × 10−7* | 6.70* | ||||
Pyrophosphoric acid | H4P2O7 | 1.2 × 10−1 | 0.91 | 7.9 × 10−3 | 2.10 | 2.0 × 10−7 | 6.70 | 4.8 × 10−10 | 9.32 |
Resorcinol | C6H4(OH)2 | 4.8 × 10−10 | 9.32 | 7.9 × 10−12 | 11.1 | ||||
Selenic acid | H2SeO4 | Strong | Strong | 2.0 × 10−2 | 1.7 | ||||
Selenious acid | H2SeO3 | 2.4 × 10−3 | 2.62 | 4.8 × 10−9 | 8.32 | ||||
Sulfuric acid | H2SO4 | Strong | Strong | 1.0 × 10−2 | 1.99 | ||||
Sulfurous acid | H2SO3 | 1.4 × 10−2 | 1.85 | 6.3 × 10−8 | 7.2 | ||||
meso-Tartaric acid | C4H6O6 | 6.8 × 10−4 | 3.17 | 1.2 × 10−5 | 4.91 | ||||
Telluric acid | H2TeO4 | 2.1 × 10−8‡ | 7.68‡ | 1.0 × 10−11‡ | 11.0‡ | ||||
Tellurous acid | H2TeO3 | 5.4 × 10−7 | 6.27 | 3.7 × 10−9 | 8.43 | ||||
Trichloroacetic acid | CCl3CO2H | 2.2 × 10−1 | 0.66 | ||||||
Trifluoroacetic acid | CF3CO2H | 3.0 × 10−1 | 0.52 | ||||||
* Measured at 20°C, not 25°C. | |||||||||
‡ Measured at 18°C, not 25°C. |
Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004).
Name | Formula | K b | pKb |
---|---|---|---|
Ammonia | NH3 | 1.8 × 10−5 | 4.75 |
Aniline | C6H5NH2 | 7.4 × 10−10 | 9.13 |
n-Butylamine | C4H9NH2 | 4.0 × 10−4 | 3.40 |
sec-Butylamine | (CH3)2CHCH2NH2 | 3.6 × 10−4 | 3.44 |
tert-Butylamine | (CH3)3CNH2 | 4.8 × 10−4 | 3.32 |
Dimethylamine | (CH3)2NH | 5.4 × 10−4 | 3.27 |
Ethylamine | C2H5NH2 | 4.5 × 10−4 | 3.35 |
Hydrazine | N2H4 | 1.3 × 10−6 | 5.9 |
Hydroxylamine | NH2OH | 8.7 × 10−9 | 8.06 |
Methylamine | CH3NH2 | 4.6 × 10−4 | 3.34 |
Propylamine | C3H7NH2 | 3.5 × 10−4 | 3.46 |
Pyridine | C5H5N | 1.7 × 10−9 | 8.77 |
Trimethylamine | (CH3)3N | 6.3 × 10−5 | 4.20 |
Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004).
Half-Reaction | E° (V) |
---|---|
Ac3+ + 3e−→ Ac | −2.20 |
Ag+ + e−→ Ag | 0.7996 |
AgBr + e−→ Ag + Br− | 0.07133 |
AgCl + e−→ Ag + Cl− | 0.22233 |
Ag2CrO4 + 2e−→ 2Ag + CrO42− | 0.4470 |
AgI + e−→ Ag + I− | −0.15224 |
Ag2S + 2e−→ 2Ag + S2− | −0.691 |
Ag2S + 2H+ + 2e−→ 2Ag + H2S | −0.0366 |
AgSCN + e−→ Ag + SCN− | 0.08951 |
Al3+ + 3e−→ Al | −1.662 |
Al(OH)4− + 3e−→ Al + 4OH− | −2.328 |
Am3+ + 3e−→ Am | −2.048 |
As + 3H+ + 3e−→ AsH3 | −0.608 |
H3AsO4 + 2H+ + 2e−→ HAsO2 + 2H2O | 0.560 |
Au+ + e−→ Au | 1.692 |
Au3+ + 3e−→ Au | 1.498 |
H3BO3 + 3H+ + 3e−→ B + 3H2O | −0.8698 |
Ba2+ + 2e−→ Ba | −2.912 |
Be2+ + 2e−→ Be | −1.847 |
Bi3+ + 3e−→ Bi | 0.308 |
BiO+ + 2H+ + 3e−→ Bi + H2O | 0.320 |
Br2(aq) + 2e−→ 2Br− | 1.0873 |
Br2(l) + 2e−→ 2Br− | 1.066 |
BrO3− + 6H+ + 5e−→ Br2 + 3H2O | 1.482 |
BrO3− + 6H+ + 6e−→ Br− + 3H2O | 1.423 |
CO2 + 2H+ + 2e−→ HCO2H | −0.199 |
Ca2+ + 2e−→ Ca | −2.868 |
Ca(OH)2 + 2e−→ Ca + 2OH− | −3.02 |
Cd2+ + 2e−→ Cd | −0.4030 |
CdSO4 + 2e−→ Cd + SO42− | −0.246 |
Cd(OH)42− + 2e−→ Cd + 4OH− | −0.658 |
Ce3+ + 3e−→ Ce | −2.336 |
Ce4− + e−→ Ce3+ | 1.72 |
Cl2(g) + 2e−→ 2Cl− | 1.35827 |
HClO + H+ + e−→ Cl2 + H2O | 1.611 |
HClO + H+ + 2e−→ Cl− + H2O | 1.482 |
ClO− + H2O + 2e−→ Cl− + 2OH− | 0.81 |
ClO3− + 6H+ + 5e−→ Cl2 + 3H2O | 1.47 |
ClO3− + 6H+ + 6e−→ Cl− + 3H2O | 1.451 |
ClO4− + 8H+ + 7e−→ Cl2 + 4H2O | 1.39 |
ClO4− + 8H+ + 8e−→ Cl− + 4H2O | 1.389 |
Co2+ + 2e−→ Co | −0.28 |
Co3+ + e−→ Co2+ | 1.92 |
Cr2+ + 2e−→ Cr | −0.913 |
Cr3+ + e−→ Cr2+ | −0.407 |
Cr3+ + 3e−→ Cr | −0.744 |
Cr2O7− + 14H+ + 6e−→ 2Cr3+ + 7H2O | 1.232 |
CrO42− + 4H2O + 3e−→ Cr(OH)3 + 5OH− | −0.13 |
Cs+ + e−→ Cs | −3.026 |
Cu+ + e−→ Cu | 0.521 |
Cu2+ + e−→ Cu+ | 0.153 |
Cu2+ + 2e−→ Cu | 0.3419 |
CuI2− + e−→ Cu + 2I− | 0.00 |
Cu2O + H2O + 2e−→ 2Cu + 2OH− | −0.360 |
Dy3+ + 3e−→ Dy | −2.295 |
Er3+ + 3e−→ Er | −2.331 |
Es3+ + 3e−→ Es | −1.91 |
Eu2+ + 2e−→ Eu | −2.812 |
Eu3+ + 3e−→ Eu | −1.991 |
F2 + 2e−→ 2F− | 2.866 |
Fe2+ + 2e−→ Fe | −0.447 |
Fe3+ + 3e−→ Fe | −0.037 |
Fe3+ + e−→ Fe2+ | 0.771 |
[Fe(CN)6]3− + e−→ [Fe(CN)6]4− | 0.358 |
Fe(OH)3 + e−→ Fe(OH)2 + OH− | −0.56 |
Fm3+ + 3e−→ Fm | −1.89 |
Fm2+ + 2e−→ Fm | −2.30 |
Ga3+ + 3e−→ Ga | −0.549 |
Gd3+ + 3e−→ Gd | −2.279 |
Ge2+ + 2e−→ Ge | 0.24 |
Ge4+ + 4e−→ Ge | 0.124 |
2H+ + 2e−→ H2 | 0.00000 |
H2 + 2e−→ 2H− | −2.23 |
2H2O + 2e−→ H2 + 2OH− | −0.8277 |
H2O2 + 2H+ + 2e−→ 2H2O | 1.776 |
Hf4+ + 4e−→ Hf | −1.55 |
Hg2+ + 2e−→ Hg | 0.851 |
2Hg2+ + 2e−→ Hg22+ | 0.920 |
Hg2Cl2 + 2e−→ 2Hg + 2Cl− | 0.26808 |
Ho2+ + 2e−→ Ho | −2.1 |
Ho3+ + 3e−→ Ho | −2.33 |
I2 + 2e−→ 2I− | 0.5355 |
I3− + 2e−→ 3I− | 0.536 |
2IO3− + 12H+ + 10e−→ I2 + 6H2O | 1.195 |
IO3− + 6H+ + 6e−→ I− + 3H2O | 1.085 |
In+ + e−→ In | −0.14 |
In3+ + 2e−→ In+ | −0.443 |
In3+ + 3e−→ In | −0.3382 |
Ir3+ + 3e−→ Ir | 1.156 |
K+ + e−→ K | −2.931 |
La3+ + 3e−→ La | −2.379 |
Li+ + e−→ Li | −3.0401 |
Lr3+ + 3e−→ Lr | −1.96 |
Lu3+ + 3e−→ Lu | −2.28 |
Md3+ + 3e−→ Md | −1.65 |
Md2+ + 2e−→ Md | −2.40 |
Mg2+ + 2e−→ Mg | −2.372 |
Mn2+ + 2e−→ Mn | −1.185 |
MnO2 + 4H+ + 2e−→ Mn2+ + 2H2O | 1.224 |
MnO4− + 8H+ + 5e−→ Mn2+ + 4H2O | 1.507 |
MnO4− + 2H2O + 3e−→ MnO2 + 4OH− | 0.595 |
Mo3+ + 3e−→ Mo | −0.200 |
N2 + 2H2O + 6H+ + 6e−→ 2NH4OH | 0.092 |
HNO2 + H+ + e−→ NO + H2O | 0.983 |
NO3− + 4H+ + 3e−→ NO + 2H2O | 0.957 |
Na+ + e−→ Na | −2.71 |
Nb3+ + 3e−→ Nb | −1.099 |
Nd3+ + 3e−→ Nd | −2.323 |
Ni2+ + 2e−→ Ni | −0.257 |
No3+ + 3e−→ No | −1.20 |
No2+ + 2e−→ No | −2.50 |
Np3+ + 3e−→ Np | −1.856 |
O2 + 2H+ + 2e−→ H2O2 | 0.695 |
O2 + 4H+ + 4e−→ 2H2O | 1.229 |
O2 + 2H2O + 2e−→ H2O2 + 2OH− | −0.146 |
O3 + 2H+ + 2e−→ O2 + H2O | 2.076 |
OsO4 + 8H+ + 8e−→ Os + 4H2O | 0.838 |
P + 3H2O + 3e−→ PH3(g) + 3OH− | −0.87 |
PO43− + 2H2O + 2e−→ HPO32− + 3OH− | −1.05 |
Pa3+ + 3e−→ Pa | −1.34 |
Pa4+ + 4e−→ Pa | −1.49 |
Pb2+ + 2e−→ Pb | −0.1262 |
PbO + H2O + 2e−→ Pb + 2OH− | −0.580 |
PbO2 + SO42− + 4H+ + 2e−→ PbSO4 + 2H2O | 1.6913 |
PbSO4 + 2e−→ Pb + SO42− | −0.3588 |
Pd2+ + 2e−→ Pd | 0.951 |
Pm3+ + 3e−→ Pm | −2.30 |
Po4+ + 4e−→ Po | 0.76 |
Pr3+ + 3e−→ Pr | −2.353 |
Pt2+ + 2e−→ Pt | 1.18 |
[PtCl4]2− + 2e−→ Pt + 4Cl− | 0.755 |
Pu3+ + 3e−→ Pu | −2.031 |
Ra2+ + 2e−→ Ra | −2.8 |
Rb+ + e−→ Rb | −2.98 |
Re3+ + 3e−→ Re | 0.300 |
Rh3+ + 3e−→ Rh | 0.758 |
Ru3+ + e−→ Ru2+ | 0.2487 |
S + 2e−→ S2− | −0.47627 |
S + 2H+ + 2e−→ H2S(aq) | 0.142 |
2S + 2e−→ S22− | −0.42836 |
H2SO3 + 4H+ + 4e−→ S + 3H2O | 0.449 |
SO42− + H2O + 2e−→ SO32− + 2OH− | −0.93 |
Sb + 3H+ + 3e−→ SbH3 | −0.510 |
Sc3+ + 3e−→ Sc | −2.077 |
Se + 2e−→ Se2− | −0.924 |
Se + 2H+ + 2e−→ H2Se | −0.082 |
SiF62− + 4e−→ Si + 6F− | −1.24 |
Sm3+ + 3e−→ Sm | −2.304 |
Sn2+ + 2e−→ Sn | −0.1375 |
Sn4+ + 2e−→ Sn2+ | 0.151 |
Sr2+ + 2e−→ Sr | −2.899 |
Ta3+ + 3e−→ Ta | −0.6 |
TcO4− + 4H+ + 3e−→ TcO2 + 2H2O | 0.782 |
TcO4− + 8H+ + 7e−→ Tc + 4H2O | 0.472 |
Tb3+ + 3e−→ Tb | −2.28 |
Te + 2e−→ Te2− | −1.143 |
Te4+ + 4e−→ Te | 0.568 |
Th4+ + 4e−→ Th | −1.899 |
Ti2+ + 2e−→ Ti | −1.630 |
Tl+ + e−→ Tl | −0.336 |
Tl3+ + 2e−→ Tl+ | 1.252 |
Tl3+ + 3e−→ Tl | 0.741 |
Tm3+ + 3e−→ Tm | −2.319 |
U3+ + 3e−→ U | −1.798 |
VO2+ + 2H+ + e−→ VO2+ + H2O | 0.991 |
V2O5 + 6H+ + 2e−→ 2VO2+ + 3H2O | 0.957 |
W2O5 + 2H+ + 2e−→ 2WO2 + H2O | −0.031 |
XeO3 + 6H+ + 6e−→ Xe + 3H2O | 2.10 |
Y3+ + 3e−→Y | −2.372 |
Yb3+ + 3e−→Yb | −2.19 |
Zn2+ + 2e−→ Zn | −0.7618 |
Zn(OH)42− + 2e−→ Zn + 4OH− | −1.199 |
Zn(OH)2 + 2e−→ Zn + 2OH− | −1.249 |
ZrO2 + 4H+ + 4e−→ Zr + 2H2O | −1.553 |
Zr4+ + 4e−→ Zr | −1.45 |
Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004).
Density: 0.99984 g/cm3 at 0°C |
0.99970 g/cm3 at 10°C |
0.99821 g/cm3 at 20°C |
0.98803 g/cm3 at 50°C |
0.95840 g/cm3 at 100°C |
Enthalpy (heat) of vaporization: 45.054 kJ/mol at 0°C |
43.990 kJ/mol at 25°C |
42.482 kJ/mol at 60°C |
40.657 kJ/mol at 100°C |
Surface tension: 74.23 J/m2 at 10°C |
71.99 J/m2 at 25°C |
67.94 J/m2 at 50°C |
58.91 J/m2 at 100°C |
Viscosity: 1.793 mPa ⋅ s at 0°C |
0.890 mPa_s at 25°C |
0.547 mPa_s at 50°C |
0.282 mPa_s at 100°C |
Ion-product constant,Kw: 1.15 × 10−15 at 0°C |
1.01 × 10−14 at 25°C |
5.31 × 10−14 at 50°C |
5.43 × 10−13 at 100°C |
Specific heat (Cs): 4.2176 J/(g-°C) at 0°C |
4.1818 J/(g-°C) at 20°C |
4.1806 J/(g-°C) at 50°C |
4.2159 J/(g-°C) at 100°C |
Vapor pressure of water (kPa) | |||||||
---|---|---|---|---|---|---|---|
T(°C) | P(kPa) | T(°C) | P(kPa) | T(°C) | P(kPa) | T(°C) | P(kPa) |
0 | 0.61129 | 30 | 4.2455 | 60 | 19.932 | 90 | 70.117 |
5 | 0.87260 | 35 | 5.6267 | 65 | 25.022 | 95 | 84.529 |
10 | 1.2281 | 40 | 7.3814 | 70 | 31.176 | 100 | 101.32 |
15 | 1.7056 | 45 | 9.5898 | 75 | 38.563 | 105 | 120.79 |
20 | 2.3388 | 50 | 12.344 | 80 | 47.373 | 110 | 143.24 |
25 | 3.1690 | 55 | 15.752 | 85 | 57.815 | 115 | 169.02 |
Vapor pressure of water (mmHg) | |||||||
---|---|---|---|---|---|---|---|
T(°C) | P(mmHg) | T(°C) | P(mmHg) | T(°C) | P(mmHg) | T(°C) | P(mmHg) |
0 | 4.585 | 30 | 31.844 | 60 | 149.50 | 90 | 525.91 |
5 | 6.545 | 35 | 42.203 | 65 | 187.68 | 95 | 634.01 |
10 | 9.211 | 40 | 55.364 | 70 | 233.84 | 100 | 759.95 |
15 | 12.793 | 45 | 71.929 | 75 | 289.24 | 105 | 905.99 |
20 | 17.542 | 50 | 92.59 | 80 | 355.32 | 110 | 1074.38 |
25 | 23.769 | 55 | 118.15 | 85 | 433.64 | 115 | 1267.74 |
Source of data: CRC Handbook of Chemistry and Physics, 84th Edition (2004).
Selected Physical Constants | |
---|---|
Atomic mass unit | 1 amu = 1.6605389 × 10−24 g |
1 g = 6.022142 × 1023 amu | |
Avogadro’s number | N = 6.022142 × 1023/mol |
Boltzmann’s constant | k = 1.380651 × 10−23 J/K |
Charge on electron | e = 1.6021765 × 10−19 C |
Faraday’s constant | F = 9.6485338 × 104 C/mol |
Gas constant | R = 0.0820575 (L atm)/(mol K) |
= 8.31447 J/(mol K) | |
Mass of electron | me = 5.485799 × 10−4 amu |
= 9.109383 × 10−28 g | |
Mass of neutron | mn = 1.0086649 amu |
= 1.6749273 × 10−24 g | |
Mass of proton | mp = 1.0072765 amu |
= 1.6726217 × 10−24 g | |
Pi | π = 3.1415927 |
Planck’s constant | h = 6.626069 × 10 −34J s |
Speed of light (in vacuum) | c = 2.99792458 × 108 m/s (exact) |
Useful Conversion Factors and Relationships | |
---|---|
Length | Energy (derived) |
Si unit: meter (m) | Si unit: joule (J) |
Mass | Pressure (derived) |
SI unit: kilogram (kg) | SI unit: pascal (Pa) |
Temperature | Volume (derived) |
Si unit: kelvin (K) | SI unit: cubic meter (m 3) |
Two systems for numbering periodic groups are shown: 1–18 is the system currently recommended by the Inernational Union of Pure and Applied Chemistry (IUPAC); an older U.S. system, in which letters designate main group elements (A) and transition elements (B), is given parentheses.
An atomic mass in brackets indicates the mass of the longest-lived isotope of an element having no stable isotopes.
Elements with atomic numbers 114 (ununquadium, 289 amu) and 116 (ununhexium, 293 amu) have been recognized by the International Union of Pure and Applied Chemistry (IUPAC). The collaborating scientists from the Joint Institute for Nuclear Research in Dubna, Russia, and Lawrence Livermore National Laboratory in California have been invited to propose names for the new elements. See http://iupac.org/publications/pac/asap/PAC-REP-10-05-01
List of Elements | |||
---|---|---|---|
Name | Symbol | Atomic Number | Atomic Mass |
Actinium | Ac | 89 | [227]* |
Aluminum | Al | 13 | 26.9815386(8) |
Americium | Am | 95 | [243]* |
Antimony | Sb | 51 | 121.760(1) |
Argon | Ar | 18 | 39.948(1) |
Arsenic | As | 33 | 74.92160(2) |
Astatine | At | 85 | [210]* |
Barium | Ba | 56 | 137.327(7) |
Berkelium | Bk | 97 | [247]* |
Beryllium | Be | 4 | 9.012182(3) |
Bismuth | Bi | 83 | 208.98040(1) |
Bohrium | Bh | 107 | [267]* |
Boron | B | 5 | 10.811(7) |
Bromine | Br | 35 | 79.904(1) |
Cadmium | Cd | 48 | 112.411(8) |
Calcium | Ca | 20 | 40.078(4) |
Californium | Cf | 98 | [251]* |
Carbon | C | 6 | 12.0107(8) |
Cerium | Ce | 58 | 140.116(1) |
Cesium | Cs | 55 | 132.9054519(2) |
Chlorine | Cl | 17 | 35.453(2) |
Chromium | Cr | 24 | 51.9961(6) |
Cobalt | Co | 27 | 58.933195(5) |
Copernicium† | Cn | 112 | [285]* |
Copper | Cu | 29 | 63.546(3) |
Curium | Cm | 96 | [247]* |
Darmstadtium | Ds | 110 | [281]* |
Dubnium | Db | 105 | [268]* |
Dysprosium | Dy | 66 | 162.500(1) |
Einsteinium | Es | 99 | [252]* |
Erbium | Er | 68 | 167.259(3) |
Europium | Eu | 63 | 151.964(1) |
Fermium | Fm | 100 | [257]* |
Fluorine | F | 9 | 18.9984032(5) |
Francium | Fr | 87 | [223]* |
Gadolinium | Gd | 64 | 157.25(3) |
Gallium | Ga | 31 | 69.723(1) |
Germanium | Ge | 32 | 72.64(1) |
Gold | Au | 79 | 196.966569(4) |
Hafnium | Hf | 72 | 178.49(2) |
Hassium | Hs | 108 | [269]* |
Helium | He | 2 | 4.002602(2) |
Holmium | Ho | 67 | 164.93032(2) |
Hydrogen | H | 1 | 1.00794(7) |
Indium | In | 49 | 114.818(3) |
Iodine | I | 53 | 126.90447(3) |
Iridium | Ir | 77 | 192.217(3) |
Iron | Fe | 26 | 55.845(2) |
Krypton | Kr | 36 | 83.798(2) |
Lanthanum | La | 57 | 138.90547(7) |
Lawrencium | Lr | 103 | [262]* |
Lead | Pb | 82 | 207.2(1) |
Lithium | Li | 3 | 6.941(2) |
Lutetium | Lu | 71 | 174.967(1) |
Magnesium | Mg | 12 | 24.3050(6) |
Manganese | Mn | 25 | 54.938045(5) |
Meitnerium | Mt | 109 | [276]* |
Mendelevium | Md | 101 | [258]* |
Mercury | Hg | 80 | 200.59(2) |
Molybdenum | Mo | 42 | 95.94(2) |
Neodymium | Nd | 60 | 144.242(3) |
Neon | Ne | 10 | 20.1797(6) |
Neptunium | Np | 93 | [237]* |
Nickel | Ni | 28 | 58.6934(2) |
Niobium | Nb | 41 | 92.90638(2) |
Nitrogen | N | 7 | 14.0067(2) |
Nobelium | No | 102 | [259]* |
Osmium | Os | 76 | 190.23(3) |
Oxygen | O | 8 | 15.9994(3) |
Palladium | Pd | 46 | 106.42(1) |
Phosphorus | P | 15 | 30.973762(2) |
Platinum | Pt | 78 | 195.084(9) |
Plutonium | Pu | 94 | [244]* |
Polonium | Po | 84 | [209]* |
Potassium | K | 19 | 39.0983(1) |
Praseodymium | Pr | 59 | 140.90765(2) |
Promethium | Pm | 61 | [145]* |
Protactinium | Pa | 91 | 231.03588(2)* |
Radium | Ra | 88 | [226]* |
Radon | Rn | 86 | [222]* |
Rhenium | Re | 75 | 186.207(1) |
Rhodium | Rh | 45 | 102.90550(2) |
Roentgenium | Rg | 111 | [280]* |
Rubidium | Rb | 37 | 85.4678(3) |
Ruthenium | Ru | 44 | 101.07(2) |
Rutherfordium | Rf | 104 | [267]* |
Samarium | Sm | 62 | 150.36(2) |
Scandium | Sc | 21 | 44.955912(6) |
Seaborgium | Sg | 106 | [271]* |
Selenium | Se | 34 | 78.96(3) |
Silicon | Si | 14 | 28.0855(3) |
Silver | Ag | 47 | 107.8682(2) |
Sodium | Na | 11 | 22.98976928(2) |
Strontium | Sr | 38 | 87.62(1) |
Sulfur | S | 16 | 32.065(5) |
Tantalum | Ta | 73 | 180.94788(2) |
Technetium | Tc | 43 | [98]* |
Tellurium | Te | 52 | 127.60(3) |
Terbium | Tb | 65 | 158.92535(2) |
Thallium | Tl | 81 | 204.3833(2) |
Thorium | Th | 90 | 232.03806(2)* |
Thulium | Tm | 69 | 168.93421(2) |
Tin | Sn | 50 | 118.710(7) |
Titanium | Ti | 22 | 47.867(1) |
Tungsten | W | 74 | 183.84(1) |
Ununhexium | Uuh | 116 | [293]* |
Ununpentium | Uup | 115 | [288]* |
Ununquadium | Uuq | 114 | [289]* |
Ununtrium | Uut | 113 | [284]* |
Uranium | U | 92 | 238.02891(3)* |
Vanadium | V | 23 | 50.9415(1) |
Xenon | Xe | 54 | 131.293(6) |
Ytterbium | Yb | 70 | 173.04(3) |
Yttrium | Y | 39 | 88.90585(2) |
Zinc | Zn | 30 | 65.409(4) |
Zirconium | Zr | 40 | 91.224(2) |
*Element has no stable isotope. A value enclosed in brackets, e.g. [209], indicates the mass number of the longest-lived isotope of the element. Three such elements (Th, Pa, and U), however, do have a characteristic terrestrial isotopic composition, and an atomic mass is given for them. An uncertainty in the last digit in the Atomic Mass column is shown by the number in parentheses; e.g., 1.00794(7) indicates ±0.00007. | |||
†Element 112 named shortly before the release of this text. Other periodic tables in this version of the text may refer to it as Ununbium (Uub). |
Source of data: Atomic weights of the elements 2001 (IUPAC Technical Report) as supplemented by the Table of Standard Atomic Weights 2005 (to be published in Pure and Applied Chemistry) on the IUPAC web site, and “Nuclear Data Sheets for A-266-294” (to be published in Nuclear Data Sheets) at http://www.nndc.bnl.gov/superheavy.pdf.
Isotope | Mass (amu) | Isotope | Mass (amu) | Isotope | Mass (amu) |
---|---|---|---|---|---|
1H | 1.007825 | 14N | 14.003074 | 208Po | 207.981246 |
2H | 2.014102 | 16O | 15.994915 | 210Po | 209.982874 |
3H | 3.016049 | 52Cr | 51.940508 | 222Rn | 222.017578 |
3He | 3.016029 | 56Fe | 55.934938 | 226Ra | 226.025410 |
4He | 4.002603 | 59Co | 58.933195 | 230Th | 230.033134 |
6Li | 6.015123 | 58Ni | 57.935343 | 234Th | 234.043601 |
7Li | 7.016005 | 60Ni | 59.930786 | 234Pa | 234.043308 |
9Be | 9.012182 | 90Rb | 89.914802 | 233U | 233.039635 |
10B | 10.012937 | 144Cs | 143.932077 | 234U | 234.040952 |
11B | 11.009305 | 206Pb | 205.974465 | 235U | 235.043930 |
12C | 12 | 207Pb | 206.975897 | 238U | 238.050788 |
13C | 13.003355 | 208Pb | 207.976652 | 239Pu | 239.052163 |
14C | 14.003242 |
Data source: G. Audi, A. H. Wapstra, and C. Thibault, The AME2003 atomic mass evaluation.
This table is provided as a reference.
Isotope | Mass (amu) | Isotope | Mass (amu) |
---|---|---|---|
8B | 8.024607 | 209Fr | 208.99592 |
40K | 39.963998 | 210Po | 209.982874 |
52Cr | 51.940508 | 212At | 211.990745 |
58Ni | 57.935343 | 214Pb | 213.999797 |
59Co | 58.933195 | 214Bi | 213.998712 |
60Co | 59.933817 | 216Fr | 216.003198 |
60Ni | 59.930786 | 199Pb | 198.972917 |
90Sr | 89.907738 | 222Rn | 222.017578 |
92Kr | 91.926156 | 226Ra | 226.025410 |
141Ba | 140.914411 | 227Ra | 227.029178 |
143Xe | 142.935110 | 228Ac | 228.031021 |
167Os | 166.971550 | 230Th | 230.033134 |
171Pt | 170.981240 | 233U | 233.039635 |
194Hg | 193.965439 | 234Th | 234.043601 |
194Tl | 193.971200 | 234Pa | 234.043308 |
199Pb | 198.972917 | 233U | 233.039635 |
199Bi | 198.977672 | 234U | 234.040952 |
206Pb | 205.974465 | 235U | 235.043930 |
207Pb | 206.975897 | 238Pa | 238.054500 |
208Pb | 207.976652 | 238U | 238.050788 |
208Bi | 207.979742 | 239Pu | 239.052163 |
208Po | 207.981246 | 245Pu | 245.067747 |
We wish to thank the Cambridge Crystallographic Data Centre (CCDC) and the Fachinformationszentrum Karlsruhe (FIZ Karlsruhe) for allowing Imagineering Media Services (IMS) to access their databases of atomic coordinates for experimentally determined three-dimensional structures. CCDC’s Cambridge Structural Database (CSD) is the world repository of small molecule crystal structures (distributed as part of the CSD System), and in FIZ Karlsruhe’s Inorganic Crystal Structure Database (ICSD) is the world’s largest inorganic crystal structure database. The coordinates of organic and organometallic compounds in CSD and inorganic and intermetallic compounds in ICSD were invaluable in ensuring the accuracy of the molecular models produced by IMS for this textbook. The authors, the publisher, and IMS gratefully acknowledge the assistance of both organizations. Any errors in the molecular models in this text are entirely the responsibility of the authors, the publisher, and IMS.
The CSD System: The Cambridge Structural Database: a quarter of a million crystal structures and rising. Allen, F.H., Acta Cryst. (2002), B58, 380–388. ConQuest: New Software for searching the Cambridge Structural Database and visualizing crystal structures. Bruno, I.J., Cole, J.C., Edgington, P.R., Kessler, M., Macrae, C.F., McCabe, P., Pearson, J., Taylor, R., Acta Cryst. (2002), B58, 389–397. IsoStar: IsoStar: A Library of Information about Nonbonded Interactions. Bruno, I.J., Cole, J.C., Lommerse, J.P.M., Rowland, R.S., Taylor, R., Verdonk, M., Journal of Computer-Aided Molecular Design (1997), 11-6, 525–537.
The Inorganic Crystal Structure Database (ICSD) is produced and owned by Fachinformationszentrum Karlsruhe (FIZ Karlsruhe) and National Institute of Standards and Technology, an agency of the U.S. Commerce Department’s Technology Administration (NIST).