chemistry of the nonmetals · chemistry of the nonmetals everything we see in the chapter-opening...

22 Chemistry of the NonmetalsEverything we see in the chapter-opening photo is composed of nonmetals. The water, of course, is H2O, and the sand is mostly SiO2. Although we cannot see it, the air contains principally N2 and O2 with much lesser amounts of other nonmetallic substances. The palm tree is also composed mostly of nonmetallic elements.

22.3 Group 8A: The Noble GAses Next, we consider the noble gases, the elements of group 8A, which exhibit very limited chemical reactivity.

22.4 Group 7A: The hAloGeNs We then explore the most electronegative elements: the halogens, group 7A.

22.5 oxyGeN We next consider oxygen, the most abundant element by mass in both Earth’s crust and the human body, and the oxide and peroxide compounds it forms.

22.1 periodic TreNds ANd chemicAl reAcTioNs We begin with a review of periodic trends and types of chemical reactions, which will help us focus on general patterns of behavior as we examine each family in the periodic table.

22.2 hydroGeN The first nonmetal we consider, hydrogen, forms compounds with most other nonmetals and with many metals.

What’s ahEad

▶ A tropicAl beAch.

In this chapter, we take a panoramic view of the descriptive chemistry of the non-metallic elements, starting with hydrogen and progressing group by group across the periodic table. We will consider how the elements occur in nature, how they are isolated from their sources, and how they are used. We will emphasize hydro-gen, oxygen, nitrogen, and carbon because these four nonmetals form many com-mercially important compounds and account for 99% of the atoms required by living cells.

As you study this descriptive chemistry, it is important to look for trends rather than trying to memorize all the facts presented. The periodic table is your most valuable tool in this task.

22.1 | Periodic trends and Chemical Reactions

Recall that we can classify elements as metals, metalloids, and nonmetals. (Section 7.6) Except for hydrogen, which is a special case, the nonmetals occupy

the upper right portion of the periodic table. This division of elements relates nicely

22.6 The oTher Group 6A elemeNTs: s, se, Te, ANd po We study the other members of group 6A (S, Se, Te, and Po), of which sulfur is the most important.

22.7 NiTroGeN We next consider nitrogen, a key component of our atmosphere. It forms compounds in which its oxidation number ranges from -3 to +5, including such important compounds as NH3 and HNO3.

22.8 The oTher Group 5A elemeNTs: p, As, sb, ANd bi Of the other members of group 5A (P, As, Sb, and Bi), we take a closer look at phosphorus—the most commercially important one and the only one that plays an important and beneficial role in biological systems.

22.9 cArboN We next focus on the inorganic compounds of carbon.

22.10 The oTher Group 4A elemeNTs: si, Ge, sn, ANd pb We then consider silicon, the element most abundant and significant of the heavier members of group 4A.

22.11 boroN Finally, we examine boron—the sole nonmetallic element of group 3A.

954 cHAPTEr 22 chemistry of the Nonmetals

to trends in the properties of the elements as summarized in ◀ Figure 22.1. Electronegativity, for example, increases as we move left to right across a period and decreases as we move down a group. The nonmetals thus have higher electronegativities than the metals. This difference leads to the formation of ionic solids in reactions between metals and nonmetals. (Sections 7.6, 8.2, and 8.4) In contrast, compounds formed between two or more nonmetals are usually molecular substances. (Sections 7.8 and 8.4)

The chemistry exhibited by the first member of a nonmetal group can differ from that of subsequent members in important ways. Two differences are particularly notable: (1) The first mem-ber is able to accommodate fewer bonded neighbors. (Section 8.7) For example, nitrogen is able to bond to a maximum of three Cl atoms, NCl3, whereas phosphorus can bond to five, PCl5. The small size of nitrogen is largely responsible for this difference. (2) The first member can more readily form p bonds. This trend is also due, in part, to size because small atoms are able to approach each other

more closely. As a result, the overlap of p orbitals, which results in the formation of p bonds, is more effective for the first element in each group (▼ Figure 22.2). More effec-tive overlap means stronger p bonds, reflected in bond enthalpies. (Section 8.8) For example, the difference between the enthalpies of the C ¬ C bond and the C “ C bond is about 270 kJ>mol (Table 8.4); this large value reflects the “strength” of a carbon– carbon p bond. On the other hand, the difference between Si ¬ Si and Si “ Si bonds is only about 100 kJ/mol, significantly lower than that for carbon, reflecting much weaker p bonding.

As we shall see, p bonds are particularly important in the chemistry of carbon, nitrogen, and oxygen, each the first member in its group. The heavier elements in these groups have a tendency to form only single bonds.

▲ Figure 22.1 trends in elemental properties.

Increasing ionization energyDecreasing atomic radiusIncreasing electronegativityDecreasing metallic character

Dec

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ioni

zati

on e

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yIn

crea

sing

ato

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rad

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Dec

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elec

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ivit

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NonmetalsMetals Metalloids

▲ Figure 22.2 P bonds in period 2 and period 3 elements.

C C Si

Smaller nucleus-to-nucleus distance,more orbital overlap,stronger π bond

Larger nucleus-to-nucleus distance,less orbital overlap,weaker π bond

Si

Of the elements Li, K, N, P, and Ne, which (a) is the most electronegative, (b) has the greatest metallic character, (c) can bond to more than four atoms in a molecule, and (d) forms p bonds most readily?

solutioNAnalyze We are given a list of elements and asked to predict several properties that can be related to periodic trends.Plan We can use Figures 22.1 and 22.2 to guide us to the answers.Solve

(a) Electronegativity increases as we proceed toward the upper right portion of the periodic table, excluding the noble gases. Thus, N is the most electronegative element of our choices.

(b) Metallic character correlates inversely with electronegativity—the less electronegative an element, the greater its metallic character. The element with the greatest metallic character is therefore K, which is closest to the lower left corner of the periodic table.

(c) Nonmetals tend to form molecular compounds, so we can narrow our choice to the three nonmetals on the list: N, P, and Ne. To form more than four bonds, an element must be able to expand its valence shell to allow more than an octet of electrons around it. Valence-shell expansion occurs for period 3 elements and below; N and Ne are both in period 2 and do not undergo valence-shell expansion. Thus, the answer is P.

(d) Period 2 nonmetals form p bonds more readily than elements in period 3 and below. There are no compounds known that contain covalent bonds to Ne. Thus, N is the element from the list that forms p bonds most readily.

samPlE ExERCisE 22.1 identifying Elemental Properties

SEcTION 22.1 Periodic Trends and chemical reactions 955

The ready ability of period 2 elements to form p bonds is an important factor in determining the elemental forms of these elements. Compare, for example, carbon and silicon. Carbon has five major crystalline allotropes: diamond, graphite, buckminster-fullerene, graphene, and carbon nanotubes. (Sections 12.7 and 12.9) Diamond is a covalent-network solid that has C ¬ C s bonds but no p bonds. Graphite, buckminster-fullerene, graphene, and carbon nanotubes have p bonds that result from the sideways overlap of p orbitals. Elemental silicon, however, exists only as a diamond-like covalent-network solid with s bonds; it has no forms analogous to graphite, buckminsterfuller-ene, graphene, or carbon nanotubes, apparently because Si ¬ Si p bonds are too weak.

We likewise see significant differences in the dioxides of carbon and silicon as a result of their relative abilities to form p bonds (▶ Figure 22.3). CO2 is a molecular substance containing C “ O double bonds, whereas SiO2 is a covalent-network solid in which four oxygen atoms are bonded to each silicon atom by single bonds, forming an extended structure that has the empirical formula SiO2.

Give it Some thoughtNitrogen is found in nature as N21g2. Would you expect phosphorus to be found in nature as P21g2? Explain.

Chemical ReactionsBecause O2 and H2O are abundant in our environment, it is particularly important to consider how these substances react with other compounds. About one-third of the re-actions discussed in this chapter involve either O2 (oxidation or combustion reactions) or H2O (especially proton-transfer reactions).

In combustion reactions (Section 3.2), hydrogen-containing compounds pro-duce H2O. Carbon-containing ones produce CO2 (unless the amount of O2 is insuf-ficient, in which case CO or even C can form). Nitrogen-containing compounds tend to form N2, although NO can form in special cases or in small amounts. A reaction illustrating these points is:

4 CH3NH21g2 + 9 O21g2 ¡ 4 CO21g2 + 10 H2O1g2 + 2 N21g2 [22.1]

The formation of H2O, CO2, and N2 reflects the high thermodynamic stability of these substances, indicated by the large bond energies for the O ¬ H, C “ O, and N ‚ N bonds (463, 799, and 941 kJ/mol, respectively). (Section 8.8)

When dealing with proton-transfer reactions, remember that the weaker a Brønsted–Lowry acid, the stronger its conjugate base. (Section 16.2) For example, H2, OH-, NH3, and CH4 are exceedingly weak proton donors that have no tendency to act as acids in water. Thus, the species formed by removing one or more protons from them are extremely strong bases. All of them react readily with water, removing protons from H2O to form OH- . Two representative reactions are:

CH3- 1aq2 + H2O1l2 ¡ CH41g2 + OH- 1aq2 [22.2]

N3 - 1aq2 + 3 H2O1l2 ¡ NH31aq2 + 3 OH- 1aq2 [22.3]

Practice Exercise 1Which description correctly describes a difference between the chemistry of oxygen and sulfur?(a) Oxygen is a nonmetal and sulfur is a metalloid. (b) Oxygen can form more than four bonds whereas sulfur cannot. (c) Sulfur has a higher electronegativity than oxygen. (d) Oxygen is better able to form p bonds than sulfur.

Practice Exercise 2Of the elements Be, C, Cl, Sb, and Cs, which (a) has the lowest electronegativity, (b) has the greatest nonmetallic character, (c) is most likely to participate in extensive p bonding, (d) is most likely to be a metalloid?

▲ Figure 22.3 comparison of the bonds in SIO2 and CO2.

CO2; C forms double bonds

Fragment of extended SiO2 lattice;Si forms only single bonds


22.2 | hydrogenThe English chemist Henry Cavendish (1731–1810) was the first to isolate hydrogen. Because the element produces water when burned in air, the French chemist Antoine Lavoisier (Figure 3.1) gave it the name hydrogen, which means “water producer” (Greek: hydro, water; gennao, to produce).

Hydrogen is the most abundant element in the universe. It is the nuclear fuel consumed by our Sun and other stars to produce energy. (Section 21.8) Although about 75% of the known mass of the universe is hydrogen, it constitutes only 0.87% of Earth’s mass. Most of the hydrogen on our planet is found associated with oxygen. Water, which is 11% hydrogen by mass, is the most abundant hydro-gen compound.

Isotopes of HydrogenThe most common isotope of hydrogen, 11H, has a nucleus consisting of a single proton. This isotope, sometimes referred to as protium,* makes up 99.9844% of naturally oc-curring hydrogen.

Two other isotopes are known: 21H, whose nucleus contains a proton and a neu-

tron, and 31H, whose nucleus contains a proton and two neutrons. The 2

1H isotope, deuterium, makes up 0.0156% of naturally occurring hydrogen. It is not radioactive and is often given the symbol D in chemical formulas, as in D2O (deuterium oxide), which is known as heavy water.

Because an atom of deuterium is about twice as massive as an atom of pro-tium, the properties of deuterium-containing substances vary somewhat from those

solutioNAnalyze We are given the reactants for two chemical equations and asked to predict the products and then balance the equations.

samPlE ExERCisE 22.2 Predicting the Products of Chemical Reactions

Predict the products formed in each of the following reactions, and write a balanced equation:(a) CH3NHNH21g2 + O21g2 ¡ ? (b) Mg3P21s2 + H2O1l2 ¡ ?

Practice Exercise 1When CaC2 reacts with water, what carbon-containing compound forms?(a) CO, (b) CO2, (c) CH4, (d) C2H2, (e) H2CO3.

Practice Exercise 2Write a balanced equation for the reaction of solid sodium hydride with water.

Plan We need to examine the reactants to see if we might recognize a reaction type. In (a) the carbon compound is reacting with O2, which suggests a combustion reaction. In (b) water reacts with an

ionic compound. The anion P3 - is a strong base and H2O is able to act as an acid, so the reactants suggest an acid–base (proton-transfer) reaction.

Solve

(a) Based on the elemental composition of the carbon com-pound, this combustion reaction should produce CO2, H2O, and N2: 2 CH3NHNH21g2 + 5 O21g2 ¡ 2 CO21g2 + 6 H2O1g2 + 2 N21g2

(b) Mg3P2 is ionic, consisting of Mg2 + and P3 - ions. The P3 - ion, like N3 - , has a strong affinity for protons and reacts with H2O to form OH- and PH3 (PH2 - , PH2

- , and PH3 are all exceedingly weak proton donors). Mg3P21s2 + 6 H2O1l2 ¡ 2 PH31g2 + 3 Mg1OH221s2

Mg1OH22 has low solubility in water and will precipitate.

*Giving unique names to isotopes is limited to hydrogen. Because of the proportionally large differences in their masses, the isotopes of H show appreciably more differences in their properties than isotopes of heavier elements.

SEcTION 22.2 Hydrogen 957

of the protium-containing analogs. For example, the normal melting and boiling points of D2O are 3.81 °C and 101.42 °C, respectively, versus 0.00 °C and 100.00 °C for H2O. Not surprisingly, the density of D2O at 25 °C 11.104 g>mL2 is greater than that of H2O 10.997 g>mL2. Replacing protium with deuterium (a process called deu-teration) can also have a profound effect on reaction rates, a phenomenon called a kinetic-isotope effect. For example, heavy water can be obtained from the electrolysis 32 H2O1l2 ¡ 2 H21g2 + O21g24 of ordinary water because the small amount of nat-urally occurring D2O in the sample undergoes electrolysis more slowly than H2O and, therefore, becomes concentrated during the reaction.

The third isotope, 31H, tritium, is radioactive, with a half-life of 12.3 yr:

31H ¡ 32He + 0

- 1e t1>2 = 12.3 yr [22.4]

Because of its short half-life, only trace quantities of tritium exist naturally. The isotope can be synthesized in nuclear reactors by neutron bombardment of lithium-6:

63Li + 1

0n ¡ 31H + 42He [22.5]

Deuterium and tritium are useful in studying reactions of compounds containing hydrogen. A compound is “labeled” by replacing one or more ordinary hydrogen atoms with deuterium or tritium at specific locations in a molecule. By comparing the locations of the label atoms in reactants and products, the reaction mechanism can often be inferred. When methyl alcohol 1CH3OH2 is placed in D2O, for example, the H atom of the O ¬ H bond exchanges rapidly with the D atoms, forming CH3OD. The H atoms of the CH3 group do not exchange. This experiment demonstrates the kinetic stability of C ¬ H bonds and reveals the speed at which the O ¬ H bond in the molecule breaks and re-forms.

Properties of HydrogenHydrogen is the only element that is not a member of any family in the periodic table. Because of its ls1 electron configuration, it is generally placed above lithium in the table. However, it is definitely not an alkali metal. It forms a positive ion much less read-ily than any alkali metal. The ionization energy of the hydrogen atom is 1312 kJ>mol, whereas that of lithium is 520 kJ>mol.

Hydrogen is sometimes placed above the halogens in the periodic table because the hydrogen atom can pick up one electron to form the hydride ion, H- , which has the same electron configuration as helium. However, the electron affinity of hydrogen, E = -73 kJ>mol, is not as large as that of any halogen. In general, hydrogen shows no closer resemblance to the halogens than it does to the alkali metals.

Elemental hydrogen exists at room temperature as a colorless, odorless, tasteless gas composed of diatomic molecules. We can call H2 dihydrogen, but it is more commonly referred to as either molecular hydrogen or simply hydrogen. Because H2 is nonpolar and has only two electrons, attractive forces between molecules are extremely weak. As a result, its melting point 1-259 °C2 and boiling point 1-253 °C2 are very low.

The H ¬ H bond enthalpy 1436 kJ>mol2 is high for a single bond. (Table 8.4) By comparison, the Cl ¬ Cl bond enthalpy is only 242 kJ>mol. Because H2 has a strong bond, most reactions involving H2 are slow at room temperature. However, the mol-ecule is readily activated by heat, irradiation, or catalysis. The activation generally pro-duces hydrogen atoms, which are very reactive. Once H2 is activated, it reacts rapidly and exothermically with a wide variety of substances.

Give it Some thoughtIf H2 is activated to produce H+, what must the other product be?

Hydrogen forms strong covalent bonds with many other elements, including oxy-gen; the O ¬ H bond enthalpy is 463 kJ>mol. The formation of the strong O ¬ H bond makes hydrogen an effective reducing agent for many metal oxides. When H2 is passed over heated CuO, for example, copper is produced:

CuO1s2 + H21g2 ¡ Cu1s2 + H2O1g2 [22.6]


When H2 is ignited in air, a vigorous reaction occurs, forming H2O. Air containing as little as 4% H2 by volume is potentially explosive. Combustion of hydrogen–oxygen mixtures is used in liquid-fuel rocket engines such as those of the Space Shuttle. The hydrogen and oxygen are stored at low temperatures in liquid form.

Production of HydrogenWhen a small quantity of H2 is needed in the laboratory, it is usually obtained by the reaction between an active metal such as zinc and a dilute strong acid such as HCl or H2SO4:

Zn1s2 + 2 H+1aq2 ¡ Zn2 + 1aq2 + H21g2 [22.7]

Large quantities of H2 are produced by reacting methane with steam at 1100 °C. We can view this process as involving two reactions:

CH41g2 + H2O1g2 ¡ CO1g2 + 3 H21g2 [22.8]

CO1g2 + H2O1g2 ¡ CO21g2 + H21g2 [22.9]

Carbon heated with water to about 1000 °C is another source of H2:

C1s2 + H2O1g2 ¡ H21g2 + CO1g2 [22.10]

This mixture, known as water gas, is used as an industrial fuel.Electrolysis of water consumes too much energy and is consequently too costly

to be used commercially to produce H2. However, H2 is produced as a by-product in the electrolysis of brine (NaCl) solutions in the course of commercial Cl2 and NaOH manufacture:

2 NaCl1aq2 + 2 H2O1l2 ¡electrolysis H21g2 + Cl21g2 + 2 NaOH1aq2 [22.11]

Give it Some thoughtWhat are the oxidation states of the H atoms in Equations 22.7–22.11?

only hydrogen and oxygen. (Figure 1.7 and Section 20.9) However, the energy required to electrolyze water must come from somewhere. If we burn fossil fuels to generate this energy, we have not advanced very far toward a true hydrogen economy. If the energy for electrolysis came instead from a hydroelectric or nuclear power plant, solar cells, or wind generators, consumption of nonrenewable energy sources and undesired production of CO2 could be avoided.

The storage of hydrogen is another technical obstacle that must be overcome in developing a hydrogen economy. Although H2(g) has a high energy density by mass, it has a low energy density by volume. Thus, storing hydrogen as a gas requires a large volume compared to the energy it delivers. There are also safety issues associated with handling and storing the gas because its combustion can be explosive. Storing hydrogen in the form of various hydride compounds such as LiAlH4 is being investigated as a means of reducing the volume and increasing the safety. One problem with this approach, however, is that such compounds have high energy density by volume but low en-ergy density by mass.

Related Exercises: 22.29, 22.30, 22.94

Closer look

the hydrogen economy

The reaction of hydrogen with oxygen is highly exothermic:

2 H21g2 + O21g2 ¡ 2 H2O1g2 ∆H = -483.6 kJ [22.12]

Because H2 has a low molar mass and a high enthalpy of com-bustion, it has a high energy density by mass. (That is, its combustion produces high energy per gram.) Furthermore, the only product of the reaction is water vapor, which means that hydrogen is environmen-tally cleaner than fossil fuels. Thus, the prospect of using hydrogen widely as a fuel is attractive.

The term “hydrogen economy” is used to describe the con-cept of delivering and using hydrogen as a fuel in place of fossil fu-els. In order to develop a hydrogen economy, it would be necessary to generate elemental hydrogen on a large scale and arrange for its transport and storage. These matters provide significant technical challenges.

▶ Figure 22.4 illustrates various sources and uses of H2 fuel. The generation of H2 through electrolysis of water is in principle the clean-est route, because this process—the reverse of Equation 22.11—produces

SEcTION 22.2 Hydrogen 959

Uses of HydrogenHydrogen is commercially important. About 5.0 * 1010 kg (50 million metric tons) is produced annually across the world. About half of the H2 produced is used to syn-thesize ammonia by the Haber process. (Section 15.2) Much of the remaining hy-drogen is used to convert high-molecular-weight hydrocarbons from petroleum into lower-molecular-weight hydrocarbons suitable for fuel (gasoline, diesel, and others) in a process known as cracking. Hydrogen is also used to manufacture methanol via the catalytic reaction of CO and H2 at high pressure and temperature:

CO1g2 + 2 H21g2 ¡ CH3OH1g2 [22.13]

Binary Hydrogen CompoundsHydrogen reacts with other elements to form three types of compounds: (1) ionic hy-drides, (2) metallic hydrides, and (3) molecular hydrides.

The ionic hydrides are formed by the alkali metals and by the heavier alka-line earths (Ca, Sr, and Ba). These active metals are much less electronegative than hydrogen. Consequently, hydrogen acquires electrons from them to form hydride ions 1H- 2:

Ca1s2 + H21g2 ¡ CaH21s2 [22.14]

The hydride ion is very basic and reacts readily with compounds having even weakly acidic protons to form H2:

H- 1aq2 + H2O1l2 ¡ H21g2 + OH- 1aq2 [22.15]

Ionic hydrides can therefore be used as convenient (although expensive) sources of H2.Calcium hydride 1CaH22 is used to inflate life rafts, weather balloons, and the like

where a simple, compact means of generating H2 is desired (Figure 22.5).

▲ Figure 22.4 the “hydrogen economy” would require hydrogen to be produced from various sources and would use hydrogen in energy-related applications.

Transport

Renewable energies

Indu

stry

Solarthermal

Naturalgas

Hydroelectric,solar photovoltaics,wind

Electrical nuclear

Thermalnuclear

Turbines,internalcombustionengines

Processes,syntheses

Thermal

OtherResidentialCom-mercial

Coal

Fuel cellengines

Internalcombustionengines

Demand

Supply

Biomass

Buildings

Renewable energies

Indu

stry

Transport

H2

Fuel cells

Electrolysis


The reaction between H- and H2O (Equation 22.15) is an acid–base reaction and a redox reaction. The H- ion, therefore, is a good base and a good reducing agent. In fact, hydrides are able to reduce O2 to OH- :

2 NaH1s2 + O21g2 ¡ 2 NaOH1s2 [22.16]

For this reason, hydrides are normally stored in an environment that is free of both moisture and air.

Metallic hydrides are formed when hydrogen reacts with transition metals. These compounds are so named because they retain their metallic properties. In many metallic hydrides, the ratio of metal atoms to hydrogen atoms is not fixed or in small whole numbers. The composition can vary within a range, depending on reaction conditions. TiH2 can be produced, for example, but preparations usually yield TiH1.8. These nonstoichiometric metallic hydrides are sometimes called inter-stitial hydrides. Because hydrogen atoms are small enough to fit between the sites occupied by the metal atoms, many metal hydrides behave like interstitial alloys.

(Section 12.3)The molecular hydrides, formed by nonmetals and metalloids, are either gases or liq-

uids under standard conditions. The simple molecular hydrides are listed in ◀ Figure 22.6, together with their standard free energies of formation, ∆G°f . (Section 19.5) In each family, the thermal stability 1measured as ∆G°f2 decreases as we move down the family. (Recall that the more stable a compound is with respect to its elements under standard conditions, the more negative ∆G°f is.)

22.3 | Group 8a: the Noble GasesThe elements of group 8A are chemically unreactive. Indeed, most of our references to these elements have been in relation to their physical properties, as when we discussed intermolecular forces. (Section 11.2) The relative inertness of these elements is due to the presence of a completed octet of valence-shell electrons (except He, which only

▲ Figure 22.5 the reaction of CaH2 with water.

Go FiGuREThis reaction is exothermic. Is the beaker on the right warmer or colder than the beaker on the left?

CaH2

H2 gas

H2O with pH indicator

Color change indicates presence of OH−

4A 5A 6A 7A

CH4(g)−50.8

NH3(g)−16.7

H2O(l)−237

HF(g)−271

SiH4(g)+56.9

PH3(g)+18.2

H2S(g)−33.0

HCl(g)−95.3

GeH4(g)+117

AsH3(g)+111

H2Se(g)+71

HBr(g)−53.2

SbH3(g)+187

H2Te(g)+138

HI(g)+1.30

▲ Figure 22.6 Standard free energies of formation of molecular hydrides. All values are kilojoules per mole of hydride.

Go FiGuREWhich is the most thermodynamically stable hydride? Which is the least thermodynamically stable?

SEcTION 22.3 Group 8A: The Noble Gases 961

has a filled 1s shell). The stability of such an arrangement is reflected in the high ioniza-tion energies of the group 8A elements. (Section 7.4)

The group 8A elements are all gases at room temperature. They are components of Earth’s atmosphere, except for radon, which exists only as a short-lived radioisotope.

(Section 21.9) Only argon is relatively abundant. (Table 18.1)Neon, argon, krypton, and xenon are used in lighting, display, and laser applica-

tions in which the atoms are excited electrically and electrons that are in a higher energy state emit light as they fall to the ground state. (Section 6.2) Argon is used as a blanketing atmosphere in electric lightbulbs. The gas conducts heat away from the fila-ment but does not react with it. Argon is also used as a protective atmosphere to prevent oxidation in welding and certain high-temperature metallurgical processes.

Helium is in many ways the most important noble gas. Liquid helium is used as a coolant to conduct experiments at very low temperatures. Helium boils at 4.2 K and 1 atm, the lowest boiling point of any substance. It is found in relatively high concen-trations in many natural-gas wells from which it is isolated.

Noble-Gas CompoundsBecause the noble gases are exceedingly stable, they react only under rigorous condi-tions. We expect the heavier ones to be most likely to form compounds because their ionization energies are lower. (Figure 7.9) A lower ionization energy suggests the possibility of sharing an electron with another atom, leading to a chemical bond. In ad-dition, because the group 8A elements (except helium) already contain eight electrons in their valence shell, formation of covalent bonds will require an expanded valence shell. Valence-shell expansion occurs most readily with larger atoms. (Section 8.7)

The first noble-gas compound was reported in 1962. This discovery caused a sen-sation because it undercut the belief that the noble-gas elements were inert. The initial study involved xenon in combination with fluorine, the element we would expect to be most reactive in pulling electron density from another atom. Since that time chemists have prepared several xenon compounds of fluorine and oxygen (▼ table 22.1). The fluorides XeF2, XeF4, and XeF6 are made by direct reaction of the elements. By varying the ratio of reactants and altering reaction conditions, one of the three compounds can be obtained. The oxygen-containing compounds are formed when the fluorides react with water as, for example,

XeF61s2 + 3 H2O1l2 ¡ XeO31aq2 + 6 HF1aq2 [22.17]

The other noble-gas elements form compounds much less readily than xenon. For many years, only one binary krypton compound, KrF2, was known with certainty, and

8A2He

10Ne

18Ar

36Kr54Xe

86Rn

table 22.1 Properties of xenon Compounds

Compound oxidation state of xe melting Point 1 °C 2 ∆H°f 1kJ ,mol 2aXeF2 +2 129 -1091g2XeF4 +4 117 -2181g2XeF6 +6 49 -2981g2XeOF4 +6 -41 to -28 +1461l2XeO3 +6 -b +4021s2XeO2F2 +6 31 +1451s2XeO4 +8 - c -aAt 25 °C, for the compound in the state indicated.bA solid; decomposes at 40 °C.cA solid; decomposes at -40 °C.


it decomposes to its elements at -10 °C. Other compounds of krypton have been isolated at very low temperatures (40 K).

solutioNAnalyze We must predict the geometrical structure given only the molecular formula.

Plan We must first write the Lewis structure for the molecule. We then count the number of electron pairs (domains) around the Xe atom and use that number and the number of bonds to predict the geometry.

Solve There are 36 valence-shell electrons (8 from xenon and 7 from each fluorine). If we make four single Xe ¬ F bonds, each fluorine has its octet satisfied. Xe then has 12 electrons in its valence shell, so we expect an octahedral arrangement of six elec-tron pairs. Two of these are nonbonded pairs. Because nonbonded pairs require more volume than bonded pairs (Section 9.2), it is reasonable to expect these nonbonded pairs to be opposite each other. The expected structure is square planar, as shown in ▶ Figure 22.7.

Comment The experimentally determined structure agrees with this prediction.

samPlE ExERCisE 22.3 Predicting a molecular structure

Use the VSEPR model to predict the structure of XeF4.

Practice Exercise 1Compounds containing the XeF3

+ ion have been characterized. Describe the electron-domain geometry and molecular geometry of this ion.(a) trigonal planar, trigonal planar; (b) tetrahedral, trigonal pyramidal; (c) trigonal bipyramidal, T shaped; (d) tetrahedral, tetrahedral; (e) octahedral, square planar.

Practice Exercise 2Describe the electron-domain geometry and molecular geometry of KrF2.

▲ Figure 22.7 Xenon tetrafluoride.

XeF F

F F

22.4 | Group 7a: the halogensThe elements of group 7A, the halogens, have the outer-electron configuration ns2np5, where n ranges from 2 through 6. The halogens have large negative electron affinities

(Section 7.5), and they most often achieve a noble-gas configuration by gaining an electron, which results in a -1 oxidation state. Fluorine, being the most electronegative element, exists in compounds only in the -1 state. The other halogens exhibit positive oxidation states up to +7 in combination with more electronegative atoms such as O. In the positive oxidation states, the halogens tend to be good oxidizing agents, readily accepting electrons.

Chlorine, bromine, and iodine are found as the halides in seawater and in salt deposits. Fluorine occurs in the minerals fluorspar 1CaF22, cryolite 1Na3AlF62, and flu-orapatite 3Ca51PO423F4.* Only fluorspar is an important commercial source of fluorine.

All isotopes of astatine are radioactive. The longest-lived isotope is astatine-210, which has a half-life of 8.1 h and decays mainly by electron capture. Because astatine is so unstable, very little is known about its chemistry.

Properties and Production of the HalogensMost properties of the halogens vary in a regular fashion as we go from fluorine to iodine (▶ table 22.2).

Under ordinary conditions the halogens exist as diatomic molecules. The molecules are held together in the solid and liquid states by dispersion forces. (Section 11.2) Because I2 is the largest and most polarizable halogen molecule, the intermolecular forces between I2 molecules are the strongest. Thus, I2 has the highest melting point and boil-ing point. At room temperature and 1 atm, I2 is a purple solid, Br2 is a red-brown liquid, and Cl2 and F2 are gases. (Figure 7.27) Chlorine readily liquefies upon compression

7A9F

17Cl

35Br

53I85At

*Minerals are solid substances that occur in nature. They are usually known by their common names rather than by their chemical names. What we know as rock is merely an aggregate of different minerals.

SEcTION 22.4 Group 7A: The Halogens 963

at room temperature and is normally stored and handled in liquid form under pressure in steel containers.

The comparatively low bond enthalpy of F2 1155 kJ>mol2 accounts in part for the extreme reactivity of elemental fluorine. Because of its high reactivity, F2 is difficult to work with. Certain metals, such as copper and nickel, can be used to contain F2 because their surfaces form a protective coating of metal fluoride. Chlorine and the heavier hal-ogens are also reactive, although less so than fluorine.

Because of their high electronegativities, the halogens tend to gain electrons from other substances and thereby serve as oxidizing agents. The oxidizing ability of the halogens, indicated by their standard reduction potentials, decreases going down the group. As a result, a given halogen is able to oxidize the halide anions below it. For example, Cl2 oxidizes Br - and I - but not F - , as seen in ▶ Figure 22.8.

table 22.2 some Properties of the halogens

Property F Cl Br i

Atomic radius 1A° 2 0.57 1.02 1.20 1.39

Ionic radius, X- 1A° 2 1.33 1.81 1.96 2.20

First ionization energy 1kJ>mol2 1681 1251 1140 1008

Electron affinity 1kJ>mol2 -328 -349 -325 -295

Electronegativity 4.0 3.0 2.8 2.5

X ¬ X single-bond enthalpy 1kJ>mol2 155 242 193 151

Reduction potential (V):12X21aq2 + e - ¡ X- 1aq2 2.87 1.36 1.07 0.54

NaF(aq) NaBr(aq) NaI(aq)

Cl2(aq) added

CCl4 layer

Noreaction

Br−oxidized

to Br2

I−oxidized

to I2

▲ Figure 22.8 reaction of Cl2 with aqueous solutions of NaF, Nabr, and Nai in the presence of carbon tetrachloride. The top liquid layer in each vial is water; the bottom liquid layer is carbon tetrachloride. The Cl21aq2, which has been added to each vial, is colorless. The brown color in the carbon tetrachloride layer indicates the presence of Br2, whereas purple indicates the presence of I2.

Go FiGuREDo Br2 and I2 appear to be more or less soluble in CCl4 than in H2O?

Write the balanced equation for the reaction, if any, between (a) I - 1aq2 and Br21l2, (b) Cl- 1aq2 and I21s2.

solutioNAnalyze We are asked to determine whether a reaction occurs when a particular halide and halogen are combined.Plan A given halogen is able to oxidize anions of the halogens below it in the periodic table. Thus, in each pair the halogen having the smaller atomic number ends up as the halide ion. If the halogen with the smaller atomic number is already the halide ion, there is no reaction. Thus, the key to determining whether a reaction occurs is locating the elements in the periodic table.Solve

(a) Br2 can oxidize (remove electrons from) the anions of the halogens below it in the periodic table. Thus, it oxidizes I - :

2 I - 1aq2 + Br21aq2 ¡ I21s2 + 2 Br - 1aq2(b) Cl- is the anion of a halogen above iodine in the periodic table. Thus, I2 cannot oxidize Cl- ;

there is no reaction.

Practice Exercise 1Which is (are) able to oxidize Cl- ?(a) F2

(b) F -

(c) Both Br2 and I2

(d) Both Br - and I -

Practice Exercise 2Write the balanced chemical equation for the reaction between Br - 1aq2 and Cl21aq2.

samPlE ExERCisE 22.4 Predicting Chemical Reactions among

the halogens


Notice in Table 22.2 that the standard reduction potential of F2 is exceptionally high. As a result, fluorine gas readily oxidizes water:

F21aq2 + H2O1l2 ¡ 2 HF1aq2 + 12 O21g2 E° = 1.80 V [22.18]

Fluorine cannot be prepared by electrolytic oxidation of aqueous solutions of fluo-ride salts because water is oxidized more readily than F - . (Section 20.9) In practice, the element is formed by electrolytic oxidation of a solution of KF in anhydrous HF.

Chlorine is produced mainly by electrolysis of either molten or aqueous sodium chloride. Both bromine and iodine are obtained commercially from brines containing the halide ions; the reaction used is oxidation with Cl2.

Uses of the HalogensFluorine is used to prepare fluorocarbons—very stable carbon–fluorine com-pounds used as refrigerants, lubricants, and plastics. Teflon® (◀ Figure 22.9) is a polymeric fluorocarbon noted for its high thermal stability and lack of chemical reactivity.

Chlorine is by far the most commercially important halogen. About 1 * 1010 kg (10 million tons) of Cl2 is produced annually in the United States. In addition, hydro-gen chloride production is about 4.0 * 109 kg (4.4 million tons) annually. About half of this chlorine finds its way eventually into the manufacture of chlorine-containing organic compounds, such as the vinyl chloride 1C2H3Cl2 used in making polyvinyl chloride (PVC) plastics. (Section 12.8) Much of the remainder is used as a bleach-ing agent in the paper and textile industries.

When Cl2 dissolves in cold dilute base, it converts into Cl- and hypochlorite, ClO - :

Cl21aq2 + 2 OH- 1aq2 ∆ Cl - 1aq2 + ClO - 1aq2 + H2O1l2 [22.19]

Sodium hypochlorite (NaClO) is the active ingredient in many liquid bleaches. Chlorine is also used in water treatment to oxidize and thereby destroy bacteria.

(Section 18.4)

Give it Some thoughtWhat is the oxidation state of Cl in each Cl species in Equation 22.19?

A common use of iodine is as KI in table salt. Iodized salt provides the small amount of iodine necessary in our diets; it is essential for the formation of thyroxin, a hormone secreted by the thyroid gland. Lack of iodine in the diet results in an enlarged thyroid gland, a condition called goiter.

The Hydrogen HalidesAll the halogens form stable diatomic molecules with hydrogen. Aqueous solutions of HCl, HBr, and HI are strong acids. The hydrogen halides can be formed by di-rect reaction of the elements. The most important means of preparing HF and HCl, however, is by reacting a salt of the halide with a strong nonvolatile acid, as in the reaction

CaF21s2 + H2SO41l2 ¡∆ 2 HF1g2 + CaSO41s2 [22.20]

Neither HBr nor HI can be prepared in this way, however, because H2SO4 oxi-dizes Br - and I - (▶ Figure 22.10). This difference in reactivity reflects the greater ease of oxidation of Br - and I - relative to F - and Cl- . These undesirable oxidations are avoided by using a nonvolatile acid, such as H3PO4, that is a weaker oxidizing agent than H2SO4.

▲ Figure 22.9 Structure of teflon®, a fluorocarbon polymer.

Go FiGuREWhat is the repeating unit in this polymer?

SEcTION 22.4 Group 7A: The Halogens 965

I2formed

Br2formed

H2SO4

NaI NaBr

▲ Figure 22.10 reaction of H2SO4 with Nai and Nabr.

Go FiGuREAre these reactions acid–base reactions or oxidation–reduction reactions?

solutioNAnalyze We are asked to write a balanced equation for the reaction between NaBr and H3PO4 to form HBr and another product.Plan As in Equation 22.20, a metathesis reaction takes place.

(Section 4.2) Let’s assume that only one H in H3PO4 reacts. (The actual number depends on the reaction conditions.) The H2PO4

- and Na+ will form NaH2PO4 as one product.Solve The balanced equation is

NaBr1s2 + H3PO41l2 ¡ NaH2PO41s2 + HBr1g2

samPlE ExERCisE 22.5 Writing a Balanced Chemical Equation

Write a balanced equation for the formation of hydrogen bromide gas from the reaction of solid sodium bromide with phosphoric acid.

Practice Exercise 1Which of the following are oxidized by H2SO4?(a) Cl- , (b) Cl- and Br - , (c) Br - and I - , (d) Cl2, (e) Br2 and I2.

Practice Exercise 2Write the balanced equation for the preparation of HI from NaI and H3PO4.

The hydrogen halides form hydrohalic acid solutions when dissolved in water. These solutions have the characteristic properties of acids, such as reactions with active metals to produce hydrogen gas. (Section 4.4) Hydrofluoric acid also reacts readily with silica 1SiO22 and with silicates to form hexafluorosilicic acid 1H2SiF62:

SiO21s2 + 6 HF1aq2 ¡ H2SiF61aq2 + 2 H2O1l2 [22.21]

Interhalogen CompoundsBecause the halogens exist as diatomic molecules, diatomic molecules made up of two dif-ferent halogen atoms exist. These compounds are the simplest examples of interhalogens, compounds, such as ClF and IF5, formed between two halogen elements.

The vast majority of the higher interhalogen compounds have a central Cl, Br, or I atom surrounded by fluorine atoms. The large size of the iodine atom allows the forma-tion of IF3, IF5, and IF7, in which the oxidation state of I is +3, +5, and +7, respectively.


With the smaller bromine and chlorine atoms, only compounds with 3 or 5 fluorines form. The only higher interhalogen compounds that do not have outer F atoms are ICl3 and ICl5; the large size of the I atom can accommodate 5 Cl atoms, whereas Br is not large enough to allow even BrCl3 to form. All of the interhalogen compounds are powerful oxidizing agents.

Oxyacids and Oxyanions▲ table 22.3 summarizes the formulas of the known oxyacids of the halogens and the way they are named*. (Section 2.8) The acid strengths of the oxyacids increase with increasing oxidation state of the central halogen atom. (Section 16.10) All the oxy-acids are strong oxidizing agents. The oxyanions, formed on removal of H+ from the oxyacids, are generally more stable than the oxyacids. Hypochlorite salts are used as bleaches and disinfectants because of the powerful oxidizing capabilities of the ClO - ion. Chlorate salts are similarly very reactive. For example, potassium chlorate is used to make matches and fireworks.

Give it Some thoughtWhich do you expect to be the stronger oxidizing agent, NaBrO3 or NaClO3?

Perchloric acid and its salts are the most stable oxyacids and oxyanions. Dilute solutions of perchloric acid are quite safe, and many perchlorate salts are stable except when heated with organic materials. When heated, however, perchlorates can become vigorous, even violent, oxidizers. Considerable caution should be exercised, therefore, when handling these substances, and it is crucial to avoid contact between perchlorates and readily oxidized material. The use of ammonium perchlorate 1NH4ClO42 as the oxidizer in the solid booster rockets for the Space Shuttle demonstrates the oxidizing power of perchlorates. The solid propellant contains a mixture of NH4ClO4 and pow-dered aluminum, the reducing agent. Each shuttle launch requires about 6 * 105 kg (700 tons) of NH4ClO4 (◀ Figure 22.11).

22.5 | oxygenBy the middle of the seventeenth century, scientists recognized that air contained a component associated with burning and breathing. That component was not iso-lated until 1774, however, when English scientist Joseph Priestley discovered oxygen. Lavoisier subsequently named the element oxygen, meaning “acid former.”

Oxygen is found in combination with other elements in a great variety of compounds—water 1H2O2, silica 1SiO22, alumina 1Al2O32, and the iron oxides

*Fluorine forms one oxyacid, HOF. Because the electronegativity of fluorine is greater than that of oxygen, we must consider fluorine to be in a -1 oxidation state and oxygen to be in the 0 oxidation state in this compound.

table 22.3 the stable oxyacids of the halogens

Formula of acid

oxidation state of halogen Cl Br i acid Name

+1 HClO HBrO HIO Hypohalous acid+3 HClO2 — — Halous acid+5 HClO3 HBrO3 HIO3 Halic acid+7 HClO4 HBrO4 HIO4 Perhalic acid

▲ Figure 22.11 launch of the Space Shuttle Columbia from the Kennedy Space center.

10 Al(s) + 6 NH4ClO4(s)

4 Al2O3(s) + 2 AlCl3(s)

+ 12 H2O(g) + 3 N2(g)

The large volume of gases produced provides thrust for the booster rockets

SEcTION 22.5 Oxygen 967

1Fe2O3, Fe3O42 are obvious examples. Indeed, oxygen is the most abundant element by mass both in Earth’s crust and in the human body. (Section 1.2) It is the oxidizing agent for the metabolism of our foods and is crucial to human life.

Properties of OxygenOxygen has two allotropes, O2 and O3. When we speak of molecular oxygen or sim-ply oxygen, it is usually understood that we are speaking of dioxygen 1O22, the normal form of the element; O3 is ozone.

At room temperature, dioxygen is a colorless, odorless gas. It condenses to a liquid at -183 °C and freezes at -218 °C. It is only slightly soluble in water (0.04 g>L, or 0.001 M at 25 °C), but its presence in water is essential to marine life.

The electron configuration of the oxygen atom is 3He42s22p4. Thus, oxygen can complete its octet of valence electrons either by picking up two electrons to form the oxide ion 1O2 - 2 or by sharing two electrons. In its covalent compounds, it tends to form either two single bonds, as in H2O, or a double bond, as in formaldehyde 1H2C “ O2. The O2 molecule contains a double bond. The bond in O2 is very strong (bond enthalpy 495 kJ>mol). Oxygen also forms strong bonds with many other ele-ments. Consequently, many oxygen-containing compounds are thermodynamically more stable than O2. In the absence of a catalyst, however, most reactions of O2 have high activation energies and thus require high temperatures to proceed at a suitable rate. Once a sufficiently exothermic reaction begins, it may accelerate rapidly, produc-ing a reaction of explosive violence.

Production of OxygenNearly all commercial oxygen is obtained from air. The normal boiling point of O2 is -183 °C, whereas that of N2, the other principal component of air, is -196 °C. Thus, when air is liquefied and then allowed to warm, the N2 boils off, leaving liquid O2 con-taminated mainly by small amounts of N2 and Ar.

In the laboratory, O2 can be obtained by heating either aqueous hydrogen peroxide or solid potassium chlorate 1KClO32:

2 KClO31s2 ¡ 2 KCl1s2 + 3 O21g2 [22.22]

Manganese dioxide 1MnO22 catalyzes both reactions.Much of the O2 in the atmosphere is replenished through photosynthesis, in which

green plants use the energy of sunlight to generate O2 (along with glucose, C6H12O6) from atmospheric CO2:

6 CO21g2 + 6 H2O1l2 ¡ C6H12O61aq2 + 6 O21g2

Uses of OxygenIn industrial use, oxygen ranks behind only sulfuric acid 1H2SO42 and nitrogen 1N22. About 3 * 1010 kg (30 million tons) of O2 is used annually in the United States. It is shipped and stored either as a liquid or in steel containers as a compressed gas. About 70% of the O2 output, however, is generated where it is needed.

Oxygen is by far the most widely used oxidizing agent in industry. Over half of the O2 produced is used in the steel industry, mainly to remove impurities from steel. It is also used to bleach pulp and paper. (Oxidation of colored compounds often gives colorless products.) Oxygen is used together with acetylene 1C2H22 in oxyacetylene welding (▶ Figure 22.12). The reaction between C2H2 and O2 is highly exothermic, producing temperatures in excess of 3000 °C.

OzoneOzone is a pale blue, poisonous gas with a sharp, irritating odor. Many people can detect as little as 0.01 ppm in air. Exposure to 0.1 to 1 ppm produces headaches, burning eyes, and irritation to the respiratory passages.

▲ Figure 22.12 Welding with an oxyacetylene torch.

2 C2H2(g) + 5 O2(g)4 CO2(g) + 2 H2O(g); ∆H° = −2510 kJ


The O3 molecule possesses p electrons that are delocalized over the three oxygen atoms. (Section 8.6) The molecule dissociates readily, forming reactive oxygen atoms:

O31g2 ¡ O21g2 + O1g2 ∆H ° = 105 kJ [22.23]

Ozone is a stronger oxidizing agent than dioxygen. Ozone forms oxides with many elements under conditions where O2 will not react; indeed, it oxidizes all the common metals except gold and platinum.

Ozone can be prepared by passing electricity through dry O2 in a flow-through apparatus. The electrical discharge causes the O2 bond to break, resulting in reactions like those described in Section 18.1. During thunderstorms, ozone is generated (and can be smelled, if you are too close) from lightning strikes:

3 O21g2 ¡electricity 2 O31g2 ∆H ° = 285 kJ [22.24]

Ozone is sometimes used to treat drinking water. Like Cl2, ozone kills bacteria and oxidizes organic compounds. The largest use of ozone, however, is in the preparation of pharmaceuticals, synthetic lubricants, and other commercially useful organic com-pounds, where O3 is used to sever carbon–carbon double bonds.

Ozone is an important component of the upper atmosphere, where it screens out ultraviolet radiation and so protects us from the effects of these high-energy rays. For this reason, depletion of stratospheric ozone is a major scientific concern. (Section 18.2) In the lower atmosphere, ozone is considered an air pollutant and is a major constituent of smog. (Section 18.2) Because of its oxidizing power, ozone damages living systems and structural materials, especially rubber.

OxidesThe electronegativity of oxygen is second only to that of fluorine. As a result, oxygen has negative oxidation states in all compounds except OF2 and O2F2. The -2 oxidation state is by far the most common. Compounds that contain oxygen in this oxidation state are called oxides.

Nonmetals form covalent oxides, most of which are simple molecules with low melting and boiling points. Both SiO2 and B2O3, however, have extended structures. Most nonmetal oxides combine with water to give oxyacids. Sulfur dioxide 1SO22, for example, dissolves in water to give sulfurous acid 1H2SO32:

SO21g2 + H2O1l2 ¡ H2SO31aq2 [22.25]

This reaction and that of SO3 with H2O to form H2SO4 are largely responsible for acid rain. (Section 18.2) The analogous reaction of CO2 with H2O to form carbonic acid 1H2CO32 causes the acidity of carbonated water.

Oxides that form acids when they react with water are called either acidic anhydrides (anhydride means “without water”) or acidic oxides. A few nonmetal oxides, especially ones with the nonmetal in a low oxidation state—such as N2O, NO, and CO—do not react with water and are not acidic anhydrides.

Give it Some thoughtWhat acid is produced by the reaction of I2O5 with water?

Most metal oxides are ionic compounds. The ionic oxides that dissolve in water form hydroxides and, consequently, are called either basic anhydrides or basic oxides. Barium oxide, for example, reacts with water to form barium hydroxide (▶ Figure 22.13). These kinds of reactions are due to the high basicity of the O2 - ion and its virtually complete hydrolysis in water:

O2 - 1aq2 + H2O1l2 ¡ 2 OH- 1aq2 [22.26]

Even those ionic oxides that are insoluble in water tend to dissolve in strong acids. Iron(III) oxide, for example, dissolves in acids:

Fe2O31s2 + 6 H+1aq2 ¡ 2 Fe3 + 1aq2 + 3 H2O1l2 [22.27]

SEcTION 22.5 Oxygen 969

This reaction is used to remove rust 1Fe2O3 # nH2O2 from iron or steel before a protec-tive coat of zinc or tin is applied.

Oxides that can exhibit both acidic and basic characters are said to be amphoteric. (Section 17.5) If a metal forms more than one oxide, the basic character of the

oxide decreases as the oxidation state of the metal increases (▼ table 22.4).

H2O(l) Ba(OH)2(aq)BaO(s) +

H2Owithindicator

Pink colorindicatesbasicsolution

▲ Figure 22.13 reaction of a basic oxide with water.

Go FiGuREIs this reaction a redox reaction?

table 22.4 acid–Base Character of Chromium oxides

oxide oxidation state of Cr Nature of oxide

CrO +2 BasicCr2O3 +3 AmphotericCrO3 +6 Acidic

▲ Figure 22.14 A self-contained breathing apparatus.

4 KO2(s) + 2 H2O(l, from breath)4 K+(aq) + 4 OH−(aq) + 3 O2(g)

2 OH−(aq) + CO2(g, from breath)H2O(l) + CO3

2−(aq)

Peroxides and SuperoxidesCompounds containing O ¬ O bonds and oxygen in the -1 oxidation state are perox-ides. Oxygen has an oxidation state of -1

2 in O2- , which is called the superoxide ion. The

most active (easily oxidized) metals (K, Rb, and Cs) react with O2 to give superoxides 1KO2, RbO2, and CsO22. Their active neighbors in the periodic table (Na, Ca, Sr, and Ba) react with O2, producing peroxides 1Na2O2, CaO2, SrO2, and BaO22. Less active metals and nonmetals produce normal oxides. (Section 7.6)

When superoxides dissolve in water, O2 is produced:

4 KO21s2 + 2 H2O1l2 ¡ 4 K+1aq2 + 4 OH- 1aq2 + 3 O21g2 [22.28]

Because of this reaction, potassium superoxide is used as an oxygen source in masks worn by rescue workers (▶ Figure 22.14). For proper breathing in toxic environments, oxygen must be generated in the mask and exhaled carbon dioxide in the mask must be eliminated. Moisture in the breath causes the KO2 to decompose to O2 and KOH, and the KOH removes CO2 from the exhaled breath:

2 OH- 1aq2 + CO21g2 ¡ H2O1l2 + CO32 - 1aq2 [22.29]


Hydrogen peroxide (◀ Figure 22.15) is the most familiar and commercially impor-tant peroxide. Pure hydrogen peroxide is a clear, syrupy liquid that melts at -0.4 °C. Concentrated hydrogen peroxide is dangerously reactive because the decomposition to water and oxygen is very exothermic:

2 H2O21l2 ¡ 2 H2O1l2 + O21g2 ∆H ° = -196.1 kJ [22.30]

This is another example of a disproportionation reaction, in which an element is simultaneously oxidized and reduced. The oxidation number of oxygen changes from -1 to -2 and 0.

Hydrogen peroxide is marketed as a chemical reagent in aqueous solutions of up to about 30% by mass. A solution containing about 3% H2O2 by mass is sold in drug-stores and used as a mild antiseptic. Somewhat more concentrated solutions are used to bleach fabrics.

The peroxide ion is a by-product of metabolism that results from the reduction of O2.The body disposes of this reactive ion with enzymes such as peroxidase and

catalase.

22.6 | the other Group 6a Elements: s, se, te, and Po

The other group 6A elements are sulfur, selenium, tellurium, and polonium. In this sec-tion, we will survey the properties of the group as a whole and then examine the chem-istry of sulfur, selenium, and tellurium. We will not discuss polonium, which has no stable isotopes and is found only in minute quantities in radium-containing minerals.

General Characteristics of the Group 6A ElementsThe group 6A elements possess the general outer-electron configuration ns2np4 with n ranging from 2 to 6. Thus, these elements attain a noble-gas electron configuration by adding two electrons, which results in a -2 oxidation state. Except for oxygen, the group 6A elements are also commonly found in positive oxidation states up to +6, and they can have expanded valence shells. Thus, we have such compounds as SF6, SeF6, and TeF6 with the central atom in the +6 oxidation state.

▼ table 22.5 summarizes some properties of the group 6A elements.

Occurrence and Production of S, Se, and TeSulfur, selenium, and tellurium can all be mined from the earth. Large underground deposits are the principal source of elemental sulfur (◀ Figure 22.16). Sulfur also oc-curs widely as sulfide 1S2 - 2 and sulfate 1SO4

2 - 2 minerals. Its presence as a minor com-ponent of coal and petroleum poses a major problem. Combustion of these “unclean”

111°

▲ Figure 22.15 Molecular structure of hydrogen peroxide. The repulsive interaction of the O ¬ H bonds with the lone-pairs of electrons on each O atom restricts the free rotation around the O ¬ O single bond.

Go FiGuREDoes H2O2 have a dipole moment?

6A8O

16S

34Se52Te

84Po

table 22.5 some Properties of the Group 6a Elements

Property o s se te

Atomic radius 1A° 2 0.66 1.05 1.21 1.38

X2 - ionic radius 1A° 2 1.40 1.84 1.98 2.21

First ionization energy 1kJ>mol2 1314 1000 941 869

Electron affinity 1kJ>mol2 -141 -200 -195 -190

Electronegativity 3.5 2.5 2.4 2.1

X ¬ X single-bond enthalpy 1kJ>mol2 146* 266 172 126

Reduction potential to H2X in acidic solution (V)

1.23 0.14 -0.40 - 0.72

*Based on O ¬ O bond energy in H2O2.

▲ Figure 22.16 Massive amounts of sulfur are extracted every year from the earth.

SEcTION 22.6 The Other Group 6A Elements: S, Se, Te, and Po 971

fuels leads to serious pollution by sulfur oxides. (Section 18.2) Much effort has been directed at removing this sulfur, and these efforts have increased the availability of sulfur.

Selenium and tellurium occur in rare minerals, such as Cu2Se, PbSe, Cu2Te, and PbTe, and as minor constituents in sulfide ores of copper, iron, nickel, and lead.

Properties and Uses of Sulfur, Selenium, and TelluriumElemental sulfur is yellow, tasteless, and nearly odorless. It is insoluble in water and exists in several allotropic forms. The thermodynamically stable form at room tempera-ture is rhombic sulfur, which consists of puckered S8 rings with each sulfur atom form-ing two bonds (Figure 7.26). Rhombic sulfur melts at 113 °C.

Most of the approximately 1 * 1010 kg (10 million tons) of sulfur produced in the United States each year is used to manufacture sulfuric acid. Sulfur is also used to vulcanize rubber, a process that toughens rubber by introducing cross-linking between polymer chains. (Section 12.8)

Selenium and tellurium do not form eight-membered rings in their elemental forms. (Section 7.8) The most stable allotropes of these elements are crystalline substances containing helical chains of atoms (▶ Figure 22.17). In all the allotropes each atom forms two bonds to its neighbors. Each atom is close to atoms in adjacent chains, and it appears that some sharing of electron pairs between these atoms occurs.

The electrical conductivity of elemental selenium is low in the dark but increases greatly upon exposure to light. This property is exploited in photoelectric cells and light meters. Photocopiers also depend on the photoconductivity of selenium. Photocopy machines contain a belt or drum coated with a film of selenium. This drum is elec-trostatically charged and then exposed to light reflected from the image being photo-copied. The charge drains from the regions where the selenium film has been made conductive by exposure to light. A black powder (the toner) sticks only to the areas that remain charged. The photocopy is made when the toner is transferred to a sheet of plain paper.

SulfidesWhen an element is less electronegative than sulfur, sulfides that contain S2 - form. Many metallic elements are found in the form of sulfide ores, such as PbS (galena) and HgS (cinnabar). A series of related ores containing the disulfide ion, S2

2 - (analogous to the peroxide ion), are known as pyrites. Iron pyrite, FeS2, occurs as golden yellow cubic crystals (▶ Figure 22.18). Because it has been occasionally mistaken for gold by miners, iron pyrite is often called fool’s gold.

One of the most important sulfides is hydrogen sulfide 1H2S2. This substance is normally prepared by action of dilute acid on iron(II) sulfide:

FeS1s2 + 2 H+1aq2 ¡ H2S1aq2 + Fe2 + 1aq2 [22.31]

One of hydrogen sulfide’s most readily recognized properties is its odor, which is most frequently encountered as the offensive odor of rotten eggs. Hydrogen sulfide is toxic but our noses can detect H2S in extremely low, nontoxic concentrations. A sulfur-containing organic molecule, such as dimethyl sulfide, 1CH322S, which is simi-larly odoriferous and can be detected by smell at a level of one part per trillion, is added to natural gas as a safety factor to give it a detectable odor.

Oxides, Oxyacids, and Oxyanions of SulfurSulfur dioxide, formed when sulfur burns in air, has a choking odor and is poisonous. The gas is particularly toxic to lower organisms, such as fungi, so it is used to sterilize dried fruit and wine. At 1 atm and room temperature, SO2 dissolves in water to pro-duce a 1.6 M solution. The SO2 solution is acidic, and we describe it as sulfurous acid 1H2SO32.

▲ Figure 22.17 portion of helical chains making up the structure of crystalline selenium.

Se

Se

Se

Se

Se

Se

Se

Se

Se

Se

Se

Se

Se

Se

▲ Figure 22.18 iron pyrite (FeS2, on the right) with gold for comparison.


Salts of SO32 - (sulfites) and HSO3

- (hydrogen sulfites or bisulfites) are well known. Small quantities of Na2SO3 or NaHSO3 are used as food additives to prevent bacterial spoilage. However, they are known to increase asthma symptoms in approxi-mately 5% of asthmatics. Thus, all food products with sulfites must now carry a warn-ing label disclosing their presence (◀ Figure 22.19).

Although combustion of sulfur in air produces mainly SO2, small amounts of SO3 are also formed. The reaction produces chiefly SO2 because the activation-energy bar-rier for oxidation to SO3 is very high unless the reaction is catalyzed. Interestingly, the SO3 by-product is used industrially to make H2SO4, which is the ultimate prod-uct of the reaction between SO3 and water. In the manufacture of sulfuric acid, SO2 is obtained by burning sulfur and then oxidized to SO3, using a catalyst such as V2O5 or platinum. The SO3 is dissolved in H2SO4 because it does not dissolve quickly in water, and then the H2S2O7 formed in this reaction, called pyrosulfuric acid, is added to water to form H2SO4:

SO31g2 + H2SO41l2 ¡ H2S2O71l2 [22.32]

H2S2O71l2 + H2O1l2 ¡ 2 H2SO41l2 [22.33]

Give it Some thoughtWhat is the net reaction of Equations 22.32 and 22.33?

Commercial sulfuric acid is 98% H2SO4. It is a dense, colorless, oily liquid that boils at 340 °C. It is a strong acid, a good dehydrating agent (▼ Figure 22.20), and a moderately good oxidizing agent.

Year after year, the production of sulfuric acid is the largest of any chemical pro-duced in the United States. About 4 * 1010 kg (40 million tons) is produced annually in this country. Sulfuric acid is employed in some way in almost all manufacturing.

Sulfuric acid is a strong acid, but only the first hydrogen is completely ionized in aqueous solution:

H2SO41aq2 ¡ H+1aq2 + HSO4- 1aq2 [22.34]

HSO4- 1aq2 ∆ H+1aq2 + SO4

2 - 1aq2 Ka = 1.1 * 10 - 2 [22.35]

▲ Figure 22.19 Food label warning of sulfites.

Table sugar (sucrose), C12H22O11

H2SO4

Pure carbon, C

▲ Figure 22.20 Sulfuric acid dehydrates table sugar to produce elemental carbon.

Go FiGuREIn this reaction, what has happened to the H and O atoms in the sucrose?

SEcTION 22.7 Nitrogen 973

Consequently, sulfuric acid forms both sulfates 1SO42- salts2 and

bisulfates (or hydrogen sulfates, HSO4- salts). Bisulfate salts are

common components of the “dry acids” used for adjusting the pH of swimming pools and hot tubs; they are also components of many toilet bowl cleaners.

The term thio indicates substitution of sulfur for oxygen, and the thiosulfate ion 1S2O3

2 -2 is formed by boiling an alkaline solu-tion of SO3

2 - with elemental sulfur:

8 SO32 - 1aq2 + S81s2 ¡ 8 S2O3

2 - 1aq2 [22.36]

The structures of the sulfate and thiosulfate ions are compared in ▶ Figure 22.21.

22.7 | NitrogenNitrogen constitutes 78% by volume of Earth’s atmosphere, where it occurs as N2 mol-ecules. Although nitrogen is a key element in living organisms, compounds of nitrogen are not abundant in Earth’s crust. The major natural deposits of nitrogen compounds are those of KNO3 (saltpeter) in India and NaNO3 (Chile saltpeter) in Chile and other desert regions of South America.

Properties of NitrogenNitrogen is a colorless, odorless, tasteless gas composed of N2 molecules. Its melting point is -210 °C, and its normal boiling point is -196 °C.

The N2 molecule is very unreactive because of the strong triple bond between nitrogen atoms (the N ‚ N bond enthalpy is 941 kJ>mol, nearly twice that for the bond in O2. (Table 8.4) When substances burn in air, they normally react with O2 but not with N2.

The electron configuration of the nitrogen atom is 3He42s22p3. The element exhibits all formal oxidation states from +5 to -3 (▶ table 22.6). The +5, 0, and -3 oxidation states are the most common and generally the most stable of these. Because nitrogen is more electronegative than all other elements except fluorine, oxygen, and chlorine, it exhibits positive oxidation states only in combination with these three elements.

Production and Uses of NitrogenElemental nitrogen is obtained in commercial quantities by fractional distillation of liquid air. About 4 * 1010 kg (40 million tons) of N2 is produced annually in the United States.

Because of its low reactivity, large quantities of N2 are used as an inert gaseous blanket to exclude O2 in food processing, manufacture of chemicals, metal fabrication, and production of electronic devices. Liquid N2 is employed as a coolant to freeze foods rapidly.

The largest use of N2 is in the manufacture of nitrogen-containing fertilizers, which provide a source of fixed nitrogen. We have previously discussed nitrogen fixation in the “Chemistry and Life” box in Section 14.7 and in the “Chemistry Put to Work” box in Section 15.2. Our starting point in fixing nitrogen is the manufacture of ammonia via the Haber process. (Section 15.2) The ammonia can then be converted into a variety of useful, simple nitrogen-containing species (Figure 22.22).

Hydrogen Compounds of NitrogenAmmonia is one of the most important compounds of nitrogen. It is a colorless, toxic gas that has a characteristic irritating odor. As noted in previous discussions, the NH3 molecule is basic 1Kb = 1.8 * 10 - 52. (Section 16.7)

2− 2−

▲ Figure 22.21 Structures of the sulfate (left) and thiosulfate (right) ions.

Go FiGuREWhat are the oxidation states of the sulfur atoms in the S2O3

2 - ion?

table 22.6 oxidation states of Nitrogen

oxidation state Examples

+5 N2O5, HNO3, NO3-

+4 NO2, N2O4

+3 HNO2, NO2- , NF3

+2 NO+1 N2O, H2N2O2, N2O2

2 - , HNF2

0 N2

-1 NH2OH, NH2F-2 N2H4

-3 NH3, NH4+, NH2

-


In the laboratory, NH3 can be prepared by the action of NaOH on an ammonium salt. The NH4

+ ion, which is the conjugate acid of NH3, transfers a proton to OH- . The resultant NH3 is volatile and is driven from the solution by mild heating:

NH4Cl1aq2 + NaOH1aq2 ¡ NH31g2 + H2O1l2 + NaCl1aq2 [22.37]

Commercial production of NH3 is achieved by the Haber process:

N21g2 + 3 H21g2 ¡ 2 NH31g2 [22.38]

About 1 * 1010 kg (10 million tons) of ammonia is produced annually in the United States. About 75% is used for fertilizer.

Hydrazine 1N2H42 is another important hydride of nitrogen. The hydrazine mol-ecule contains an N ¬ N single bond (◀ Figure 22.23). Hydrazine is quite poisonous. It can be prepared by the reaction of ammonia with hypochlorite ion 1OCl- 2 in aqueous solution:

2 NH31aq2 + OCl- 1aq2 ¡ N2H41aq2 + Cl- 1aq2 + H2O1l2 [22.39]

The reaction involves several intermediates, including chloramine 1NH2Cl2. The poi-sonous NH2Cl bubbles out of solution when household ammonia and chlorine bleach (which contains OCl- ) are mixed. This reaction is one reason for the frequently cited warning not to mix bleach and household ammonia.

Pure hydrazine is a strong and versatile reducing agent. The major use of hydra-zine and compounds related to it, such as methylhydrazine (Figure 22.23), is as rocket fuel.

▲ Figure 22.22 Sequence of conversion of N2 into common nitrogen compounds.

Go FiGuREIn which of these species is the oxidation number of nitrogen +3?

N2(dinitrogen)

NH3(ammonia)

NO(nitric oxide)

NO2(nitrogen dioxide)

HNO3(nitric acid)

NO3−

(nitrate salts)NO2

−

(nitrite salts)

N2H4(hydrazine)

NH4+

(ammonium salts)

▲ Figure 22.23 hydrazine (top, N2H4) and methylhydrazine (bottom, CH3NHNH2).

Go FiGuREIs the N ¬ N bond length in these molecules shorter or longer than the N ¬ N bond length in N2?

samPlE ExERCisE 22.6 Writing a Balanced Equation

Hydroxylamine 1NH2OH2 reduces copper(II) to the free metal in acid solutions. Write a balanced equation for the reaction, assuming that N2 is the oxidation product.

solutioNAnalyze We are asked to write a balanced oxidation–reduction equation in which NH2OH is converted to N2 and Cu2 + is converted to Cu.Plan Because this is a redox reaction, the equation can be balanced by the method of half-reactions discussed in Section 20.2. Thus, we begin with two half-reactions, one involving the NH2OH and N2 and the other involving Cu2 + and Cu.Solve The unbalanced and incomplete Cu2 + 1aq2 ¡ Cu1s2half-reactions are NH2OH1aq2 ¡ N21g2

Cu2 + 1aq2 + 2 e - ¡ Cu1s2

SEcTION 22.7 Nitrogen 975

Oxides and Oxyacids of NitrogenNitrogen forms three common oxides: N2O (nitrous oxide), NO (nitric oxide), and NO2 (nitrogen dioxide). It also forms two unstable oxides that we will not discuss, N2O3 (dinitrogen trioxide) and N2O5 (dinitrogen pentoxide).

Nitrous oxide 1N2O2 is also known as laughing gas because a person becomes giddy after inhaling a small amount. This colorless gas was the first substance used as a general anesthetic. It is used as the compressed gas propellant in several aerosols and foams, such as in whipped cream. It can be prepared in the laboratory by carefully heat-ing ammonium nitrate to about 200 °C:

NH4NO31s2 ¡∆ N2O1g2 + 2 H2O1g2 [22.40]

Nitric oxide (NO) is also a colorless gas but, unlike N2O, it is slightly toxic. It can be prepared in the laboratory by reduction of dilute nitric acid, using copper or iron as a reducing agent:

3 Cu1s2 + 2 NO3- 1aq2 + 8 H+1aq2 ¡ 3 Cu2 + 1aq2 + 2 NO1g2 + 4 H2O1l2 [22.41]

Nitric oxide is also produced by direct reaction of N2 and O2 at high temperatures. This reaction is a significant source of nitrogen oxide air pollutants. (Section 18.2) The direct combination of N2 and O2 is not used for commercial production of NO, how-ever, because the yield is low, the equilibrium constant Kp at 2400 K being only 0.05.

(Section 15.7, “Chemistry Put to Work: Controlling Nitric Oxide Emissions”)The commercial route to NO (and hence to other oxygen-containing compounds

of nitrogen) is via the catalytic oxidation of NH3:

4 NH31g2 + 5 O21g2 ¡Pt catalyst

850 °C 4 NO1g2 + 6 H2O1g2 [22.42]

This reaction is the first step in the Ostwald process, by which NH3 is converted com-mercially into nitric acid 1HNO32.

When exposed to air, nitric oxide reacts readily with O2 (▼ Figure 22.24):

2 NO1g2 + O21g2 ¡ 2 NO21g2 [22.43]

Practice Exercise 1In power plants, hydrazine is used to prevent corrosion of the metal parts of steam boilers by the O2 dissolved in the water. The hydrazine reacts with O2 in water to give N2 and H2O. Write a balanced equation for this reaction.

Practice Exercise 2Methylhydrazine, N2H3CH31l2, is used with the oxidizer dinitrogen tetroxide, N2O41l2, to power the steering rockets of the Space Shuttle orbiter. The reaction of these two substances produces N2, CO2, and H2O. Write a balanced equation for this reaction.

Balancing these equations as described in Section 20.2 gives 2 NH2OH1aq2 ¡ N21g2 + 2 H2O1l2 + 2 H+1aq2 + 2 e -

Adding these half-reactions gives the balanced equation: Cu2 + 1aq2 + 2 NH2OH1aq2 ¡ Cu1s2 + N21g2 + 2 H2O1l2 + 2 H+1aq2

▲ Figure 22.24 Formation of No2(g) as No(g) combines with o2(g) in the air.

O2 in air

NO2(g)

NO(g)


When dissolved in water, NO2 forms nitric acid:

3 NO21g2 + H2O1l2 ¡ 2 H+1aq2 + 2 NO3- 1aq2 + NO1g2 [22.44]

Nitrogen is both oxidized and reduced in this reaction, which means it disproportion-ates. The NO can be converted back into NO2 by exposure to air (Equation 22.43) and thereafter dissolved in water to prepare more HNO3.

NO is an important neurotransmitter in the human body. It causes the muscles that line blood vessels to relax, thus allowing an increased passage of blood (see the “Chemistry and Life”).

Nitrogen dioxide 1NO22 is a yellow-brown gas (Figure 22.24). Like NO, it is a major constituent of smog. (Section 18.2) It is poisonous and has a choking odor. As dis-cussed in Section 15.1, NO2 and N2O4 exist in equilibrium:

2 NO21g2 ∆ N2O41g2 ∆H ° = -58 kJ [22.45]

The two common oxyacids of nitrogen are nitric acid 1HNO32 and nitrous acid 1HNO22 (◀ Figure 22.25). Nitric acid is a strong acid. It is also a powerful oxidizing agent, as indicated by the standard reduction potential in the reaction

NO3- 1aq2 + 4 H+1aq2 + 3 e- ¡ NO1g2 + 2 H2O1l2 E° = +0.96 V [22.46]

Concentrated nitric acid attacks and oxidizes most metals except Au, Pt, Rh, and Ir.About 7 * 109 kg (8 million tons) of nitric acid is produced annually in the United

States. Its largest use is in the manufacture of NH4NO3 for fertilizers. It is also used in the production of plastics, drugs, and explosives. Among the explosives made from nitric acid are nitroglycerin, trinitrotoluene (TNT), and nitrocellulose. The following reaction occurs when nitroglycerin explodes:

4 C3H5N3O91l2 ¡ 6 N21g2 + 12 CO21g2 + 10 H2O1g2 + O21g2 [22.47]

All the products of this reaction contain very strong bonds and are gases. As a result, the reaction is very exothermic, and the volume of the products is far larger than the volume occupied by the reactant. Thus, the expansion resulting from the heat gener-ated by the reaction produces the explosion. (Section 8.8, “Explosives and Alfred Nobel”)

Nitrous acid is considerably less stable than HNO3 and tends to dispropor-tionate into NO and HNO3. It is normally made by action of a strong acid, such as H2SO4, on a cold solution of a nitrite salt, such as NaNO2. Nitrous acid is a weak acid 1Ka = 4.5 * 10 - 42.

Give it Some thoughtWhat are the oxidation numbers of the nitrogen atoms in(a) nitric acid(b) nitrous acid

▲ Figure 22.25 Structures of nitric acid (top) and nitrous acid (bottom).

Go FiGuREWhich is the shortest NO bond in these two molecules?

Chemistry and life

Nitroglycerin, Nitric oxide, and heart Disease

During the 1870s, an interesting observation was made in Alfred Nobel’s dynamite factories. Workers who suffered from heart disease that caused chest pains when they exerted themselves found relief from the pains during the workweek. It quickly became apparent that nitroglycerin, present in the air of the factory, acted to enlarge blood vessels. Thus, this powerfully explosive chemical became a stan-dard treatment for angina pectoris, the chest pains accompanying heart failure. It took more than 100 yrs to discover that nitroglyc-erin was converted in the vascular smooth muscle into NO, which is the chemical agent actually causing dilation of the blood vessels.

In 1998, the Nobel Prize in Physiology or Medicine was awarded to Robert F. Furchgott, Louis J. Ignarro, and Ferid Murad for their discoveries of the detailed pathways by which NO acts in the car-diovascular system. It was a sensation that this simple, common air pollutant could exert important functions in mammals, including humans.

As useful as nitroglycerin is to this day in treating angina pec-toris, it has a limitation in that prolonged administration results in development of tolerance, or desensitization, of the vascular muscle to further vasorelaxation by nitroglycerin. The bioactivation of nitro-glycerin is the subject of active research in the hope that a means of circumventing desensitization can be found.

H N

O

O O

H NO O

SEcTION 22.8 The Other Group 5A Elements: P, As, Sb, and Bi 977

22.8 | the other Group 5a Elements: P, as, sb, and Bi

Of the other group 5A elements—phosphorus, arsenic, antimony, and bismuth—phosphorus has a central role in several aspects of biochemistry and environmental chemistry.

General Characteristics of the Group 5A ElementsThe group 5A elements have the outer-shell electron configuration ns2np3, with n rang-ing from 2 to 6. A noble-gas configuration is achieved by adding three electrons to form the -3 oxidation state. Ionic compounds containing X3 - ions are not common, how-ever. More commonly, the group 5A element acquires an octet of electrons via covalent bonding and oxidation numbers ranging from -3 to +5.

Because of its lower electronegativity, phosphorus is found more frequently in positive oxidation states than is nitrogen. Furthermore, compounds in which phos-phorus has the +5 oxidation state are not as strongly oxidizing as the corresponding compounds of nitrogen. Compounds in which phosphorus has a -3 oxidation state are much stronger reducing agents than are the corresponding nitrogen compounds.

Some properties of the group 5A elements are listed in ▼ table 22.7. The general pattern is similar to what we saw with other groups: Size and metallic character increase as atomic number increases in the group.

The variation in properties among group 5A elements is more striking than that seen in groups 6A and 7A. Nitrogen at the one extreme exists as a gaseous diatomic molecule, clearly nonmetallic. At the other extreme, bismuth is a reddish white, metallic-looking substance that has most of the characteristics of a metal.

The values listed for X ¬ X single-bond enthalpies are not reliable because it is dif-ficult to obtain such data from thermochemical experiments. However, there is no doubt about the general trend: a low value for the N ¬ N single bond, an increase at phosphorus, and then a gradual decline to arsenic and antimony. From observations of the elements in the gas phase, it is possible to estimate the X ‚ X triple-bond enthalpies. Here, we see a trend that is different from that for the X ¬ X single bond. Nitrogen forms a much stron-ger triple bond than the other elements, and there is a steady decline in the triple-bond enthalpy down through the group. These data help us to appreciate why nitrogen alone of the group 5A elements exists as a diatomic molecule in its stable state at 25 °C. All the other elements exist in structural forms with single bonds between the atoms.

Occurrence, Isolation, and Properties of PhosphorusPhosphorus occurs mainly in the form of phosphate minerals. The principal source of phosphorus is phosphate rock, which contains phosphate principally as Ca31PO422. The element is produced commercially by the reduction of calcium phosphate with carbon in the presence of SiO2:

2 Ca31PO4221s2 + 6 SiO21s2 + 10 C1s2 ¡1500 °C P41g2 + 6 CaSiO31l2 + 10 CO1g2 [22.48]

5A7N

15P

33As

51Sb83Bi

table 22.7 Properties of the Group 5a Elements

Property N P as sb Bi

Atomic radius 1A° 2 0.71 1.07 1.19 1.39 1.48

First ionization energy 1kJ>mol2 1402 1012 947 834 703

Electron affinity 1kJ>mol2 7 0 -72 -78 -103 -91

Electronegativity 3.0 2.1 2.0 1.9 1.9

X ¬ X single-bond enthalpy 1kJ>mol2* 163 200 150 120 —

X ‚ X triple-bond enthalpy 1kJ>mol2 941 490 380 295 192

*Approximate values only.


The phosphorus produced in this fashion is the allotrope known as white phosphorus. This form distills from the reaction mixture as the reaction proceeds.

Phosphorus exists in several allotropic forms. White phosphorus consists of P4 tet-rahedra (◀ Figure 22.26). The bond angles in this molecule, 60°, are unusually small, so there is much strain in the bonding, which is consistent with the high reactivity of white phosphorus. This allotrope bursts spontaneously into flames if exposed to air. When heated in the absence of air to about 400 °C, white phosphorus is converted to a more stable allotrope known as red phosphorus, which does not ignite on contact with air. Red phosphorus is also considerably less poisonous than the white form. We will denote elemental phosphorus as simply P(s).

Phosphorus HalidesPhosphorus forms a wide range of compounds with the halogens, the most important

of which are the trihalides and pentahalides. Phosphorus trichloride 1PCl32 is com-mercially the most significant of these compounds and is used to prepare a wide

variety of products, including soaps, detergents, plastics, and insecticides.Phosphorus chlorides, bromides, and iodides can be made by direct oxidation

of elemental phosphorus with the elemental halogen. PCl3, for example, which is a liquid at room temperature, is made by passing a stream of dry chlorine gas over white or red phosphorus:

2 P1s2 + 3 Cl21g2 ¡ 2 PCl31l2 [22.49]

If excess chlorine gas is present, an equilibrium is established between PCl3 and PCl5. PCl31l2 + Cl21g2 ∆ PCl51s2 [22.50]

The phosphorus halides react readily with water, and most fume in air because of reaction with water vapor. In the presence of excess water, the products are the corre-sponding phosphorus oxyacid and hydrogen halide:

PBr31l2 + 3 H2O1l2 ¡ H3PO31aq2 + 3 HBr1aq2 [22.51]

PCl51l2 + 4 H2O1l2 ¡ H3PO41aq2 + 5 HCl1aq2 [22.52]

Give it Some thoughtWhich oxyacid is produced when PF3 reacts with water?

Oxy Compounds of PhosphorusProbably the most significant phosphorus compounds are those in which the ele-ment is combined with oxygen. Phosphorus(III) oxide 1P4O62 is obtained by allow-ing white phosphorus to oxidize in a limited supply of oxygen. When oxidation takes place in the presence of excess oxygen, phosphorus(V) oxide 1P4O102 forms. This compound is also readily formed by oxidation of P4O6. These two oxides represent the two most common oxidation states for phosphorus, +3 and +5. The structural relationship between P4O6 and P4O10 is shown in ▶ Figure 22.27. Notice the resem-blance these molecules have to the P4 molecule (Figure 22.27); all three substances have a P4 core.

Phosphorus(V) oxide is the anhydride of phosphoric acid 1H3PO42, a weak tri-protic acid. In fact, P4O10 has a very high affinity for water and is consequently used as a drying agent. Phosphorus(III) oxide is the anhydride of phosphorous acid 1H3PO32, a weak diprotic acid (▶ Figure 22.28).*

▲ Figure 22.26 White and red phosphorus. Despite the fact that both contain only phosphorus atoms, these two forms of phosphorus differ greatly in reactivity. The white allotrope, which reacts violently with oxygen, must be stored under water so that it is not exposed to air. The much less reactive red form does not need to be stored this way.

White phosphorusRed phosphorus

*Note that the element phosphorus 1FOS # for # us2 has a -us suffix, whereas the first word in the name phosphorous 1fos # FOR # us2 acid has an -ous suffix.

SEcTION 22.8 The Other Group 5A Elements: P, As, Sb, and Bi 979

PO

▲ Figure 22.27 Structures of P4O6 (top) and P4O10 (bottom).

Go FiGuREHow do the electron domains about P in P4O6 differ from those about P in P4O10?

One characteristic of phosphoric and phosphorous acids is their tendency to undergo condensation reactions when heated. (Section 12.8) For example, two H3PO4 molecules are joined by the elimination of one H2O molecule to form H4P2O7:

These atoms are eliminated as H2O

P

O

OH

HO OH P

O

OH

OH+ P

O

OH

HO O P

O

OH

OH +HO H2O

[22.53]

Phosphoric acid and its salts find their most important uses in detergents and fertilizers. The phosphates in detergents are often in the form of sodium tripolyphosphate 1Na5P3O102.

The phosphate ions “soften” water by binding their oxygen groups to the metal ions that contribute to the hardness of water. This keeps the metal ions from interfer-ing with the action of the detergent. The phosphates also keep the pH above 7 and thus prevent the detergent molecules from becoming protonated.

Most mined phosphate rock is converted to fertilizers. The Ca31PO422 in phos-phate rock is insoluble 1Ksp = 2.0 * 10 - 292. It is converted to a soluble form for use in fertilizers by treatment with sulfuric or phosphoric acid. The reaction with phosphoric acid yields Ca1H2PO422:

Ca31PO4221s2 + 4 H3PO41aq2 ¡ 3 Ca2+1aq2 + 6 H2PO4- 1aq2 [22.54]

Although the solubility of Ca1H2PO422 allows it to be assimilated by plants, it also allows it to be washed from the soil and into bodies of water, thereby contributing to water pollution. (Section 18.4)

Phosphorus compounds are important in biological systems. The element occurs in phosphate groups in RNA and DNA, the molecules responsible for the control of protein biosynthesis and transmission of genetic information. It also occurs in adenosine triphos-phate (ATP), which stores energy in biological cells and has the following structure:

P

O

P

O

O− O−

OO CH2OP

O

O−

−O

C

H

C

H

H

C

OH

H

C

OH

O

Adenosine

NH2

N CN

NNCH

CHC

C

The P ¬ O ¬ P bond of the end phosphate group is broken by hydrolysis with water, forming adenosine diphosphate (ADP):

P

O

O− O−

−O P

O

O O

O−

P

O

ATP

ADP

O Adenosine +

P

O

O− O−

P

O

O Adenosine +O P

O

O−

−O

H2O

HO OH

[22.55]

This H not acidic because P–H bond is nonpolar

▲ Figure 22.28 Structures of H3PO4 (top) and H3PO3 (bottom).


This reaction releases 33 kJ of energy under standard conditions, but in the living cell, the Gibbs free energy change for the reaction is closer to -57 kJ>mol. The concen-tration of ATP inside a living cell is in the range of 1–10 mM, which means a typical human metabolizes her or his body mass of ATP in one day! ATP is continually made from ADP and continually converted back to ADP, releasing energy that can be har-nessed by other cellular reactions.

Chemistry and life

Arsenic in Drinking Water

Arsenic, in the form of its oxides, has been known as a poison for centu-ries. The current Environmental Protection Agency (EPA) standard for arsenic in public water supplies is 10 ppb (equivalent to 10 mg>L). Most regions of the United States tend to have low to moderate (2–10 ppb) groundwater arsenic levels (▼ Figure 22.29). The western region tends to have higher levels, coming mainly from natural geological sources in the area. Estimates, for example, indicate that 35% of water-supply wells in Arizona have arsenic concentrations above 10 ppb.

The problem of arsenic in drinking water in the United States is dwarfed by the problem in other parts of the world—especially in Bangladesh, where the problem is tragic. Historically, surface water

▲ Figure 22.29 Geographic distribution of arsenic in groundwater.

At least 25% of samples exceed 5 ppm AsAt least 25% of samples exceed 3 ppm AsAt least 25% of samples exceed 1 ppm As

EXPLANATION

At least 25% of samples exceed 50 ppm AsAt least 25% of samples exceed 10 ppm As

Insuf�cient data

AlaskaPuerto

RicoHawaii

sources in that country have been contaminated with micro-organisms, causing significant health problems. During the 1970s, international agencies, headed by the United Nations Children’s Fund (UNICEF), began investing millions of dol-lars of aid money in Bangladesh for wells to provide “clean” drinking water. Unfortunately, no one tested the well water for the presence of arsenic; the problem was not discovered until the 1980s. The result has been the biggest outbreak of mass poisoning in history. Up to half of the country’s esti-mated 10 million wells have arsenic concentrations above 50 ppb.

In water, the most common forms of arsenic are the arsenate ion and its protonated hydrogen anions (AsO4

3 - , HAsO42 - , and H2AsO4

- ) and the arsenite ion and its protonated forms 1AsO3

3 - , HAsO32 - , H2AsO3

- , and H3AsO32. These species are collectively referred to by the oxidation number of the arsenic as arsenic(V) and arsenic(III), respectively. Arsenic(V) is more prevalent in oxygen-rich (aerobic) surface waters, whereas arsenic(III) is more likely to occur in oxygen-poor (anaerobic) groundwaters.

One of the challenges in determining the health effects of arsenic in drinking waters is the different chemistry of arsenic(V) and arsenic(III), as well as the different concen-trations required for physiological responses in different in-dividuals. In Bangladesh, skin lesions were the first sign of the arsenic problem. Statistical studies correlating arsenic levels with the occurrence of disease indicate a lung and blad-der cancer risk arising from even low levels of arsenic.

The current technologies for removing arsenic per-form most effectively when treating arsenic in the form of arsenic(V), so water treatment strategies require preoxida-tion of the drinking water. Once in the form of arsenic(V), there are a number of possible removal strategies. For exam-ple, Fe3 + can be added to precipitate FeAsO4, which is then removed by filtration.

22.9 | CarbonCarbon constitutes only 0.027% of Earth’s crust. Although some carbon occurs in elemental form as graphite and diamond, most is found in combined form. Over half occurs in carbonate compounds, and carbon is also found in coal, petroleum, and natu-ral gas. The importance of the element stems in large part from its occurrence in all living organisms: Life as we know it is based on carbon compounds.

Elemental Forms of CarbonWe have seen that carbon exists in several allotropic crystalline forms: graphite, dia-mond, fullerenes, carbon nanotubes, and graphene. Fullerenes, nanotubes, and graphene are discussed in Chapter 12; here we focus on graphite and diamond.

SEcTION 22.9 carbon 981

Graphite is a soft, black, slippery solid that has a metallic luster and conducts elec-tricity. It consists of parallel sheets of sp2@hybridized carbon atoms held together by dispersion forces. (Section 12.7) Diamond is a clear, hard solid in which the carbon atoms form an sp3@hybridized covalent network. (Section 12.7) Diamond is denser than graphite (d = 2.25 g>cm3 for graphite; d = 3.51 g>cm3 for diamond). At approx-imately 100,000 atm at 3000 °C, graphite converts to diamond. In fact, almost any carbon-containing substance, if put under sufficiently high pressure, forms diamonds; scientists at General Electric in the 1950s used peanut butter to make diamonds. About 3 * 104 kg of industrial-grade diamonds are synthesized each year, mainly for use in cutting, grinding, and polishing tools.

Graphite has a well-defined crystalline structure, but it also exists in two com-mon amorphous forms: carbon black and charcoal. Carbon black is formed when hydrocarbons are heated in a very limited supply of oxygen, such as in this methane reaction:

CH41g2 + O21g2 ¡ C1s2 + 2 H2O1g2 [22.56]

Carbon black is used as a pigment in black inks; large amounts are also used in making automobile tires.

Charcoal is formed when wood is heated strongly in the absence of air. Charcoal has an open structure, giving it an enormous surface area per unit mass. “Activated charcoal,” a pulverized form of charcoal whose surface is cleaned by heating with steam, is widely used to adsorb molecules. It is used in filters to remove offensive odors from air and colored or bad-tasting impurities from water.

Oxides of CarbonCarbon forms two principal oxides: carbon monoxide (CO) and carbon dioxide 1CO22. Carbon monoxide is formed when carbon or hydrocarbons are burned in a limited sup-ply of oxygen:

2 C1s2 + O21g2 ¡ 2 CO1g2 [22.57]

CO is a colorless, odorless, tasteless gas that is toxic because it binds to hemoglobin in the blood and thus interferes with oxygen transport. Low-level poisoning results in headaches and drowsiness; high-level poisoning can cause death.

Carbon monoxide is unusual in that it has a nonbonding pair of electrons on car-bon: ∶C ‚ O∶. It is isoelectronic with N2, so you might expect CO to be equally unre-active. Moreover, both substances have high bond energies (1072 kJ>mol for C ‚ O and 941 kJ>mol for N ‚ N). Because of the lower nuclear charge on carbon (compared with either N or O), however, the carbon nonbonding pair is not held as strongly as that on N or O. Consequently, CO is better able to function as a Lewis base than is N2; for example, CO can coordinate its nonbonding pair to the iron of hemoglobin, displacing O2, but N2 cannot. In addition, CO forms a variety of covalent compounds, known as metal carbonyls, with transition metals. Ni1CO24, for example, is a volatile, toxic compound formed by warming metallic nickel in the presence of CO. The forma-tion of metal carbonyls is the first step in the transition-metal catalysis of a variety of reactions of CO.

Carbon monoxide has several commercial uses. Because it burns readily, forming CO2, it is employed as a fuel:

2 CO1g2 + O21g2 ¡ 2 CO21g2 ∆H ° = -566 kJ [22.58]

Carbon monoxide is an important reducing agent, widely used in metallurgical opera-tions to reduce metal oxides, such as the iron oxides:

Fe3O41s2 + 4 CO1g2 ¡ 3 Fe1s2 + 4 CO21g2 [22.59]

Carbon dioxide is produced when carbon-containing substances are burned in excess oxygen, such as in this reaction involving ethanol:

C2H5OH1l2 + 3 O21g2 ¡ 2 CO21g2 + 3 H2O1g2 [22.60]


It is also produced when many carbonates are heated:

CaCO31s2 ¡∆ CaO1s2 + CO21g2 [22.61]

In the laboratory, CO2 can be produced by the action of acids on carbonates (◀ Figure 22.31):

CO32 - 1aq2 + 2 H+1aq2 ¡ CO21g2 + H2O1l2 [22.62]

Carbon dioxide is a colorless, odorless gas. It is a minor component of Earth’s atmo-sphere but a major contributor to the greenhouse effect. (Section 18.2) Although it is not toxic, high concentrations of CO2 increase respiration rate and can cause suf-focation. It is readily liquefied by compression. When cooled at atmospheric pressure, however, CO2 forms a solid rather than liquefying. The solid sublimes at atmospheric pressure at -78 °C. This property makes solid CO2, known as dry ice, valuable as a refrigerant. About half of the CO2 consumed annually is used for refrigeration. The other major use of CO2 is in the production of carbonated beverages. Large quanti-ties are also used to manufacture washing soda 1Na2CO3 # 10 H2O2, used to precipitate metal ions that interfere with the cleansing action of soap, and baking soda 1NaHCO32. Baking soda is so named because this reaction occurs during baking:

NaHCO31s2 + H+1aq2 ¡ Na+1aq2 + CO21g2 + H2O1l2 [22.63]

The H+1aq2 is provided by vinegar, sour milk, or the hydrolysis of certain salts. The bubbles of CO2 that form are trapped in the baking dough, causing it to rise.

Give it Some thoughtYeast are living organisms that make bread rise in the absence of baking soda and acid. What must the yeast be producing to make bread rise?

Chemistry Put to Work

carbon Fibers and composites

The properties of graphite are anisotropic; that is, they differ in dif-ferent directions through the solid. Along the carbon planes, graphite possesses great strength because of the number and strength of the carbon–carbon bonds in this direction. The bonds between planes are relatively weak, however, making graphite weak in that direction.

Fibers of graphite can be prepared in which the carbon planes are aligned to varying extents parallel to the fiber axis. These fibers are lightweight (density of about 2 g>cm3) and chemically quite unreac-tive. The oriented fibers are made by first slowly pyrolyzing (decom-posing by action of heat) organic fibers at about 150 to 300 °C. These fibers are then heated to about 2500 °C to graphitize them (convert amorphous carbon to graphite). Stretching the fiber during pyrolysis helps orient the graphite planes parallel to the fiber axis. More amor-phous carbon fibers are formed by pyrolysis of organic fibers at lower temperatures (1200 to 400 °C). These amorphous materials, com-monly called carbon fibers, are the type most often used in commercial materials.

Composite materials that take advantage of the strength, sta-bility, and low density of carbon fibers are widely used. Compos-ites are combinations of two or more materials. These materials are present as separate phases and are combined to form structures that take advantage of certain desirable properties of each component. In carbon composites, the graphite fibers are often woven into a fabric that is embedded in a matrix that binds them into a solid structure.

The fibers transmit loads evenly throughout the matrix. The finished composite is thus stronger than any one of its components.

Carbon composite materials are used widely in a number of appli-cations, including high-performance graphite sports equipment such as tennis racquets, golf clubs, and bicycle wheels (▼ Figure 22.30). Heat-resistant composites are required for many aerospace applica-tions, where carbon composites now find wide use.

▲ Figure 22.30 carbon composites in commercial products.

▲ Figure 22.31 CO2 formation from the reaction between an acid and calcium carbonate in rock.

CaCO3

Strong acid CO2(g)

SEcTION 22.9 carbon 983

Carbonic Acid and CarbonatesCarbon dioxide is moderately soluble in H2O at atmospheric pressure. The resulting solution is moderately acidic because of the formation of carbonic acid 1H2CO32:

CO21aq2 + H2O1l2 ∆ H2CO31aq2 [22.64]

Carbonic acid is a weak diprotic acid. Its acidic character causes carbonated beverages to have a sharp, slightly acidic taste.

Although carbonic acid cannot be isolated, hydrogen carbonates (bicarbonates) and carbonates can be obtained by neutralizing carbonic acid solutions. Partial neutral-ization produces HCO3

- , and complete neutralization gives CO32 - . The HCO3

- ion is a stronger base than acid 1Kb = 2.3 * 10 - 8; Ka = 5.6 * 10 - 112. The carbonate ion is much more strongly basic 1Kb = 1.8 * 10 - 42.

The principal carbonate minerals are calcite 1CaCO32, magnesite 1MgCO32, dolo-mite 3MgCa1CO3224, and siderite 1FeCO32. Calcite is the principal mineral in lime-stone and the main constituent of marble, chalk, pearls, coral reefs, and the shells of marine animals such as clams and oysters. Although CaCO3 has low solubility in pure water, it dissolves readily in acidic solutions with evolution of CO2:

CaCO31s2 + 2 H+1aq2 ∆ Ca2 + 1aq2 + H2O1l2 + CO21g2 [22.65]

Because water containing CO2 is slightly acidic (Equation 22.64), CaCO3 dissolves slowly in this medium:

CaCO31s2 + H2O1l2 + CO21g2 ¡ Ca2 + 1aq2 + 2 HCO3- 1aq2 [22.66]

This reaction occurs when surface waters move underground through limestone deposits. It is the principal way Ca2 + enters groundwater, producing “hard water.” If the limestone deposit is deep enough underground, dissolution of the limestone pro-duces a cave.

One of the most important reactions of CaCO3 is its decomposition into CaO and CO2 at elevated temperatures (Equation 22.61). About 2 * 1010 kg (20 million tons) of calcium oxide, known as lime or quicklime, is produced in the United States annually. Because calcium oxide reacts with water to form Ca1OH22, it is an important commer-cial base. It is also important in making mortar, the mixture of sand, water, and CaO used to bind bricks, blocks, or rocks together. Calcium oxide reacts with water and CO2 to form CaCO3, which binds the sand in the mortar:

CaO1s2 + H2O1l2 ¡ Ca2 + 1aq2 + 2 OH- 1aq2 [22.67]

Ca2 + 1aq2 + 2 OH- 1aq2 + CO21aq2 ¡ CaCO31s2 + H2O1l2 [22.68]

CarbidesThe binary compounds of carbon with metals, metalloids, and certain nonmetals are called carbides. The more active metals form ionic carbides, and the most common of these contain the acetylide ion 1C2

2 - 2. This ion is isoelectronic with N2, and its Lewis structure, 3:C ‚ C:42 - , has a carbon–carbon triple bond. The most important ionic carbide is calcium carbide 1CaC22, produced by the reduction of CaO with carbon at high temperature:

2 CaO1s2 + 5 C1s2 ¡ 2 CaC21s2 + CO21g2 [22.69]

The carbide ion is a very strong base that reacts with water to form acetylene 1H ¬ C ‚ C ¬ H2:

CaC21s2 + 2 H2O1l2 ¡ Ca1OH221aq2 + C2H21g2 [22.70]

Calcium carbide is therefore a convenient solid source of acetylene, which is used in welding (Figure 22.13).

Interstitial carbides are formed by many transition metals. The carbon atoms occupy open spaces (interstices) between the metal atoms in a manner analogous to the interstitial hydrides. (Section 22.2) This process generally makes the metal harder.


Tungsten carbide (WC), for example, is very hard and very heat-resistant and, thus, used to make cutting tools.

Covalent carbides are formed by boron and silicon. Silicon carbide (SiC), known as Carborundum®, is used as an abrasive and in cutting tools. Almost as hard as diamond, SiC has a diamond-like structure with alternating Si and C atoms.

22.10 | the other Group 4a Elements: si, Ge, sn, and Pb

The trend from nonmetallic to metallic character as we go down a family is strikingly evident in group 4A. Carbon is a nonmetal; silicon and germanium are metalloids; tin and lead are metals. In this section, we consider a few general characteristics of group 4A and then look more thoroughly at silicon.

General Characteristics of the Group 4A ElementsThe group 4A elements possess the outer-shell electron configuration ns2np2. The elec-tronegativities of the elements are generally low (▼ table 22.8); carbides that formally contain C4 - ions are observed only in the case of a few compounds of carbon with very active metals. Formation of 4+ ions by electron loss is not observed for any of these ele-ments; the ionization energies are too high. The +4 oxidation state is common, how-ever, and is found in the vast majority of the compounds of the group 4A elements. The +2 oxidation state is found in the chemistry of germanium, tin, and lead, however, and it is the principal oxidation state for lead. Carbon, except in highly unusual examples, forms a maximum of four bonds. The other members of the family are able to form more than four bonds. (Section 8.7)

Table 22.8 shows that the strength of a bond between two atoms of a given ele-ment decreases as we go down group 4A. Carbon–carbon bonds are quite strong. Carbon, therefore, has a striking ability to form compounds in which carbon atoms are bonded to one another in extended chains and rings, which accounts for the large number of organic compounds that exist. Other elements can form chains and rings, but these bonds are far less important in the chemistries of these other elements. The Si ¬ Si bond strength 1226 kJ>mol2, for example, is much lower than the Si ¬ O bond strength 1386 kJ>mol2. As a result, the chemistry of silicon is dominated by the forma-tion of Si ¬ O bonds, and Si ¬ Si bonds play a minor role.

Occurrence and Preparation of SiliconSilicon is the second most abundant element, after oxygen, in Earth’s crust. It occurs in SiO2 and in an enormous variety of silicate minerals. The element is obtained by the reduction of molten silicon dioxide with carbon at high temperature:

SiO21l2 + 2 C1s2 ¡ Si1l2 + 2 CO1g2 [22.71]

Elemental silicon has a diamond-like structure. Crystalline silicon is a gray metal-lic-looking solid that melts at 1410 °C. The element is a semiconductor, as we saw in Chapters 7 and 12, and is used to make solar cells and transistors for computer chips. To be used as a semiconductor, it must be extremely pure, possessing less than 10 - 7%

4A6C

14Si

32Ge

50Sn82Pb

table 22.8 some Properties of the Group 4a Elements

Property C si Ge sn Pb

Atomic radius 1A° 2 0.76 1.11 1.20 1.39 1.46

First ionization energy 1kJ>mol2 1086 786 762 709 716

Electronegativity 2.5 1.8 1.8 1.8 1.9

X ¬ X single-bond enthalpy 1kJ>mol2 348 226 188 151 —

SEcTION 22.10 The Other Group 4A Elements: Si, Ge, Sn, and Pb 985

(1 ppb) impurities. One method of purification is to treat the element with Cl2 to form SiCl4, a volatile liquid that is purified by fractional distillation and then converted back to elemental silicon by reduction with H2:

SiCl41g2 + 2 H21g2 ¡ Si1s2 + 4 HCl1g2 [22.72]

The process known as zone refining can further purify the element (▶ Figure 22.30). As a heated coil is passed slowly along a silicon rod, a narrow band of the element is melted. As the molten section is swept slowly along the length of the rod, the impurities concen-trate in this section, following it to the end of the rod. The purified top portion of the rod crystallizes as 99.999999999% pure silicon.

SilicatesSilicon dioxide and other compounds that contain silicon and oxygen make up over 90% of Earth’s crust. In silicates, a silicon atom is surrounded by four oxygens and silicon is found in its most common oxidation state, +4. The orthosilicate ion, SiO4

4 - , is found in very few silicate minerals, but we can view it as the “building block” for many mineral structures. As ▼ Figure 22.33 shows, adjacent tetrahedra are linked by a common oxygen atom. Two tetrahedra joined in this way, called the disilicate ion, contain two Si atoms and seven O atoms. Silicon and oxygen are in the +4 and -2 oxidation states, respectively, in all silicates, so the overall charge of any silicate ion must be consistent with these oxidation states. For example, the charge on Si2O7 is 1221+42 + 1721-22 = -6; it is the Si2O7

6 - ion.In most silicate minerals, silicate tetrahedra are linked together to form chains,

sheets, or three-dimensional structures. We can connect two vertices of each tetrahedron to two other tetrahedra, for example, leading to an infinite chain with an g O ¬ Si ¬ O ¬ Si g backbone as shown in Figure 22.33(b). Notice that each silicon in this structure has two unshared (terminal) oxygens and two shared (bridging) oxygens. The stoichiometry is then 2112 + 211>22 = 3 oxygens per silicon. Thus, the formula unit for this chain is SiO3

2 - . The mineral enstatite 1MgSiO32 has this kind of structure, consisting of rows of single-strand silicate chains with Mg2 + ions between the strands to balance charge.

In Figure 22.33(c), each silicate tetrahedron is linked to three others, forming an infinite sheet structure. Each silicon in this structure has one unshared oxygen and three shared oxygens. The stoichiometry is then 1112 + 31½2 = 2 ½ oxygens per silicon. The simplest formula of this sheet is Si2O5

2 - . The mineral talc, also known as talcum powder, has the formula Mg31Si2O5221OH22 and is based on this sheet

Molten section

Silicon rod

Inert atmosphere

As heating coil slowly moves down, impurities concen-trate in molten section, leaving behind ultrapure Si

▲ Figure 22.32 Zone-refining apparatus for the production of ultrapure silicon.

Go FiGuREWhat limits the range of temperatures you can use for zone refining of silicon?

▲ Figure 22.33 Silicate chains and sheets.

SiO

O

O

O

SiO32−

formulaunit

Silicate ion

(a)

Fragment of silicate chain Fragment of silicate sheet

(b) (c)

4−

Si2O52−

formulaunit


structure. The Mg2 + and OH- ions lie between the silicate sheets. The slippery feel of talcum powder is due to the silicate sheets sliding relative to one another.

Numerous minerals are based on silicates, and many are useful as clays, ceram-ics, and other materials. A few silicates have harmful effects on human health, the best-known example being asbestos, a general term applied to a group of fibrous sili-cate minerals. The structure of these minerals is either chains of silicate tetrahedra or sheets formed into rolls. The result is that the minerals have a fibrous character (◀ Figure 22.34). Asbestos minerals were once widely used as thermal insulation, especially in high-temperature applications, because of the great chemical stability of the silicate structure. In addition, the fibers can be woven into asbestos cloth, which was used for fireproof curtains and other applications. However, the fibrous structure of asbestos minerals poses a health risk because the fibers readily penetrate soft tissues, such as the lungs, where they can cause diseases, including cancer. The use of asbestos as a common building material has therefore been discontinued.

When all four vertices of each SiO4 tetrahedron are linked to other tetrahedra, the structure extends in three dimensions. This linking of the tetrahedra forms quartz 1SiO22. Because the structure is locked together in a three-dimensional array much like diamond (Section 12.7), quartz is harder than strand- or sheet-type silicates.

▲ Figure 22.34 Serpentine asbestos.

GlassQuartz melts at approximately 1600 °C, forming a tacky liquid. In the course of melt-ing, many silicon–oxygen bonds are broken. When the liquid cools rapidly, silicon– oxygen bonds are re-formed before the atoms are able to arrange themselves in a regular fashion. An amorphous solid, known as quartz glass or silica glass, results. Many sub-stances can be added to SiO2 to cause it to melt at a lower temperature. The common glass used in windows and bottles, known as soda-lime glass, contains CaO and Na2O

solutioNAnalyze A mineral is described that has a sheet silicate structure with Mg2 + and OH- ions to balance charge and 1.5 Mg for each 1 Si. We are asked to write the empirical formula for the mineral.Plan As shown in Figure 22.33(c), the silicate sheet structure has the simplest formula Si2O5

2 - . We first add Mg2 + to give the proper Mg:Si ratio. We then add OH- ions to obtain a neutral compound.Solve The observation that the Mg:Si ratio equals 1.5 is consistent with three Mg2 + ions per Si2O5

2 - unit. The addition of three Mg2 + ions would make Mg31Si2O524 + . In order to achieve charge balance in the mineral, there must be four OH- per Si2O5

2 - . Thus, the formula of chrysotile is Mg31Si2O521OH24. Since this is not reducible to a simpler formula, this is the empirical formula.

samPlE ExERCisE 22.7 determining an Empirical Formula

The mineral chrysotile is a noncarcinogenic asbestos mineral that is based on the sheet struc-ture shown in Figure 22.33(c). In addition to silicate tetrahedra, the mineral contains Mg2 + and OH- ions. Analysis of the mineral shows that there are 1.5 Mg atoms per Si atom. What is the empirical formula for chrysotile?

Practice Exercise 1In the mineral beryl, six silicate tetrahedra are connected to form a ring as shown here. The negative charge of this polyanion is balanced by Be2 + and Al3 + cations. If elemental analysis gives a Be:Si ratio of 1:2 and an Al:Si ratio of 1:3 what is the empirical formula of beryl: (a) Be2Al3Si6O19, (b) Be3Al2 (SiO4)6, (c) Be3Al2Si6O18, (d) BeAl2Si6O15?

Practice Exercise 2The cyclosilicate ion consists of three silicate tetrahedra linked together in a ring. The ion contains three Si atoms and nine O atoms. What is the overall charge on the ion?

Beryl

SEcTION 22.11 Boron 987

in addition to SiO2 from sand. The CaO and Na2O are produced by heating two inex-pensive chemicals, limestone 1CaCO32 and soda ash 1Na2CO32, which decompose at high temperatures:

CaCO31s2 ¡ CaO1s2 + CO21g2 [22.73]

Na2CO31s2 ¡ Na2O1s2 + CO21g2 [22.74]

Other substances can be added to soda-lime glass to produce color or to change the properties of the glass in various ways. The addition of CoO, for example, produces the deep blue color of “cobalt glass.” Replacing Na2O with K2O results in a harder glass that has a higher melting point. Replacing CaO with PbO results in a denser “lead crystal” glass with a higher refractive index. Lead crystal is used for decorative glassware; the higher refractive index gives this glass a particularly sparkling appearance. Addition of nonmetal oxides, such as B2O3 and P4O10, which form network structures related to the silicates, also changes the properties of the glass. Adding B2O3 creates a “boro-silicate” glass with a higher melting point and a greater ability to withstand tempera-ture changes. Such glasses, sold commercially under trade names such as Pyrex® and Kimax®, are used where resistance to thermal shock is important, such as in laboratory glassware or coffeemakers.

SiliconesSilicones consist of O ¬ Si ¬ O chains in which the two remaining bonding positions on each silicon are occupied by organic groups such as CH3:

O O OSi

H3C CH3

Si

H3C CH3 H3C CH3

OSi . . .. . .

Depending on chain length and degree of cross-linking, silicones can be either oils or rubber-like materials. Silicones are nontoxic and have good stability toward heat, light, oxygen, and water. They are used commercially in a wide variety of prod-ucts, including lubricants, car polishes, sealants, and gaskets. They are also used for waterproofing fabrics. When applied to a fabric, the oxygen atoms form hydrogen bonds with the molecules on the surface of the fabric. The hydrophobic (water-repelling) organic groups of the silicone are then left pointing away from the surface as a barrier.

Give it Some thoughtDistinguish among the substances silicon, silica, and silicone.

22.11 | BoronBoron is the only group 3A element that can be considered nonmetallic and thus is our final element in this chapter. The element has an extended network struc-ture with a melting point 12300 °C2 that is intermediate between the melting points of carbon 13550 °C2 and silicon 11410 °C2. The electron configuration of boron is 3He42s22p1.

In the family of compounds called boranes, the molecules contain only boron and hydrogen. The simplest borane is BH3. This molecule contains only six valence electrons and is therefore an exception to the octet rule. As a result, BH3 reacts with itself to form diborane 1B2H62. This reaction can be viewed as a Lewis acid–base reac-tion in which one B ¬ H bonding pair of electrons in each BH3 molecule is donated to the other. As a result, diborane is an unusual molecule in which hydrogen atoms form a bridge between two B atoms (▶ Figure 22.35). Such hydrogens, called bridging hydrogens, exhibit interesting chemical reactivity, which you may learn about in a more advanced chemistry course.

3A5B

13Al

31Ga

49In81Tl

▲ Figure 22.35 the structure of diborane 1B2H62.


Sharing hydrogen atoms between the two boron atoms compensates somewhat for the deficiency in valence electrons around each boron. Nevertheless, diborane is an extremely reactive molecule, spontaneously flammable in air in a highly exothermic reaction:

B2H61g2 + 3 O21g2 ¡ B2O31s2 + 3 H2O1g2 ∆H ° = -2030 kJ [22.75]

Boron and hydrogen form a series of anions called borane anions. Salts of the boro-hydride ion 1BH4

- 2 are widely used as reducing agents. For example, sodium borohy-dride 1NaBH42 is a commonly used reducing agent for certain organic compounds.

Give it Some thoughtRecall that the hydride ion is H- . What is the oxidation state of boron in sodium borohydride?

The only important oxide of boron is boric oxide 1B2O32. This substance is the anhydride of boric acid, which we may write as H3BO3 or B1OH23. Boric acid is so weak an acid 1Ka = 5.8 * 10 - 102 that solutions of H3BO3 are used as an eyewash. Upon heating, boric acid loses water by a condensation reaction similar to that described for phosphorus in Section 22.8:

4 H3BO31s2 ¡ H2B4O71s2 + 5 H2O1g2 [22.76]

The diprotic acid H2B4O7 is tetraboric acid. The hydrated sodium salt Na2B4O7 #10 H2O, called borax, occurs in dry lake deposits in California and can also be prepared from other borate minerals. Solutions of borax are alkaline, and the substance is used in various laundry and cleaning products.

solutioN(a) The BrF2

+ ion has 7 + 2172 - 1 = 20 valence-shell electrons. The Lewis structure for the ion is

Br FF+

Because there are four electron domains around the central Br atom, the resulting electron domain geometry is tetrahedral

(Section 9.2). Because bonding pairs of electrons occupy two of these domains, the molecular geometry is bent:

+ +

The BrF4- ion has 7 + 4172 + 1 = 36 electrons, leading to the

Lewis structure

The interhalogen compound BrF3 is a volatile, straw-colored liquid. The compound exhibits appre-ciable electrical conductivity because of autoionization (“solv” refers to BrF3 as the solvent):

2 BrF31l2 ∆ BrF2+1solv2 + BrF4

- 1solv2(a) What are the molecular structures of the BrF2

+ and BrF4- ions?

(b) The electrical conductivity of BrF3 decreases with increasing temperature. Is the autoionization process exothermic or endothermic?

(c) One chemical characteristic of BrF3 is that it acts as a Lewis acid toward fluoride ions. What do we expect will happen when KBr is dissolved in BrF3?

Br FF

F

F

−

Because there are six electron domains around the central Br atom in this ion, the electron-domain geometry is octahedral. The two nonbonding pairs of electrons are located opposite each other on the octahedron, leading to a square-planar molecular geometry:

−−

Putting Concepts togethersamPlE iNtEGRativE ExERCisE

chapter Summary and Key Terms 989

Chapter summary and Key termsthe perioDic tAble AND cheMicAl reActioNS (SectioN 22.1) The periodic table is useful for organizing and remembering the descriptive chemistry of the elements. Among elements of a given group, size increases with increasing atomic number, and electronega-tivity and ionization energy decrease. Nonmetallic character parallels electronegativity, so the most nonmetallic elements are found in the upper right portion of the periodic table.

Among the nonmetallic elements, the first member of each group differs dramatically from the other members; it forms a maximum of four bonds to other atoms and exhibits a much greater tendency to form p bonds than the heavier elements in its group.

Because O2 and H2O are abundant in our world, we focus on two important and general reaction types as we discuss the nonmet-als: oxidation by O2 and proton-transfer reactions involving H2O or aqueous solutions.

hyDroGeN (SectioN 22.2) Hydrogen has three isotopes: protium 11

1H2, deuterium 121H2, and tritium 13

1H2. Hydrogen is not a member of any particular periodic group, although it is usually placed above lithium. The hydrogen atom can either lose an electron, forming H+, or gain one, forming H- (the hydride ion). Because the H ¬ H bond is relatively strong, H2 is fairly unreactive unless activated by heat or a catalyst. Hydrogen forms a very strong bond to oxygen, so the reac-tions of H2 with oxygen-containing compounds usually lead to the formation of H2O. Because the bonds in CO and CO2 are even stron-ger than the O ¬ H bond, the reaction of H2O with carbon or certain organic compounds leads to the formation of H2. The H+1aq2 ion is able to oxidize many metals, forming H21g2. The electrolysis of water also forms H21g2.

The binary compounds of hydrogen are of three general types: ionic hydrides (formed by active metals), metallic hydrides (formed by transition metals), and molecular hydrides (formed by nonmetals). The ionic hydrides contain the H- ion; because this ion is extremely basic, ionic hydrides react with H2O to form H2 and OH- .

Group 8A: the Noble GASeS AND Group 7A: hAloGeNS (SectioNS 22.3 AND 22.4) The noble gases (group 8A) exhibit a very limited chemical reactivity because of the exceptional stability of their electron configurations. The xenon fluorides and oxides and KrF2 are the best-established compounds of the noble gases.

The halogens (group 7A) occur as diatomic molecules. All except fluorine exhibit oxidation states varying from -1 to +7. Fluorine is the most electronegative element, so it is restricted to the oxidation states 0 and -1. The oxidizing power of the element (the tendency to form the -1 oxidation state) decreases as we proceed down the group.

The hydrogen halides are among the most useful compounds of these elements; these gases dissolve in water to form the hydrohalic

acids, such as HCl(aq). Hydrofluoric acid reacts with silica. The interhalogens are compounds formed between two different halogen elements. Chlorine, bromine, and iodine form a series of oxyacids, in which the halogen atom is in a positive oxidation state. These com-pounds and their associated oxyanions are strong oxidizing agents.

oXyGeN AND the other Group 6A eleMeNtS (SectioNS 22.5 AND 22.6) Oxygen has two allotropes, O2 and O3 (ozone). Ozone is unstable compared to O2, and it is a stronger oxidizing agent than O2. Most reactions of O2 lead to oxides, compounds in which oxygen is in the -2 oxidation state. The soluble oxides of nonmet-als generally produce acidic aqueous solutions; they are called acidic anhydrides or acidic oxides. In contrast, soluble metal oxides produce basic solutions and are called basic anhydrides or basic oxides. Many metal oxides that are insoluble in water dissolve in acid, accompanied by the formation of H2O. Peroxides contain O ¬ O bonds and oxygen in the -1 oxidation state. Peroxides are unstable, decomposing to O2 and oxides. In such reactions peroxides are simultaneously oxidized and reduced, a process called disproportionation. Superoxides contain the O2

- ion in which oxygen is in the - 12 oxidation state.Sulfur is the most important of the other group 6A elements. It

has several allotropic forms; the most stable one at room tempera-ture consists of S8 rings. Sulfur forms two oxides, SO2 and SO3, vand both are important atmospheric pollutants. Sulfur trioxide is the anhydride of sulfuric acid, the most important sulfur compound and the most-produced industrial chemical. Sulfuric acid is a strong acid and a good dehydrating agent. Sulfur forms several oxyanions as well, including the SO3

2 - (sulfite), SO42 - (sulfate), and S2O3

2 - (thiosul-fate) ions. Sulfur is found combined with many metals as a sulfide, in which sulfur is in the -2 oxidation state. These compounds often react with acids to form hydrogen sulfide 1H2S2, which smells like rotten eggs.

NitroGeN AND the other Group 5A eleMeNtS (SectioNS 22.7 AND 22.8) Nitrogen is found in the atmosphere as N2 mol-ecules. Molecular nitrogen is chemically very stable because of the strong N ‚ N bond. Molecular nitrogen can be converted into ammo-nia via the Haber process. Once the ammonia is made, it can be con-verted into a variety of different compounds that exhibit nitrogen oxidation states ranging from -3 to +5. The most important indus-trial conversion of ammonia is the ostwald process, in which ammonia is oxidized to nitric acid 1HNO32.

Nitrogen has three important oxides: nitrous oxide 1N2O2, nitric oxide (NO), and nitrogen dioxide 1NO22. Nitrous acid 1HNO22 is a weak acid; its conjugate base is the nitrite ion 1NO2

- 2. Another impor-tant nitrogen compound is hydrazine 1N2H42.

Phosphorus is the most important of the remaining group 5A elements. It occurs in nature as phosphate minerals. Phosphorus has

(b) The observation that conductivity decreases as temperature increases indicates that there are fewer ions present in the solu-tion at the higher temperature. Thus, increasing the temperature causes the equilibrium to shift to the left. According to Le Châte-lier’s principle, this shift indicates that the reaction is exothermic as it proceeds from left to right. (Section 15.7)

(c) A Lewis acid is an electron-pair acceptor. (Section 16.11) The fluoride ion has four valence-shell electron pairs and can act as a Lewis base (an electron-pair donor). Thus, we can envision the following reaction occurring:

Br FFF −F

Br

F− BrF3 BrF4−

F

FF F

−

+

+


visualizing Concepts

22.1 (a) One of these structures is a stable compound; the other is not. Identify the stable compound, and explain why it is sta-ble. Explain why the other compound is not stable. (b) What is the geometry around the central atoms of the stable com-pound? [Section 22.1]

H

H

H

H

SiSi

H

H

H

H

CC

22.2 (a) Identify the type of chemical reaction represented by the following diagram. (b) Place appropriate charges on the spe-cies on both sides of the equation. (c) Write the chemical equation for the reaction. [Section 22.1]

+ +

22.3 Which of the following species (there may be more than one) is/are likely to have the structure shown here: (a) XeF4, (b) BrF4

+, (c) SiF4, (d) TeCl4, (e) HClO4? The colors do not reflect atom identities.) [Sections 22.3, 22.4, 22.6, and 22.10]

22.4 You have two glass bottles, one containing oxygen and one filled with nitrogen. How could you determine which one is which? [Sections 22.5 and 22.7]

22.5 Write the molecular formula and Lewis structure for each of the following oxides of nitrogen: [Section 22.7]

several allotropes, including white phosphorus, which consists of P4 tetrahedra. In reaction with the halogens, phosphorus forms tri-halides PX3 and pentahalides PX5. These compounds undergo hydro-lysis to produce an oxyacid of phosphorus and HX.

Phosphorus forms two oxides, P4O6 and P4O10. Their corre-sponding acids, phosphorous acid and phosphoric acid, undergo condensation reactions when heated. Phosphorus compounds are important in biochemistry and as fertilizers.

cArboN AND the other Group 4A eleMeNtS (SectioNS 22.9 AND 22.10) The allotropes of carbon include diamond, graph-ite, fullerenes, carbon nanotubes, and graphene. Amorphous forms of graphite include charcoal and carbon black. Carbon forms two com-mon oxides, CO and CO2. Aqueous solutions of CO2 produce the weak diprotic acid carbonic acid 1H2CO32, which is the parent acid of hydrogen carbonate and carbonate salts. Binary compounds of carbon are called carbides. Carbides may be ionic, interstitial, or covalent. Cal-cium carbide 1CaC22 contains the strongly basic acetylide ion 1C2

2 - 2, which reacts with water to form acetylene.

The other group 4A elements show great diversity in physical and chemical properties. Silicon, the second most abundant element, is a semi-conductor. It reacts with Cl2 to form SiCl4, a liquid at room temperature, a reaction that is used to help purify silicon from its native minerals. Sili-con forms strong Si ¬ O bonds and therefore occurs in a variety of silicate minerals. Silica is SiO2; silicates consist of SiO4 tetrahedra, linked together at their vertices to form chains, sheets, or three-dimensional structures. The most common three-dimensional silicate is quartz 1SiO22. Glass is an amorphous (noncrystalline) form of SiO2. Silicones contain O ¬ Si ¬ O chains with organic groups bonded to the Si atoms. Like silicon, germa-nium is a metalloid; tin and lead are metallic.

boroN (SectioN 22.11) Boron is the only group 3A element that is a nonmetal. It forms a variety of compounds with hydrogen called boron hydrides, or boranes. Diborane 1B2H62 has an unusual structure with two hydrogen atoms that bridge between the two boron atoms. Boranes react with oxygen to form boric oxide 1B2O32, in which boron is in the +3 oxidation state. Boric oxide is the anhydride of boric acid 1H3BO32. Boric acid readily undergoes condensation reactions.

learning outcomes after studying this chapter, you should be able to:

• Use periodic trends to explain the basic differences between the elements of a group or period. (Section 22.1)

• Explain two ways in which the first element in a group differs from subsequent elements in the group. (Section 22.1)

• Be able to determine electron configurations, oxidation numbers, and molecular shapes of elements and compounds. (Sections 22.2–22.11)

• Know the sources of the common nonmetals, how they are obtained, and how they are used. (Sections 22.2–22.11)

• Understand how phosphoric and phosphorous acids undergo con-densation reactions. (Section 22.8)

• Explain how the bonding and structures of silicates relate to their chemical formulas and properties. (Section 22.10)

Exercises

Exercises 991

22.6 Which property of the group 6A elements might be the one de-picted in the graph shown here: (a) electronegativity, (b) first ionization energy, (c) density, (d) X ¬ X single-bond enthalpy, (e) electron affinity? Explain your answer. [Sections 22.5 and 22.6]

O

S

Te

Po

Se

22.7 The atomic and ionic radii of the first three group 6A elements are

Ionicradius(Å)

Atomicradius(Å)

1.20 1.98

Se Se2−

1.05 1.84

S S2−

0.66 1.40

O O2−

(a) Explain why the atomic radius increases in moving downward in the group. (b) Explain why the ionic radii are larger than the atomic radii. (c) Which of the three anions would you expect to be the strongest base in water? Explain. [Sections 22.5 and 22.6]

22.8 Which property of the third-row nonmetallic elements might be the one depicted below: (a) first ionization energy, (b) atomic radius, (c) electronegativity, (d) melting point, (e) X ¬ X single-bond enthalpy? Explain both your choice and why the other choices would not be correct. [Sections 22.3, 22.4, 22.6, 22.8, and 22.10]

Si

P

Cl

Ar

S

22.9 Which of the following compounds would you expect to be the most generally reactive, and why? (Each corner in these structures represents a CH2 group.) [Section 22.8]

OO O

22.10 (a) Draw the Lewis structures for at least four species that have the general formula

X Yn

where X and Y may be the same or different, and n may have a value from +1 to -2. (b) Which of the compounds is likely to be the strongest Brønsted base? Explain. [Sections 22.1, 22.7, and 22.9]

Periodic trends and Chemical Reactions (section 22.1)

22.11 Identify each of the following elements as a metal, nonmetal, or metalloid: (a) phosphorus, (b) strontium, (c) manganese, (d) selenium, (e) sodium, (f) krypton.

22.12 Identify each of the following elements as a metal, nonmetal, or metalloid: (a) gallium, (b) molybdenum, (c) tellurium, (d) arsenic, (e) xenon, (f) ruthenium.

22.13 Consider the elements O, Ba, Co, Be, Br, and Se. From this list, select the element that (a) is most electronegative, (b) ex-hibits a maximum oxidation state of +7, (c) loses an electron most readily, (d) forms p bonds most readily, (e) is a transi-tion metal, (f) is a liquid at room temperature and pressure.

22.14 Consider the elements Li, K, Cl, C, Ne, and Ar. From this list, select the element that (a) is most electronegative, (b) has the greatest metallic character, (c) most readily forms a positive ion, (d) has the smallest atomic radius, (e) forms p bonds most readily, (f) has multiple allotropes.

22.15 Explain the following observations: (a) The highest fluoride compound formed by nitrogen is NF3, whereas phosphorus readily forms PF5. (b) Although CO is a well-known com-pound, SiO does not exist under ordinary conditions. (c) AsH3 is a stronger reducing agent than NH3.

22.16 Explain the following observations: (a) HNO3 is a stronger oxidizing agent than H3PO4. b) Silicon can form an ion with six fluorine atoms, SiF6

2 - , whereas carbon is able to bond to a maximum of four, CF4. (c) There are three compounds formed by carbon and hydrogen that contain two carbon at-oms each (C2H2, C2H4, and C2H6), whereas silicon forms only one analogous compound 1Si2H62.

22.17 Complete and balance the following equations:(a) NaOCH31s2 + H2O1l2 ¡(b) CuO1s2 + HNO31aq2 ¡(c) WO31s2 + H21g2 ¡∆

(d) NH2OH1l2 + O21g2 ¡(e) Al4C31s2 + H2O1l2 ¡

22.18 Complete and balance the following equations:(a) Mg3N21s2 + H2O1l2 ¡(b) C3H7OH1l2 + O21g2 ¡(c) MnO21s2 + C1s2 ¡∆

(d) AlP1s2 + H2O1l2 ¡(e) Na2S1s2 + HCl1aq2 ¡


hydrogen, the Noble Gases, and the halogens (sections 22.2, 22.3, and 22.4)

22.19 (a) Give the names and chemical symbols for the three isotopes of hydrogen. (b) List the isotopes in order of decreasing natural abundance. (c) Which hydrogen isotope is radioactive? (d) Write the nuclear equation for the radioactive decay of this isotope.

22.20 Are the physical properties of H2O different from D2O? Explain. 22.21 Give a reason why hydrogen might be placed along with the

group 1A elements of the periodic table. 22.22 What does hydrogen have in common with the halogens?

Explain. 22.23 Write a balanced equation for the preparation of H2 using (a) Mg

and an acid, (b) carbon and steam, (c) methane and steam. 22.24 List (a) three commercial means of producing H2, (b) three

industrial uses of H2. 22.25 Complete and balance the following equations:

(a) NaH1s2 + H2O1l2 ¡(b) Fe1s2 + H2SO41aq2 ¡(c) H21g2 + Br21g2 ¡(d) Na1l2 + H21g2 ¡(e) PbO1s2 + H21g2 ¡

22.26 Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chap-ter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magne-sium oxide and hydrogen. (c) Manganese(IV) oxide is re-duced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.

22.27 Identify the following hydrides as ionic, metallic, or molecu-lar: (a) BaH2, (b) H2Te, (c) TiH1.7.

22.28 Identify the following hydrides as ionic, metallic, or molecular: (a) B2H6, (b) RbH, (c) Th4H1.5.

22.29 Describe two characteristics of hydrogen that are favorable for its use as a general energy source in vehicles.

22.30 The H2>O2 fuel cell converts elemental hydrogen and oxygen into water, producing, theoretically, 1.23 V. What is the most sustainable way to obtain hydrogen to run a large number of fuel cells? Explain.

22.31 Why does xenon form stable compounds with fluorine, whereas argon does not?

22.32 A friend tells you that the “neon” in neon signs is a compound of neon and aluminum. Can your friend be correct? Explain.

22.33 Write the chemical formula for each of the following, and indi-cate the oxidation state of the halogen or noble-gas atom in each: (a) calcium hypobromite, (b) bromic acid, (c) xenon trioxide, (d) perchlorate ion, (e) iodous acid, (f) iodine pentafluoride.

22.34 Write the chemical formula for each of the following com-pounds, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) chlorate ion, (b) hydroiodic acid, (c) iodine trichloride, (d) sodium hypochlorite, (e) perchloric acid, (f) xenon tetrafluoride.

22.35 Name the following compounds and assign oxidation states to the halogens in them: (a) Fe1ClO323, (b) HClO2, (c) XeF6, (d) BrF5, (e) XeOF4, (f) HIO3.

22.36 Name the following compounds and assign oxidation states to the halogens in them: (a) KClO3, (b) Ca1IO322, (c) AlCl3, (d) HBrO3, (e) H5IO6, (f) XeF4.

22.37 Explain each of the following observations: (a) At room tempera-ture I2 is a solid, Br2 is a liquid, and Cl2 and F2 are both gases. (b) F2 cannot be prepared by electrolytic oxidation of aqueous F - solutions. (c) The boiling point of HF is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxi-dizing power in the order F2 7 Cl2 7 Br2 7 I2.

22.38 Explain the following observations: (a) For a given oxidation state, the acid strength of the oxyacid in aqueous solution de-creases in the order chlorine 7 bromine 7 iodine. (b) Hy-drofluoric acid cannot be stored in glass bottles. (c) HI cannot be prepared by treating NaI with sulfuric acid. (d) The interh-alogen ICl3 is known, but BrCl3 is not.

oxygen and the other Group 6a Elements (sections 22.5 and 22.6)

22.39 Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form O2 and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, PbS(s), reacts with ozone to form PbSO41s2 and O21g2. (d) When heated in air, ZnS(s) is converted to ZnO. (e) Potassium peroxide reacts with CO21g2 to give potassium carbonate and O2. (f) Oxygen is converted to ozone in the upper atmosphere.

22.40 Complete and balance the following equations:(a) CaO1s2 + H2O1l2 ¡(b) Al2O31s2 + H+1aq2 ¡(c) Na2O21s2 + H2O1l2 ¡(d) N2O31g2 + H2O1l2 ¡(e) KO21s2 + H2O1l2 ¡(f) NO1g2 + O31g2 ¡

22.41 Predict whether each of the following oxides is acidic, basic, amphoteric, or neutral: (a) NO2, (b) CO2, (c) Al2O3, (d) CaO.

22.42 Select the more acidic member of each of the following pairs: (a) Mn2O7 and MnO2, (b) SnO and SnO2, (c) SO2 and SO3, (d) SiO2 and SO2, (e) Ga2O3 and In2O3, (f) SO2 and SeO2.

22.43 Write the chemical formula for each of the following com-pounds, and indicate the oxidation state of the group 6A element in each: (a) selenous acid, (b) potassium hydrogen sulfite, (c) hydrogen telluride, (d) carbon disulfide, (e) cal-cium sulfate, (f) cadmium sulfide, (g) zinc telluride.

22.44 Write the chemical formula for each of the following com-pounds, and indicate the oxidation state of the group 6A ele-ment in each: (a) sulfur tetrachloride, (b) selenium trioxide, (c) sodium thiosulfate, (d) hydrogen sulfide, (e) sulfuric acid, (f) sulfur dioxide, (g) mercury telluride.

22.45 In aqueous solution, hydrogen sulfide reduces (a) Fe3 + to Fe2 + , (b) Br2 to Br - , (c) MnO4

- to Mn2 + , (d) HNO3 to NO2. In all cases, under appropriate conditions, the product is elemental sul-fur. Write a balanced net ionic equation for each reaction.

22.46 An aqueous solution of SO2 reduces (a) aqueous KMnO4 to MnSO41aq2, (b) acidic aqueous K2Cr2O7 to aqueous Cr3 + , (c) aqueous Hg21NO322 to mercury metal. Write balanced equa-tions for these reactions.

22.47 Write the Lewis structure for each of the following species, and indicate the structure of each: (a) SeO3

2 - ; (b) S2Cl2; (c) chlorosulfonic acid, HSO3Cl (chlorine is bonded to sulfur).

22.48 The SF5- ion is formed when SF41g2 reacts with fluoride salts

containing large cations, such as CsF(s). Draw the Lewis structures for SF4 and SF5

- , and predict the molecular structure of each.

Exercises 993

22.49 Write a balanced equation for each of the following reactions: (a) Sulfur dioxide reacts with water. (b) Solid zinc sulfide reacts with hydrochloric acid. (c) Elemental sulfur reacts with sulfite ion to form thiosulfate. (d) Sulfur trioxide is dissolved in sulfuric acid.

22.50 Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction prod-ucts, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on alumi-num selenide. (b) Sodium thiosulfate is used to remove excess Cl2 from chlorine-bleached fabrics. The thiosulfate ion forms SO4

2 - and elemental sulfur, while Cl2 is reduced to Cl- .

Nitrogen and the other Group 5a Elements (sections 22.7 and 22.8)

22.51 Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) sodium nitrite, (b) ammonia, (c) nitrous oxide, (d) sodium cyanide, (e) nitric acid, (f) nitrogen dioxide, (g) nitrogen, (h) boron nitride.

22.52 Write the chemical formula for each of the following com-pounds, and indicate the oxidation state of nitrogen in each: (a) nitric oxide, (b) hydrazine, (c) potassium cyanide, (d) so-dium nitrite, (e) ammonium chloride, (f) lithium nitride.

22.53 Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) HNO2, (b) N3

- , (c) N2H5+, (d) NO3

- . 22.54 Write the Lewis structure for each of the following species,

describe its geometry, and indicate the oxidation state of the nitrogen: (a) NH4

+, (b) NO2- , (c) N2O, (d) NO2.

22.55 Complete and balance the following equations:(a) Mg3N21s2 + H2O1l2 ¡(b) NO1g2 + O21g2 ¡(c) N2O51g2 + H2O1l2 ¡(d) NH31aq2 + H+1aq2 ¡(e) N2H41l2 + O21g2 ¡

Which ones of these are redox reactions? 22.56 Write a balanced net ionic equation for each of the follow-

ing reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid re-acts with sulfur with formation of nitrogen dioxide. (c) Con-centrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming NF3. (e) Hydrazine reduces CrO4

2 - to Cr1OH24- in

base (hydrazine is oxidized to N2). 22.57 Write complete balanced half-reactions for (a) oxidation of

nitrous acid to nitrate ion in acidic solution, (b) oxidation of N2 to N2O in acidic solution.

22.58 Write complete balanced half-reactions for (a) reduction of nitrate ion to NO in acidic solution, (b) oxidation of HNO2 to NO2 in acidic solution.

22.59 Write a molecular formula for each compound, and indicate the oxidation state of the group 5A element in each formula: (a) phosphorous acid, (b) pyrophosphoric acid, (c) antimony trichloride, (d) magnesium arsenide, (e) diphosphorus pent-oxide, (f) sodium phosphate.

22.60 Write a chemical formula for each compound or ion, and in-dicate the oxidation state of the group 5A element in each for-mula: (a) phosphate ion, (b) arsenous acid, (c) antimony(III) sulfide, (d) calcium dihydrogen phosphate, (e) potassium phosphide, (f) gallium arsenide.

22.61 Account for the following observations: (a) Phosphorus forms a pentachloride, but nitrogen does not. (b) H3PO2 is a monoprotic acid. (c) Phosphonium salts, such as PH4Cl, can be formed under anhydrous conditions, but they cannot be made in aqueous solution. (d) White phosphorus is more re-active than red phosphorus.

22.62 Account for the following observations: (a) H3PO3 is a di-protic acid. (b) Nitric acid is a strong acid, whereas phos-phoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) So-lutions of Na3PO4 are quite basic.

22.63 Write a balanced equation for each of the following reactions: (a) preparation of white phosphorus from calcium phosphate, (b) hydrolysis of PBr3, (c) reduction of PBr3 to P4 in the gas phase, using H2.

22.64 Write a balanced equation for each of the following reactions: (a) hydrolysis of PCl5, (b) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, (c) reaction of P4O10 with water.

Carbon, the other Group 4a Elements, and Boron (sections 22.9, 22.10, and 22.11)

22.65 Give the chemical formula for (a) hydrocyanic acid, (b) nickel tetracarbonyl, (c) barium bicarbonate, (d) calcium acetylide, (e) potassium carbonate.

22.66 Give the chemical formula for (a) carbonic acid, (b) sodium cyanide, (c) potassium hydrogen carbonate, (d) acetylene, (e) iron pentacarbonyl.

22.67 Complete and balance the following equations:(a) ZnCO31s2 ¡∆

(b) BaC21s2 + H2O1l2 ¡(c) C2H21g2 + O21g2 ¡(d) CS21g2 + O21g2 ¡(e) Ca1CN221s2 + HBr1aq2 ¡

22.68 Complete and balance the following equations:(a) CO21g2 + OH- 1aq2 ¡(b) NaHCO31s2 + H+1aq2 ¡(c) CaO1s2 + C1s2 ¡∆

(d) C1s2 + H2O1g2 ¡∆

(e) CuO1s2 + CO1g2 ¡ 22.69 Write a balanced equation for each of the following reac-

tions: (a) Hydrogen cyanide is formed commercially by pass-ing a mixture of methane, ammonia, and air over a catalyst at 800 °C. Water is a by-product of the reaction. (b) Baking soda reacts with acids to produce carbon dioxide gas. (c) When barium carbonate reacts in air with sulfur dioxide, barium sulfate and carbon dioxide form.

22.70 Write a balanced equation for each of the following reactions: (a) Burning magnesium metal in a carbon dioxide atmo-sphere reduces the CO2 to carbon. (b) In photosynthesis, so-lar energy is used to produce glucose 1C6H12O62 and O2 from carbon dioxide and water. (c) When carbonate salts dissolve in water, they produce basic solutions.

22.71 Write the formulas for the following compounds, and indi-cate the oxidation state of the group 4A element or of boron in each: (a) boric acid, (b) silicon tetrabromide, (c) lead(II) chloride, (d) sodium tetraborate decahydrate (borax), (e) bo-ric oxide, (f) germanium dioxide.


22.72 Write the formulas for the following compounds, and indicate the oxidation state of the group 4A element or of boron in each: (a) silicon dioxide, (b) germanium tetrachloride, (c) sodium boro-hydride, (d) stannous chloride, (e) diborane, (f) boron trichloride.

22.73 Select the member of group 4A that best fits each description: (a) has the lowest first ionization energy, (b) is found in oxi-dation states ranging from -4 to +4, (c) is most abundant in Earth’s crust.

22.74 Select the member of group 4A that best fits each description: (a) forms chains to the greatest extent, (b) forms the most ba-sic oxide, (c) is a metalloid that can form 2+ ions.

22.75 (a) What is the characteristic geometry about silicon in all sil-icate minerals? (b) Metasilicic acid has the empirical formula H2SiO3. Which of the structures shown in Figure 22.34 would you expect metasilicic acid to have?

22.76 Speculate as to why carbon forms carbonate rather than sili-cate analogs.

22.77 (a) Determine the number of calcium ions in the chemical for-mula of the mineral hardystonite, CaxZn1Si2O72. (b) Determine the number of hydroxide ions in the chemical formula of the mineral pyrophyllite , Al21Si2O5221OH2x.

22.78 (a) Determine the number of sodium ions in the chemi-cal formula of albite, NaxAlSi3O8. (b) Determine the num-ber of hydroxide ions in the chemical formula of tremolite, Ca2Mg51Si4O11221OH2x.

22.79 (a) How does the structure of diborane 1B2H62 differ from that of ethane 1C2H62? (b) Explain why diborane adopts the geom-etry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as “hydridic”?

22.80 Write a balanced equation for each of the following reac-tions: (a) Diborane reacts with water to form boric acid and molecular hydrogen. (b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid. (c) Boron oxide dissolves in water to give a solution of boric acid.

additional Exercises 22.81 Indicate whether each of the following statements is true

or false (a) H21g2 and D21g2 are allotropic forms of hydro-gen. (b) ClF3 is an interhalogen compound. (c) MgO(s) is an acidic anhydride. (d) SO21g2 is an acidic anhydride. (e) 2 H3PO41l2 S H4P2O71l2+H2O1g2 is an example of a con-densation reaction. (f) Tritium is an isotope of the element hydrogen. (g) 2SO21g2+O21g2 S 2SO31g2 is an example of a disproportionation reaction.

22.82 Although the ClO4- and IO4

- ions have been known for a long time, BrO4

- was not synthesized until 1965. The ion was synthe-sized by oxidizing the bromate ion with xenon difluoride, produc-ing xenon, hydrofluoric acid, and the perbromate ion. (a) Write the balanced equation for this reaction. (b) What are the oxidation states of Br in the Br-containing species in this reaction?

22.83 Write a balanced equation for the reaction of each of the following compounds with water: (a) SO21g2, (b) Cl2O71g2, (c) Na2O21s2, (d) BaC21s2, (e) RbO21s2 (f) Mg3N21s2, (g) NaH(s).

22.84 What is the anhydride for each of the following acids: (a) H2SO4, (b) HClO3, (c) HNO2, (d) H2CO3, (e) H3PO4?

22.85 Hydrogen peroxide is capable of oxidizing (a) hydrazine to N2 and H2O, (b) SO2 to SO4

2 - , (c) NO2- to NO3

- , (d) H2S1g2 to S(s), (e) Fe2 + to Fe3 + . Write a balanced net ionic equation for each of these redox reactions.

22.86 Explain why SO2 can be used as a reducing agent but SO3 cannot. 22.87 A sulfuric acid plant produces a considerable amount of heat.

This heat is used to generate electricity, which helps reduce operating costs. The synthesis of H2SO4 consists of three main

chemical processes: (a) oxidation of S to SO2, (b) oxidation of SO2 to SO3, (c) the dissolving of SO3 in H2SO4 and its reac-tion with water to form H2SO4. If the third process produces 130 kJ>mol, how much heat is produced in preparing a mole of H2SO4 from a mole of S? How much heat is produced in preparing 5000 pounds of H2SO4?

22.88 (a) What is the oxidation state of P in PO43 - and of N in NO3

- ? (b) Why doesn’t N form a stable NO4

3 - . ion analogous to P? 22.89 (a) The P4, P4O6 and P4O10 molecules have a common struc-

tural feature of four P atoms arranged in a tetrahedron (Figures 22.27 and 22.28). Does this mean that the bonding between the P atoms is the same in all these cases? Explain. (b) Sodium trimetaphosphate 1Na3P3O92 and sodium tetram-etaphosphate 1Na4P4O122 are used as water-softening agents. They contain cyclic P3O9

3 - and P4O124 - ions, respectively.

Propose reasonable structures for these ions. 22.90 Ultrapure germanium, like silicon, is used in semiconductors.

Germanium of “ordinary” purity is prepared by the high-tem-perature reduction of GeO2 with carbon. The Ge is converted to GeCl4 by treatment with Cl2 and then purified by distillation; GeCl4 is then hydrolyzed in water to GeO2 and reduced to the elemental form with H2. The element is then zone refined. Write a balanced chemical equation for each of the chemical transfor-mations in the course of forming ultrapure Ge from GeO2.

22.91 (a) Determine the charge of the aluminosilicate ion whose composition is AlSi3O10. (b) Using Figure 22.33, propose a reasonable description of the structure of this aluminosilicate.

integrative Exercises [22.92] (a) How many grams of H2 can be stored in 100.0 kg of the

alloy FeTi if the hydride FeTiH2 is formed? (b) What volume does this quantity of H2 occupy at STP? (c) If this quantity of hydrogen was combusted in air to produce liquid water, how much energy could be produced?

[22.93] Using the thermochemical data in Table 22.1 and Appendix C, calculate the average Xe ¬ F bond enthalpies in XeF2, XeF4, and XeF6, respectively. What is the significance of the trend in these quantities?

22.94 Hydrogen gas has a higher fuel value than natural gas on a mass basis but not on a volume basis. Thus, hydrogen is not competitive with natural gas as a fuel transported long dis-tances through pipelines. Calculate the heats of combustion of H2 and CH4 (the principal component of natural gas) (a) per mole of each, (b) per gram of each, (c) per cubic meter of each at STP. Assume H2O1l2 as a product.

22.95 Using ∆G°f for ozone from Appendix C, calculate the equi-

librium constant for Equation 22.24 at 298.0 K, assuming no electrical input.

Designing an Experiment 995

22.96 The solubility of Cl2 in 100 g of water at STP is 310 cm3. Assume that this quantity of Cl2 is dissolved and equilibrated as follows:

Cl21aq2 + H2O ∆ Cl- 1aq2 + HClO1aq2 + H+1aq2 (a) If the equilibrium constant for this reaction is 4.7 * 10 - 4,

calculate the equilibrium concentration of HClO formed. (b) What is the pH of the final solution?

[22.97] When ammonium perchlorate decomposes thermally, the products of the reaction are N21g2, O21g2, H2O1g2, and HCl(g). (a) Write a balanced equation for the reaction. (Hint: You might find it easier to use fractional coefficients for the products.) (b) Calculate the enthalpy change in the reaction per mole of NH4ClO4. The standard enthalpy of formation of NH4ClO41s2 is -295.8 kJ. (c) When NH4ClO41s2 is employed in solid-fuel booster rockets, it is packed with powdered aluminum. Given the high temperature needed for NH4ClO41s2 decomposition and what the products of the reaction are, what role does the aluminum play? (d) Calculate the volume of all the gases that would be produced at STP, assuming complete reaction of one pound of ammonium perchlorate.

22.98 The dissolved oxygen present in any highly pressurized, high-temperature steam boiler can be extremely corrosive to its metal parts. Hydrazine, which is completely miscible with water, can be added to remove oxygen by reacting with it to form nitrogen and water. (a) Write the balanced equa-tion for the reaction between gaseous hydrazine and oxygen. (b) Calculate the enthalpy change accompanying this reaction. (c) Oxygen in air dissolves in water to the extent of 9.1 ppm at 20 °C at sea level. How many grams of hydrazine are required to react with all the oxygen in 3.0 * 104 L (the volume of a small swimming pool) under these conditions?

22.99 One method proposed for removing SO2 from the flue gases of power plants involves reaction with aqueous H2S. Elemental sul-fur is the product. (a) Write a balanced chemical equation for the reaction. (b) What volume of H2S at 27 °C and 760 torr would be required to remove the SO2 formed by burning 2.0 tons of coal containing 3.5% S by mass? (c) What mass of elemental sulfur is produced? Assume that all reactions are 100% efficient.

22.100 The maximum allowable concentration of H2S1g2 in air is 20 mg per kilogram of air (20 ppm by mass). How many grams of FeS would be required to react with hydrochloric acid to produce this concentration at 1.00 atm and 25 °C in an average room measuring 12 ft * 20 ft * 8 ft? (Under these conditions, the average molar mass of air is 29.0 g>mol.)

22.101 The standard heats of formation of H2O1g2, H2S1g2, H2Se1g2, and H2Te1g2 are -241.8, -20.17, +29.7, and +99.6 kJ>mol, respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous at-oms are 248, 277, 227, and 197 kJ>mol of atoms for O, S, Se,

and Te, respectively. The enthalpy for dissociation of H2 is 436 kJ>mol. Calculate the average H ¬ O, H ¬ S, H ¬ Se, and H ¬ Te bond enthalpies, and comment on their trend.

22.102 Manganese silicide has the empirical formula MnSi and melts at 1280 °C. It is insoluble in water but does dissolve in aque-ous HF. (a) What type of compound do you expect MnSi to be: metallic, molecular, covalent-network, or ionic? (b) Write a likely balanced chemical equation for the reaction of MnSi with concentrated aqueous HF.

[22.103] Chemists tried for a long time to make molecular compounds containing silicon–silicon double bonds; they finally succeeded in 1981. The trick is having large, bulky R groups on the silicon atoms to make R2Si “ SiR2 compounds. What experiments could you do to prove that a new compound has a silicon–silicon double bond rather than a silicon–silicon single bond?

22.104 Hydrazine has been employed as a reducing agent for metals. Using standard reduction potentials, predict whether the fol-lowing metals can be reduced to the metallic state by hydra-zine under standard conditions in acidic solution: (a) Fe2 + , (b) Sn2 + , (c) Cu2 + , (d) Ag+, (e) Cr3 + , (f) Co3 + .

22.105 Both dimethylhydrazine, 1CH322NNH2, and methylhydra-zine, CH3NHNH2, have been used as rocket fuels. When dini-trogen tetroxide 1N2O42 is used as the oxidizer, the products are H2O, CO2, and N2. If the thrust of the rocket depends on the volume of the products produced, which of the sub-stituted hydrazines produces a greater thrust per gram total mass of oxidizer plus fuel? (Assume that both fuels generate the same temperature and that H2O1g2 is formed.)

22.106 Carbon forms an unusual unstable oxide of formula C3O2, called carbon suboxide. Carbon suboxide is made by using P2O5 to dehydrate the dicarboxylic acid called malonic acid, which has the formula HOOC ¬ CH2 ¬ COOH. (a) Write a balanced reaction for the production of carbon suboxide from malonic acid. (b) How many grams of carbon suboxide could be made from 20.00 g of malonic acid? (c) Suggest a Lewis structure for C3O2. (Hint: The Lewis structure of malonic acid suggests which atoms are connected to which.) (d) By using the information in Table 8.5, predict the C ¬ C and C ¬ O bond lengths in C3O2. (e) Sketch the Lewis structure of a product that could result by the addition of 2 mol of H2 to 1 mol of C3O2.

22.107 Borazine, 1BH231NH23, is an analog of C6H6, benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohy-dride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemi-cal equations for the production of borazine using both syn-thetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from 2.00 L of ammonia at STP, assuming diborane is in excess?

designing an ExperimentYou are given samples of five substances. At room temperature, three are colorless gases, one is a colorless liquid, and one is a white solid. You are told that the substances are NF3, PF3, PCl3, PF5, and PCl5. Let’s design experiments to determine which substance is which, using concepts from this and earlier chapters. (a) Assuming that you don’t have access to either the internet or to a handbook of chemistry (as is the case during your exams!), design experiments that would allow you to identify the substances. (b) How might you proceed differently if you had access to data from the internet? (c) Which of the substances could undergo reaction to add more atoms around the central atom? What types of reactions might you choose to test this hypothesis? (d) Based on what you know about intermolecular forces, which of the substances is likely the solid?

23Transition Metals and Coordination ChemistryThe colors of our world are beautiful, but to a chemist they are also informative—providing insights into the structure and bonding of matter. Compounds of the transition metals constitute an important group of colored substances. Some of them are used in pigments; others produce the colors in glass and precious gems. The use of vivid green, yellow, and

23.3 Common Ligands in Coordination Chemistry We examine some common geometries found in coordination complexes and how the geometries relate to coordination numbers.

23.4 nomenCLature and isomerism in Coordination Chemistry We introduce the nomenclature used for coordination compounds. We see that coordination compounds exhibit isomerism, in which two compounds have the same composition but different structures, and then look at two types: structural isomers and stereoisomers.

23.1 the transition metaLs We examine the physical properties, electron configurations, oxidation states, and magnetic properties of the transition metals.

23.2 transition-metaL CompLexes We introduce the concepts of metal complexes and ligands and provide a brief history of the development of coordination chemistry.

WhaT’s ahead

blue colors in the paintings of impressionists like Monet, Cezanne, and van Gogh was made possible by the development of synthetic pigments in the 1800s. Three such pig-ments used extensively by the impressionists were cobalt blue, CoAl2O4, chrome yellow, PbCrO4, and emerald green, Cu4(CH3COO)2(AsO2)6. In each case, the presence of a transition metal ion is directly responsible for the color of the pigment—Co2+ in cobalt blue, Cr6+ in chrome yellow, and Cu2+ in emerald green.

The color of a given transition-metal compound depends upon not only the transition-metal ion but also upon the identity and geometry of the ions and/or molecules that surround it. To appreciate the importance of the transition-metal ion surroundings, consider the colors of the minerals azurite, Cu3(CO3)2(OH)2, and mala-chite, Cu2(CO3)(OH)2 (Figure 23.1). Both minerals contain Cu2+ ions surrounded by carbonate and hydroxide ions, but subtle differences in the local surroundings of the Cu2+ ion lead to the contrasting colors of these two minerals. In this chapter, we will learn why transition-metal compounds are colored and how changes in the identity and environment of the transition-metal ion lead to changes in color.

▶ WATER LILLIES AND THE JAPANESE BRIDGE, was painted by Claude Monet in 1899.

23.6 CrystaL-FieLd theory We explore how crystal-field theory allows us to explain some of the interesting spectral and magnetic properties of coordination compounds.

23.5 CoLor and magnetism in Coordination Chemistry We discuss color and magnetism in coordination compounds, emphasizing the visible portion of the electromagnetic spectrum and the notion of complementary colors. We then see that many transition-metal complexes are paramagnetic because they contain unpaired electrons.

998 chapter 23 transition Metals and coordination chemistry

Transition metals and their compounds are important for much more than their color. For example, they are used as catalysts and as magnets, and they play an impor-tant role in biology.

In earlier chapters, we saw that metal ions can function as Lewis acids, forming covalent bonds with molecules and ions functioning as Lewis bases. (Section 16.11) We have encountered many ions and compounds that result from such interactions, such as 3Ag(NH3)24+ in Section 17.5 and hemoglobin in Section 13.6. In this chapter, we focus on the rich and important chemistry associated with such complex assemblies of metal ions surrounded by molecules and ions. Metal compounds of this kind are called coordination compounds, and the branch of chemistry that focuses on them is called coordination chemistry.

23.1 | The Transition MetalsThe part of the periodic table in which the d orbitals are being filled as we move left to right across a row is the home of the transition metals (◀ Figure 23.2). (Section 6.8)

With some exceptions (e.g., platinum, gold), metallic elements are found in nature as solid inorganic compounds called minerals. Notice from ▼ Table 23.1 that minerals are identified by common names rather than chemical names.

Most transition metals in minerals have oxidation states ranging from +1 to +4. To obtain a pure metal from its min-eral, various chemical processes must be performed to reduce the metal to the 0 oxidation state. Metallurgy is the sci-ence and technology of extracting metals from their natu-ral sources and preparing them for practical use. It usually involves several steps: (1) mining, that is, removing the relevant ore (a mixture of minerals) from the ground, (2) concentrat-ing the ore or otherwise preparing it for further treatment, (3) reducing the ore to obtain the free metal, (4) purifying the metal, and (5) mixing it with other elements to modify its properties. This last process produces an alloy, a metallic material composed of two or more elements. (Section 12.3)

Physical PropertiesSome physical properties of the period 4 (also known as “first-row”) transition metals are listed in ▶ Table 23.2. The properties of the heavier transition metals vary similarly across periods 5 and 6.

▲ Figure 23.1 Crystals of blue azurite and green malachite. These semiprecious stones were ground up and used as pigments in the Middle Ages and Renaissance, but they were eventually replaced with blue and green pigments that have superior chemical stability.

Table 23.1 Principal Mineral sources of some Transition Metals

Metal Mineral Mineral Composition

Chromium Chromite FeCr2O4

Cobalt Cobaltite CoAsSCopper Chalcocite Cu2S

Chalcopyrite CuFeS2

Malachite Cu2CO3(OH)2

Iron Hematite Fe2O3

Magnetite Fe3O4

Manganese Pyrolusite MnO2

Mercury Cinnabar HgSMolybdenum Molybdenite MoS2

Titanium Rutile TiO2

Ilmenite FeTiO3

Zinc Sphalerite ZnS

▲ Figure 23.2 The position of the transition metals in the periodic table. They are the B groups in periods 4, 5, and 6. The short lived, radioactive transition metals from period 7 are not shown.

21Sc

23V

41Nb

73Ta

39Y

22Ti

40Zr

72Hf

3B3

4B4

5B5

6B6

7B7

1B11

2B128 9

24Cr

42Mo

74W

25Mn

43Tc

75Re

71Lu

26Fe

44Ru

76Os

28Ni

46Pd

78Pt

29Cu

47Ag

79Au

30Zn

48Cd

80Hg

27Co

10

45Rh

77Ir

8B

section 23.1 the transition Metals 999

▶ Figure 23.3 shows the atomic radius observed in close-packed metallic structures as a function of group number.* The trends seen in the graph are a result of two competing forces. On the one hand, increasing effective nuclear charge favors a decrease in radius as we move left to right across each period. (Section 7.2) On the other hand, the metallic bonding strength increases until we reach the middle of each period and then decreases as we fill antibonding orbitals. (Section 12.4) As a general rule, a bond shortens as it becomes stronger. (Section 8.8) For groups 3B through 6B, these two effects work cooperatively and the result is a marked decrease in radius. In ele-ments to the right of group 6B, the two effects counteract each other, reducing the decrease and eventually leading to an increase in radius.

Give It Some ThoughtWhich element has the largest bonding atomic radius: Sc, Fe, or Au?

In general, atomic radii increase as we move down in a given group in the periodic table because of the increasing principal quantum num-ber of the outer-shell electrons. ( Section 7.3) Note in Figure 23.2, however, that once we move beyond the group 3B elements, the period 5 and period 6 transition elements in a given group have virtually the same radii. In group 5B, for example, tantalum in period 6 has virtually the same radius as niobium in period 5. This interesting and important effect has its origin in the lanthanide series, elements 57 through 70. The filling of 4f orbitals through the lanthanide elements (Figure 6.31) causes a steady increase in the effective nuclear charge, producing a size decrease, called the lanthanide contraction, that just offsets the increase we expect as we go from period 5 transition metals to period 6. Thus, the period 5 and period 6 transition metals in each group have about the same radii and similar chemical properties. For example, the chemical properties of the group 4B metals zirconium (period 5) and hafnium (period 6) are remarkably similar. These two metals always occur together in nature, and they are very difficult to separate.

Electron Configurations and Oxidation StatesTransition metals owe their location in the periodic table to the filling of the d subshells, as you saw in Figure 6.31. Many of the chemical and physical properties of transition metals result from the unique characteristics of the d orbitals. For a given transition-metal atom, the valence (n - 1)d orbitals are smaller than the corresponding valence ns and np orbitals. In quantum mechanical terms, the (n - 1)d orbital wave functions drop off more rapidly as we move away from the nucleus than do the ns and np orbital

Table 23.2 Properties of the Period 4 Transition Metals

Group 3B 4B 5B 6B 7B 8B 1B 2B

element: sc Ti V Cr Mn Fe Co Ni Cu Zn

Ground state electron configuration 3d14s2 3d24s2 3d34s2 3d54s1 3d54s2 3d64s2 3d74s2 3d84s2 3d104s1 3d104s2

First ionization energy (kJ/mol) 631 658 650 653 717 759 758 737 745 906Metallic radius (Å) 1.64 1.47 1.35 1.29 1.37 1.26 1.25 1.25 1.28 1.37

Density (g/cm3) 3.0 4.5 6.1 7.9 7.2 7.9 8.7 8.9 8.9 7.1

Melting point (°C) 1541 1660 1917 1857 1244 1537 1494 1455 1084 420Crystal structure* hcp hcp bcc bcc ** bcc hcp fcc fcc hcp

*Abbreviations for crystal structures are hcp = hexagonal close packed, fcc = face centered cubic, bcc = body centered cubic. (Section 12.3)

**Manganese has a more complex crystal structure.

*Note that the radii defined in this way, often referred to as metallic radii, differ somewhat from the bonding atomic radii defined in Section 7.3.

3B 4B 5B 6B 7B 8B 1B 2B1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

Group

Ato

mic

rad

ius

(Å)

Period 4 (Sc–Zn)Period 5 (Y–Cd)Period 6 (Lu–Hg)

▲ Figure 23.3 Radii of transition metals as a function of group number.

Go FiGureDoes the variation in radius of the transition metals follow the same trend as the effective nuclear charge on moving from left to right across the periodic table?


wave functions. This characteristic feature of the d orbitals limits their interaction with orbitals on neighboring atoms, but not so much that they are insensitive to surround-ing atoms. As a result, electrons in these orbitals behave sometimes like valence elec-trons and sometimes like core electrons. The details depend on location in the periodic table and the atom’s environment.

When transition metals are oxidized, they lose their outer s electrons before they lose electrons from the d subshell. (Section 7.4) The electron configuration of Fe is 3Ar43d 64s2, for example, whereas that of Fe2+ is 3Ar43d6. Formation of Fe3+ requires loss of one 3d electron, giving 3Ar43d 5. Most transition-metal ions contain partially occupied d subshells, which are responsible in large part for three characteristics:

1. Transition metals often have more than one stable oxidation state. 2. Many transition-metal compounds are colored, as shown in ▲ Figure 23.4. 3. Transition metals and their compounds often exhibit magnetic properties.

◀ Figure 23.5 shows the common nonzero oxidation states for the period 4 transition metals. The +2 oxidation state, which is common for most transition metals, is due to the loss of the two outer 4s electrons. This oxidation state is found for all these elements except Sc, where the 3+ ion with an [Ar] configuration is particularly stable.

Oxidation states above +2 are due to successive losses of 3d electrons. From Sc through Mn the maximum oxidation state increases from +3 to +7, equaling in each case the total number of 4s plus 3d electrons in the atom. Thus, manganese has a maximum oxidation state of 2 + 5 = +7. As we move to the right beyond Mn in Figure 23.5, the maximum oxidation state decreases. This decrease is due in part to the attraction of d orbital electrons to the nucleus, which increases faster than the attraction of the s orbital elec-trons to the nucleus as we move left to right across the periodic table. In other words, in each period the d electrons become more corelike as the atomic

number increases. By the time we get to zinc, it is not possible to remove electrons from the 3d orbitals through chemical oxidation.

▲ Figure 23.4 Aqueous solutions of transition metal ions. Left to right: Co2+, Ni2+, Cu2+, and Zn2+. The counterion is nitrate in all cases.

Go FiGureIn which transition-metal ion of this group are the 3d orbitals completely filled?

Sc Ti V Cr Mn Fe Co Ni Cu Zn

3B 4B 5B

Most frequently seen

Less common

6B 7B 8B 1B 2B

0

+2

+4

+6

+8

▲ Figure 23.5 Nonzero oxidation states of the period 4 transition metals.

Go FiGureFor which of the ions shown in this figure are the 4s orbitals empty? For which ions are the 3d orbitals empty?

section 23.1 the transition Metals 1001

In the transition metals of periods 5 and 6, the increased size of the 4d and 5d orbitals makes it possible to attain maximum oxidation states as high as +8, which is achieved in RuO4 and OsO4. In general, the maximum oxidation states are found only when the metals are combined with the most electronegative elements, especially O, F, and in some cases Cl.

Give It Some ThoughtWhy does Ti5+ not exist?

MagnetismThe spin an electron possesses gives the electron a magnetic mo-ment, a property that causes the electron to behave like a tiny magnet. In a diamagnetic solid, defined as one in which all the electrons in the solid are paired, the spin-up and spin-down elec-trons cancel one another. (Section 9.8) Diamagnetic sub-stances are generally described as being nonmagnetic, but when a diamagnetic substance is placed in a magnetic field, the motions of the electrons cause the substance to be very weakly repelled by the magnet. In other words, these supposedly nonmagnetic substances do show some very faint magnetic character in the presence of a magnetic field.

A substance in which the atoms or ions have one or more unpaired electrons is paramagnetic. (Section 9.8) In a para-magnetic solid, the electrons on one atom or ion do not influence the unpaired electrons on neighboring atoms or ions. As a result, the magnetic moments on the atoms or ions are randomly oriented and constantly changing direction, as shown in ▶ Figure 23.6(a). When a paramagnetic substance is placed in a magnetic field, how-ever, the magnetic moments tend to align parallel to one another, producing a net attractive interaction with the magnet. Thus, unlike a diamagnetic substance, which is weakly repulsed by a magnetic field, a paramagnetic substance is attracted to a magnetic field.

When you think of a magnet, you probably picture a simple iron magnet. Iron exhibits ferromagnetism, a form of magnetism much stronger than paramagnetism. Ferromagnetism arises when the unpaired electrons of the atoms or ions in a solid are influenced by the orientations of the electrons in neighboring atoms or ions. The most stable (lowest-energy) arrangement is when the spins of electrons on adjacent atoms or ions are aligned in the same direction, as in Figure 23.6(b). When a ferromagnetic solid is placed in a magnetic field, the electrons tend to align strongly in a direction parallel to the magnetic field. The attraction to the magnetic field that results may be as much as one million times stronger than that for a paramagnetic substance.

When a ferromagnet is removed from an external magnetic field, the interac-tions between the electrons cause the ferromagnetic substance to maintain a magnetic moment. We then refer to it as a permanent magnet (▶ Figure 23.7).

The only ferromagnetic transition metals are Fe, Co, and Ni, but many alloys also exhibit ferromagnetism, which is in some cases stronger than the ferromagnetism of the pure metals. Particularly powerful ferromagnetism is found in compounds contain-ing both transition metals and lanthanide metals. Two of the most important examples are SmCo5 and Nd2Fe14B.

Two additional types of magnetism involving ordered arrangements of unpaired electrons are depicted in Figure 23.6. In materials that exhibit antiferromagnetism [Figure 23.6(c)], the unpaired electrons on a given atom or ion align so that their spins are oriented in the direction opposite the spin direction on neighboring atoms.

Antiferromagnetic; spins align in opposite directions and cancel each other

Ferrimagnetic; unequal spins align in opposite directions but do not completely cancel each other

(a) (b)

(c) (d)

Paramagnetic; spins random; spins do align if in a magnetic field

Ferromagnetic; spins aligned parallel to each other

▲ Figure 23.6 The relative orientation of electron spins in various types of magnetic substances.

Go FiGureDescribe how the representation shown for the paramagnetic material would change if the material were placed in a magnetic field.

▲ Figure 23.7 A permanent magnet. Permanent magnets are made from ferromagnetic and ferrimagnetic materials.


This means that the spin-up and spin-down electrons cancel each other. Examples of antiferromagnetic substances are chromium metal, FeMn alloys, and such transition-metal oxides as Fe2O3, LaFeO3, and MnO.

A substance that exhibits ferrimagnetism [Figure 23.6(d)] has both ferromagnetic and antiferromagnetic characteristics. Like an antiferromagnet, the unpaired electrons align so that the spins in adjacent atoms or ions point in opposite directions. However, unlike an antiferromagnet, the net magnetic moments of the spin-up electrons are not fully canceled by the spin-down electrons. This can happen because the magnetic cen-ters have different numbers of unpaired electrons (NiMnO3), because the number of magnetic sites aligned in one direction is larger than the number aligned in the other direction (Y3Fe5O12), or because both these conditions apply (Fe3O4). Because the mag-netic moments do not cancel, the properties of ferrimagnetic materials are similar to the properties of ferromagnetic materials.

Give It Some ThoughtHow do you think spin–spin interactions of unpaired electrons on adjacent atoms in a substance are affected by the interatomic distance?

Ferromagnets, ferrimagnets, and antiferromagnets all become paramagnetic when heated above a critical temperature. This happens when the thermal energy is sufficient to overcome the forces determining the spin directions of the electrons. This tempera-ture is called the Curie temperature, TC, for ferromagnets and ferrimagnets and the Néel temperature, TN, for antiferromagnets.

23.2 | Transition-Metal ComplexesThe transition metals occur in many interesting and important molecular forms. Spe-cies that are assemblies of a central transition-metal ion bonded to a group of sur-rounding molecules or ions, such as 3Ag(NH3)24+ and 3Fe(H2O)643+, are called metal complexes, or merely complexes.* If the complex carries a net charge, it is generally called a complex ion. (Section 17.5) Compounds that contain complexes are known as coordination compounds.

The molecules or ions that bond to the metal ion in a complex are known as ligands (from the Latin word ligare, “to bind”). There are two NH3 ligands bonded to Ag+ in the complex ion 3Ag(NH3)24+, for instance, and six H2O ligands bonded to Fe3+ in 3Fe(H2O)643+. Each ligand functions as a Lewis base and so donates a pair of electrons to form the ligand–metal bond. (Section 16.11) Thus, every ligand has at least one unshared pair of valence electrons. Four of the most frequently encountered ligands,

O

H

H N

H

H

H Cl C N −−

illustrate that most ligands are either polar molecules or anions. In forming a complex, the ligands are said to coordinate to the metal.

Give It Some ThoughtIs the interaction between an ammonia ligand and a metal cation a Lewis acid–base interaction? If so, which species acts as the Lewis acid?

*Most of the coordination compounds we examine in this chapter contain transition-metal ions, although ions of other metals can also form complexes.

section 23.2 transition-Metal complexes 1003

The Development of Coordination Chemistry: Werner’s TheoryBecause compounds of the transition metals are beautifully colored, the chemistry of these elements fascinated chemists even before the periodic table was introduced. Dur-ing the late 1700s through the 1800s, the many coordination compounds that were iso-lated and studied had properties that were puzzling in light of the bonding theories prevailing at the time. ▼ Table 23.3, for example, lists a series of CoCl39NH3 com-pounds that have strikingly different colors. Note that the third and fourth species have different colors even though the originally assigned formula was the same for both, CoCl3 # 4 NH3.

The modern formulations of the compounds in Table 23.3 are based on various lines of experimental evidence. For example, all four compounds are strong electro-lytes (Section 4.1) but yield different numbers of ions when dissolved in water. Dissolving CoCl3 # 6 NH3 in water yields four ions per formula unit (3Co(NH3)643+ plus three Cl- ions), whereas CoCl3 # 5 NH3 yields only three ions per formula unit (3Co(NH3)5Cl42+ and two Cl- ions). Furthermore, the reaction of the compounds with excess aqueous silver nitrate leads to the precipitation of different amounts of AgCl(s). When CoCl3 # 6 NH3 is treated with excess AgNO3(aq), 3 mol of AgCl(s) precipitate per mole of complex, which means all three Cl- ions in the complex can react to form AgCl(s). By contrast, when CoCl3 # 5 NH3 is treated with excess AgNO3(aq), only 2 mol of AgCl(s) precipitate per mole of complex, telling us that one of the Cl- ions in the complex does not react. These results are summarized in Table 23.3.

In 1893 the Swiss chemist Alfred Werner (1866–1919) proposed a theory that suc-cessfully explained the observations in Table 23.3. In a theory that became the basis for understanding coordination chemistry, Werner proposed that any metal ion exhibits both a primary valence and a secondary valence. The primary valence is the oxidation state of the metal, which is +3 for the complexes in Table 23.3. (Section 4.4) The secondary valence is the number of atoms bonded to the metal ion, which is also called the coordination number. For these cobalt complexes, Werner deduced a coordina-tion number of 6 with the ligands in an octahedral arrangement around the Co3+ ion.

Werner’s theory provided a beautiful explanation for the results in Table 23.3. The NH3 molecules are ligands bonded to the Co3+ ion (through the nitrogen atom as we will see later); if there are fewer than six NH3 molecules, the remaining ligands are Cl- ions. The central metal and the ligands bound to it constitute the coordination sphere of the complex.

In writing the chemical formula for a coordination compound, Werner suggested using square brackets to signify the makeup of the coordination sphere in any given compound. He therefore proposed that CoCl3 # 6 NH3 and CoCl3 # 5 NH3 are better written as 3Co(NH3)64Cl3 and 3Co(NH3)5Cl4Cl2, respectively. He further proposed that the chloride ions that are part of the coordination sphere are bound so tightly that they do not dissociate when the complex is dissolved in water. Thus, dissolving 3Co(NH3)5Cl4Cl2 in water produces a 3Co(NH3)5Cl42+ ion and two Cl- ions.

Werner’s ideas also explained why there are two forms of CoCl3 # 4 NH3. Using Werner’s postulates, we write the formula as 3Co(NH3)4Cl24Cl. As shown in Figure 23.8, there are two ways to arrange the ligands in the 3Co(NH3)4Cl24+ complex, called the cis and trans forms. In the cis form, the two chloride ligands occupy adjacent vertices of the octahedral arrangement. In trans@3Co(NH3)4Cl24+ the two chlorides are opposite

Table 23.3 Properties of some ammonia Complexes of Cobalt(iii)

original Formulation Color

ions per Formula unit

“Free” Cl− ions per Formula unit Modern Formulation

CoCl3 # 6 NH3 Orange 4 3 3Co(NH3)64Cl3

CoCl3 # 5 NH3 Purple 3 2 3Co(NH3)5Cl4Cl2

CoCl3 # 4 NH3 Green 2 1 trans@3Co(NH3)4Cl24ClCoCl3 # 4 NH3 Violet 2 1 cis@3Co(NH3)4Cl24Cl


each other. It is this difference in positions of the Cl ligands that leads to two com-pounds, one violet and one green.

The insight Werner provided into the bonding in coordination compounds is even more remarkable when we realize that his theory predated Lewis’s ideas of covalent bonding by more than 20 years! Because of his tremendous contributions to coordina-tion chemistry, Werner was awarded the 1913 Nobel Prize in Chemistry.

▲ Figure 23.8 Isomers of [Co(NH3)4Cl2]+. The cis isomer is violet, and the trans isomer is green.

Go FiGureIs there another way to arrange the chloride ions in the 3Co(NH3)4Cl24+ ion besides the two shown in this figure?

Two Cl on same side of metal ion

Two Cl on opposite sides of metal ion

trans isomercis isomer

+ +

Palladium(II) tends to form complexes with coordination number 4. A compound has the composition PdCl2 # 3 NH3. (a) Write the formula for this compound that best shows the co-ordination structure. (b) When an aqueous solution of the compound is treated with excess AgNO3(aq), how many moles of AgCl(s) are formed per mole of PdCl2 # 3 NH3?

soluTioNAnalyze We are given the coordination number of Pd(II) and a chemical formula indicating that the complex contains NH3 and Cl-. We are asked to determine (a) which ligands are attached to Pd(II) in the compound and (b) how the compound behaves toward AgNO3 in aqueous solution.Plan (a) Because of their charge, the Cl- ions can be either in the coordination sphere, where they are bonded directly to the metal, or outside the coordination sphere, where they are bonded ionically to the complex. The electrically neutral NH3 ligands must be in the coordination sphere, if we assume four ligands bonded to the Pd(II) ion. (b) Any chlorides in the coordination sphere do not precipitate as AgCl.Solve

(a) By analogy to the ammonia complexes of cobalt(III) shown in Figure 23.7, we predict that the three NH3 are ligands attached to the Pd(II) ion. The fourth ligand around Pd(II) is one chloride ion. The second chloride ion is not a ligand; it serves only as a counterion (a nonco-ordinating ion that balances charge) in the compound. We conclude that the formula show-ing the structure best is 3Pd(NH3)3Cl4Cl.

saMPle exerCise 23.1 identifying the Coordination sphere

of a Complex

section 23.2 transition-Metal complexes 1005

The Metal–Ligand BondThe bond between a ligand and a metal ion is a Lewis acid–base interaction.

(Section 16.11) Because the ligands have available pairs of electrons, they can func-tion as Lewis bases (electron-pair donors). Metal ions (particularly transition-metal ions) have empty valence orbitals, so they can act as Lewis acids (electron-pair accep-tors). We can picture the bond between the metal ion and ligand as the result of their sharing a pair of electrons initially on the ligand:

Ag+(aq) + 2 H(aq) (aq)AgN

H

H

N

H

H

HN

H

H

H

+

[23.1]

The formation of metal–ligand bonds can markedly alter the properties we observe for the metal ion. A metal complex is a distinct chemical species that has physical and chemical properties different from those of the metal ion and ligands from which it is formed. As one example, ▼ Figure 23.9 shows the color change that occurs when aqueous solutions of NCS- (colorless) and Fe3+ (yellow) are mixed, forming 3Fe(H2O)5NCS42+.

(b) Because only the non-ligand Cl- can react, we expect to produce 1 mol of AgCl(s) per mole of complex. The balanced equation is

3Pd(NH3)3Cl4Cl(aq) + AgNO3(aq) ¡ 3Pd(NH3)3Cl4NO3(aq) + AgCl(s)

This is a metathesis reaction (Section 4.2) in which one of the cations is the 3Pd(NH3)3Cl4+ complex ion.

Practice exercise 1When the compound RhCl3 # 4 NH3 is dissolved in water and treated with excess AgNO3(aq) one mole of AgCl(s) is formed for every mole of RhCl3 # 4 NH3. What is the correct way to write the formula of this compound? (a) 3Rh(NH3)4Cl34, (b) 3Rh(NH3)4Cl24Cl, (c) 3Rh(NH3)4Cl4Cl2, (d) 3Rh(NH3)44Cl3, (e) 3RhCl34(NH3)4.

Practice exercise 2Predict the number of ions produced per formula unit when the compound CoCl2 # 6 H2O dissolves in water to form an aqueous solution.

▲ Figure 23.9 Reaction of Fe3+(aq) and NCS−(aq).

Go FiGureWrite a balanced chemical equation for the reaction depicted in this figure.

Red [Fe(H2O)5NCS]2+

forms

NH4NCS(aq)solution

[Fe(H2O)6]3+(aq)solution


Complex formation can also significantly change other properties of metal ions, such as their ease of oxidation or reduction. Silver ion, for example, is readily reduced in water,

Ag+(aq) + e- ¡ Ag(s) E° = +0.799 V [23.2]

but the 3Ag(CN)24- ion is not so easily reduced because complexation by CN- ions stabilizes silver in the +1 oxidation state:

3Ag(CN)24-(aq) + e- ¡ Ag(s) + 2 CN-(aq) E° = -0.31 V [23.3]

Hydrated metal ions are complexes in which the ligand is water. Thus, Fe3+(aq) consists largely of 3Fe(H2O)643+. (Section 16.11) It is important to realize that ligands can undergo reaction. For example, we saw in Figure 16.16 that a water mol-ecule in 3Fe(H2O)643+(aq) can be deprotonated to yield 3Fe(H2O)5OH42+(aq) and H+(aq). The iron ion retains its oxidation state; the coordinated hydroxide ligand, with a 1- charge, reduces the complex charge to 2+ . Ligands can also be displaced from the coordination sphere by other ligands, if the incoming ligands bind more strongly to the metal ion than the original ones. For example, ligands such as NH3, NCS-, and CN- can replace H2O in the coordination sphere of metal ions.

Charges, Coordination Numbers, and GeometriesThe charge of a complex is the sum of the charges on the metal and on the ligands. In 3Cu(NH3)44SO4 we can deduce the charge on the complex ion because we know that the charge of the sulfate ion is 2- . Because the compound is electrically neutral, the complex ion must have a 2+ charge, 3Cu(NH3)442+. We can then use the charge of the complex ion to deduce the oxidation number of copper. Because the NH3 ligands are uncharged molecules, the oxidation number of copper must be +2:

+2 + 4(0) = +2

[Cu(NH3)4]2+

soluTioNAnalyze We are given the chemical formula of a coordination com-pound and asked to determine the oxidation number of its metal atom.Plan To determine the oxidation number of Rh, we need to figure out what charges are contributed by the other groups. The overall charge is zero, so the oxidation number of the metal must balance the charge due to the rest of the compound.Solve The NO3 group is the nitrate anion, which has a 1- charge. The NH3 ligands carry zero charge, and the Cl is a coordinated chloride ion, which has a 1- charge. The sum of all the charges must be zero:

+ 5(0) + (−1) + 2(−1) = 0

[Rh(NH3)5Cl](NO3)2

The oxidation number of rhodium, x, must therefore be +3.

saMPle exerCise 23.2 determining the oxidation Number of a Metal in a Complex

What is the oxidation number of the metal in 3Rh(NH3)5Cl4(NO3)2?

Practice exercise 1In which of the following compounds does the transition-metal have the highest oxidation number? (a) 3Co(NH3)4Cl2], (b) K23PtCl64, (c) Rb33MoO3F34, (d) Na3Ag(CN)24, (e) K43Mn(CN)64.Practice exercise 2What is the charge of the complex formed by a platinum(II) metal ion surrounded by two ammonia molecules and two bromide ions?

Recall that the number of atoms directly bonded to the metal atom in a complex is the coordination number of the complex. Thus, the silver ion in 3Ag(NH3)24+ has a coordination number of 2, and the cobalt ion has a coordination number of 6 in all four complexes in Table 23.3.

section 23.3 common Ligands in coordination chemistry 1007

Some metal ions have only one observed coordination number. The coordina-tion number of chromium(III) and cobalt(III) is invariably 6, for example, and that of platinum(II) is always 4. For most metals, however, the coordination number is different for different ligands. In these complexes, the most common coordination numbers are 4 and 6.

The coordination number of a metal ion is often influenced by the relative sizes of the metal ion and the ligands. As the ligand gets larger, fewer of them can coordinate to the metal ion. Thus, iron(III) is able to coordinate to six fluorides in 3FeF643- but to only four chlorides in 3FeCl44-. Ligands that transfer substantial negative charge to the metal also produce reduced coordination numbers. For example, six ammonia molecules can coordinate to nickel(II), forming 3Ni(NH3)642+, but only four cyanide ions can coordinate to this ion, forming 3Ni(CN)442-.

The most common coordination geometries for coordination com-plexes are shown in ▶ Figure 23.10. Complexes in which the coordina-tion number is 4 have two geometries—tetrahedral and square planar. The tetrahedral geometry is the more common of the two and is espe-cially prevalent among nontransition metals. The square planar geom-etry is characteristic of transition-metal ions with eight d electrons in the valence shell, such as platinum(II) and gold(III). Complexes with a coordination number of 6 almost always have an octahedral geometry. Even though the octahedron can be drawn as a square with one ligand above and another below the plane, all six vertices are equivalent.

23.3 | Common ligands in Coordination Chemistry

The ligand atom that binds to the central metal ion in a coordination complex is called the donor atom of the ligand. Ligands having only one donor atom are called monodentate ligands (from the Latin, meaning “one-toothed”). These ligands are able to occupy only one site in a coordina-tion sphere. Ligands having two donor atoms are bidentate ligands (“two-toothed”), and those having three or more donor atoms are polydentate ligands (“many-toothed”). In both bidentate and polydentate species, the multiple donor atoms can simultaneously bond to the metal ion, thereby occupying two or more sites in a coordination sphere. Table 23.4 gives ex-amples of all three types of ligands.

Because they appear to grasp the metal between two or more donor atoms, bidentate and polydentate ligands are also known as chelating agents (pronounced “KEE-lay-ting”; from the Greek chele, “claw”).

One common chelating agent is the bidentate ligand ethylenediamine, denoted en:

CH2

H2N NH2

CH2

in which each donor nitrogen atom has one nonbonding electron pair. These donor atoms are sufficiently far apart to allow both of them to bond to the metal ion in adja-cent positions. The 3Co(en)343 + complex ion, which contains three ethylenediamine ligands in the octahedral coordination sphere of cobalt(III), is shown in Figure 23.11. Notice that in the image on the right the en is written in a shorthand notation as two nitrogen atoms connected by an arc.

The ethylenediaminetetraacetate ion, 3EDTA44-, is an important polydentate ligand that has six donor atoms. It can wrap around a metal ion using all six donor atoms, as shown in Figure 23.12, although it sometimes binds to a metal using only five of its donor atoms.

Give It Some ThoughtBoth H2O and ethylenediamine (en) have two nonbonding pairs of electrons. Why cannot water act as a bidentate ligand?

−Cl

ClCl

ClFe

Tetrahedral geometry

Square planar geometry

Octahedral geometry

2−Cl

ClCl

ClPt

3−

Fe

F

F

FF

F

F

A tetrahedron has four triangular faces and four equivalent vertices.

The metal and all four ligands lie in the same plane.

An octahedron has eight triangular faces and six equivalent vertices.

▲ Figure 23.10 Common geometries of coordination complexes. In complexes having coordination number 4, the geometry is typically either tetrahedral or square planar. In complexes having coordination number 6, the geometry is nearly always octahedral.

Go FiGureIn the drawings on the right-hand side, what does the solid wedge connecting atoms represent? What does the dashed wedge connecting atoms represent?


In general, the complexes formed by chelating ligands (that is, bidentate and poly-dentate ligands) are more stable than the complexes formed by related monodentate ligands. The equilibrium formation constants for 3Ni(NH3)642+ and 3Ni(en)342+ illus-trate this observation:

3Ni(H2O)642+(aq) + 6 NH3(aq) ∆ 3Ni(NH3)642+(aq) + 6 H2O(l) Kf = 1.2 * 109 [23.4]

3Ni(H2O)642+(aq) + 3 en(aq) ∆ 3Ni(en)342+(aq) + 6 H2O(l) Kf = 6.8 * 1017 [23.5]

Table 23.4 some Common ligands

ligand Type examples

C

O 2−

Carbonate ion

4−

Ethylenediaminetetraacetate ion (EDTA4−)

O OC

C

O

O

CO

OO

O2−

Oxalate ion

CH2H2C

NH2H2N

Ethylenediamine (en)

CH2H2C

NH

CH2CH2

NH2

N CH2 CH2 N

H2N

Diethylenetriamine

Bidentate

NH3 Ammonia

H2O Water

Chloride ion

F −

Cl −

Fluoride ion

Thiocyanate ion

N[ ]C −

N ]C

CH2

C

O

O CH2

C

O

CH2

C

O

CH2 O

O

5−

Triphosphate ion

P

O

O

O O P

O

O

O P

O

O

O

or[ S −

H[ ]O

O

−

ON ]or

[ −

Cyanide ion

Nitrite ion

Hydroxide ionMonodentate

Polydentate

Bipyridine(bipy or bpy)

N N

Ortho-phenanthroline(o-phen)

NN

▲ Figure 23.11 The 3Co(en)3 43+ ion. The ligand is ethylenediamine.

[Co(en)3]3+

3+

Co

H2N

H2C

H2C

H2N

H2C

NH2 H2N

NH2

NH2

CH2

CH2CH2

3+

=

▲ Figure 23.12 The complex ion 3Co(EDTA) 4−. The ligand is the polydentate ethylenediaminetetraacetate ion, whose full representation is given in Table 23.2. This representation shows how the two N and four O donor atoms coordinate to cobalt.

−


Although the donor atom is nitrogen in both instances, 3Ni(en)342+ has a formation constant that is more than 108 times larger than that of 3Ni(NH3)642+. This trend of generally larger formation constants for bidentate and polydentate ligands, known as the chelate effect, is examined in the “A Closer Look” essay on page 1010.

Chelating agents are often used to prevent one or more of the customary reactions of a metal ion without removing the ion from solution. For example, a metal ion that interferes with a chemical analysis can often be complexed and its interference thereby removed. In a sense, the chelating agent hides the metal ion. For this reason, scientists sometimes refer to these ligands as sequestering agents.

Phosphate ligands, such as sodium tripolyphosphate, Na53OPO2OPO2OPO34, are used to sequester Ca2+ and Mg2+ ions in hard water so that these ions cannot interfere with the action of soap or detergents.

Chelating agents are used in many prepared foods, such as salad dressings and fro-zen desserts, to complex trace metal ions that catalyze decomposition reactions. Che-lating agents are used in medicine to remove toxic heavy metal ions that have been ingested, such as Hg2+, Pb2+, and Cd2+. One method of treating lead poisoning, for example, is to administer Na2Ca(EDTA). The EDTA chelates the lead, allowing it to be removed from the body via urine.

Give It Some ThoughtCobalt(III) has a coordination number of 6 in all its complexes. Is the carbonate ion a monodentate or bidentate ligand in the 3Co(NH3)4(CO3)4 + ion?

Metals and Chelates in Living SystemsTen of the 29 elements known to be necessary for human life are transition metals. (Section 2.7, “Elements Required by Living Organisms”) These ten elements—V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Mo, and Cd—form complexes with a variety of groups present in biological systems.

Although our bodies require only small quantities of metals, deficiencies can lead to serious illness. A deficiency of manganese, for example, can lead to convulsive disorders. Some epilepsy patients have been helped by the addition of manganese to their diets.

Among the most important chelating agents in nature are those derived from the porphine molecule (▶ Figure 23.13). This molecule can coordinate to a metal via its four nitrogen donor atoms. Once porphine bonds to a metal ion, the two H atoms on the nitrogens are displaced to form complexes called porphyrins. Two important por-phyrins are hemes, in which the metal ion is Fe(II), and chlorophylls, with a Mg(II) central ion.

Figure 23.14 shows a schematic structure of myoglobin, a pro-tein that contains one heme group. Myoglobin is a globular protein, one that folds into a compact, roughly spherical shape. Myoglobin is found in the cells of skeletal muscle, particularly in seals, whales, and porpoises. It stores oxygen in cells, one molecule of O2 per myo-globin, until it is needed for metabolic activities. Hemoglobin, the protein that transports oxygen in human blood, is made up of four heme-containing subunits, each of which is very similar to myoglo-bin. One hemoglobin can bind up to four O2 molecules.

In both myoglobin and hemoglobin, the iron is coordinated to the four nitrogen atoms of a porphyrin and to a nitrogen atom from the protein chain (Figure 23.15). In hemoglobin, the sixth position around the iron is occupied either by O2 (in oxyhemo-globin, the bright red form) or by water (in deoxyhemoglobin, the purplish red form). (The oxy form is the one shown in Figure 23.15.)

NH

NHN

N

N

NMg2+

N

NCH3

CH3

OCH3

CH2

H3C

H3C

H3C

CH3 CH3 CH3

CH3

OO

2 O O

N

NFe2+

N

N

O OH OHO

Heme b

Chlorophyll a

Porphine

▲ Figure 23.13 Porphine and two porphyrins, heme b and chlorophyll a. Fe(II) and Mg(II) ions replace the two blue H atoms in porphine and bond with all four nitrogens in heme b and chlorophyll a, respectively.

Go FiGureWhat is the coordination number of the metal ion in heme b? In chlorophyll a?


Carbon monoxide is poisonous because the equilibrium binding constant of human hemoglobin for CO is about 210 times greater than that for O2. As a result, a relatively small quantity of CO can inactivate a substantial fraction of the hemoglo-bin in the blood by displacing the O2 molecule from the heme-containing subunit. For example, a person breathing air that contains only 0.1% CO takes in enough CO after a few hours to convert up to 60% of the hemoglobin (Hb) into COHb, thereby reducing the blood’s normal oxygen-carrying capacity by 60%.

Under normal conditions, a nonsmoker breathing unpolluted air has about 0.3 to 0.5% COHb in her or his blood. This amount arises mainly from the production of small quantities of CO in the course of normal body chemistry and from the small amount of CO present in clean air. Exposure to higher concentrations of CO causes the COHb level to increase, which in turn leaves fewer Hb sites to which O2 can bind. If the level of COHb becomes too high, oxygen transport is effectively shut down and death occurs. Because CO is colorless and odorless, CO poisoning occurs with very lit-tle warning. Improperly ventilated combustion devices, such as kerosene lanterns and stoves, thus pose a potential health hazard.

▲ Figure 23.14 Myoglobin. This ribbon diagram does not show most of the atoms.

Heme

Heme

Protein (globin)

OO

N

HN

▲ Figure 23.15 Coordination sphere of the hemes in oxymyoglobin and oxyhemoglobin.

Go FiGureWhat is the coordination number of iron in the heme unit shown here? What is the identity of the donor atoms?

ligands also causes the 3Cu(H2O)2(NH3)242+ ion to be more rigid, which is probably the reason ∆S° is slightly negative.

We can use Equation 19.20, ∆G° = -RT ln K, to calcu-late the equilibrium constant of the reaction at 27 °C. The result, K = 3.1 * 107, tells us that the equilibrium lies far to the right, fa-voring replacement of H2O by NH3. For this equilibrium, therefore, the enthalpy change, ∆H ° = -46 kJ, is large enough and negative enough to overcome the entropy change, ∆S° = -8.4 J>K.

Now let’s use a single bidentate ethylenediamine (en) ligand in our substitution reaction:

3Cu(H2O)442+(aq) + en(aq) ∆3Cu(H2O)2(en)42+(aq) + 2 H2O(l)

∆H° = -54 kJ; ∆S° = +23 J>K; ∆G° = -61 kJ

The en ligand binds slightly more strongly to the Cu2+ ion than two NH3 ligands, so the enthalpy change here (-54 kJ) is slightly more negative than for 3Cu(H2O)2(NH3)242+(-46 kJ). There is a big difference in the

a Closer look

Entropy and the Chelate Effect

We learned in Section 19.5 that chemical processes are favored by positive entropy changes and by negative enthalpy changes. The special stability associated with the formation of chelates, called the chelate effect, can be explained by comparing the entropy changes that occur with monoden-tate ligands with the entropy changes that occur with polydentate ligands.

We begin with the reaction in which two H2O ligands of the square-planar Cu(II) complex 3Cu(H2O)442+ are replaced by mono-dentate NH3 ligands at 27 °C:

3Cu(H2O)442+(aq) + 2 NH3(aq) ∆ 3Cu(H2O)2(NH3)242+(aq) + 2 H2O(l)

∆H° = -46 kJ; ∆S° = -8.4 J>K; ∆G° = -43 kJThe thermodynamic data tell us about the relative abilities of

H2O and NH3 to serve as ligands in this reaction. In general, NH3 binds more tightly to metal ions than does H2O, so this substitution reaction is exothermic (∆H 6 0). The stronger bonding of the NH3


The chlorophylls, which are porphyrins that contain Mg(II) (Figure 23.13), are the key components in the conversion of solar energy into forms that can be used by living organisms. This process, called photosynthesis, occurs in the leaves of green plants:

6 CO2(g) + 6 H2O(l) ¡ C6H12O6(aq) + 6 O2(g) [23.6]

The formation of 1 mol of glucose, C6H12O6, requires the absorption of 48 mol of photons from sunlight or other sources of light. Chlorophyll-containing pig-ments in the leaves of plants absorb the photons. Figure 23.13 shows that the chlorophyll molecule has a series of alternating, or conjugated, double bonds in the ring surrounding the metal ion. This system of conjugated double bonds makes it possible for chlorophyll to absorb light strongly in the visible region of the spectrum. As ▶ Figure 23.16 shows, chlorophyll is green because it absorbs red light (maximum absorption at 655 nm) and blue light (maximum absorption at 430 nm) and transmits green light.

Photosynthesis is nature’s solar energy–conversion machine, and thus all living systems on Earth depend on photosynthesis for continued existence.

Give It Some ThoughtWhat property of the porphine ligand makes it possible for chlorophyll to play a role in plant photosynthesis?

∆H ° = (-54 kJ) - (-46 kJ) = -8 kJ ∆S° = (+23 J>K) - (-8.4 J>K) = +31 J>K ∆G° = (-61 kJ) - (-43 kJ) = -18 kJ

Notice that at 27 °C, the entropic contribution (-T∆S°) to the free-energy change, ∆G° = ∆H ° - T∆S° (Equation 19.12), is nega-tive and greater in magnitude than the enthalpic contribution (∆H °). The equilibrium constant for the NH39en reaction, 1.4 * 103, shows that the replacement of NH3 by en is thermodynamically favorable.

The chelate effect is important in biochemistry and molecu-lar biology. The additional thermodynamic stabilization provided by entropy effects helps stabilize biological metal–chelate com-plexes, such as porphyrins, and can allow changes in the oxidation state of the metal ion while retaining the structural integrity of the complex.

Related Exercises: 23.31, 23.32, 23.96, 23.98

entropy change, however: ∆S° is -8.4 J>K for the NH3 reaction but +23 J>K for the en reaction. We can explain the positive ∆S° value using concepts discussed in Section 19.3. Because a single en ligand occupies two coordination sites, two molecules of H2O are released when one en ligand bonds. Thus, there are three product molecules in the reaction but only two reactant molecules. The greater number of product molecules leads to the positive entropy change for the equilibrium.

The slightly more negative value of ∆H ° for the en reaction (-54 kJ versus -46 kJ) coupled with the positive entropy change leads to a much more negative value of ∆G° (–61 kJ for en, -43 kJ for NH3) and thus a larger equilibrium constant: K = 4.2 * 1010.

We can combine our two equations using Hess’s law (Sec-tion 5.6) to calculate the enthalpy, entropy, and free-energy changes that occur for en to replace ammonia as ligands on Cu(II):

3Cu(H2O)2(NH3)242+(aq) + en(aq) ∆ 3Cu(H2O)2(en)42+(aq) + 2 NH3(aq)

400 500 600 700Wavelength (nm)

Lig

ht a

bsor

ptio

n

▲ Figure 23.16 The absorption of sunlight by chlorophyll.

Go FiGureWhich peak in this curve corresponds to the lowest-energy transition by an electron in a chlorophyll molecule?

Chemistry and life

The Battle for Iron In Living Systems

Because living systems have difficulty assimilating enough iron to satisfy their nutritional needs, iron-deficiency anemia is a common problem in humans. Chlorosis, an iron deficiency in plants that makes leaves turn yellow, is also commonplace.

Living systems have difficulty assimilating iron because most iron compounds found in nature are not very soluble in water. Micro-organisms have adapted to this problem by secreting an iron-binding compound, called a siderophore, that forms an extremely stable wa-ter-soluble complex with iron(III). One such complex is ferrichrome (Figure 23.17). The iron-binding strength of a siderophore is so great that it can extract iron from iron oxides.

When ferrichrome enters a living cell, the iron it carries is re-moved through an enzyme-catalyzed reaction that reduces the strongly bonding iron(III) to iron(II), which is only weakly complexed by the siderophore (Figure 23.18). Microorganisms thus acquire iron by excreting a siderophore into their immediate environment and then taking the resulting iron complex into the cell.

In humans, iron is assimilated from food in the intestine. A pro-tein called transferrin binds iron and transports it across the intesti-nal wall to distribute it to other tissues in the body. The normal adult body contains about 4 g of iron. At any one time, about 3 g of this iron is in the blood, mostly in the form of hemoglobin. Most of the remain-der is carried by transferrin.

A bacterium that infects the blood requires a source of iron if it is to grow and reproduce. The bacterium excretes a siderophore into the


23.4 | Nomenclature and isomerism in Coordination Chemistry

When complexes were first discovered, they were named after the chemist who originally prepared them. A few of these names persist, as, for example, with the dark red substance NH43Cr(NH3)2(NCS)44, which is still known as Reinecke’s salt. Once the structures of complexes were more fully understood, it became possible to name them in a more system-atic manner. Let’s use two substances to illustrate how coordination compounds are named:

[Co(NH3)5Cl]Cl2 Pentaamminechlorocobalt(III)

Cation

Cobalt in+3 oxidationstate

Cl−

ligand5 NH3ligands

Chloride

Anion

Na2[MoOCl4] Sodium

Cation

Molybdenumin +4 oxidationstate

4 Cl−

ligandsOxide, O2−

ligand

Tetrachlorooxomolybdate(IV)

Anion

H

CH

CO

H

CH3

CH3CH2

CH2 CH2

CH2

CH2

CH2

Fe3+

CH2CH2

CH2

CH3

O

O

ON

N

CO

C

C

N

CN

HC

H

C

H

H

O

H

H

O

C

C

C

C

NH

O

NH

H

H

O

ON

H

N

HC

NO

O

C

C

blood to compete with transferrin for iron. The equilibrium constants for forming the iron complex are about the same for transferrin and siderophores. The more iron available to the bacterium, the more rap-idly it can reproduce and thus the more harm it can do.

Several years ago, New Zealand clinics regularly gave iron sup-plements to infants soon after birth. However, the incidence of certain bacterial infections was eight times higher in treated than in untreated infants. Presumably, the presence of more iron in the blood than abso-lutely necessary makes it easier for bacteria to obtain the iron needed for growth and reproduction.

In the United States, it is common medical practice to supplement infant formula with iron sometime during the first year of life. How-ever, iron supplements are not necessary for infants who breast-feed because breast milk contains two specialized proteins, lactoferrin and transferrin, which provide sufficient iron while denying its availability to bacteria. Even for infants fed with infant formulas, supplementing with iron during the first several months of life may be ill-advised.

For bacteria to continue to multiply in the blood, they must synthesize new supplies of siderophores. Synthesis of siderophores in bacteria slows, however, as the temperature is increased above the normal body temperature of 37 °C and stops completely at 40 °C. This suggests that fever in the presence of an invading microbe is a mecha-nism used by the body to deprive bacteria of iron.

Related Exercise: 23.74

▲ Figure 23.17 Ferrichrome.

Fe3+

Fe3+

e−

Fe3+

Fe2+

Ferrichrome

Siderophore

Cell wall

▲ Figure 23.18 The iron-transport system of a bacterial cell.

section 23.4 nomenclature and isomerism in coordination chemistry 1013

1. In naming complexes that are salts, the name of the cation is given before the name of the anion. Thus, in 3Co(NH3)5Cl4Cl2 we name the 3Co(NH3)5Cl42+ cat-ion and then the Cl-.

2. In naming complex ions or molecules, the ligands are named before the metal. Ligands are listed in alphabetical order, regardless of their charges. Prefixes that give the number of ligands are not considered part of the ligand name in determining alphabetical order. Thus, the 3Co(NH3)5Cl42+ ion is pentaamminechlorocobalt(III). (Be careful to note, however, that the metal is written first in the chemical formula.)

3. The names of anionic ligands end in the letter o, but electrically neutral li-gands ordinarily bear the name of the molecules (▲ Table 23.5). Special names are used for H2O (aqua), NH3 (ammine), and CO (carbonyl). For example, 3Fe(CN)2(NH3)2(H2O)24+ is the diamminediaquadicyanoiron(III) ion.

4. Greek prefixes (di-, tri-, tetra-, penta-, hexa-) are used to indicate the number of each kind of ligand when more than one is present. If the ligand contains a Greek prefix (for example, ethylenediamine) or is polydentate, the alternate prefixes bis-, tris-, tetrakis-, pentakis-, and hexakis- are used and the ligand name is placed in paren-theses. For example, the name for 3Co(en)34Br3 is tris(ethylenediamine)-cobalt(III) bromide.

5. If the complex is an anion, its name ends in -ate. The compound K43Fe(CN)64 is potassium hexacyanoferrate(II), for example, and the ion 3CoCl442 - is tetrachlorocobaltate(II) ion.

6. The oxidation number of the metal is given in parentheses in Roman numerals following the name of the metal.Three examples for applying these rules are3Ni(NH3)64Br2 Hexaamminenickel(II) bromide3Co(en)2(H2O)(CN)4Cl2 Aquacyanobis(ethylenediamine)cobalt(III) chlorideNa23MoOCl44 Sodium tetrachlorooxomolybdate(IV)

saMPle exerCise 23.3 Naming Coordination Compounds

Name the compounds (a) 3Cr(H2O)4Cl24Cl, (b) K43Ni(CN)44.soluTioNAnalyze We are given the chemical formulas for two coordination compounds and assigned the task of naming them.Plan To name the complexes, we need to determine the ligands in the complexes, the names of the ligands, and the oxidation state of the metal ion. We then put the information together following the rules listed in the text.Solve

(a) The ligands are four water molecules—tetraaqua—and two chloride ions—dichloro. By assigning all the oxidation numbers we know for this molecule, we see that the oxida-tion number of Cr is +3:

+3 + 4(0) + 2(−1) + (−1) = 0

[Cr(H2O)4Cl2]Cl

Table 23.5 some Common ligands and Their Names

ligand Name in Complexes ligand Name in Complexes

Azide, N3- Azido Oxalate, C2O4

2- Oxalato

Bromide, Br- Bromo Oxide, O2- Oxo

Chloride, Cl- Chloro Ammonia, NH3 AmmineCyanide, CN- Cyano Carbon monoxide, CO CarbonylFluoride, F - Fluoro Ethylenediamine, en EthylenediamineHydroxide, OH- Hydroxo Pyridine, C5H5N Pyridine

Carbonate, CO32- Carbonato Water, H2O Aqua


IsomerismWhen two or more compounds have the same composition but a different arrangement of atoms, we call them isomers. (Section 2.9) Here we consider two main kinds of isomers in coordination compounds: structural isomers (which have different bonds) and stereoisomers (which have the same bonds but different ways in which the ligands occupy the space around the metal center). Each of these classes also has subclasses, as shown in ▼ Figure 23.19.

Structural IsomerismMany types of structural isomerism are known in coordination chemistry, including the two named in Figure 23.19: linkage isomerism and coordination-sphere isomerism. Linkage isomerism is a relatively rare but interesting type that arises when a particular ligand is capable of coordinating to a metal in two ways. The nitrite ion, NO2

-, for ex-ample, can coordinate to a metal ion through either its nitrogen or one of its oxygens

Thus, we have chromium(III). Finally, the anion is chloride. The name of the compound is tetraaquadichlorochromium(III) chloride.

(b) The complex has four cyanide ion ligands, CN-, which means tetracyano, and the oxidation state of the nickel is zero:

Because the complex is an anion, the metal is indicated as nickelate(0). Putting these parts together and naming the cation first, we have potassium tetracyanonickelate(0).

Practice exercise 1What is the name of the compound 3Rh(NH3)4Cl24Cl? (a) Rhodium(III) tetraamminedichloro chloride, (b) Tetraammoniadichlororhodium(III) chloride, (c) Tetraamminedichlororhodium(III) chloride, (d) Tetraamminetrichlororhodium(III), (e) Tetraamminedichlororhodium(II) chloride.

Practice exercise 2Name the compounds (a) 3Mo(NH3)3Br34NO3, (b) (NH4)23CuBr44. (c) Write the formula for sodium diaquabis(oxalato)ruthenate(III).

4(+1) + 0 + 4(−1) = 0

K4[Ni(CN)4]

▲ Figure 23.19 Forms of isomerism in coordination compounds.

Do the molecules havethe same atoms?

Do the same ligands bind to the metal ion?

Do the molecules havethe same bonds (atoms

linked to the samepartners)?

Structuralisomers Stereoisomers

Coordinationsphere isomers

Linkageisomers

Notisomers YESNO

YESNO

YESNO

Are the molecules non-superimposablemirror images?

Geometricalisomers

Opticalisomers

YESNO


(▶ Figure 23.20). When it coordinates through the nitrogen atom, the NO2- ligand is

called nitro; when it coordinates through the oxygen atom, it is called nitrito and is gen-erally written ONO-. The isomers shown in Figure 23.20 have different properties. The nitro isomer is yellow, for example, whereas the nitrito isomer is red.

Another ligand capable of coordinating through either of two donor atoms is thiocy-anate, SCN-, whose potential donor atoms are N and S.

Give It Some ThoughtCan the ammonia ligand engage in linkage isomerism? Explain.

Coordination-sphere isomers are isomers that differ in which species in the complex are ligands and which are outside the coordination sphere in the solid. For example, three isomers have the formula CrCl3(H2O)6. When the ligands are six H2O and the chloride ions are in the crystal lattice (as counterions), we have the violet compound 3Cr(H2O)64Cl3. When the ligands are five H2O and one Cl-, with the sixth H2O and the two Cl- out in the lattice, we have the green compound 3Cr(H2O)5Cl4Cl2 # H2O. The third isomer, 3Cr(H2O)4Cl24Cl # 2 H2O, is also a green compound. In the two green compounds, either one or two water molecules have been displaced from the coordination sphere by chloride ions. The displaced H2O molecules occupy a site in the crystal lattice.

StereoisomerismStereoisomers have the same chemical bonds but differ-ent spatial arrangements. In the square-planar complex 3Pt(NH3)2Cl24, for example, the chloro ligands can be either adjacent to or opposite each other (▼ Figure 23.21). (We saw an earlier example of this type of isomerism in the cobalt complex of Figure 23.8, and we will return to that complex in a moment.) This form of stereoisomerism, in which the arrangement of the atoms is different but the same bonds are present, is called geometric isomerism. The isomer on the left in

Nitrito isomerBonding via ligand O atom

Nitro isomerBonding via ligand N atom

2+2+

▲ Figure 23.20 Linkage isomerism.

Go FiGureWhat is the chemical formula and name for each of the complex ions in this figure?

cisCl ligands adjacent to each other

NH3 ligands adjacent to each other

transCl ligands on opposite sides of central atom

NH3 ligands on opposite sides of central atom

= Pt= Cl = H= N

▲ Figure 23.21 Geometric isomerism.

Go FiGureWhich of these isomers has a nonzero dipole moment?


Figure 23.21, with like ligands in adjacent positions, is the cis isomer, and the isomer on the right, with like ligands across from one another, is the trans isomer.

Geometric isomers generally have different physical properties and may also have markedly different chemical reactivities. For example, cis@3Pt(NH3)2Cl24, also called cisplatin, is effective in the treatment of testicular, ovarian, and certain other cancers, whereas the trans isomer is ineffective. This is because cisplatin forms a chelate with two nitrogens of DNA, displacing the chloride ligands. The chloride ligands of the trans isomer are too far apart to form the N-Pt-N chelate with the nitrogen donors in DNA.

Geometric isomerism is also possible in octahedral complexes when two or more different ligands are present, as in the cis and trans tetraamminedichlorocobalt(III) ion in Figure 23.8. Because all the corners of a tetrahedron are adjacent to one another, cis–trans isomerism is not observed in tetrahedral complexes.

The Lewis structure :C ‚ O: indicates that the CO molecule has two lone pairs of electrons. When CO binds to a transition-metal atom, it nearly always does so by using the C lone pair. How many geometric isomers are there for tetracarbonyldichloroiron(II)?

soluTioNAnalyze We are given the name of a complex containing only monodentate ligands, and we must determine the number of isomers the complex can form.Plan We can count the number of ligands to determine the coordination number of the Fe and then use the coordination number to predict the geometry. We can then either make a series of drawings with ligands in different positions to determine the number of isomers or deduce the number of isomers by analogy to cases we have discussed.Solve The name indicates that the complex has four carbonyl (CO) ligands and two chloro (Cl-) ligands, so its formula is Fe(CO)4Cl2. The complex therefore has a coordination number of 6, and we can assume an octahedral geometry. Like 3Co(NH3)4Cl24+ (Figure 23.8), it has four li-gands of one type and two of another. Consequently, there are two isomers possible: one with the Cl- ligands across the metal from each other, trans@3Fe(CO)4Cl24, and one with the Cl- ligands adjacent to each other, cis@3Fe(CO)4Cl24.Comment It is easy to overestimate the number of geometric isomers. Sometimes different ori-entations of a single isomer are incorrectly thought to be different isomers. If two structures can be rotated so that they are equivalent, they are not isomers of each other. The problem of iden-tifying isomers is compounded by the difficulty we often have in visualizing three-dimensional molecules from their two-dimensional representations. It is sometimes easier to determine the number of isomers if we use three-dimensional models.

Practice exercise 1Which of the following molecules does not have a geometric isomer?

[MX2Y2] [MX4Y2] [MX3Y3][MX3Y]

(a) (b) (c) (d)

Practice exercise 2How many isomers exist for the square-planar molecule 3Pt(NH3)2ClBr4?

saMPle exerCise 23.4 determining the Number of Geometric isomers

The second type of stereoisomerism listed in Figure 23.19 is optical isomerism. Optical isomers, called enantiomers, are mirror images that cannot be superimposed on each other. They bear the same resemblance to each other that your left hand bears to your right hand. If you look at your left hand in a mirror, the image is identical to your right hand (▶ Figure 23.22). No matter how hard you try, however, you cannot


superimpose your two hands on each other. An example of a complex that exhibits this type of isomerism is the 3Co(en)343 + ion. Figure 23.22 shows the two enantiomers of this complex and their mirror-image relationship. Just as there is no way that we can twist or turn our right hand to make it look identical to our left, so also there is no way to rotate one of these enantiomers to make it identical to the other. Molecules or ions that are not superimposable on their mirror image are said to be chiral (pronounced KY-rul).

Mirror Mirror

Left hand Mirror image of left handis identical to right hand

Enantiomers of [Co(en)3]3+

Co CoN N

▲ Figure 23.22 Optical isomerism.

Does either cis-3Co(en)2Cl24+ or trans@3Co(en)2Cl24- have optical isomers?

soluTioNAnalyze We are given the chemical formula for two geometric isomers and asked to determine whether either one has optical isomers. Because en is a bidentate ligand, we know that both com-plexes are octahedral and both have coordination number 6.Plan We need to sketch the structures of the cis and trans isomers and their mirror images. We can draw the en ligand as two N atoms connected by an arc. If the mirror image cannot be super-imposed on the original structure, the complex and its mirror image are optical isomers.Solve The trans isomer of 3Co(en)2Cl24+ and its mirror image are:

Cl

Cl

Co

N

NN

N

Cl

Cl

Co

N

NN

N

Mirror

Notice that the mirror image of the isomer is identical to the original. Consequently trans-3Co(en)2Cl24+ does not exhibit optical isomerism.The mirror image of the cis isomer cannot be superimposed on the original:

Cl

ClCo

N

N

N

N

Mirror

Cl

ClCo

N

N

N

N

saMPle exerCise 23.5 Predicting Whether a Complex has

optical isomers


The properties of two optical isomers differ only if the isomers are in a chiral environment—that is, an environment in which there is a sense of right- and left- handedness. A chiral enzyme, for example, might catalyze the reaction of one opti-cal isomer but not the other. Consequently, one optical isomer may produce a specific physiological effect in the body, with its mirror image producing either a different effect or none at all. Chiral reactions are also extremely important in the synthesis of pharma-ceuticals and other industrially important chemicals.

Optical isomers are usually distinguished from each other by their interac-tion with plane-polarized light. If light is polarized—for example, by being passed through a sheet of polarizing film—the electric-field vector of the light is confined to a single plane (▼ Figure 23.23). If the polarized light is then passed through a solu-tion containing one optical isomer, the plane of polarization is rotated either to the right or to the left. The isomer that rotates the plane of polarization to the right is dextrorotatory; it is the dextro, or d, isomer (Latin dexter, “right”). Its mirror image rotates the plane of polarization to the left; it is levorotatory and is the levo, or l, isomer (Latin laevus, “left”). The 3Co(en)343 + isomer on the right in Figure 23.22 is found experimentally to be the l isomer of this ion. Its mirror image is the d isomer. Because of their effect on plane-polarized light, chiral molecules are said to be optically active.

Give It Some ThoughtWhat is the similarity and what is the difference between the d and l isomers of a compound?

When a substance with optical isomers is prepared in the laboratory, the chemical environment during the synthesis is not usually chiral. Consequently, equal amounts of the two isomers are obtained, and the mixture is said to be racemic. A racemic mixture does not rotate polarized light because the rotatory effects of the two isomers cancel each other.

Thus, the two cis structures are optical isomers (enantiomers). We say that cis-3Co(en)2Cl24+ is a chiral complex.

Practice exercise 1Which of the following complexes has optical isomers? (a) Tetrahedral 3CdBr2Cl242-, (b) Octahedral 3CoCl4(en)42-, (c) Octahedral 3Co(NH3)4Cl242+, (d) Tetrahedral 3Co(NH3)BrClI4-.

Practice exercise 2Does the square-planar complex ion 3Pt(NH3)(N3)ClBr4- have optical isomers? Explain your answer.

▲ Figure 23.23 Using polarized light to detect optical activity.

Lightsource

Polarizeraxis

Polarizedlight

Rotatedpolarized light

Angle ofrotation of

plane ofpolarization

Analyzer

Optically activesolution

PolarizingfilmUnpolarized

light

section 23.5 color and Magnetism in coordination chemistry 1019

23.5 | Color and Magnetism in Coordination Chemistry

Studies of the colors and magnetic properties of transition-metal complexes have played an important role in the development of modern models for metal–ligand bonding. We discussed the various types of magnetic behavior of the transition metals in Section 23.1, and we discussed the interaction of radiant energy with matter in Section 6.3. Let’s briefly examine the significance of these two properties for transition-metal complexes before we develop a model for metal–ligand bonding.

ColorIn Figure 23.4, we saw the diverse range of colors seen in salts of transition-metal ions and their aqueous solutions. In general, the color of a complex depends on the identity of the metal ion, on its oxidation state, and on the ligands bound to it. ▼ Figure 23.24, for instance, shows how the pale blue color characteristic of 3Cu(H2O)442+ changes to deep blue-violet as NH3 ligands replace the H2O ligands to form 3Cu(NH3)442+.

For a substance to have color we can see, it must absorb some portion of the spec-trum of visible light. (Section 6.1) Absorption happens, however, only if the energy needed to move an electron in the substance from its ground state to an excited state corresponds to the energy of some portion of the visible spectrum. (Section 6.3) Thus, the particular energies of radiation a substance absorbs dictate the color we see for the substance.

When an object absorbs some portion of the visible spectrum, the color we per-ceive is the sum of the unabsorbed portions, which are either reflected or transmitted by the object and strike our eyes. (Opaque objects reflect light, and transparent ones transmit it.) If an object absorbs all wavelengths of visible light, none reaches our eyes and the object appears black. If it absorbs no visible light, it is white if opaque or color-less if transparent. If it absorbs all but orange light, the orange light is what reaches our eye and therefore is the color we see.

[Cu(H2O)4]2+(aq) [Cu(NH3)4]

2+(aq)NH3(aq)

▲ Figure 23.24 The color of a coordination complex changes when the ligand changes.

Go FiGureIs the equilibrium binding constant of ammonia for Cu(II) likely to be larger or smaller than that of water for Cu(II)?


An interesting phenomenon of vision is that we also perceive an orange color when an object absorbs only the blue portion of the visible spectrum and all the other col-ors strike our eyes. This is because orange and blue are complementary colors, which means that the removal of blue from white light makes the light look orange (and the removal of orange makes the light look blue).

Complementary colors can be determined with an artist’s color wheel, which shows complementary colors on opposite sides (▲ Figure 23.25).

The amount of light absorbed by a sample as a function of wavelength is known as the sample’s absorption spectrum. The visible absorption spectrum of a transparent sam-ple can be determined using a spectrometer, as described in the “A Closer Look” box on page 582. The absorption spectrum of the ion 3Ti(H2O)643 + is shown in ▶ Figure 23.26. The absorption maximum is at 500 nm, but the graph shows that much of the yellow, green, and blue light is also absorbed. Because the sample absorbs all of these colors, what we see is the unabsorbed red and violet light, which we perceive as purple (the color pur-ple is classified as a tertiary color located between red and violet on an artists color wheel).

▲ Figure 23.25 Two ways of perceiving the color orange. An object appears orange either when it reflects orange light to the eye (left), or when it transmits to the eye all colors except blue, the complement of orange (middle). Complementary colors lie opposite to each other on an artist’s color wheel (right).

650 nm 580 nm

560 nm

490 nm430 nm

750 nm400 nm

O

B

V

R Y

G

Only blue light absorbed; eye perceives orange, blue’s complementary colorEye perceives orange since

only orange light reflected

soluTioNAnalyze We need to relate the color absorbed by a complex (red) to the color observed for the complex.

Plan For an object that absorbs only one color from the visible spec-trum, the color we see is complementary to the color absorbed. We can use the color wheel of Figure 23.25 to determine the complemen-tary color.

Solve From Figure 23.25, we see that green is complementary to red, so the complex appears green.

Comment As noted in Section 23.2, this green complex was one of those that helped Werner establish his theory of coordina-tion (Table 23.3). The other geometric isomer of this complex, cis@3Co(NH3)4Cl24+, absorbs yellow light and therefore appears violet.

Practice exercise 1A solution containing a certain transition-metal complex ion has the absorption spectrum shown here.

saMPle exerCise 23.6 relating Color absorbed to Color observed

The complex ion trans-3Co(NH3)4Cl24+ absorbs light primarily in the red region of the visible spectrum (the most intense absorption is at 680 nm). What is the color of the complex ion?


Ab

sorb

ance

What color would you expect a solution containing this ion to be? (a) violet, (b) blue, (c) green, (d) orange, (e) red.

Practice exercise 2A certain transition-metal complex ion absorbs at 695 nm. Which color is this ion most likely to be—blue, yellow, green, or red?

section 23.6 crystal-field theory 1021

Magnetism of Coordination CompoundsMany transition-metal complexes exhibit paramagnetism, as described in Sections 9.8 and 23.1. In such compounds the metal ions possess some number of unpaired elec-trons. It is possible to experimentally determine the number of unpaired electrons per metal ion from the measured degree of paramagnetism, and experiments reveal some interesting comparisons.

Compounds of the complex ion 3Co(CN)643- have no unpaired electrons, for example, but compounds of the 3CoF643- ion have four unpaired electrons per metal ion. Both complexes contain Co(III) with a 3d6 electron configuration. (Section 7.4) Clearly, there is a major difference in the ways in which the electrons are arranged in these two cases. Any successful bonding theory must explain this difference, and we present such a theory in the next section.

Give It Some ThoughtWhat is the electron configuration for (a) the Co atom and (b) the Co3+ ion? How many unpaired electrons does each possess? (See Section 7.4 to review electron configurations of ions.)

23.6 | Crystal-field TheoryScientists have long recognized that many of the magnetic properties and colors of transition-metal complexes are related to the presence of d electrons in the metal cat-ion. In this section, we consider a model for bonding in transition-metal complexes, crystal-field theory, that accounts for many of the observed properties of these sub-stances.* Because the predictions of crystal-field theory are essentially the same as those obtained with more advanced molecular-orbital theories, crystal-field theory is an excellent place to start in considering the electronic structure of coordination compounds.

The attraction of a ligand to a metal ion is essentially a Lewis acid–base interaction in which the base—that is, the ligand—donates a pair of electrons to an empty orbital on the metal ion (▶ Figure 23.27). Much of the attractive interaction between the metal ion and the ligands is due, however, to the electrostatic forces between the positive charge on the metal ion and negative charges on the ligands. An ionic ligand, such as Cl- or SCN-, experiences the usual cation–anion attraction. When the ligand is a neu-tral molecule, as in the case of H2O or NH3, the negative ends of these polar molecules, which contain an unshared electron pair, are directed toward the metal ion. In this case, the attractive interaction is of the ion–dipole type. (Section 11.2) In either case, the ligands are attracted strongly toward the metal ion. Because of the metal–ligand elec-trostatic attraction, the energy of the complex is lower than the combined energy of the separated metal ion and ligands.

Although the metal ion is attracted to the ligand electrons, the metal ion’s d elec-trons are repulsed by the ligands. Let’s examine this effect more closely, specifically the case in which the ligands form an octahedral array around a metal ion that has coordi-nation number 6.

In crystal-field theory, we consider the ligands to be negative points of charge that repel the negatively charged electrons in the d orbitals of the metal ion. The energy diagram in Figure 23.28 shows how these ligand point charges affect the energies of the d orbitals. First we imagine the complex as having all the ligand point charges uniformly distributed on the surface of a sphere centered on the metal ion. The average energy of the metal ion’s d orbitals is raised by the presence of this uniformly charged sphere. Hence, the energies of all five d orbitals are raised by the same amount.


Abs

orba

nce

Blue, green, yellow absorbed; a mixture of violet and red light travel to the eye, solution appears purple

▲ Figure 23.26 The color of [Ti(H2O)6]3+ A solution containing the 3Ti(H2O)643+ ion appears purple because, as its visible absorption spectrum shows, the solution does not absorb light from the violet and red ends of the spectrum. That unabsorbed light is what reaches our eyes.

Go FiGureHow would this absorbance spectrum change if you decreased the concentration of the 3Ti(H2O)643+ in solution?

*The name crystal field arose because the theory was first developed to explain the properties of solid crystal-line materials. The theory applies equally well to complexes in solution, however.

M Ln+

▲ Figure 23.27 Metal–ligand bond formation. The ligand acts as a Lewis base by donating its nonbonding electron pair to an empty orbital on the metal ion. The bond that results is strongly polar with some covalent character.


This energy picture is only a first approximation, however, because the ligands are not distributed uniformly on a spherical surface and, therefore, do not approach the metal ion equally from every direction. Instead, we envision the six ligands approach-ing along x-, y-, and z-axes, as shown on the right in Figure 23.28. This arrangement of ligands is called an octahedral crystal field. Because the metal ion’s d orbitals have different orientations and shapes, they do not all experience the same repulsion from the ligands and, therefore, do not all have the same energy under the influence of the octahedral crystal field. To see why, we must consider the shapes of the d orbitals and how their lobes are oriented relative to the ligands.

Figure 23.28 shows that the lobes of the dz2 and dx2 - y2 orbitals are directed along the x-, y-, and z-axes and so point directly toward the ligand point charges. In the dxy, dxz, and dyz orbitals, however, the lobes are directed between the axes and so do not point directly toward the charges. The result of this difference in orientation—dx2 - y2 and dz2 lobes point directly toward the ligand charges; dxy, dxz, and dyz lobes do not—is that the energy of the dx2 - y2 and dz2 orbitals is higher than the energy of the dxy, dxz, and dyz orbitals. This difference in energy is represented by the red boxes in the energy diagram of Figure 23.28.

It might seem like the energy of the dx2 - y2 orbital should be different from that of the dz2 orbital because the dx2 - y2 has four lobes pointing at ligands and the dz2 has only two lobes pointing at ligands. However, the dz2 orbital does have electron density in the xy plane, represented by the ring encircling the point where the two lobes meet. More advanced calculations show that two orbitals do indeed have the same energy in the pres-ence of the octahedral crystal field.

Because their lobes point directly at the negative ligand charges, electrons in the metal ion’s dz2 and dx2 - y2 orbitals experience stronger repulsions than those in the dxy, dxz, and dyz orbitals. As a result, the energy splitting shown in Figure 23.28 occurs. The three lower-energy d orbitals are called the t2 set of orbitals, and the two higher-energy ones are called the e set.* The energy gap ∆ between the two sets is often called the crystal-field splitting energy.

Ene

rgy

Free metal ion

t2 set (dxy, dxz, dyz)

e set (dz2, dx2−y2)

∆

Metal ion plusligands (negativepoint chargeswith sphericalsymmetry)

In octahedralcrystal �eld

y

x

z

y

x

z

y

x

z

y

x

z

y

x

z

+

−

+

− −

−

−

−

−

− −

−

−

−

−

− −

−

−

−

−

− −

−

−

−

−

− −

−

−

−

−

dz2 dx2−y2

dxy dxz dyz

▲ Figure 23.28 Energies of d orbitals in a free metal ion, a spherically symmetric crystal field, and an octahedral crystal field.

Go FiGureWhich d orbitals have lobes that point directly toward the ligands in an octahedral crystal field?

*The labels t2 for the dxy, dxz, and dyz orbitals and e for the dz2 and dx22y2 orbitals come from the application of a branch of mathematics called group theory to crystal-field theory. Group theory can be used to analyze the effects of symmetry on molecular properties.


Crystal-field theory helps us account for the colors observed in transition-metal complexes. The energy gap ∆ between the e and t2 sets of d orbitals is of the same order of magnitude as the energy of a photon of visible light. It is therefore possible for a transition-metal complex to absorb visible light that excites an electron from a lower-energy (t2) d orbital into a higher-energy (e) d orbital. In 3Ti(H2O)643+, for example, the Ti(III) ion has an 3Ar43d1 electron configuration. (Recall from Section 7.4 that when determining the electron configurations of transition-metal ions, we remove the s elec-trons first.) Ti(III) is thus called a d1 ion. In the ground state of 3Ti(H2O)643+, the single 3d electron resides in an orbital in the t2 set (▶ Figure 23.29). Absorption of 495-nm light excites this electron up to an orbital in the e set, generating the absorption spec-trum shown in Figure 23.26. Because this transition involves exciting an electron from one set of d orbitals to the other, we call it a d-d transition. As noted earlier, the absorp-tion of visible radiation that produces this d-d transition causes the 3Ti(H2O)643+ ion to appear purple.

Give It Some ThoughtWhy are compounds of Ti(IV) colorless?

The magnitude of the crystal-field splitting energy and, consequently, the color of a complex depend on both the metal and the ligands. For example, we saw in Figure 23.4 that the color of 3M(H2O)642+ complexes changes from reddish-pink when the metal ion is Co2+, to green for Ni2+, to pale blue for Cu2+. If we change the ligands in the 3Ni(H2O)642+ ion the color also changes. 3Ni(NH3)642+ has a blue-violet color, while 3Ni(en)342+ is purple (▼ Figure 23.30). In a ranking called the spectrochemical series, ligands are arranged in order of their abilities to increase splitting energy, as in this abbreviated list:

::Increasing ∆ ¡

Cl- 6 F- 6 H2O 6 NH3 6 en 6 NO2-(N@bonded) 6 CN-

Light

of495 nmE

nerg

y e

t2

▲ Figure 23.29 The d-d transition in [Ti(H2O)6]3+ is produced by the absorption of 495-nm light.

Go FiGureHow would you calculate the energy gap between the t2 and e orbitals from this diagram?

Ene

rgy

[Ni(H2O)6]2+

Max absorption in visible = 720 nm

[Ni(NH3)6]2+


t2

ee

∆ ∆

t2

[Ni(en)3]2+


e

∆

t2

▲ Figure 23.30 Effect of ligand on crystal-field splitting. The greater the crystal-field strength of the ligand, the greater the energy gap ∆ it causes between the t2 and e sets of the metal ion’s orbitals. This shifts the wavelength of the absorption maximum to shorter values.

Go FiGureIf you were to use a ligand L that was a stronger field ligand than ethylenediamine, what color would you expect the 3NiL642+ complex ion to have?


The magnitude of ∆ increases by roughly a factor of 2 from the far left to the far right of the spectrochemical series. Ligands at the low@∆ end of the spectrochemical series are termed weak-field ligands; those at the high@∆ end are termed strong-field ligands.

Let’s take a closer look at the colors and crystal field splitting as we vary the ligands for the series of Ni2+ complexes discussed above. Because the Ni atom has an 3Ar43d84s2 electron configuration, Ni2+ has the configuration 3Ar43d8 and therefore is a d8 ion. The t2 set of orbitals holds six electrons, two in each orbital, while the last two electrons go into the e set of orbitals. Consistent with Hund’s rule, each e orbital holds one electron and both electrons have the same spin. (Section 6.8)

As the ligand changes from H2O to NH3 to ethylenediamine the spectrochemical series tells us that the crystal field, ∆, exerted by the six ligands should increase. When there is more than one electron in the d orbitals, interactions between the electrons make the absorption spectra more complicated than the spectrum shown for 3Ti(H2O)643+ in Figure 23.26, which complicates the task of relating changes in ∆ with color. With d8 ions like Ni2+ three peaks are observed in the absorption spectra. Fortunately, for Ni2+ complexes we can simplify the analysis because only one of these three peaks falls in the visible region of the spectrum.* Because the energy separation ∆ is increasing, the wavelength of the absorp-tion peak should shift to a shorter wavelength. (Section 6.3) In the case of 3Ni(H2O)642+ the absorption peak in the visible part of the spectrum reaches a maximum near 720 nm, in the red region of the spectrum. So the complex ion takes the complementary color—green. For 3Ni(NH3)642+ the absorption peak reaches its maximum at 570 nm near the bound-ary between orange and yellow. The resulting color of the complex ion is a mixture of the complementary colors—blue and violet. Finally for 3Ni(en)342+ the peak shifts to an even shorter wavelength, 540 nm, which lies near the boundary between green and yellow. The resulting color purple is a mixture of the complimentary colors red and violet.

Electron Configurations in Octahedral ComplexesCrystal-field theory helps us understand the magnetic properties and some important chemical properties of transition-metal ions. From Hund’s rule, we expect electrons to always occupy the lowest-energy vacant orbitals first and to occupy a set of degenerate (same-energy) orbitals one at a time with their spins parallel. (Section 6.8) Thus, if we have a d1, d2, or d3 octahedral complex, the electrons go into the lower-energy t2 orbitals, with their spins parallel. When a fourth electron must be added, we have the two choices shown in ▼ Figure 23.31: The electron can either go into an e orbital, where

▲ Figure 23.31 Two possibilities for adding a fourth electron to a d3 octahedral complex. Whether the fourth electron goes into a t2 orbital or into an e orbital depends on the relative energies of the crystal-field splitting energy and the spin-pairing energy.

t2

e

t2

et2

e

Ene

rgy

Spin pairing energy larger than ∆

d3 ion d4 ion

∆ larger than spin pairing energy

*The other two peaks fall in the infrared (IR) and ultraviolet (UV) regions of the spectrum. For 3Ni(H2O)642+ the IR peak is found at 1176 nm and the UV peak at 388 nm.


it will be the sole electron in the orbital, or become the second electron in a t2 orbital. Because the energy difference between the t2 and e sets is the splitting energy ∆, the energy cost of going into an e orbital rather than a t2 orbital is also ∆. Thus, the goal of filling lowest-energy available orbitals first is met by putting the electron in a t2 orbital.

There is a penalty for doing this, however, because the electron must now be paired with the electron already occupying the orbital. The difference between the energy required to pair an electron in an occupied orbital and the energy required to place that electron in an empty orbital is called the spin-pairing energy. The spin-pairing energy arises from the fact that the electrostatic repulsion between two electrons that share an orbital (and so must have opposite spins) is greater than the repulsion between two electrons that are in different orbitals and have parallel spins.

In coordination complexes, the nature of the ligands and the charge on the metal ion often play major roles in determining which of the two electron arrangements shown in Figure 23.31 is used. In 3CoF643 - and 3Co(CN)643 - , both ligands have a 1- charge. The F- ion, however, is on the low end of the spectrochemical series, so it is a weak-field ligand. The CN- ion is on the high end and so is a strong-field ligand, which means it produces a larger energy gap ∆ than the F- ion. The splittings of the d-orbital energies in these two complexes are compared in ▶ Figure 23.32.

Cobalt(III) has an 3Ar43d6 electron configuration, so both complexes in Figure 23.32 are d6 complexes. Let’s imagine that we add these six electrons one at a time to the d orbitals of the 3CoF643 - ion. The first three go into the t2 orbitals with their spins parallel. The fourth electron could pair up in one of the t2 orbitals. The F- ion is a weak-field ligand, however, and so the energy gap ∆ between the t2 set and the e set is small. In this case, the more stable arrangement is the fourth electron in one of the e orbitals. By the same energy argument, the fifth electron goes into the other e orbital. With all five d orbitals containing one electron, the sixth must pair up, and the energy needed to place the sixth electron in a t2 orbital is less than that needed to place it in an e orbital. We end up with four t2 electrons and two e electrons.

Figure 23.32 shows that the crystal-field splitting energy ∆ is much larger in the 3Co(CN)643 - complex. In this case, the spin-pairing energy is smaller than ∆, so the lowest-energy arrangement is the six electrons paired in the t2 orbitals.

The 3CoF643 - complex is a high-spin complex; that is, the electrons are arranged so that they remain unpaired as much as possible. The 3Co(CN)643 - ion is a low-spin complex; that is, the electrons are arranged so that they remain paired as much as possible while still following Hund’s rule. These two electronic arrangements can be readily dis-tinguished by measuring the magnetic properties of the complex. Experiments show that 3CoF643 - has four unpaired electrons and 3Co(CN)643 - has none. The absorption spec-trum also shows peaks corresponding to the different values of ∆ in these two complexes.

Give It Some ThoughtIn octahedral complexes, for which d electron configurations is it possible to have high-spin and low-spin arrangements with different numbers of unpaired electrons?

In the transition metal ions of periods 5 and 6 (which have 4d and 5d valence elec-trons), the d orbitals are larger than in the period 4 ions (which have only 3d electrons). Thus, ions from periods 5 and 6 interact more strongly with ligands, resulting in a larger crystal-field splitting. Consequently, metal ions in periods 5 and 6 are invariably low spin in an octahedral crystal field.

▲ Figure 23.32 High-spin and low-spin complexes. The high-spin 3CoF643- ion has a weak-field ligand and so a small ∆ value. The low-spin 3Co(CN)643- ion has a strong-field ligand and so a large ∆ value.

∆∆

Ene

rgy

[CoF6]3−

High-spincomplex

Low-spincomplex

t2

t2

e

e

[Co(CN)6]3−

∆ less than spin-pairing energy, electrons singly occupy each e orbital before pairing in the t2 orbitals

∆ greater than spin-pairing energy, t2 orbitals are filled with electrons before occupying the e orbitals

saMPle exerCise 23.7 The spectrochemical series, Crystal Field splitting, Color, and Magnetism

The compound hexaamminecobalt(III) chloride is diamagnetic and orange in color with a single absorption peak in its visible absorption spectrum. (a) What is the electron configuration of the cobalt(III) ion? (b) Is 3Co(NH3)643+ a high-spin complex or a low-spin complex? (c) Estimate the wavelength where you expect the absorption of light to reach a maximum? (d) What color and magnetic behavior would you predict for the complex ion 3Co(en)343+?


Tetrahedral and Square-Planar ComplexesThus far we have considered crystal-field theory only for complexes having an octahe-dral geometry. When there are only four ligands in a complex, the geometry is gener-ally tetrahedral, except for the special case of d8 metal ions, which we will discuss in a moment.

The crystal-field splitting of d orbitals in tetrahedral complexes differs from that in octahedral complexes. Four equivalent ligands can interact with a central metal ion most effectively by approaching along the vertices of a tetrahedron. In this geometry the lobes of the two e orbitals point toward the edges of the tetrahedron, exactly in between the ligands (▶ Figure 23.33). This orientation keeps the dx2 - y2 and dz2 as far from the ligand point charges as possible. Consequently, these two d orbitals experi-ence less repulsion from the ligands and lie at lower energy than the other three d orbit-als. The three t2 orbitals do not point directly at the ligand point charges, but they do come closer to the ligands than the e set and as a result they experience more repulsion and are higher in energy. As we see in Figure 23.33, the splitting of d orbitals in a tet-rahedral geometry is the opposite of what we find for the octahedral geometry, namely the e orbitals are now below the t2 orbitals. The crystal-field splitting energy ∆ is much smaller for tetrahedral complexes than it is for comparable octahedral complexes, in part because there are fewer ligand point charges in the tetrahedral geometry, and in part because neither set of orbitals have lobes that point directly at the ligand point charges. Calculations show that for the same metal ion and ligand set, ∆ for the tetrahe-dral complex is only four-ninths as large as for the octahedral complex. For this reason,

soluTioNAnalyze We are given the color and magnetic behavior of an octahe-dral complex containing Co with a +3 oxidation number. We need to use this information to determine its electron configuration, its spin state (low-spin or high-spin), and the color of light it absorbs. In part (d), we must use the spectrochemical series to predict how its proper-ties will change if NH3 is replaced by ethylenediamine (en).Plan (a) From the oxidation number and the periodic table we can determine the number of valence electrons for Co(III) and from that we can determine the electron configuration. (b) The magnetic be-havior can be used to determine whether this compound is a low-spin or high-spin complex. (c) Since there is a single peak in the visible absorption spectrum the color of the compound should be comple-mentary to the color of light that is absorbed most strongly. (d) Eth-ylenediamine is a stronger field ligand than NH3 so we expect a larger ∆ for 3Co(en)343+ than for 3Co(NH3)643+.Solve

(a) Co has an electron configuration of 3Ar44s23d7 and Co3+ has three fewer electrons than Co. Because transition-metal ions always lose their valence shell s electrons, the electron configura-tion of Co3+ is 3Ar43d6.

(b) There are six valence electrons in the d orbitals. The filling of the t2 and e orbitals for both high-spin and low-spin complexes is shown below. Because the compound is diamagnetic we know all of the electrons must be paired up, which allows us to determine that 3Co(NH3)643+ is a low-spin complex.

Low spin

t2

e

∆

High spin

t2

e

∆

(c) We are told that the compound is orange and has a single absorp-tion peak in the visible region of the spectrum. The compound must therefore absorb the complementary color of orange, which is blue. The blue region of the spectrum ranges from approxi-mately 430 nm to 490 nm. As an estimate we assume that the complex ion absorbs somewhere in the middle of the blue region, near 460 nm.

(d) Ethylenediammine is higher in the spectrochemical series than ammonia. Therefore, we expect a larger ∆ for 3Co(en)343+. Because ∆ was already greater than the spin-pairing energy for 3Co(NH3)643+, we expect 3Co(en)343+ to be a low-spin complex as well, with a d6 configuration, so it will also be diamagnetic. The wavelength at which the complex absorbs light will shift to higher energy. If we assume a shift in the absorption maxi-mum from blue to violet, the color of the complex will become yellow.

Practice exercise 1Which of the following octahedral complex ions will have the fewest number of unpaired electrons? (a) 3Cr(H2O)643+, (b) 3V(H2O)643+, (c) 3FeF643-, (d) 3RhCl643-, (e) 3Ni(NH3)642+.

Practice exercise 2Consider the colors of the ammonia complexes of Co3+ given in Table 23.3. Based on the change in color would you expect 3Co(NH3)5Cl42+ to have a larger or smaller value of ∆ than 3Co(NH3)643+? Is this prediction consistent with the spectrochemical series?


all tetrahedral complexes are high spin; the crystal-field splitting energy is never large enough to overcome the spin-pairing energies.

In a square-planar complex, four ligands are arranged about the metal ion such that all five species are in the xy plane. The resulting energy levels of the d orbitals are illustrated in ▶ Figure 23.34. Note in particular that the dz2 orbital is considerably lower in energy than the dx2 - y2 orbital. To understand why this is so, recall from Figure 23.28 that in an octahedral field the dz2 orbital of the metal ion interacts with the ligands positioned above and below the xy plane. There are no ligands in these two positions in a square-planar complex, which means that the dz2 orbital experiences less repulsion and so remains in a lower-energy, more stable state.

Square-planar complexes are characteristic of metal ions with a d8 electron con-figuration. They are nearly always low spin, with the eight d electrons spin-paired to form a diamagnetic complex. This pairing leaves the dx2 - y2 orbital empty. Such an elec-tronic arrangement is particularly common among the d8 ions of periods 5 and 6, such as Pd2+, Pt2+, Ir+, and Au3+.

Give It Some ThoughtWhy is the energy of the dxz and d yz orbitals in a square-planar complex lower than that of the dxy orbital?

▲ Figure 23.33 Energies of the d orbitals in a tetrahedral crystal field. The splitting of the e and t2 sets of orbitals is inverted with respect to the splitting associated with an octahedral crystal field. The crystal field splitting energy ∆ is much smaller than it is in an octahedral crystal field.

e set (dz2, dx2−y2)

t2 set (dxy, dxz, dyz)

∆

In sphericalcrystal �eld

In tetrahedralcrystal �eld

x x x

z z z

y y

x

z

y

x

z

y

y

dz2 dx2−y2

dxy dxz dyz

Ene

rgy

Square planar

dxz, dyz

dxy

d 2

dx2−y2

▲ Figure 23.34 Energies of the d orbitals in a square-planar crystal field.

Go FiGureFor which d orbital(s) do the lobes point directly at the ligands in a square-planar crystal field?

soluTioNAnalyze We are given two complexes containing Ni2+ and their mag-netic properties. We are given two molecular geometry choices and asked to use crystal-field splitting diagrams from the text to deter-mine which complex has which geometry.

saMPle exerCise 23.8 Populating d orbitals in Tetrahedral and square-Planar Complexes

Nickel(II) complexes in which the metal coordination number is 4 can have either square-planar or tetrahedral geometry. 3NiCl442- is paramagnetic, and 3Ni(CN)442- is diamagnetic. One of these complexes is square planar, and the other is tetrahedral. Use the relevant crystal-field splitting dia-grams in the text to determine which complex has which geometry.

Plan We need to determine the number of d electrons in Ni2+ and then use Figure 23.33 for the tetrahedral complex and Figure 23.34 for the square-planar complex.


Crystal-field theory can be used to explain many observations in addition to those we have discussed. The theory is based on electrostatic interactions between ions and atoms, which essentially means ionic bonds. Many lines of evidence show, however, that the bonding in complexes must have some covalent character. Therefore, molecular- orbital theory (Sections 9.7 and 9.8) can also be used to describe the bonding in complexes, although the application of molecular-orbital theory to coordination compounds is beyond the scope of our discussion. Crystal-field theory, although not entirely accurate in all details, provides an adequate and useful first description of the electronic structure of complexes.

Solve Nickel(II) has the electron configuration 3Ar43d8. Tetrahedral complexes are always high spin, and square-planar complexes are al-most always low spin. Therefore, the population of the d electrons in the two geometries is

dx2−y2

dz2

Tetrahedral

dxz, dyz

dxy

Square planar

t2 (dxy, dyz, dxz)

e (dx2−y2, dz2)

The tetrahedral complex has two unpaired electrons, and the square-planar complex has none. We know from Section 23.1 that the tetrahedral com-plex must be paramagnetic and the square planar must be diamagnetic. Therefore, 3NiCl442- is tetrahedral, and 3Ni(CN)442- is square planar.Comment Nickel(II) forms octahedral complexes more frequently than square-planar ones, whereas d8 metals from periods 5 and 6 tend to favor square-planar coordination.

Practice exercise 1How many unpaired electrons do you predict for the tetrahedral 3MnCl442 - ion? (a) 1, (b) 2, (c) 3, (d) 4, (e) 5.

Practice exercise 2Are there any diamagnetic tetrahedral complexes containing tran-sition metal ions with partially filled d orbitals? If so what electron count(s) leads to diamagnetism?

KMnO4 K2CrO4 KClO4

▲ Figure 23.35 The colors of compounds can arise from charge-transfer transitions. KMnO4 and K2CrO4 are colored due to ligand-to-metal charge-transfer transitions in their anions. The perchlorate anion in KClO4 has no occupied d orbitals and its charge-transfer transition is at higher energy, corresponding to ultraviolet absorption; therefore it appears white.

of Mn(VII). Because Mn(VII) has a d0 electron configuration, the ab-sorption cannot be due to a d-d transition because there are no d elec-trons to excite! That does not mean, however, that the d orbitals are not involved in the transition. The excitation in the MnO4

- ion is due to a charge-transfer transition, in which an electron on one oxygen ligand is excited into a vacant d orbital on the Mn ion (▶ Figure 23.36). In essence, an electron is transferred from a ligand to the metal, so this transition is called a ligand-to-metal charge-transfer (LMCT) transition.

An LMCT transition is also responsible for the color of the CrO4

2-, which is a d0 Cr(VI) complex.Also shown in Figure 23.35 is a salt of the perchlorate ion (ClO4

-). Like MnO4

-, ClO4- is tetrahedral and has its central atom in the +7 ox-

idation state. However, because the Cl atom does not have low-lying

a Closer look

Charge-Transfer Color

In the laboratory portion of your course, you have probably seen many colorful transition-metal compounds, including those shown in ▼ Figure 23.35. Many of these compounds are colored because of d-d transitions. Some colored complexes, however, including the violet permanganate ion, MnO4

-, and the yellow chromate ion, CrO42-, derive

their color from a different type of excitation involving the d orbitals.The permanganate ion strongly absorbs visible light, with a maxi-

mum absorption at 565 nm. Because violet is complementary to yellow, this strong absorption in the yellow portion of the visible spectrum is responsible for the violet color of salts and solutions of the ion. What is happening during this absorption of light? The MnO4

- ion is a complex


t2 set

e set

Filled ligand orbitals

Empty Mn 3d orbitals

Ene

rgy

▲ Figure 23.36 Ligand-to-metal charge-transfer transition in MnO4–.

As shown by the blue arrow, an electron is excited from a nonbonding pair on O into one of the empty d orbitals on Mn.

d orbitals, exciting a Cl electron requires a more energetic photon than does MnO4

-. The first absorption for ClO4- is in the ultraviolet

portion of the spectrum, so all the visible light is transmitted and the salt appears white.

Other complexes exhibit charge-transfer excitations in which an electron from the metal atom is excited to an empty orbital on a ligand. Such an excitation is called a metal-to-ligand charge-transfer (MLCT) transition.

Charge-transfer transitions are generally more intense than d-d transitions. Many metal-containing pigments used for oil painting, such as cadmium yellow (CdS), chrome yellow (PbCrO4), and red ochre (Fe2O3), have intense colors because of charge-transfer transitions.

Related Exercises: 23.82, 23.83

soluTioN(a) The complex formed by coordination of one oxalate ion is octahedral:

(b) Because the oxalate ion has a charge of 2- , the net charge of a complex with three oxalate anions and one Co2+ ion is 4- . Therefore, the coordination compound has the formula Na43Co(C2O4)34.

(c) There is only one geometric isomer. The complex is chiral, however, in the same way the 3Co(en)343+ complex is chiral (Figure 23.22). The two mirror images are not superimposable, so there are two enantiomers:

4− 4−

saMPle iNTeGraTiVe exerCise Putting Concepts Together

The oxalate ion has the Lewis structure shown in Table 23.4. (a) Show the geometry of the com-plex formed when this ion complexes with cobalt(II) to form 3Co(C2O4)(H2O)44. (b) Write the formula for the salt formed when three oxalate ions complex with Co(II), assuming that the charge-balancing cation is Na+. (c) Sketch all the possible geometric isomers for the cobalt complex formed in part (b). Are any of these isomers chiral? Explain. (d) The equilibrium constant for the formation of the cobalt(II) complex produced by coordination of three oxalate anions, as in part (b), is 5.0 * 109, and the equilibrium constant for formation of the cobalt(II) complex with three mol-ecules of ortho-phenanthroline (Table 23.4) is 9 * 1019. From these results, what conclusions can you draw regarding the relative Lewis base properties of the two ligands toward cobalt(II)? (e) Using the approach described in Sample Exercise 17.14, calculate the concentration of free aqueous Co(II) ion in a solution initially containing 0.040 M oxalate (aq) and 0.0010 M Co2+(aq).


Chapter summary and Key TermsTHE TRANSITION METALS (SECTION 23.1) Metallic elements are obtained from minerals, which are solid inorganic compounds found in nature. Metallurgy is the science and technology of extracting metals from the earth and processing them for further use. Transition met-als are characterized by incomplete filling of the d orbitals. The pres-ence of d electrons in transition elements leads to multiple oxidation states. As we proceed through the transition metals in a given row of the periodic table, the attraction between the nucleus and the valence electrons increases more markedly for d electrons than for s electrons. As a result, the later transition elements in a period tend to have lower oxidation states.

The atomic and ionic radii of period 5 transition metals are larger than those of period 4 metals. The transition metals of periods 5 and 6 have comparable atomic and ionic radii and are also similar in other properties. This similarity is due to the lanthanide contraction.

The presence of unpaired electrons in valence orbitals leads to mag-netic behavior in transition metals and their compounds. In ferromagnetic, ferrimagnetic, and antiferromagnetic substances, the unpaired electron spins on atoms in a solid are affected by spins on neighboring atoms. In a ferromagnetic substance the spins all point in the same direction. In an antiferromagnetic substance the spins point in opposite directions and cancel one another. In a ferrimagnetic substance the spins point in oppo-site directions but do not fully cancel. Ferromagnetic and ferrimagnetic substances are used to make permanent magnets.

TRANSITION-METAL COMPLExES (SECTION 23.2) Coordination compounds are substances that contain metal complexes. Metal com-plexes contain metal ions bonded to several surrounding anions or molecules known as ligands. The metal ion and its ligands make up the coordination sphere of the complex. The number of atoms attached to the metal ion is the coordination number of the metal ion. The most common coordination numbers are 4 and 6; the most common coor-dination geometries are tetrahedral, square planar, and octahedral.

COMMON LIGANDS IN COORDINATION CHEMISTRy (SECTION 23.3) Ligands that occupy only one site in a coordination sphere are called

monodentate ligands. The atom of the ligand that bonds to the metal ion is the donor atom. Ligands that have two donor atoms are bidentate ligands. Polydentate ligands have three or more donor atoms. Bidentate and polydendate ligands are also called chelating agents. In general, chelating agents form more stable complexes than do related mono-dentate ligands, an observation known as the chelate effect. Many biologically important molecules, such as the porphyrins, are com-plexes of chelating agents. A related group of plant pigments known as chlorophylls are important in photosynthesis, the process by which plants use solar energy to convert CO2 and H2O into carbohydrates.

NOMENCLATURE AND ISOMERISM IN COORDINATION CHEMISTRy (SECTION 23.4) In naming coordination compounds, the number and type of ligands attached to the metal ion are specified, as is the oxidation state of the metal ion. Isomers are compounds with the same composition but different arrangements of atoms and therefore differ-ent properties. Structural isomers differ in the bonding arrangements of the ligands. Linkage isomerism occurs when a ligand can coordinate to a metal ion through different donor atoms. Coordination-sphere isomers contain different ligands in the coordination sphere. Stereoisomers are isomers with the same chemical bonding arrangements but different spatial arrangements of ligands. The most common forms of stereo-isomerism are geometric isomerism and optical isomerism. Geometric isomers differ from one another in the relative locations of donor atoms in the coordination sphere; the most common are cis–trans isomers. Geometric isomers differ from one another in their chemical and physical properties. Optical isomers are nonsuperimposable mirror images of each other. Optical isomers, or enantiomers, are chiral, mean-ing that they have a specific “handedness” and differ only in the pres-ence of a chiral environment. Optical isomers can be distinguished from one another by their interactions with plane-polarized light; solutions of one isomer rotate the plane of polarization to the right (dextrorotatory), and solutions of its mirror image rotate the plane to the left (levorotatory). Chiral molecules, therefore, are optically active. A 50950 mixture of two optical isomers does not rotate plane-polarized light and is said to be racemic.

(d) The ortho-phenanthroline ligand is bidentate, like the oxalate ligand, so they both exhibit the chelate effect. Thus, we conclude that ortho-phenanthroline is a stronger Lewis base toward Co2+ than oxalate. This conclusion is consistent with what we learned about bases in Section 16.7, that nitrogen bases are generally stronger than oxygen bases. (Recall, for example, that NH3 is a stronger base than H2O.)

(e) The equilibrium we must consider involves 3 mol of oxalate ion (represented as Ox2 - ).Co2+(aq) + 3 Ox2-(aq) ∆ 3Co(Ox)344-(aq)

The formation-constant expression is

Kf =33Co(Ox)344-43Co2+43Ox2-43

Because Kf is so large, we can assume that essentially all the Co2+ is converted to the oxalato complex. Under that assumption, the final concentration of 3Co(Ox)344- is 0.0010 M and that of oxalate ion is 3Ox2-4 = (0.040) - 3(0.0010) = 0.037 M (three Ox2- ions react with each Co2+ ion). We then have

3Co2+4 = xM, 3Ox2-4 ≅ 0.037 M, 33Co(Ox)344-4 ≅ 0.0010 MInserting these values into the equilibrium-constant expression, we have

Kf =(0.0010)x(0.037)3 = 5 * 109

Solving for x, we obtain 4 * 10-9 M. From this, we see that the oxalate has complexed all but a tiny fraction of the Co2+ in solution.

exercises 1031

23.1 The three graphs below show the variation in radius, effective nuclear charge, and maxi-mum oxidation state for the transition metals of period 4. In each part below identify which property is being plotted. [Section 23.1]

COLOR AND MAGNETISM IN COORDINATION CHEMISTRy (SECTION 23.5) A substance has a particular color because it either reflects or transmits light of that color or absorbs light of the comple-mentary color. The amount of light absorbed by a sample as a function of wavelength is known as its absorption spectrum. The light absorbed provides the energy to excite electrons to higher-energy states.

It is possible to determine the number of unpaired electrons in a complex from its degree of paramagnetism. Compounds with no unpaired electrons are diamagnetic.

CRySTAL-FIELD THEORy (SECTION 23.6) Crystal-field theory suc-cessfully accounts for many properties of coordination compounds, including their color and magnetism. In crystal-field theory, the interaction between metal ion and ligand is viewed as electrostatic. Because some d orbitals point directly at the ligands whereas others point between them, the ligands split the energies of the metal d orbit-als. For an octahedral complex, the d orbitals are split into a lower-energy set of three degenerate orbitals (the t2 set) and a higher-energy set of two degenerate orbitals (the e set). Visible light can cause a

d-d transition, in which an electron is excited from a lower-energy d orbital to a higher-energy d orbital. The spectrochemical series lists ligands in order of their ability to increase the split in d-orbital ener-gies in octahedral complexes.

Strong-field ligands create a splitting of d-orbital energies that is large enough to overcome the spin-pairing energy. The d electrons then preferentially pair up in the lower-energy orbitals, producing a low-spin complex. When the ligands exert a weak crystal field, the split-ting of the d orbitals is small. The electrons then occupy the higher-energy d orbitals in preference to pairing up in the lower-energy set, producing a high-spin complex.

Crystal-field theory also applies to tetrahedral and square- planar complexes, which leads to different d-orbital splitting patterns. In a tetrahedral crystal field, the splitting of the d orbitals results in a higher-energy t2 set and a lower-energy e set, the opposite of the octa-hedral case. The splitting in a tetrahedral crystal field is much smaller than that in an octahedral crystal field, so tetrahedral complexes are always high-spin complexes.

learning outcomes after studying this chapter, you should be able to:

• Describe the periodic trends in radii and oxidation states of the transition-metal ions, including the origin and effect of the lantha-nide contraction. (Section 23.1)

• Determine the oxidation number and number of d electrons for metal ions in complexes. (Section 23.2)

• Identify common ligands and distinguish between chelating and nonchelating ligands. (Section 23.3)

• Name coordination compounds given their formula and write their formula given their name. (Section 23.4)

• Recognize and draw the geometric isomers of a complex. (Section 23.4)

• Recognize and draw the optical isomers of a complex. (Section 23.4)

• Use crystal-field theory to explain the colors and to determine the number of unpaired electrons in a complex. (Sections 23.5 and 23.6)

exercisesVisualizing Concepts

23.2 Draw the structure for Pt(en)Cl2 and use it to answer the following questions: (a) What is the coordination number for platinum in this complex? (b) What is the coordination geometry? (c) What is the oxidation state of the platinum? (d) How many unpaired electrons are there? [Sections 23.2 and 23.6]

Sc Ti V Cr Mn Fe Co Ni Cu Zn Sc Ti V Cr Mn Fe Co Ni Cu Zn(a) (b)

Sc Ti V Cr Mn Fe Co Ni Cu Zn(c)


23.8 Which of these crystal-field splitting diagrams represents: (a) a weak-field octahedral complex of Fe3+, (b) a strong-field octahedral complex of Fe3+, (c) a tetrahedral complex of Fe3+, (d) a tetrahedral complex of Ni2+? (The diagrams do not indi-cate the relative magnitudes of ∆.) [Section 23.6]

(1) (4)(2) (3)

23.9 In the linear crystal field shown here, the negative charges are on the z-axis. Using Figure 23.28 as a guide, predict which of the following choices most accurately describes the splitting of the d orbitals in a linear crystal field? [Section 23.6]

(a) (b) (c) (d) (e)

23.10 Two Fe(II) complexes are both low spin but have different li-gands. A solution of one is green and a solution of the other is red. Which solution is likely to contain the complex that has the stronger-field ligand? [Section 23.6]

23.3 Draw the Lewis structure for the ligand shown here. (a) Which atoms can serve as donor atoms? Classify this ligand as monodentate, bidentate, or polydentate. (b) How many of these ligands are needed to fill the coordination sphere in an octahedral complex? [Section 23.2]

NH2CH2CH2NHCH2CO2−

23.4 The complex ion shown here has a 1- charge. Name the complex ion. [Section 23.4]

= Pt

= Cl

= H

= N

23.5 There are two geometric isomers of octahedral complexes of the type MA3X3, where M is a metal and A and X are mono-dentate ligands. Of the complexes shown here, which are identical to (1) and which are the geometric isomers of (1)? [Section 23.4]

(1) (2) (5)(4)(3)

23.6 Which of the complexes shown here are chiral? [Section 23.4]

= NH2CH2CH2NH2 = NH3

(1) (4)(3)(2)

= Cl= Cr

23.7 The solutions shown here each have an absorption spectrum with a single absorption peak like that shown in Figure 23.26. What color does each solution absorb most strongly? [Section 23.5]

exercises 1033

Co(NH3)6Cl3. (e) What are the most common coordination numbers for metal complexes?

23.23 A complex is written as NiBr2 # 6 NH3. (a) What is the oxi-dation state of the Ni atom in this complex? (b) What is the likely coordination number for the complex? (c) If the com-plex is treated with excess AgNO3(aq), how many moles of AgBr will precipitate per mole of complex?

23.24 Crystals of hydrated chromium(III) chloride are green, have an empirical formula of CrCl3 # 6 H2O, and are highly solu-ble, (a) Write the complex ion that exists in this compound. (b) If the complex is treated with excess AgNO3(aq), how many moles of AgCl will precipitate per mole of CrCl3 # 6 H2O dis-solved in solution? (c) Crystals of anhydrous chromium(III) chloride are violet and insoluble in aqueous solution. The co-ordination geometry of chromium in these crystals is octahe-dral as is almost always the case for Cr3+. How can this be the case if the ratio of Cr to Cl is not 1:6?

23.25 Indicate the coordination number and the oxidation number of the metal for each of the following complexes:(a) Na23CdCl44(b) K23MoOCl44(c) 3Co(NH3)4Cl24Cl(d) 3Ni(CN)543 -

(e) K33V(C2O4)34(f) 3Zn(en)24Br2

23.26 Indicate the coordination number and the oxidation number of the metal for each of the following complexes:(a) K33Co(CN)64(b) Na23CdBr44(c) 3Pt(en)34(ClO4)4

(d) 3Co(en)2(C2O4)4+

(e) NH43Cr(NH3)2(NCS)44(f) 3Cu(bipy)2I]I

Common ligands in Coordination Chemistry (section 23.3)

23.27 (a) What is the difference between a monodentate ligand and a bidentate ligand? (b) How many bidentate ligands are nec-essary to fill the coordination sphere of a six-coordinate com-plex? (c) You are told that a certain molecule can serve as a tridentate ligand. Based on this statement, what do you know about the molecule?

23.28 For each of the following polydentate ligands, determine (i) the maximum number of coordination sites that the li-gand can occupy on a single metal ion and (ii) the number and type of donor atoms in the ligand: (a) ethylenediamine (en), (b) bipyridine (bipy), (c) the oxalate anion (C2O4

2-), (d) the 2- ion of the porphine molecule (Figure 23.13), (e) 3EDTA44-.

23.29 Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies:(a) 3Co(NH3)4(o@phen)]Cl3

(b) 3Cr(C2O4)(H2O)44Br

The Transition Metals (section 23.1)

23.11 The lanthanide contraction explains which of the following periodic trends? (a) The atomic radii of the transition met-als first decrease and then increase when moving horizontally across each period. (b) When forming ions the transi-tion metals lose their valence s orbitals before their valence d orbitals. (c) The radii of the period 5 transition metals (Y–Cd) are very similar to the radii of the period 6 transition metals (Lu–Hg).

23.12 Which periodic trend is responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7B and 8B? (a) The number of valence electrons reaches a maximum at group 8B. (b) The effective nuclear charge increases on moving left across each period. (c) The radii of the transition-metal elements reaches a mini-mum for group 8B and as the size of the atoms decreases it becomes easier to remove electrons.

23.13 For each of the following compounds determine the electron configuration of the transition metal ion. (a) TiO, (b) TiO2, (c) NiO, (d) ZnO.

23.14 Among the period 4 transition metals (Sc–Zn), which elements do not form ions where there are partially filled 3d orbitals?

23.15 Write out the ground-state electron configurations of (a) Ti3+, (b) Ru2+, (c) Au3+, (d) Mn4+.

23.16 How many electrons are in the valence d orbitals in these transition-metal ions? (a) Co3+, (b) Cu+, (c) Cd2+, (d) Os3+.

23.17 Which type of substance is attracted by a magnetic field, a diamagnetic substance or a paramagnetic substance?

23.18 Which type of magnetic material cannot be used to make permanent magnets, a ferromagnetic substance, an antiferro-magnetic substance, or a ferrimagnetic substance?

23.19 What kind of magnetism is exhibited by this diagram:

Put in vertical magnetic field

Magnetic field

23.20 The most important oxides of iron are magnetite, Fe3O4, and hematite, Fe2O3. (a) What are the oxidation states of iron in these compounds? (b) One of these iron oxides is ferrimag-netic, and the other is antiferromagnetic. Which iron oxide is likely to show which type of magnetism? Explain.

Transition-Metal Complexes (section 23.2)

23.21 (a) Using Werner’s definition of valence, which property is the same as oxidation number, primary valence or secondary valence? (b) What term do we normally use for the other type of valence? (c) Why can the NH3 molecule serve as a ligand but the BH3 molecule cannot?

23.22 (a) What is the meaning of the term coordination number as it applies to metal complexes? (b) Give an example of a ligand that is neutral and one that is negatively charged. (c) Would you expect ligands that are positively charged to be common? Explain. (d) What type of chemical bonding is characteristic of coordination compounds? Illustrate with the compound


(d) potassium diaquatetrabromovanadate(III)(e) bis(ethylenediamine)zinc(II) tetraiodomercurate(II)

23.36 Write the formula for each of the following compounds, be-ing sure to use brackets to indicate the coordination sphere:(a) tetraaquadibromomanganese(III) perchlorate(b) bis(bipyridyl)cadmium(II) chloride(c) potassium tetrabromo(ortho-phenanthroline)-cobaltate (III)(d) cesium diamminetetracyanochromate(III)(e) tris(ethylenediamine)rhodium(III) tris(oxalato)-

cobaltate(III) 23.37 Write the names of the following compounds, using the stan-

dard nomenclature rules for coordination complexes:(a) 3Rh(NH3)4Cl24Cl(b) K23TiCl64(c) MoOCl4

(d) 3Pt(H2O)4(C2O4)]Br2

23.38 Write names for the following coordination compounds:(a) 3Cd(en)Cl24(b) K43Mn(CN)64(c) 3Cr(NH3)5(CO3)]Cl(d) 3Ir(NH3)4(H2O)24(NO3)3

23.39 Consider the following three complexes: (Complex 1) 3Co(NH3)4Br24Cl (Complex 2) 3Pd(NH3)2(ONO)24 (Complex 3) 3V(en)2Cl24+, Which of the three complexes can have (a) geometric isomers,

(b) linkage isomers, (c) optical isomers, (d) coordination-sphere isomers?

23.40 Consider the following three complexes: (Complex 1) 3Co(NH3)5SCN42+

(Complex 2) 3Co(NH3)3Cl342+

(Complex 3) CoClBr # 5NH3

Which of the three complexes can have (a) geometric isomers, (b) linkage isomers, (c) optical isomers, (d) coordination-sphere isomers?

23.41 A four-coordinate complex MA2B2 is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?

23.42 Consider an octahedral complex MA3B3. How many geomet-ric isomers are expected for this compound? Will any of the isomers be optically active? If so, which ones?

23.43 Sketch all the possible stereoisomers of (a) tetrahedral 3Cd(H2O)2Cl24, (b) square-planar 3IrCl2(PH3)24-, (c) octahe-dral 3Fe(o@phen)2Cl24+.

23.44 Sketch all the possible stereoisomers of (a) 3Rh(bipy) (o-phen)243+, (b) 3Co(NH3)3(bipy)Br42+, (c) square-planar 3Pd(en)(CN)24.

Color and Magnetism in Coordination Chemistry; Crystal-Field Theory (sections 23.5 and 23.6)

23.45 (a) If a complex absorbs light at 610 nm, what color would you expect the complex to be? (b) What is the energy in Joules

(c) 3Cr(EDTA)(H2O)4-

(d) 3Zn(en)24(ClO4)2

23.30 Indicate the likely coordination number of the metal in each of the following complexes:(a) 3Rh(bipy)34(NO3)3

(b) Na33Co(C2O4)2Cl24(c) 3Cr(o@phen)34(CH3COO)3

(d) Na23Co(EDTA)Br4 23.31 (a) What is meant by the term chelate effect? (b) What ther-

modynamic factor is generally responsible for the chelate effect? (c) Why are polydentate ligands often called sequester-ing agents?

23.32 Pyridine (C5H5N), abbreviated py, is the molecule

N

(a) Why is pyridine referred to as a monodentate ligand? (b) For the equilibrium reaction

3Ru(py)4(bipy)42+ + 2 py ∆ 3Ru(py)642+ + bipy

what would you predict for the magnitude of the equilibrium constant? Explain your answer.

23.33 True or false? The following ligand can act as a bidentate ligand?

N

N

23.34 When silver nitrate is reacted with the molecular base ortho-phenanthroline, colorless crystals form that contain the transition-metal complex shown below. (a) What is the co-ordination geometry of silver in this complex? (b) Assuming no oxidation or reduction occurs during the reaction, what is charge of the complex shown here? (c) Do you expect any nitrate ions will be present in the crystal? (d) Write a formula for the compound that forms in this reaction. (e) Use the ac-cepted nomenclature to write the name of this compound.

N N

NN

Ag

Nomenclature and isomerism in Coordination Chemistry (section 23.4)

23.35 Write the formula for each of the following compounds, be-ing sure to use brackets to indicate the coordination sphere:(a) hexaamminechromium(III) nitrate(b) tetraamminecarbonatocobalt(III) sulfate(c) dichlorobis(ethylenediamine)platinum(IV) bromide

additional exercises 1035

for the 3Ni(bipy)342+ ion? (b) Based on these data, where would you put bipy in the spectrochemical series?

23.55 Give the number of (valence) d electrons associated with the cen-tral metal ion in each of the following complexes: (a) K33TiCl64, (b) Na33Co(NO2)64, (c) 3Ru(en)34Br3, (d) 3Mo(EDTA)]ClO4, (e) K33ReCl64.

23.56 Give the number of (valence) d electrons associated with the central metal ion in each of the following complexes: (a) K33Fe(CN)64, (b) 3Mn(H2O)64(NO3)2, (c) Na3Ag(CN)24, (d) 3Cr(NH3)4Br24ClO4, (e) 3Sr(EDTA)42 - .

23.57 A classmate says, “A weak-field ligand usually means the complex is high spin.” Is your classmate correct? Explain.

23.58 A classmate says, “A strong-field ligand means that the ligand binds strongly to the metal ion.” Is your classmate correct? Explain.

23.59 For each of the following metals, write the electronic config-uration of the atom and its 2+ ion: (a) Mn, (b) Ru, (c) Rh. Draw the crystal-field energy-level diagram for the d orbitals of an octahedral complex, and show the placement of the d electrons for each 2+ ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

23.60 For each of the following metals, write the electronic config-uration of the atom and its 3+ ion: (a) Fe, (b) Mo, (c) Co. Draw the crystal-field energy-level diagram for the d orbit-als of an octahedral complex, and show the placement of the d electrons for each 3+ ion, assuming a weak-field complex. How many unpaired electrons are there in each case?

23.61 Draw the crystal-field energy-level diagrams and show the placement of d electrons for each of the following: (a) 3Cr(H2O)642+ (four unpaired electrons), (b) 3Mn(H2O)642+ (high spin), (c) 3Ru(NH3)5(H2O)42+ (low spin), (d) 3IrCl642- (low spin), (e) 3Cr(en)343+, (f) 3NiF644-.

23.62 Draw the crystal-field energy-level diagrams and show the placement of electrons for the following com-plexes: (a) 3VCl643-, (b) 3FeF643- (a high-spin complex), (c) 3Ru(bipy)343+ (a low-spin complex), (d) 3NiCl442- (tetra-hedral), (e) 3PtBr642-, (f) 3Ti(en)342+.

23.63 The complex 3Mn(NH3)642+ contains five unpaired electrons. Sketch the energy-level diagram for the d orbitals, and indi-cate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

23.64 The ion 3Fe(CN)643- has one unpaired electron, whereas 3Fe(NCS)643- has five unpaired electrons. From these results, what can you conclude about whether each complex is high spin or low spin? What can you say about the placement of NCS- in the spectrochemical series?

of a photon with a wavelength of 610 nm? (c) What is the energy of this absorption in kJ>mol?

23.46 (a) A complex absorbs photons with an energy of 4.51 * 10-19 J. What is the wavelength of these photons? (b) If this is the only place in the visible spectrum where the complex absorbs light, what color would you expect the complex to be?

23.47 Identify each of the following coordination complexes as either diamagnetic or paramagnetic:(a) 3ZnCl442-

(b) 3Pd(NH3)2Cl24(c) 3V(H2O)643+

(d) 3Ni(en)342+

23.48 Identify each of the following coordination complexes as either diamagnetic or paramagnetic:(a) 3Ag(NH3)24+

(b) square planar 3Cu(NH3)442+

(c) 3Ru(bipy)342+

(d) 3CoCl442-

23.49 In crystal-field theory, ligands are modeled as if they are point negative charges. What is the basis of this assumption, and how does it relate to the nature of metal–ligand bonds?

23.50 The lobes of which d orbitals point directly between the li-gands in (a) octahedral geometry, (b) tetrahedral geometry?

23.51 (a) Sketch a diagram that shows the definition of the crys-tal-field splitting energy (∆) for an octahedral crystal field. (b) What is the relationship between the magnitude of ∆ and the energy of the d-d transition for a d1 complex? (c) Calculate ∆ in kJ>mol if a d1 complex has an absorption maximum at 545 nm.

23.52 As shown in Figure 23.26, the d-d transition of 3Ti(H2O)643+ produces an absorption maximum at a wavelength of about 500 nm. (a) What is the magnitude of ∆ for 3Ti(H2O)643+ in kJ>mol? (b) How would the magnitude of ∆ change if the H2O ligands in 3Ti(H2O)643+ were replaced with NH3 ligands?

23.53 The colors in the copper-containing minerals malachite (green) and azurite (blue) come from a single d-d transition in each compound. (a) What is the electron configuration of the copper ion in these minerals? (b) Based on their colors in which compound would you predict the crystal field splitting ∆ is larger?

23.54 The color and wavelength of the absorption maximum for 3Ni(H2O)642+, 3Ni(NH3)642+, and 3Ni(en)342+ are given in Figure 23.30. The absorption maximum for the 3Ni(bipy)342+ ion occurs at about 520 nm. (a) What color would you expect

additional exercises 23.65 The Curie temperature is the temperature at which a ferro-

magnetic solid switches from ferromagnetic to paramagnetic, and for nickel, the Curie temperature is 354 °C. Knowing this, you tie a string to two paper clips made of nickel and hold the paper clips near a permanent magnet. The magnet attracts the paper clips, as shown in the photograph on the left. Now you heat one of the paper clips with a cigarette lighter, and the clip drops (right photograph). Explain what happened.


Tri�uoromethylacetylacetonate

(tfac)CF3 CH3C C

C

H

O O

−

Sketch all possible isomers for the complex with three tfac li-gands on cobalt(III). (You can use the symbol to repre-sent the ligand.)

23.74 Which transition metal atom is present in each of the fol-lowing biologically important molecules: (a) hemoglobin, (b) chlorophylls, (c) siderophores.

23.75 Carbon monoxide, CO, is an important ligand in coordina-tion chemistry. When CO is reacted with nickel metal the product is 3Ni(CO)44, which is a toxic, pale yellow liquid. (a) What is the oxidation number for nickel in this com-pound? (b) Given that 3Ni(CO)44 is diamagnetic molecule with a tetrahedral geometry, what is the electron configuration of nickel in this compound? (c) Write the name for 3Ni(CO)44 using the nomenclature rules for coordination compounds.

23.76 Some metal complexes have a coordination number of 5. One such complex is Fe(CO)5, which adopts a trigonal bi-pyramidal geometry (see Figure 9.8). (a) Write the name for Fe(CO)5, using the nomenclature rules for coordination compounds. (b) What is the oxidation state of Fe in this com-pound? (c) Suppose one of the CO ligands is replaced with a CN- ligand, forming 3Fe(CO)4(CN)4-. How many geometric isomers would you predict this complex could have?

23.77 Which of the following objects is chiral: (a) a left shoe, (b) a slice of bread, (c) a wood screw, (d) a molecular model of Zn(en)Cl2, (e) a typical golf club?

23.78 The complexes 3V(H2O)643+ and 3VF643- are both known. (a) Draw the d-orbital energy-level diagram for V(III) octa-hedral complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy?

[23.79] One of the more famous species in coordination chemistry is the Creutz–Taube complex:

(NH3)5RuN NRu(NH3)5

5+

It is named for the two scientists who discovered it and ini-tially studied its properties. The central ligand is pyrazine, a planar six-membered ring with nitrogens at opposite sides. (a) How can you account for the fact that the complex, which has only neutral ligands, has an odd overall charge? (b) The metal is in a low-spin configuration in both cases. Assuming octahedral coordination, draw the d-orbital energy-level dia-gram for each metal. (c) In many experiments the two metal ions appear to be in exactly equivalent states. Can you think of a reason that this might appear to be so, recognizing that electrons move very rapidly compared to nuclei?

23.80 Solutions of 3Co(NH3)642+, 3Co(H2O)642+ (both octahedral), and 3CoCl442- (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in octahedral ones, assign a color to each complex.

23.66 Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

23.67 Based on the molar conductance values listed here for the series of platinum(IV) complexes, write the formula for each complex so as to show which ligands are in the coordina-tion sphere of the metal. By way of example, the molar con-ductances of 0.050 M NaCl and BaCl2 are 107 ohm-1 and 197 ohm-1, respectively.

23.68 (a) A compound with formula RuCl3 # 5 H2O is dissolved in water, forming a solution that is approximately the same color as the solid. Immediately after forming the solution, the addition of excess AgNO3(aq) forms 2 mol of solid AgCl per mole of complex. Write the formula for the compound, showing which ligands are likely to be present in the coordi-nation sphere. (b) After a solution of RuCl3 # 5 H2O has stood for about a year, addition of AgNO3(aq) precipitates 3 mol of AgCl per mole of complex. What has happened in the ensu-ing time?

23.69 Sketch the structure of the complex in each of the following compounds and give the full compound name:(a) cis@3Co(NH3)4(H2O)24(NO3)2

(b) Na23Ru(H2O)Cl54(c) trans@NH43Co(C2O4)2(H2O)24(d) cis@3Ru(en)2Cl24

23.70 (a) Which complex ions in Exercise 23.69 have a mirror plane? (b) Will any of the complexes have optical isomers?

23.71 The molecule dimethylphosphinoethane 3(CH3)2PCH2CH2P(CH3)2, which is abbreviated dmpe] is used as a ligand for some complexes that serve as catalysts. A complex that con-tains this ligand is Mo(CO)4(dmpe). (a) Draw the Lewis structure for dmpe, and compare it with ethylenediamine as a coordinating ligand. (b) What is the oxidation state of Mo in Na23Mo(CN)2(CO)2(dmpe)]? (c) Sketch the structure of the 3Mo(CN)2(CO)2(dmpe)42- ion, including all the possible isomers.

23.72 Although the cis configuration is known for 3Pt(en)Cl24, no trans form is known. (a) Explain why the trans compound is not possible. (b) Would NH2CH2CH2CH2CH2NH2 be more likely than en (NH2CH2CH2NH2) to form the trans com-pound? Explain.

23.73 The acetylacetone ion forms very stable complexes with many metallic ions. It acts as a bidentate ligand, coordinating to the metal at two adjacent positions. Suppose that one of the CH3 groups of the ligand is replaced by a CF3 group, as shown here:

Complex

Molar Conductance (ohm−1)* of 0.050 M solution

Pt(NH3)6Cl4 523

Pt(NH3)4Cl4 228

Pt(NH3)3Cl4 97

Pt(NH3)2Cl4 0

KPt(NH3)Cl5 108

*The ohm is a unit of resistance; conductance is the inverse of resistance.

integrative exercises 1037

model of complexes that may have played a role in the origin of life. (a) Sketch the structure of the complex. (b) The com-plex is isolated as a sodium salt. Write the complete name of this salt. (c) What is the oxidation state of Fe in this complex? How many d electrons are associated with the Fe in this com-plex? (d) Would you expect this complex to be high spin or low spin? Explain.

[23.86] When Alfred Werner was developing the field of coordina-tion chemistry, it was argued by some that the optical activ-ity he observed in the chiral complexes he had prepared was because of the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral com-plex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) com-plex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthe-size the complex you design, but we will not worry about that for now.)

23.87 Generally speaking, for a given metal and ligand, the stabil-ity of a coordination compound is greater for the metal in the +3 rather than in the +2 oxidation state (for metals that form stable +3 ions in the first place). Suggest an explanation, keeping in mind the Lewis acid–base nature of the metal– ligand bond.

23.88 Many trace metal ions exist in the blood complexed with amino acids or small peptides. The anion of the amino acid glycine (gly),

H2NCH2C O−

O

can act as a bidentate ligand, coordinating to the metal through nitrogen and oxygen atoms. How many isomers are possible for (a) 3Zn(gly)24 (tetrahedral), (b) 3Pt(gly)24 (square planar), (c) 3Co(gly)34 (octahedral)? Sketch all possible iso-mers. Use the symbol to represent the ligand.

23.89 The coordination complex 3Cr(CO)64 forms colorless, dia-magnetic crystals that melt at 90 °C. (a) What is the oxida-tion number of chromium in this compound? (b) Given that 3Cr(CO)64 is diamagnetic, what is the electron configuration of chromium in this compound? (c) Given that 3Cr(CO)64 is colorless, would you expect CO to be a weak-field or strong-field ligand? (d) Write the name for 3Cr(CO)64 using the no-menclature rules for coordination compounds.

23.81 Oxyhemoglobin, with an O2 bound to iron, is a low-spin Fe(II) complex; deoxyhemoglobin, without the O2 molecule, is a high-spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many un-paired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of O2 in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15-min-ute exposure to air containing 400 ppm of CO causes about 10% of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and O2 to hemoglobin? (e) CO is a strong-field ligand. What color might you expect carboxy-hemoglobin to be?

[23.82] Consider the tetrahedral anions VO43- (orthovanadate ion),

CrO42- (chromate ion), and MnO4

- (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit d-d tran-sitions? Explain. (c) As mentioned in “A Closer Look” on charge-transfer color, the violet color of MnO4

- is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in MnO4

- oc-curs at a wavelength of 565 nm. The CrO4

2- ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for MnO4

-? Explain. (e) The VO43- ion

is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spec-trum? Explain your reasoning.

23.83 (a) Given the colors observed for VO43- (orthovanadate ion),

CrO42- (chromate ion), and MnO4

- (permanganate ion) (see Exercise 23.82), what can you say about how the energy sepa-ration between the ligand orbitals and the empty d orbitals changes as a function of the oxidation state of the transition metal at the center of the tetrahedral anion?

[23.84] The red color of ruby is due to the presence of Cr(III) ions at octahedral sites in the close-packed oxide lattice of Al2O3. Draw the crystal-field splitting diagram for Cr(III) in this environ-ment. Suppose that the ruby crystal is subjected to high pres-sure. What do you predict for the variation in the wavelength of absorption of the ruby as a function of pressure? Explain.

23.85 In 2001, chemists at SUNY-Stony Brook succeeded in synthe-sizing the complex trans@3Fe(CN)4(CO)242-, which could be a

integrative exercises [23.90] Metallic elements are essential components of many impor-

tant enzymes operating within our bodies. Carbonic anhy-drase, which contains Zn2+ in its active site, is responsible for rapidly interconverting dissolved CO2 and bicarbonate ion, HCO3

-. The zinc in carbonic anhydrase is tetrahedrally co-ordinated by three neutral nitrogen-containing groups and a water molecule. The coordinated water molecule has a pKa of 7.5, which is crucial for the enzyme’s activity. (a) Draw the ac-tive site geometry for the Zn(II) center in carbonic anhydrase, just writing “N” for the three neutral nitrogen ligands from the protein. (b) Compare the pKa of carbonic anhydrase’s ac-tive site with that of pure water; which species is more acidic?

(c) When the coordinated water to the Zn(II) center in car-bonic anhydrase is deprotonated, what ligands are bound to the Zn(II) center? Assume the three nitrogen ligands are unaffected. (d) The pKa of 3Zn(H2O)642+ is 10. Suggest an explanation for the difference between this pKa and that of carbonic anhydrase. (e) Would you expect carbonic anhy-drase to have a deep color, like hemoglobin and other metal-ion containing proteins do? Explain.

23.91 Two di f ferent compounds have the formulat ion CoBr(SO4) # 5 NH3. Compound A is dark violet, and com-pound B is red-violet. When compound A is treated with AgNO3(aq), no reaction occurs, whereas compound B


then treated with sulfate ion to precipitate Ca2+ as calcium sulfate. The Mg2+ was then titrated with 18.7 mL of 0.0104 M 3EDTA44-. Calculate the concentrations of Mg2+ and Ca2+ in the hard water in mg>L.

23.97 Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does O2, as indicated by these approximate standard free-energy changes in blood:

Hb + O2 ¡ HbO2 ∆G° = -70 kJHb + CO ¡ HbCO ∆G° = -80 kJ

Using these data, estimate the equilibrium constant at 298 K for the equilibrium

HbO2 + CO ∆ HbCO + O2

[23.98] The molecule methylamine (CH3NH2) can act as a monoden-tate ligand. The following are equilibrium reactions and the thermochemical data at 298 K for reactions of methylamine and en with Cd2+(aq):

Cd2+(aq) + 4 CH3NH2(aq) ∆ 3Cd(CH3NH2)442+(aq)∆H ° = -57.3 kJ; ∆S° = -67.3 J>K; ∆G° = -37.2 kJ

Cd2+(aq) + 2 en(aq) ∆ 3Cd(en)242+(aq)∆H ° = -56.5 kJ; ∆S° = +14.1 J>K; ∆G° = -60.7 kJ

(a) Calculate ∆G° and the equilibrium constant K for the fol-lowing ligand exchange reaction:

3Cd(CH3NH2)442+(aq) + 2 en(aq) ∆3Cd(en)242+(aq) + 4 CH3NH2(aq)

Based on the value of K in part (a), what would you conclude about this reaction? What concept is demonstrated? (b) Deter-mine the magnitudes of the enthalpic (∆H °) and the entropic (-T∆S°) contributions to ∆G° for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the “A Closer Look” box on the chelate effect, predict the sign of ∆H° for the following hypothetical reaction:

3Cd(CH3NH2)442+(aq) + 4 NH3(aq) ∆3Cd(NH3)442+(aq) + 4 CH3NH2(aq)

23.99 The value of ∆ for the 3CrF643- complex is 182 kJ>mol. Calculate the expected wavelength of the absorption corre-sponding to promotion of an electron from the lower-energy to the higher-energy d-orbital set in this complex. Should the complex absorb in the visible range?

[23.100] A Cu electrode is immersed in a solution that is 1.00 M in 3Cu(NH3)442+ and 1.00 M in NH3. When the cathode is a standard hydrogen electrode, the emf of the cell is found to be +0.08 V. What is the formation constant for 3Cu(NH3)442+?

[23.101] The complex 3Ru(EDTA)(H2O)4- undergoes substitution reactions with several ligands, replacing the water molecule with the ligand. In all cases, the ruthenium stays in the +3 oxidation state and the ligands use a nitrogen donor atom to bind to the metal.

3Ru(EDTA)(H2O)4- + L ¡ 3Ru(EDTA)L4- + H2O

The rate constants for several ligands are as follows:

reacts with AgNO3(aq) to form a white precipitate. When compound A is treated with BaCl2(aq), a white precipi-tate is formed, whereas compound B has no reaction with BaCl2(aq). (a) Is Co in the same oxidation state in these com-plexes? (b) Explain the reactivity of compounds A and B with AgNO3(aq) and BaCl2(aq). (c) Are compounds A and B iso-mers of one another? If so, which category from Figure 23.19 best describes the isomerism observed for these complexes? (d) Would compounds A and B be expected to be strong elec-trolytes, weak electrolytes, or nonelectrolytes?

23.92 A manganese complex formed from a solution containing potassium bromide and oxalate ion is purified and analyzed. It contains 10.0% Mn, 28.6% potassium, 8.8% carbon, and 29.2% bromine by mass. The remainder of the compound is oxygen. An aqueous solution of the complex has about the same electrical conductivity as an equimolar solution of K43Fe(CN)64. Write the formula of the compound, using brackets to denote the manganese and its coordination sphere.

23.93 The E° values for two low-spin iron complexes in acidic solu-tion are as follows:

3Fe(o@phen)343+(aq) + e- ∆Fe(o@phen)342+(aq) E° = 1.12 V

3Fe(CN)643-(aq) + e- ∆Fe(CN)644-(aq) E° = 0.36 V

(a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which com-plex, 3Fe(o@phen)343+ or 3Fe(CN)643-, is more difficult to re-duce? (c) Suggest an explanation for your answer to (b).

23.94 A palladium complex formed from a solution containing bro-mide ion and pyridine, C5H5N (a good electron-pair donor), is found on elemental analysis to contain 37.6% bromine, 28.3% carbon, 6.60% nitrogen, and 2.37% hydrogen by mass. The compound is slightly soluble in several organic solvents; its solutions in water or alcohol do not conduct electricity. It is found experimentally to have a zero dipole moment. Write the chemical formula, and indicate its probable structure.

23.95 (a) In early studies it was observed that when the complex 3Co(NH3)4Br24Br was placed in water, the electrical conductivity of a 0.05 M solution changed from an initial value of 191 ohm-1 to a final value of 374 ohm-1 over a period of an hour or so. Sug-gest an explanation for the observed results. (See Exercise 23.67 for relevant comparison data.) (b) Write a balanced chemical equation to describe the reaction. (c) A 500-mL solution is made up by dissolving 3.87 g of the complex. As soon as the solution is formed, and before any change in conductivity has occurred, a 25.00-mL portion of the solution is titrated with 0.0100 M AgNO3 solution. What volume of AgNO3 solution do you ex-pect to be required to precipitate the free Br-(aq)? (d) Based on the response you gave to part (b), what volume of AgNO3 so-lution would be required to titrate a fresh 25.00-mL sample of 3Co(NH3)4Br24Br after all conductivity changes have occurred?

23.96 The total concentration of Ca2+ and Mg2+ in a sample of hard water was determined by titrating a 0.100-L sample of the wa-ter with a solution of EDTA4-. The EDTA4- chelates the two cations:

Mg2+ + 3EDTA44- ¡ 3Mg(EDTA)42-

Ca2+ + 3EDTA44- ¡ 3Ca(EDTA)42-

It requires 31.5 mL of 0.0104 M 3EDTA44- solution to reach the end point in the titration. A second 0.100-L sample was

ligand, l k (M−1s−1)

Pyridine 6.3 * 103

SCN- 2.7 * 102

CH3CN 3.0 * 10

Design an experiment 1039

a single concerted step. Which of these two mechanisms is more consistent with the data? Explain. (b) What do the re-sults suggest about the relative donor ability of the nitrogens of the three ligands toward Ru(III)? (c) Assuming that the complexes are all low spin, how many unpaired electrons are in each?

(a) One possible mechanism for this substitution reaction is that the water molecule dissociates from the Ru(III) in the rate-determining step, and then the ligand L binds to Ru(III) in a rapid second step. A second possible mecha-nism is that L approaches the complex, begins to form a new bond to the Ru(III), and displaces the water molecule, all in

design an experimentFollowing a procedure found in a scientific paper you go into the lab and attempt to pre-pare crystals of dichlorobis(ethylenediamine)cobalt(III) chloride. The paper states that this compound can be made by reacting CoCl2 # 6 H2O, an excess of ethylenediamine, O2 from the air (which acts as an oxidizing agent), water, and concentrated hydrochloric acid. At the end of the reaction you filter off the solution and are left with a green, crystalline product. (a) What experiment(s) could you perform to confirm that you have prepared 3CoCl2(en)24Cl and not 3Co(en)34Cl3? (b) How could you verify that cobalt was present as Co3+ and deter-mine the spin state of the cobalt complex in your product? (c) How many geometric isomers exist for 3CoCl2(en)24Cl? How could you determine if your product contains a single geometric isomer or a mixture of geometric isomers? (d) If the product does contain a single geometric isomer how would you determine which one was present? (Hint: You may find the information in Table 23.3 helpful.)

24 The Chemistry of Life: Organic and Biological ChemistryWe are all familiar with how chemical substances can influence our health and behavior. Aspirin, also known as acetylsalicylic acid, relieves aches and pains. Cocaine, whose full chemical name is methyl (1R,2R,3S,5S)- 3-(benzoyloxy)-8-methyl-8-azabicyclo[3.2.1]octane-2-carboxylate, is a plant-derived substance that is used in clinical situations as an anesthetic, but also is used illegally to experience extreme euphoria (a “high”).

24.3 Alkenes, Alkynes, And AromAtic HydrocArbons We next explore hydrocarbons with one or more C “ C bonds, called alkenes, and those with one or more C ‚ C bonds, called alkynes. Aromatic hydrocarbons have at least one planar ring with delocalized p electrons.

24.4 orgAnic FunctionAl groups We recognize that a central organizing principle of organic chemistry is the functional group, a group of atoms at which most of the compound’s chemical reactions occur.

24.1 generAl cHArActeristics oF orgAnic molecules We begin with a review of the structures and properties of organic compounds.

24.2 introduction to HydrocArbons We consider hydrocarbons, compounds containing only C and H, including the hydrocarbons called alkanes, which contain only single C ¬ C bonds. We also look at isomers, compounds with identical compositions but different molecular structures.

WhaT’s ahead

▶ BRAIN ON COCAINE. These positron-emitting tomography (PET) scans of the human brain show how rapidly glucose is metabolized in different parts of the brain. The top row of images show the brain of a normal person; the bottom two rows of images show the brain of a person who has taken cocaine, after 10 days and after 100 days (red = high glucose metabolism, yellow = medium, blue = low). Notice that glucose metabolism is suppressed in the brain of the person who has taken cocaine.

Understanding how these molecules exert their effects, and developing new molecules that can target disease and pain, is an enormous part of the modern chemical enterprise. This chapter is about the molecules, composed mainly of carbon, hydrogen, oxygen, and nitrogen, that bridge chemistry and biology.

More than 16 million carbon-containing compounds are known. Chemists make thousands of new compounds every year, about 90% of which contain carbon. The study of compounds whose molecules contain carbon constitutes the branch of chem-istry known as organic chemistry. This term arose from the eighteenth-century belief that organic compounds could be formed only by living (that is, organic) systems. This idea was disproved in 1828 by the German chemist Friedrich Wöhler when he synthe-sized urea 1H2NCONH22, an organic substance found in the urine of mammals, by heating ammonium cyanate 1NH4OCN2, an inorganic (nonliving) substance.

24.8 cArboHydrAtes We observe that carbohydrates are sugars and polymers of sugars used primarily as fuel by organisms (glucose) or as structural support in plants (cellulose).

24.9 lipids We find that lipids are a large class of molecules used primarily for energy storage in organisms.

24.10 nucleic Acids We learn that nucleic acids are polymers of nucleotides that contain an organism’s genetic information. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are polymers composed of nucleotides.

24.5 cHirAlity in orgAnic cHemistry We learn that compounds with nonsuperimposable mirror images are chiral and that chirality plays important roles in organic and biological chemistry.

24.6 introduction to biocHemistry We introduce the chemistry of living organisms, known as biochemistry, biological chemistry, or chemical biology. Important classes of compounds that occur in living systems are proteins, carbohydrates, lipids, and nucleic acids.

24.7 proteins We learn that proteins are polymers of amino acids linked with amide (also called peptide) bonds. Proteins are used by organisms for structural support, as molecular transporters and as catalysts for biochemical reactions.

1042 CHAPteR 24 the Chemistry of Life: Organic and Biological Chemistry

The study of the chemistry of living species is called biological chemistry, chemical biology, or biochemistry. In this chapter, we present some of the elementary aspects of both organic chemistry and biochemistry.

24.1 | General Characteristics of Organic Molecules

What is it about carbon that leads to the tremendous diversity in its compounds and allows it to play such crucial roles in biology and society? Let’s consider some general features of organic molecules and, as we do, review principles we learned in earlier chapters.

The Structures of Organic MoleculesBecause carbon has four valence electrons (3He42s22p2), it forms four bonds in virtu-ally all its compounds. When all four bonds are single bonds, the electron pairs are disposed in a tetrahedral arrangement. (Section 9.2) In the hybridization model, the carbon 2s and 2p orbitals are then sp3 hybridized. (Section 9.5) When there is one double bond, the arrangement is trigonal planar (sp2 hybridization). With a triple bond, it is linear (sp hybridization). Examples are shown in ▼ Figure 24.1.

Almost every organic molecule contains C ¬ H bonds. Because the valence shell of H can hold only two electrons, hydrogen forms only one covalent bond. As a result, hydrogen atoms are always located on the surface of organic molecules whereas the C ¬ C bonds form the backbone, or skeleton, of the molecule, as in the propane molecule:

C C CH H

H

H

H

H

H

H

Tetrahedral4 single bonds

sp3 hybridization

Trigonal planar2 single bonds1 double bond

sp2 hybridization

Linear1 single bond1 triple bond

sp hybridization

180°

120°109.5°

▲ Figure 24.1 Carbon geometries. The three common geometries around carbon are tetrahedral as in methane 1CH42, trigonal planar as in formaldehyde 1CH2O2, and linear as in acetonitrile 1CH3CN2. Notice that in all cases each carbon atom forms four bonds.

GO FiGureWhat is the geometry around the bottom carbon atom in acetonitrile?

seCtION 24.1 General Characteristics of Organic Molecules 1043

The Stabilities of Organic SubstancesCarbon forms strong bonds with a variety of elements, especially H, O, N, and the halo-gens. (Section 8.8) Carbon also has an exceptional ability to bond to itself, form-ing a variety of molecules made up of chains or rings of carbon atoms. Most reactions with low or moderate activation energy (Section 14.5) begin when a region of high electron density on one molecule encounters a region of low electron density on an-other molecule. The regions of high electron density may be due to the presence of a multiple bond or to the more electronegative atom in a polar bond. Because of their strength (the C ¬ C single bond enthalpy is 348 kJ>mol, the C ¬ H bond enthalpy is 413 kJ>mol Table 8.4) and lack of polarity, both C ¬ C single bonds and C ¬ H bonds are relatively unreactive. To better understand the implications of these facts, consider ethanol:

C C OH H

H

H

H

H

The differences in the electronegativity values of C (2.5) and O (3.5) and of O and H (2.1) indicate that the C ¬ O and O ¬ H bonds are quite polar. Thus, many reac-tions of ethanol involve these bonds while the hydrocarbon portion of the molecule remains intact. A group of atoms such as the C ¬ O ¬ H group, which determines how an organic molecule reacts (in other words, how the molecule functions), is called a functional group. The functional group is the center of reactivity in an organic molecule.

Give It Some ThoughtWhich bond is most likely to be the location of a chemical reaction: C “ N, C ¬ C, or C ¬ H?

Solubility and Acid–Base Properties of Organic SubstancesIn most organic substances, the most prevalent bonds are carbon–carbon and carbon–hydrogen, which are not polar. For this reason, the overall polarity of organic molecules is often low, which makes them generally soluble in nonpolar solvents and not very soluble in water. (Section 13.3) Organic molecules that are soluble in polar solvents are those that have polar groups on the molecule surface, such as glucose and ascorbic acid (▶ Figure 24.2). Organic molecules that have a long, nonpolar part bonded to a polar, ionic part, such as the stearate ion shown in Figure 24.2, function as surfactants and are used in soaps and detergents. (Section 13.6) The nonpolar part of the mol-ecule extends into a nonpolar medium such as grease or oil, and the polar part extends into a polar medium such as water.

Many organic substances contain acidic or basic groups. The most impor-tant acidic organic substances are the carboxylic acids, which bear the functional group ¬ COOH. (Sections 4.3 and 16.10) The most important basic organic sub-stances are amines, which bear the ¬ NH2, ¬ NHR, or ¬ NR2 groups, where R is an organic group made up of carbon and hydrogen atoms. (Section 16.7)

As you read this chapter, you will find many concept links ( ) to related mate-rials in earlier chapters. We strongly encourage you to follow these links and review the earlier material. Doing so will enhance your understanding and appreciation of organic chemistry and biochemistry.

Glucose (C6H12O6)

Ascorbic acid (HC6H7O6)

Stearate (C17H35COO−)

−

▲ Figure 24.2 Organic molecules that are soluble in polar solvents.

GO FiGureHow would replacing OH groups on ascorbic acid with CH3 groups affect the substance’s solubility in (a) polar solvents and (b) nonpolar solvents?


24.2 | introduction to hydrocarbonsBecause carbon compounds are so numerous, it is convenient to organize them into families that have structural similarities. The simplest class of organic compounds is the hydrocarbons, compounds composed of only carbon and hydrogen. The key struc-tural feature of hydrocarbons (and of most other organic substances) is the presence of stable carbon–carbon bonds. Carbon is the only element capable of forming stable, extended chains of atoms bonded through single, double, or triple bonds.

Hydrocarbons can be divided into four types, depending on the kinds of carbon–carbon bonds in their molecules. ▼ Table 24.1 shows an example of each type.

Alkanes contain only single C ¬ C bonds. Alkenes, also known as olefins, contain at least one C “ C double bond, and alkynes contain at least one C ‚ C triple bond. In aromatic hydrocarbons the carbon atoms are connected in a planar ring structure, joined by both s and delocalized p bonds between carbon atoms. (Section 8.6) Benzene 1C6H62 is the best-known example of an aromatic hydrocarbon.

Each type of hydrocarbon exhibits different chemical behaviors, as we will see shortly. The physical properties of all four types, however, are similar in many ways. Because hydrocarbon molecules are relatively nonpolar, they are almost completely insoluble in water but dissolve readily in nonpolar solvents. Their melting points and boiling points are determined by dispersion forces. (Section 11.2) As a result, hydro-carbons of very low molecular weight, such as C2H61bp = -89 °C2, are gases at room temperature; those of moderate molecular weight, such as C6H141bp = 69 °C2, are liquids; and those of high molecular weight, such as C22H461mp = 44 °C2, are solids.

▶ Table 24.2 lists the ten simplest alkanes. Many of these substances are familiar because they are used so widely. Methane is a major component of natural gas. Pro-pane is the major component of bottled gas used for home heating and cooking in areas where natural gas is not available. Butane is used in disposable lighters and in fuel can-isters for gas camping stoves and lanterns. Alkanes with 5 to 12 carbon atoms per mol-ecule are used to make gasoline. Notice that each succeeding compound in Table 24.2 has an additional CH2 unit.

Table 24.1 The Four hydrocarbon Types with Molecular examples

Alkane Ethane CH3CH3

Ethylene CH2 CH2Alkene

Alkyne

Aromatic

Acetylene CH CH

Benzene C6H6

C C

H

HHH

HH

C CHH

HH

C CH H

1.54 Å

120°

180°

1.34 Å

1.21 Å

122°

109.5°

1.39 Å

CH HC

C

CH H

HH

C

C

Type Example

seCtION 24.2 Introduction to Hydrocarbons 1045

The formulas for the alkanes given in Table 24.2 are written in a notation called condensed structural formulas. This notation reveals the way in which atoms are bonded to one another but does not require drawing in all the bonds. For example, the struc-tural formula and the condensed structural formulas for butane 1C4H102 are

C C C C

H3C CH2 CH2 CH3

CH3CH2CH2CH3

H H

H

H

H

H

H

H

H

H

or

Give It Some ThoughtHow many C ¬ H and C ¬ C bonds are formed by the middle carbon atom of propane?

Structures of AlkanesAccording to the VSEPR model, the molecular geometry about each carbon atom in an alkane is tetrahedral. (Section 9.2) The bonding may be described as involv-ing sp3@hybridized orbitals on the carbon, as pictured in ▶ Figure 24.3 for methane.

(Section 9.5)Rotation about a carbon–carbon single bond is relatively easy and occurs rapidly

at room temperature. To visualize such rotation, imagine grasping either methyl group of the propane molecule in ▶ Figure 24.4 and rotating the group relative to the rest of the molecule. Because motion of this sort occurs rapidly in alkanes, a long-chain alkane molecule is constantly undergoing motions that cause it to change its shape, something like a length of chain that is being shaken.

Structural IsomersThe alkanes in Table 24.2 are called straight-chain or linear hydrocarbons because all the carbon atoms are joined in a continuous chain. Alkanes consisting of four or more carbon atoms can also form branched chains, and when they do, they are called branched-chain hydrocarbons. (The branches in organic molecules are often called side chains.) Table 24.3, for example, shows all the straight-chain and branched-chain al-kanes containing four and five carbon atoms.

Compounds that have the same molecular formula but different bonding arrange-ments (and hence different structures) are called structural isomers. Thus, C4H10 has two structural isomers and C5H12 has three. The structural isomers of a given alkane differ slightly from one another in physical properties, as the melting and boiling points in Table 24.3 indicate.

Table 24.2 First Ten Members of the straight-Chain alkane series

Molecular Formula Condensed structural Formula Name

Boiling Point 1 °C 2

CH4 CH4 Methane -161C2H6 CH3CH3 Ethane -89C3H8 CH3CH2CH3 Propane -44C4H10 CH3CH2CH2CH3 Butane -0.5C5H12 CH3CH2CH2CH2CH3 Pentane 36C6H14 CH3CH2CH2CH2CH2CH3 Hexane 68C7H16 CH3CH2CH2CH2CH2CH2CH3 Heptane 98C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 Octane 125C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 Nonane 151C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 Decane 174

▲ Figure 24.3 Bonds about carbon in methane. This tetrahedral molecular geometry is found around all carbons in alkanes.

109.5°

s orbital of H

sp3 orbital of C

▲ Figure 24.4 Rotation about a C ¬ C bond occurs easily and rapidly in all alkanes.

Table 24.1 The Four hydrocarbon Types with Molecular examples


The number of possible structural isomers increases rapidly with the number of carbon atoms in the alkane. There are 18 isomers with the molecular formula C8H18, for example, and 75 with the molecular formula C10H22.

Give It Some ThoughtWhat evidence can you cite to support the fact that although isomers have the same molecular formula they are in fact different compounds?

Nomenclature of AlkanesIn the first column of Table 24.3, the names in parentheses are called the common names. The common name of the isomer with no branches begins with the letter n (in-dicating the “normal” structure). When one CH3 group branches off the major chain, the common name of the isomer begins with iso-, and when two CH3 groups branch off, the common name begins with neo-. As the number of isomers grows, however, it becomes impossible to find a suitable prefix to denote each isomer by a common name. The need for a systematic means of naming organic compounds was recognized

−138Butane(n-butane)

2-Methylpropane(isobutane)

Pentane(n-pentane)

2-Methylbutane(isopentane)

2,2-Dimethylpropane(neopentane)

HH H

H

CH

H

CH

H

CH

H

C CH3CH2CH2CH3

−159

H

H

C

H

H

C

H

CH H

H

H C H

CH3

CH3

CH3CH

−130

H

H

CH

H

CH

H

CH

H

CH

H

CH H CH3CH2CH2CH2CH3

−160H

H

H

H

C

H

HH

C

H

H

C

H C

H

C H CH3 CH2CH CH3

CH3

−16

−0.5

−12

+36

+28

+9H

H

H

CC

H

HH

H

H

C HH

CH

CH

CH3

CH3

CH3

CH3C

Systematic Name(Common Name) Structural Formula

Condensed StructuralFormula

Space-�llingModel

MeltingPoint (°C)

BoilingPoint (°C)

Table 24.3 isomers of C4h10 and C5h12


as early as 1892, when an organization called the International Union of Chemistry met in Geneva to formulate rules for naming organic substances. Since that time the task of updating the rules for naming compounds has fallen to the International Union of Pure and Applied Chemistry (IUPAC). Chemists everywhere, regardless of their nationality, subscribe to a common system for naming compounds.

The IUPAC names for the isomers of butane and pentane are the ones given first in Table 24.3. These systematic names, as well as those of other organic compounds, have three parts to them:

Whatsubstituents?

How manycarbons?

What family?

pre�x base suf�x

The following steps summarize the procedures used to name alkanes, which all have names ending with -ane. We use a similar approach to write the names of other organic compounds.

1. Find the longest continuous chain of carbon atoms, and use the name of this chain (given in Table 24.2) as the base name. Be careful in this step because the longest chain may not be written in a straight line, as in the following structure:

CHCH3 CH3

CH3CH2 CH2CH2

12

3 4 5 6

2-Methylhexane

Because the longest continuous chain contains six C atoms, this isomer is named as a substituted hexane. Groups attached to the main chain are called substituents because they are substituted in place of an H atom on the main chain. In this molecule, the CH3 group not enclosed by the blue outline is the only substituent in the molecule.

2. Number the carbon atoms in the longest chain, beginning with the end nearest a substituent. In our example, we number the C atoms beginning at the upper right because that places the CH3 substituent on C2 of the chain. (If we had num-bered from the lower right, the CH3 would be on C5.) The chain is numbered from the end that gives the lower number to the substituent position.

3. Name each substituent. A substituent formed by removing an H atom from an alkane is called an alkyl group. Alkyl groups are named by replacing the -ane end-ing of the alkane name with -yl. The methyl group 1CH32, for example, is derived from methane 1CH42 and the ethyl group 1C2H52 is derived from ethane 1C2H62. ▶ Table 24.4 lists six common alkyl groups.

4. Begin the name with the number or numbers of the carbon or carbons to which each substituent is bonded. For our compound, the name 2-methylhexane indi-cates the presence of a methyl group on C2 of a hexane (six-carbon) chain.

5. When two or more substituents are present, list them in alphabetical order. The presence of two or more of the same substituent is indicated by the prefixes di- (two), tri- (three), tetra- (four), penta- (five), and so forth. The prefixes are ignored in determining the alphabetical order of the substituents:

CH2CH3

CH3

CH3

CH3 CH3

CH3

CH2CH

CH CH

CH

1

2

34

5 6

3-Ethyl-2,4,5-trimethylheptane

7

Table 24.4 Condensed structural Formulas and Common Names for several alkyl Groups

Group Name

CH3 ¬ MethylCH3CH2 ¬ EthylCH3CH2CH2 ¬ PropylCH3CH2CH2CH2 ¬ Butyl

CH3

CH3

HC

Isopropyl

CH3

CH3

CCH3

tert-Butyl


sOLuTiONAnalyze We are given the condensed structural formula of an alkane and asked to give its name.Plan Because the hydrocarbon is an alkane, its name ends in -ane. The name of the parent hydrocarbon is based on the longest continu-ous chain of carbon atoms. Branches are alkyl groups, named after the number of C atoms in the branch and located by counting C atoms along the longest continuous chain.Solve The longest continuous chain of C atoms extends from the upper left CH3 group to the lower left CH3 group and is seven C atoms long:

CH35CH2

6CH27CH3

4CH

2CH21CH3 CH3

3CH

The parent compound is thus heptane. There are two methyl groups branching off the main chain. Hence, this compound is a dimethyl-heptane. To specify the location of the two methyl groups, we must number the C atoms from the end that gives the lower two numbers to the carbons bearing side chains. This means that we should start numbering at the upper left carbon. There is a methyl group on C3 and one on C4. The compound is thus 3,4-dimethylheptane.

saMPLe exerCise 24.1 Naming alkanes

Give the systematic name for the following alkane:

CH3 CH2

CH2CH3

CH

CH2CH3 CH3CH

Practice exercise 1What is the proper name of this compound?

CH3 CH3CH2 C

CH2

CH3

CH3

(a) 3-ethyl-3-methylbutane, (b) 2-ethyl-2-methylbutane,(c) 3,3-dimethylpentane, (d) isoheptane,(e) 1,2-dimethyl-neopentane.

Practice exercise 2Name the following alkane:

CH3 CH3

CH2

CH3

CH

CH

CH3

sOLuTiONAnalyze We are given the systematic name for a hydrocarbon and asked to write its condensed structural formula.Plan Because the name ends in -ane, the compound is an alkane, meaning that all the carbon–carbon bonds are single bonds. The parent hydrocarbon is pentane, indicating five C atoms (Table 24.2). There are two alkyl groups specified, an ethyl group (two carbon atoms, C2H5) and a methyl group (one carbon atom, CH3). Counting from left to right along the five-carbon chain, the name tells us that the ethyl group is attached to C3 and the methyl group is attached to C2.Solve We begin by writing five C atoms attached by single bonds. These represent the backbone of the parent pentane chain:

C ¬ C ¬ C ¬ C ¬ CWe next place a methyl group on the second C and an ethyl group on the third C of the chain. We then add hydrogens to all the other C atoms to make four bonds to each carbon:

CH3 CH2

CH2CH3

CH3CH CH

CH3

saMPLe exerCise 24.2 Writing Condensed structural Formulas

Write the condensed structural formula for 3-ethyl-2-methylpentane.

The formula can be written more concisely as

CH3CH1CH32CH1C2H52CH2CH3

where the branching alkyl groups are indicated in parentheses.

Practice exercise 1How many hydrogen atoms are in 2,2-dimethylhexane?(a) 6, (b) 8, (c) 16, (d) 18, (e) 20.

Practice exercise 2Write the condensed structural formula for 2,3-dimethylhexane.


CycloalkanesAlkanes that form rings, or cycles, are called cycloalkanes. As ▼ Figure 24.5 illus-trates, cycloalkane structures are sometimes drawn as line structures, which are poly-gons in which each corner represents a CH2 group. This method of representation is similar to that used for benzene rings. (Section 8.6) (Remember from our ben-zene discussion that in aromatic structures each vertex represents a CH group, not a CH2 group.)

Carbon rings containing fewer than five carbon atoms are strained because the C ¬ C ¬ C bond angles must be less than the 109.5 ° tetrahedral angle. The amount of strain increases as the rings get smaller. In cyclopropane, which has the shape of an equilateral triangle, the angle is only 60 °; this molecule is therefore much more reactive than propane, its straight-chain analog.

Give It Some ThoughtAre the C ¬ C bonds cyclopropane weaker than those in cyclohexane?

Reactions of AlkanesBecause they contain only C ¬ C and C ¬ H bonds, most alkanes are relatively unreac-tive. At room temperature, for example, they do not react with acids, bases, or strong oxidizing agents. Their low chemical reactivity, as noted in Section 24.1, is due primarily to the strength and lack of polarity of C ¬ C and C ¬ H bonds.

Alkanes are not completely inert, however. One of their most commercially important reactions is combustion in air, which is the basis of their use as fuels. (Section 3.2) For example, the complete combustion of ethane proceeds according to this highly exothermic reaction:

2 C2H61g2 + 7 O21g2 ¡ 4 CO21g2 + 6 H2O1l2 ∆H° = -2855 kJ

▲ Figure 24.5 Condensed structural formulas and line structures for three cycloalkanes.

GO FiGureThe general formula for straight-chain alkanes is CnH2n +2. What is the general formula for cycloalkanes?

Cyclopentane

CH2

CH2H2C

H2C

Cyclohexane

CH2CH2 CH2

CH2

CH2CH2

H2C

H2C

Cyclopropane

H2C CH2

Each vertex represents one CH2 group

Five vertices = �ve CH2 groups

Three vertices = three CH2 groups


Chemistry Put to Work

Gasoline

Petroleum, or crude oil, is a mixture of hydrocarbons plus smaller quantities of other organic compounds containing nitrogen, oxygen, or sulfur. The tremendous demand for petroleum to meet the world’s energy needs has led to the tapping of oil wells in such forbidding places as the North Sea and northern Alaska.

The usual first step in the refining, or processing, of petroleum is to separate it into fractions on the basis of boiling point (▼ Table 24.5). Because gasoline is the most commercially important of these fractions, various processes are used to maximize its yield.

Gasoline is a mixture of volatile alkanes and aromatic hydrocar-bons. In a traditional automobile engine, a mixture of air and gasoline vapor is compressed by a piston and then ignited by a spark plug. The burning of the gasoline should create a strong, smooth expansion of gas, forcing the piston outward and imparting force along the drive-shaft of the engine. If the gas burns too rapidly, the piston receives a single hard slam rather than a strong, smooth push. The result is a “knocking” or “pinging” sound and a reduction in the efficiency with which energy produced by the combustion is converted to work.

The octane number of a gasoline is a measure of its resistance to knocking. Gasolines with high octane numbers burn more smoothly and are thus more effective fuels (▶ Figure 24.6). Branched al-kanes and aromatic hydrocarbons have higher octane numbers than straight-chain alkanes. The octane number of gasoline is obtained by comparing its knocking characteristics with those of isooctane (2,2,4-trimethylpentane) and heptane. Isooctane is assigned an oc-tane number of 100, and heptane is assigned 0. Gasoline with the same

knocking characteristics as a mixture of 91% isooctane and 9% hep-tane, for instance, is rated as 91 octane.

The gasoline obtained by fractionating petroleum (called straight-run gasoline) contains mainly straight-chain hydrocarbons and has an octane number around 50. To increase its octane rating, it is subjected to a process called reforming, which converts the straight-chain alkanes into branched-chain ones.

Cracking is used to produce aromatic hydrocarbons and to convert some of the less-volatile fractions of petroleum into compounds suitable for use as automobile fuel. In cracking, the hydrocarbons are mixed with a catalyst and heated to 4009500 °C. The catalysts used are either clay min-erals or synthetic Al2O39SiO2 mixtures. In addition to forming molecules more suitable for gasoline, cracking results in the formation of such low-molecular-weight hydrocarbons as ethylene and propene. These substances are used in a variety of reactions to form plastics and other chemicals.

Adding compounds called either antiknock agents or octane en-hancers increases the octane rating of gasoline. Until the mid-1970s the principal antiknock agent was tetraethyl lead, 1C2H524Pb. It is no longer used, however, because of the environmental hazards of lead and because it poisons catalytic converters. (Section 14.7, “Cata-lytic Converters”) Aromatic compounds such as toluene 1C6H5CH32 and oxygenated hydrocarbons such as ethanol 1CH3CH2OH2 are now generally used as antiknock agents.

Related Exercises: 24.19 and 24.20

Table 24.5 hydrocarbon Fractions from Petroleum

Fractionsize range of Molecules

Boiling-Point range 1 °C 2 uses

Gas C1 to C5 -160 to 30 Gaseous fuel, production of H2

Straight-run gasoline

C5 to C12 30 to 200 Motor fuel

Kerosene, fuel oil

C12 to C18 180 to 400 Diesel fuel, furnace fuel, cracking

Lubricants C16 and up 350 and up LubricantsParaffins C20 and up Low-melting

solidsCandles, matches

Asphalt C36 and up Gummy residues

Surfacing roads

▲ Figure 24.6 Octane rating. The octane rating of gasoline measures its resistance to knocking when burned in an engine. The octane rating of the gasoline in the foreground is 89.

24.3 | alkenes, alkynes, and aromatic hydrocarbons

Because alkanes have only single bonds, they contain the largest possible number of hydro-gen atoms per carbon atom. As a result, they are called saturated hydrocarbons. Alkenes, alkynes, and aromatic hydrocarbons contain multiple carbon–carbon bonds (double,

seCtION 24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons 1051

triple, or delocalized p bonds). As a result, they contain less hydrogen than an alkane with the same number of carbon atoms. Collectively, they are called unsaturated hydrocarbons. On the whole, unsaturated molecules are more reactive than saturated ones.

AlkenesAlkenes are unsaturated hydrocarbons that contain at least one C “ C bond. The sim-plest alkene is CH2 “ CH2, called ethene (IUPAC) or ethylene (common name), which plays important roles as a plant hormone in seed germination and fruit ripening. The next member of the series is CH3 ¬ CH “ CH2, called propene or propylene. Alkenes with four or more carbon atoms have several isomers. For example, the alkene C4H8 has the four structural isomers shown in ▲ Figure 24.7. Notice both their structures and their names.

The names of alkenes are based on the longest continuous chain of carbon atoms that contains the double bond. The chain is named by changing the ending of the name of the corresponding alkane from -ane to -ene. The compound on the left in Figure 24.7, for example, has a double bond as part of a three-carbon chain; thus, the parent alkene is propene.

The location of the double bond along an alkene chain is indicated by a prefix number that designates the double-bond carbon atom that is nearest an end of the chain. The chain is always numbered from the end that brings us to the double bond sooner and hence gives the smallest-numbered prefix. In propene, the only possible location for the double bond is between the first and second carbons; thus, a prefix indicating its location is unnecessary. For butene (Figure 24.7), there are two possible positions for the double bond, either after the first carbon (1-butene) or after the sec-ond carbon (2-butene).

Give It Some ThoughtHow many distinct locations are there for a double bond in a five-carbon linear chain?

If a substance contains two or more double bonds, the location of each is indi-cated by a numerical prefix, and the ending of the name is altered to identify the number of double bonds: diene (two), triene (three), and so forth. For example, CH2 “ CH ¬ CH2 ¬ CH “ CH2 is 1,4-pentadiene.

The two isomers on the right in Figure 24.7 differ in the relative locations of their methyl groups. These two compounds are geometric isomers, compounds that have the same molecular formula and the same groups bonded to one another

C

Methylpropenebp −7 °C

C

HCH3

CH33 4 3

2 1 2 1

HC

1-Butenebp −6 °C

CHH CH3

CH3CH3CH2 CH34CH31

1

4H

C

cis-2-Butenebp +4 °C

C

HH H3 2

C

trans-2-Butenebp +1 °C

C3 2

H

Red numbers mark longest chain containing C=C

Methyl group branching off longest chain

No methyl group branching off

Methyl groups on same side of double bond

Methyl groups on opposite sides of double bond

▲ Figure 24.7 The alkene C4H8 has four structural isomers.

GO FiGureHow many isomers are there for propene, C3H6?


but differ in the spatial arrangement of these groups. (Section 23.4) In the cis isomer the two methyl groups are on the same side of the double bond, whereas in the trans isomer they are on opposite sides. Geometric isomers possess distinct physical properties and can differ significantly from each other in their chemical behavior.

Geometric isomerism in alkenes arises because, unlike the C ¬ C bond, the C “ C bond resists twisting. Recall from Section 9.6 that the double bond between two carbon atoms consists of a s and a p bond. ▲ Figure 24.8 shows a cis alkene. The carbon– carbon bond axis and the bonds to the hydrogen atoms and to the alkyl groups (des-ignated R) are all in a plane, and the p orbitals that form the p bond are perpendicular to that plane. As Figure 24.8 shows, rotation around the carbon–carbon double bond requires the p bond to be broken, a process that requires considerable energy (about 250 kJ>mol). Because rotation does not occur easily around the carbon–carbon bond, the cis and trans isomers of an alkene cannot readily interconvert and, therefore, exist as distinct compounds.

R R R

RH H

R H

H RH

H

Overlapping p orbitals create one π bond

Rotation has destroyed orbital overlap

Rotation around double bond requires considerable energy to break π bond

cis isomer trans isomerThis end of molecule rotated 90°

▲ Figure 24.8 Geometric isomers exist because rotation about a carbon–carbon double bond requires too much energy to occur at ordinary temperatures.

Draw all the structural and geometric isomers of pentene, C5H10, that have an unbranched hydrocarbon chain.

sOLuTiONAnalyze We are asked to draw all the isomers (both structural and geometric) for an alkene with a five-carbon chain.Plan Because the compound is named pentene and not pentadiene or pentatriene, we know that the five-carbon chain contains only one carbon–carbon double bond. Thus, we begin by plac-ing the double bond in various locations along the chain, remembering that the chain can be numbered from either end. After finding the different unique locations for the double bond, we consider whether the molecule can have cis and trans isomers.Solve There can be a double bond after either the first carbon (1-pentene) or second carbon (2-pentene). These are the only two possibilities because the chain can be numbered from either end. Thus, what we might erroneously call 3-pentene is actually 2-pentene, as seen by numbering the carbon chain from the other end:

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

5 4 3 2 1

5 4 3 2 1

renumbered as

renumbered as

C C C C C

C C C C C

C C C C C

C C C C C

C C C C C

C C C C C

saMPLe exerCise 24.3 drawing isomers


AlkynesAlkynes are unsaturated hydrocarbons containing one or more C ‚ C bonds. The sim-plest alkyne is acetylene 1C2H22, a highly reactive molecule. When acetylene is burned in a stream of oxygen in an oxyacetylene torch, the flame reaches about 3200 K. Be-cause alkynes in general are highly reactive, they are not as widely distributed in nature as alkenes; alkynes, however, are important intermediates in many industrial processes.

Alkynes are named by identifying the longest continuous chain containing the tri-ple bond and modifying the ending of the name of the corresponding alkane from -ane to -yne, as shown in Sample Exercise 24.4.

Because the first C atom in 1-pentene is bonded to two H atoms, there are no cis–trans isomers. There are cis and trans isomers for 2-pentene, however. Thus, the three isomers for pentene are

CH2H

CH3

CH3

CH3 CH3CHCH2 CH2 CH2 CH3

HH

CH2

trans-2-Pentene

1-Pentene

cis-2-PenteneC C

C C

H

(You should convince yourself that cis-3-pentene is identical to cis-2-pentene and trans-3-pentene is identical to trans-2-pentene. However, cis-2-pentene and trans-2-pentene are the correct names because they have smaller numbered prefixes.)

Practice exercise 1Which compound does not exist?(a) 1,2,3,4,5,6,7-octaheptaene, (b) cis-2-butane, (c) trans-3-hexene,(d) 1-propene, (e) cis-4-decene.

Practice exercise 2How many straight-chain isomers are there of hexene, C6H12?

Name the following compounds:

C C

C CH

HH

CHCH3CH2CH2(a) (b)CH3

CH3

CH3CH2CH2CH

CH2CH2CH3

sOLuTiONAnalyze We are given the condensed structural formulas for an alkene and an alkyne and asked to name the compounds.Plan In each case, the name is based on the number of carbon atoms in the longest continu-ous carbon chain that contains the multiple bond. In the alkene, care must be taken to indicate whether cis–trans isomerism is possible and, if so, which isomer is given.Solve

(a) The longest continuous chain of carbons that contains the double bond is seven carbons long, so the parent hydrocarbon is heptene. Because the double bond begins at carbon 2 (numbering from the end closer to the double bond), we have 2-heptene. With a methyl group at carbon atom 4, we have 4-methyl-2-heptene. The geometrical configuration at the double bond is cis (that is, the alkyl groups are bonded to the double bond on the same side). Thus, the full name is 4-methyl-cis-2-heptene.

saMPLe exerCise 24.4 Naming unsaturated hydrocarbons


Addition Reactions of Alkenes and AlkynesThe presence of carbon–carbon double or triple bonds in hydrocarbons markedly in-creases their chemical reactivity. The most characteristic reactions of alkenes and al-kynes are addition reactions, in which a reactant is added to the two atoms that form the multiple bond. A simple example is the addition of elemental bromine to ethylene to produce 1,2,-dibromoethane:

H2C CH2 Br2

Br

H2C

Br

CH2+

[24.1]

The p bond in ethylene is broken and the electrons that formed the bond are used to form two s bonds to the two bromine atoms. The s bond between the carbon atoms is retained.

Addition of H2 to an alkene converts it to an alkane:

CH3CH “ CHCH3 + H2 ¡Ni, 500 °C CH3CH2CH2CH3 [24.2]

The reaction between an alkene and H2, referred to as hydrogenation, does not occur readily at ordinary temperatures and pressures. One reason for the lack of reactivity of H2 toward alkenes is the stability of the H2 bond. To promote the reaction, the reaction temperature must be raised 1500 °C2, and a catalyst (such as Ni) is used to assist in rup-turing the H ¬ H bond. We write such conditions over the reaction arrow to indicate they must be present in order for the reaction to occur. The most widely used catalysts are finely divided metals on which H2 is adsorbed. (Section 14.7)

Hydrogen halides and water can also add to the double bond of alkenes, as in these reactions of ethylene: CH2 “ CH2 + HBr ¡ CH3CH2Br [24.3]

CH2 “ CH2 + H2O ¡H2SO4 CH3CH2OH [24.4]

The addition of water is catalyzed by a strong acid, such as H2SO4.The addition reactions of alkynes resemble those of alkenes, as shown in these

examples:

CH3C

CH3

CH3

CCH3 Cl2

Cl

ClC C

2-Butyne trans-2,3-Dichloro-2-butene

+

[24.5]

CH3C CCH3 2 Cl2 CH3 CH3

Cl

Cl

C

Cl

Cl

C

2-Butyne 2,2,3,3-Tetrachlorobutane

+

[24.6]

—¡

(b) The longest continuous chain containing the triple bond has six carbons, so this compound is a derivative of hexyne. The triple bond comes after the first carbon (numbering from the right), making it 1-hexyne. The branch from the hexyne chain contains three carbon atoms, making it a propyl group. Because this substituent is located on C3 of the hexyne chain, the molecule is 3-propyl-1-hexyne.

Practice exercise 1If a compound has two carbon–carbon triple bonds and one carbon–carbon double bond, what class of compound is it?(a) an eneyne, (b) a dieneyne, (c) a trieneyne, (d) an enediyne, (e) an enetriyne.

Practice exercise 2Draw the condensed structural formula for 4-methyl-2-pentyne.


sOLuTiONAnalyze We are asked to predict the compound formed when a par-ticular alkene undergoes hydrogenation (reaction with H2) and to write the condensed structural formula of the product.

Plan To determine the condensed structural formula of the product, we must first write the condensed structural formula or Lewis struc-ture of the reactant. In the hydrogenation of the alkene, H2 adds to the double bond, producing an alkane.

Solve The name of the starting compound tells us that we have a chain of five C atoms with a double bond at one end (position 1) and a methyl group on C3:

CH2 CH2CH CH CH3

CH3

saMPLe exerCise 24.5 Predicting the Product of an addition reaction

Write the condensed structural formula for the product of the hydrogenation of 3-methyl-1-pentene.

Hydrogenation—the addition of two H atoms to the carbons of the double bond—leads to the following alkane:

CH2CH CH3

CH3

CH3 CH2

Comment The longest chain in this alkane has five carbon atoms; the product is therefore 3-methylpentane.

Practice exercise 1What product is formed from the hydrogenation of 2-methylpro-pene? (a) propane, (b) butane, (c) 2-methylbutane, (d) 2-methyl-propane, (e) 2-methylpropyne.

Practice exercise 2Addition of HCl to an alkene forms 2-chloropropane. What is the alkene?

The second, faster step is addition of Br- to the positively charged carbon. The bromide ion donates a pair of electrons to the carbon, forming the C ¬ Br bond:

CH3CH CH3CH CH2CH3 CH3CHCH2CH3

Br Br

CH2CH3 Br+

δ−

++ −

δ

CH3CH CH3CH CH2CH3 CH3CHCH2CH3

Br Br

CH2CH3 Br+

δ−

++ −

δ

[24.8]

Because the rate-determining step involves both the alkene and the acid, the rate law for the reaction is second order, first order in the alkene and first order in HBr:

Rate = -∆3CH3CH “ CHCH34

∆t= k3CH3CH “ CHCH343HBr4

[24.9]

The energy profile for the reaction is shown in Figure 24.9. The first energy maximum represents the transition state in the first step, and the second maximum represents the transition state in the second step. The energy minimum represents the energies of the intermediate species, CH3C

+

H ¬ CH2CH3 and Br-.

a Closer Look

Mechanism of Addition Reactions

As the understanding of chemistry has grown, chemists have ad-vanced from simply cataloging reactions that occur to explaining how they occur, by drawing the individual steps of a reaction based upon experimental and theoretical evidence. The sum of these steps is termed a reaction mechanism. (Section 14.6)

The addition reaction between HBr and an alkene, for in-stance, is thought to proceed in two steps. In the first step, which is rate determining (Section 14.6), HBr attacks the electron-rich double bond, transferring a proton to one of the double-bond car-bons. In the reaction of 2-butene with HBr, for example, the first step is

CH3CH CH3CH CHCH3 CH3CH CH2CH3

H

Br

CHCH3 HBr Brδ+

δ−

++

−+

CH3CH CH3CH CHCH3 CH3CH CH2CH3

H

Br

CHCH3 HBr Brδ+

δ−

++

−+ [24.7]

The electron pair that formed the p bond is used to form the new C ¬ H bond.


Aromatic HydrocarbonsThe simplest aromatic hydrocarbon, benzene 1C6H62, is shown in ▼ Figure 24.10 along with some other aromatic hydrocarbons. Benzene is the most important aromatic hy-drocarbon, and most of our discussion focuses on it.

To show electron movement in reactions like these, chemists of-ten use curved arrows pointing in the direction of electron flow. For the addition of HBr to 2-butene, for example, the shifts in electron positions are shown as

CH3 CHCH3CH CH3slow CH3C

H

C

H

H

CH3 CH3C

H

C

H

H

fastBrH Br

Br

++

+ −

CH3 CHCH3CH CH3slow CH3C

H

C

H

H

CH3 CH3C

H

C

H

H

fastBrH Br

Br

++

+ −

Reaction pathway

ReactantsIntermediates

First transitionstate Second transition

state

ProductCH3CH CHCH3 CH3CHCH2CH3

CH3CH CH2CH3

+ HBr

Br

+ Br−

Ene

rgy

+

▲ Figure 24.9 Energy profile for addition of HBr to 2-butene. The two maxima tell you that this is a two-step mechanism.

GO FiGureWhat features of an energy profile allow you to distinguish between an intermediate and a transition state?

▲ Figure 24.10 Line formulas and common names of several aromatic compounds. The aromatic rings are represented by hexagons with a circle inscribed inside to denote delocalized p bonds. Each vertex represents a carbon atom. Each carbon is bound to three other atoms—either three carbons, or two carbons and a hydrogen—so that each carbon has the requisite four bonds.

Benzene AnthraceneNaphthalene Toluene(Methylbenzene)

CH3

Pyrene

Stabilization of p Electrons by DelocalizationIf you draw one Lewis structure for benzene, you draw a ring that contains three CC double bonds and three single CC bonds. (Section 8.6) You might therefore expect benzene to resemble the alkenes and to be highly reactive. Benzene and the other aro-matic hydrocarbons, however, are much more stable than alkenes because the p elec-trons are delocalized in the p orbitals. (Section 9.6)

We can estimate the stabilization of the p electrons in benzene by comparing the energy required to form cyclohexane by adding hydrogen to benzene, to cyclohexene (one double bond) and to 1,4-cyclohexadiene (two double bonds):

∆H° = −208 kJ/mol3 H2+

H2+ ∆H° −120 kJ/mol=

2 H2+ ∆H° −232 kJ/mol=


From the second and third reactions, it appears that the energy required to hydro-genate each double bond is roughly 118 kJ/mol for each bond. Benzene contains the equivalent of three double bonds. We might expect, therefore, the energy required to hydrogenate benzene to be about 3 times -118, or -354 kJ>mol, if benzene behaved as though it were “cyclohexatriene,” that is, if it behaved as though it had three iso-lated double bonds in a ring. Instead, the energy released is 146 kJ less than this, indicating that benzene is more stable than would be expected for three double bonds. The difference of 146 kJ>mol between the expected heat of hydrogenation 1-354 kJ>mol2 and the observed heat of hydrogenation, 1-208 kJ>mol2 is due to stabilization of the p electrons through delocalization in the p orbitals that extend around the ring. Chemists call this stabilization energy the resonance energy.

Substitution ReactionsAlthough aromatic hydrocarbons are unsaturated, they do not readily undergo ad-dition reactions. The delocalized p bonding causes aromatic compounds to behave quite differently from alkenes and alkynes. Benzene, for example, does not add Cl2 or Br2 to its double bonds under ordinary conditions. In contrast, aromatic hydro-carbons undergo substitution reactions relatively easily. In a substitution reaction, one hydrogen atom of the molecule is removed and replaced (substituted) by an-other atom or group of atoms. When benzene is warmed in a mixture of nitric and sulfuric acids, for example, one of the benzene hydrogens is replaced by the nitro group, NO2:

HNO3 H2O

NO2H2SO4+ +

Benzene Nitrobenzene

[24.10]

More vigorous treatment results in substitution of a second nitro group into the molecule:

HNO3

NO2

+ H2O

NO2

NO2

H2SO4 +

[24.11]

There are three isomers of benzene that contain two nitro groups—the 1,2- or ortho- isomer, the 1,3- or meta-isomer, and the 1,4- or para-isomer of dinitro-benzene:

NO2

NO2

NO2

NO2

NO2

NO2

ortho-Dinitrobenzene1,2-Dinitrobenzene

mp 118 °C

meta-Dinitrobenzene1,3-Dinitrobenzene

mp 90 °C

para-Dinitrobenzene1,4-Dinitrobenzene

mp 174 °C

12

34

5

61

2

34

5

61

2

34

5

6


In the reaction of Equation 24.11, the principal product is the meta isomer.Bromination of benzene, carried out with FeBr3 as a catalyst, is another substitu-

tion reaction:

Br2 HBr

Br

+ +FeBr3

Benzene Bromobenzene

[24.12]

In a similar substitution reaction, called the Friedel–Crafts reaction, alkyl groups can be substituted onto an aromatic ring by reacting an alkyl halide with an aromatic compound in the presence of AlCl3 as a catalyst:

CH3CH2

CH2CH3

Cl HCl+ +AlCl3

Benzene Ethylbenzene

[24.13]

Give It Some ThoughtWhen the aromatic hydrocarbon naphthalene, shown in Figure 24.10, reacts with nitric and sulfuric acids, two compounds containing one nitro group are formed. Draw the structures of these two compounds.

24.4 | Organic Functional GroupsThe C “ C double bonds of alkenes and C ‚ C triple bonds of alkynes are just two of many functional groups in organic molecules. As noted earlier, these functional groups each undergo characteristic reactions, and the same is true of all other func-tional groups. Each kind of functional group often undergoes the same kinds of reac-tions in every molecule, regardless of the size and complexity of the molecule. Thus, the chemistry of an organic molecule is largely determined by the functional groups it contains.

▶ Table 24.6 lists the most common functional groups. Notice that, except for C “ C and C ‚ C, they all contain either O, N, or a halogen atom, X.

We can think of organic molecules as being composed of functional groups bonded to one or more alkyl groups. The alkyl groups, which are made of C ¬ C and C ¬ H single bonds, are the less reactive portions of the molecules. In describing gen-eral features of organic compounds, chemists often use the designation R to represent any alkyl group: methyl, ethyl, propyl, and so on. Alkanes, for example, which contain no functional group, are represented as R ¬ H. Alcohols, which contain the functional group ¬ OH, are represented as R ¬ OH. If two or more different alkyl groups are present in a molecule, we designate them R, R′, R″, and so forth.

AlcoholsAlcohols are compounds in which one or more hydrogens of a parent hydro-carbon have been replaced by the functional group ¬ OH, called either the

seCtION 24.4 Organic Functional Groups 1059

O

C C C C

C C C C

C

OC

C

C OC C

H H

HH

H H

H OC H

O

O

O

O

NC

C C

C

H

N

HC

O

C

O O

C

O

C

O

C

O

C

O

O

C

O

C

O

HC

X

(X halogen)

C

H

HH

H

H

H

H

H

H

H H

H

H

H

NC

H

H

H C

H

H

H HC

H

H

C

H

H

H C

H

H

H

HC

H

H

C

H

H

C

H H

H

H

HHH

HC

C H

N

Cl

Alkene -ene Ethene (Ethylene)

Alkyne -yneEthyne (Acetylene)

Alcohol -ol Methanol (Methyl alcohol)

Ether ether Dimethyl ether

Alkyl halideor haloalkane

-ide

Amine -amine Ethylamine

Aldehyde -al Ethanal (Acetaldehyde)

Ketone -one Propanone (Acetone)

Carboxylic acid

-oic acid Ethanoic acid (Acetic acid)

Ester -oate Methyl ethanoate (Methyl acetate)

Amide -amide Ethanamide (Acetamide)

Chloromethane (Methyl chloride)

FunctionalGroup

CompoundType

Suf�x orPre�x

StructuralFormula

Ball-and-stickModel

Example

Systematic Name(common name)

Table 24.6 Common Functional Groups


hydroxyl group or the alcohol group. Note in ▲ Figure 24.11 that the name for an alco-hol ends in -ol. The simple alcohols are named by changing the last letter in the name of the corresponding alkane to -ol—for example, ethane becomes ethanol. Where neces-sary, the location of the OH group is designated by a numeric prefix that indicates the number of the carbon atom bearing the OH group.

The O ¬ H bond is polar, so alcohols are more soluble in polar solvents than are hydrocarbons. The ¬ OH functional group can also participate in hydrogen bond-ing. As a result, the boiling points of alcohols are higher than those of their parent alkanes.

◀ Figure 24.12 shows several commercial products that consist entirely or in large part of an alcohol.

The simplest alcohol, methanol (methyl alcohol), has many industrial uses and is produced on a large scale by heating carbon monoxide and hydrogen under pressure in the presence of a metal oxide catalyst:

CO1g2 + 2 H21g2 ¡200 - 300 atm

400 °C CH3OH1g2 [24.14]

Because methanol has a very high octane rating as an automobile fuel, it is used as a gasoline additive and as a fuel in its own right.

Ethanol (ethyl alcohol, C2H5OH) is a product of the fermentation of carbohydrates such as sugars and starches. In the absence of air, yeast cells convert these carbohy-drates into ethanol and CO2:

C6H12O61aq2 ¡yeast 2 C2H5OH1aq2 + 2 CO21g2 [24.15]

In the process, the yeast cells derive energy necessary for growth. This reaction is carried out under carefully controlled conditions to produce beer, wine, and other beverages in which ethanol (called just “alcohol” in everyday language) is the active ingredient.

The simplest polyhydroxyl alcohol (an alcohol containing more than one OH group) is 1,2-ethanediol (ethylene glycol, HOCH2CH2OH), the major ingredient in automobile antifreeze. Another common polyhydroxyl alcohol is 1,2,3-propanetriol 1glycerol, HOCH2CH1OH2CH2OH2, a viscous liquid that dissolves readily in water and is used in cosmetics as a skin softener and in foods and candies to keep them moist.

Phenol is the simplest compound with an OH group attached to an aromatic ring. One of the most striking effects of the aromatic group is the greatly increased acidity of the OH group. Phenol is about 1 million times more acidic in water than a nonaromatic alcohol. Even so, it is not a very strong acid 1Ka = 1.3 * 10-102. Phenol is used indus-trially to make plastics and dyes, and as a topical anesthetic in throat sprays.

▲ Figure 24.11 Condensed structural formulas of six important alcohols. Common names are given in blue.

2-PropanolIsopropyl alcohol;

rubbing alcohol

Phenol

OH

OH

CH3 CH

2-Methyl-2-propanolt-Butyl alcohol

1,2-EthanediolEthylene glycol

OH

CH3 CH3CH3

CH3

C

1,2,3-PropanetriolGlycerol; glycerin

2,15-dimethyl-14-(1,5-dimethylhexyl)tetracyclo[8.7.0.02,7.011,15]heptacos-

7-en-5-olCholesterol

OH OH OH

CH2 CH

HO

CH2

OH

CH2

OH

CH2

H3C

H3C

H3C

CH3

CH3H H

▲ Figure 24.12 Everyday alcohols. Many of the products we use every day—from rubbing alcohol to hair spray and antifreeze—are composed either entirely or mainly of alcohols.


Cholesterol, shown in Figure 24.11, is a biochemically important alcohol. The OH group forms only a small component of this molecule, so cholesterol is only slightly soluble in water (2.6 g/L of H2O). Cholesterol is a normal and essential component of our bodies; when present in excessive amounts, however, it may precipitate from solu-tion. It precipitates in the gallbladder to form crystalline lumps called gallstones. It may also precipitate against the walls of veins and arteries and thus contribute to high blood pressure and other cardiovascular problems.

EthersCompounds in which two hydrocarbon groups are bonded to one oxygen are called ethers. Ethers can be formed from two molecules of alcohol by eliminating a molecule of water. The reaction is catalyzed by sulfuric acid, which takes up water to remove it from the system:

CH3CH2 ¬ OH + H ¬ OCH2CH3 ¡H2SO4 CH3CH2 ¬ O ¬ CH2CH3 + H2O[24.16]

A reaction in which water is eliminated from two substances is called a condensation reaction. (Sections 12.8 and 22.8)

Both diethyl ether and the cyclic ether tetrahydrofuran, shown below, are common solvents for organic reactions. Diethyl ether was formerly used as an anesthetic (known simply as “ether” in that context), but it had significant side effects.

Diethyl ether Tetrahydrofuran (THF)

CH3CH2 CH2CH3CH2

CH2

CH2

CH2O

O

Aldehydes and KetonesSeveral of the functional groups listed in Table 24.6 contain the carbonyl group, C “ O. This group, together with the atoms attached to its carbon, defines several im-portant functional groups that we consider in this section.

In aldehydes, the carbonyl group has at least one hydrogen atom attached:

Formaldehyde AcetaldehydeMethanal

O O

CH HEthanal

C HCH3

In ketones, the carbonyl group occurs at the interior of a carbon chain and is therefore flanked by carbon atoms:

Propanone

O

C2-Butanone

CH3 CH3

O

CCH3 CH2CH3

Acetone Methyl ethyl ketoneTestosterone

O

H3C

H3COH

C

The systematic names of aldehydes contain -al and that ketone names contain -one. Notice that testosterone has both alcohol and ketone groups; the ketone functional group dominates the molecular properties. Therefore, testosterone is considered a ketone first and an alcohol second, and its name reflects its ketone properties.


Many compounds found in nature contain an aldehyde or ketone functional group. Vanilla and cinnamon flavorings are naturally occurring aldehydes. Two iso-mers of the ketone carvone impart the characteristic flavors of spearmint leaves and caraway seeds.

Ketones are less reactive than aldehydes and are used extensively as solvents. Ace-tone, the most widely used ketone, is completely miscible with water, yet it dissolves a wide range of organic substances.

Carboxylic Acids and EstersCarboxylic acids contain the carboxyl functional group, often written COOH.

(Section 16.10) These weak acids are widely distributed in nature and are com-mon in citrus fruits. (Section 4.3) They are also important in the manufacture of polymers used to make fibers, films, and paints. ▼ Figure 24.13 shows the formulas of several carboxylic acids.

The common names of many carboxylic acids are based on their historical ori-gins. Formic acid, for example, was first prepared by extraction from ants; its name is derived from the Latin word formica, “ant.”

Carboxylic acids can be produced by oxidation of alcohols. Under appropriate conditions, the aldehyde may be isolated as the first product of oxidation, as in the sequence

Ethanol

O

CH3CH2OH (O)Acetaldehyde

CH3CH H2O+ +

[24.17]

O

Acetaldehyde

CH3CH

O

Acetic acid

CH3COH(O)+

[24.18]

where (O) represents any oxidant that can provide oxygen atoms. The air oxidation of ethanol to acetic acid is responsible for causing wines to turn sour, producing vinegar.

Methanoic acidFormic acid

Citric acid

Acetylsalicylic acidAspirin

CH2 CH2

H

CH3

CH3

OHLactic acid

C

O

C

O

OHC

O

C

O

O

CHCH3 OH

OHOH

C

Ethanoic acidAcetic acid

Phenyl methanoic acidBenzoic acid

HO

C

O

HO C

O

OH

C OH

O

C OH

O C

O

▲ Figure 24.13 Structural formulas of common carboxylic acids. The monocarboxylic acids are generally referred to by their common names, given in blue type.

GO FiGureWhich of these substances have both a carboxylic acid functional group and an alcohol functional group?


The oxidation processes in organic compounds are related to those oxidation reac-tions we studied in Chapter 20. Instead of counting electrons, the number of C ¬ O bonds is usually considered to show the extent of oxidation of similar compounds. For example, methane can be oxidized to methanol, then formaldehye (methanal), then formic acid (methanoic acid):

Formaldehyde

O

CH HFormic acid

O

CH O HMethanol

H

H

CH OH

Methane

H

H

CH H

From methanol to formic acid, the number of C ¬ O bonds increases from 0 to 3 (double bonds are counted as two). If you were to calculate the oxidation state of car-bon in these compounds, it would range from -4 in methane (if H’s are counted as +1) to +2 in formic acid, which is consistent with carbon being oxidized. The ultimate oxi-dation product of any organic compound, then, is CO2, which is indeed the product of combustion reactions of carbon-containing compounds (CO2 has 4 C ¬ O bonds, and C has the oxidation state +4).

Give It Some ThoughtWhat chemical process is happening when formic acid is converted back to methane?

Aldehydes and ketones can be prepared by controlled oxidation of alcohols. Com-plete oxidation results in formation of CO2 and H2O, as in the burning of methanol:

CH3OH1g2 + 32 O21g2 ¡ CO21g2 + 2 H2O1g2

Controlled partial oxidation to form other organic substances, such as aldehydes and ketones, is carried out by using various oxidizing agents, such as air, hydrogen peroxide 1H2O22, ozone 1O32, and potassium dichromate 1K2Cr2O72.

Give It Some ThoughtWrite the condensed structural formula for the acid that would result from oxidation of the alcohol

CHOHCH2

CH2CH2

CH2

Acetic acid can also be produced by the reaction of methanol with carbon monox-ide in the presence of a rhodium catalyst:

+

O

C OHCOCH3OHcatalyst

CH3 [24.19]

This reaction is not an oxidation; it involves, in effect, the insertion of a carbon mon-oxide molecule between the CH3 and OH groups. A reaction of this kind is called carbonylation.

Carboxylic acids can undergo condensation reactions with alcohols to form esters:

O

C

Acetic acid Ethanol Ethyl acetate

HO O H2OCH2CH3 CH3

O

CCH3 CH2CH3OH ++

[24.20]


Esters are compounds in which the H atom of a carboxylic acid is replaced by a carbon-containing group:

O

C O C

Give It Some ThoughtWhat is the difference between an ether and an ester?

The name of any ester consists of the name of the group contributed by the alco-hol followed by the name of the group contributed by the carboxylic acid, with the -ic replaced by -ate. For example, the ester formed from ethyl alcohol, CH3CH2OH, and butyric acid, CH31CH222 COOH, is

OCH2CH3CH3CH2CH2C

O

Ethyl butyrate

Notice that the chemical formula generally has the group originating from the acid written first, which is opposite of the way the ester is named. Another example is iso-amyl acetate, the ester formed from acetic acid and isoamyl alcohol. Isoamyl acetate smells like bananas or pears.

O C CH3(CH3)2CHCH2CH2

O

Isoamyl Acetate

Many esters such as isoamyl acetate have pleasant odors and are largely respon-sible for the pleasing aromas of fruit.

An ester treated with an acid or a base in aqueous solution is hydrolyzed; that is, the molecule is split into an alcohol and a carboxylic acid or its anion:

O

CMethyl propionate

CH3CH2

CH3CH2

CH3 OHO

O

C OPropionate Methanol

CH3OH

−

−

+

+

[24.21]

The hydrolysis of an ester in the presence of a base is called saponification, a term that comes from the Latin word for soap, sapon. Naturally occurring esters include fats and oils, and in making soap an animal fat or a vegetable oil is boiled with a strong base. The resultant soap consists of a mixture of salts of long-chain carboxylic acids (called fatty acids), which form during the saponification reaction.


In a basic aqueous solution, esters react with hydroxide ion to form the salt of the carboxylic acid and the alcohol from which the ester is constituted. Name each of the following esters, and indicate the products of their reaction with aqueous base.

C

O

C

O

(a) OCH2CH3 O(b) CH3CH2CH2

sOLuTiONAnalyze We are given two esters and asked to name them and to predict the products formed when they undergo hydrolysis (split into an alcohol and carboxylate ion) in basic solution.Plan Esters are formed by the condensation reaction between an alcohol and a carboxylic acid. To name an ester, we must analyze its structure and determine the identities of the alcohol and acid from which it is formed. We can identify the alcohol by adding an OH to the alkyl group attached to the O atom of the carboxyl (COO) group. We can identify the acid by adding an H to the O atom of the carboxyl group. We have learned that the first part of an ester name indicates the alcohol portion and the second indicates the acid portion. The name conforms to how the ester undergoes hydrolysis in base, reacting with base to form an alcohol and a carboxylate anion.Solve

(a) This ester is derived from ethanol 1CH3CH2OH2 and benzoic acid 1C6H5COOH2. Its name is therefore ethyl benzoate. The net ionic equation for reaction of ethyl benzoate with hydroxide ion is

C OCH2CH3(aq)

HOCH2CH3(aq)

OH−(aq)

O

C

O

O−(aq)

+

+

The products are benzoate ion and ethanol.(b) This ester is derived from phenol 1C6H5OH2 and butanoic acid (commonly called butyric

acid) 1CH3CH2CH2COOH2. The residue from the phenol is called the phenyl group. The ester is therefore named phenyl butyrate or phenyl butanoate. The net ionic equation for the reaction of phenyl butyrate with hydroxide ion is

CH3CH2CH2C (aq)

(aq)

O

O

CH3CH2CH2C

O

OH−(aq)

HOO−(aq)

+

+

The products are butyrate ion and phenol.

Practice exercise 1For the generic ester RC(O)OR’, which bond will hydrolyze under basic conditions?(a) the R ¬ C bond (b) the C “ O bond (c) the C ¬ O bond (d) the O ¬ R> bond(e) more than one of the above

Practice exercise 2Write the condensed structural formula for the ester formed from propyl alcohol and propi-onic acid.

saMPLe exerCise 24.6 Naming esters and Predicting hydrolysis

Products


Cocaine Morphine Codeine

O

OO

HO

HO HO

O

O

N

N

H3CCH3

CH3

O

O

N CH3

H3C

Amines and AmidesAmines are compounds in which one or more of the hydrogens of ammonia 1NH32 are replaced by an alkyl group:

(CH3)3NCH3CH2NH2 NH2

PhenylamineAniline

TrimethylamineEthylamine

Amines are the most common organic bases. (Section 16.7) As we saw in the Chemistry Put to Work box in Section 16.8, many pharmaceutically active compounds are complex amines:

An amine with at least one H bonded to N can undergo a condensation reaction with a carboxylic acid to form an amide, which contains the carbonyl group 1C “ O2 attached to N (Table 24.6):

CH3C OH H N(CH3)2 N(CH3)2 H2O

O

CH3C

O

+ + [24.22]

We may consider the amide functional group to be derived from a carboxylic acid with an NRR′, NH2 or NHR′ group replacing the OH of the acid, as in these examples:

CH3C

O O

NH2

CH3C

O

NH

C

EthanamideAcetamide

PhenylmethanamideBenzamide

NH2

N-(4-hydroxyphenyl)ethanamideAcetaminophen

OH

The amide linkage

O

CR R′N

H

where R and R′ are organic groups, is the key functional group in proteins, as we will see in Section 24.7.

seCtION 24.5 Chirality in Organic Chemistry 1067

24.5 | Chirality in Organic ChemistryA molecule possessing a nonsuperimposable mirror image is termed chiral (Greek cheir, “hand”). (Section 23.4) Compounds containing carbon atoms with four dif-ferent attached groups are inherently chiral. A carbon atom with four different attached groups is called a chiral center. For example, consider 2-bromopentane:

CH3 CH2CH2CH3C

H

Br

All four groups attached to C2 are different, making that carbon a chiral center. ▼ Figure 24.14 illustrates the nonsuperimposable mirror images of this molecule. Imagine moving the molecule shown to the left of the mirror over to the right of the mirror. If you then turn it in every possible way, you will conclude that it cannot be superimposed on the molecule shown to the right of the mirror. Nonsuperimposable mirror images are called either optical isomers or enantiomers. (Section 23.4) Organic chemists use the labels R and S to distinguish the two forms. We need not go into the rules for deciding on the labels.

Mirror

▲ Figure 24.14 The two enantiomeric forms of 2-bromopentane. The mirror-image isomers are not superimposable on each other.

GO FiGureIf you replace Br with CH3, will the compound be chiral?

▲ Figure 24.15 (R)-Albuterol. This compound, which acts as a bronchodilator in patients with asthma, is one member of an enantiomer pair. The other member, (S)-albuterol, has the OH group pointing down, and does not have the same physiological effect.

HOH2C

HO

OHHN

The two members of an enantiomer pair have identical physical properties and identical chemical properties when they react with nonchiral reagents. Only in a chiral environment do they behave differently from each other. One interesting property of chiral substances is that their solutions may rotate the plane of polarized light, as explained in Section 23.4.

Chirality is common in organic compounds. It is often not observed, however, because when a chiral substance is synthesized in a typical reaction, the two enantio-mers are formed in precisely the same quantity. The resulting mixture is called a race-mic mixture, and it does not rotate the plane of polarized light because the two forms rotate the light to equal extents in opposite directions. (Section 23.4)

Many drugs are chiral compounds. When a drug is administered as a racemic mixture, often only one enantiomer has beneficial results. The other is often inert, or nearly so, or may even have a harmful effect. For example, the drug (R)-albuterol (▶ Figure 24.15) is a bronchodilator used to relieve the symptoms of asthma. The enantiomer (S)-albuterol is not only ineffective as a bronchodilator but also actually counters the effects of (R)-albuterol. As another example, the nonsteroidal analgesic ibuprofen is a chiral molecule usually sold as the racemic mixture. However, a prep-aration consisting of just the more active enantiomer, (S)-ibuprofen (Figure 24.16), relieves pain and reduces inflammation more rapidly than the racemic mixture. For this reason, the chiral version of the drug may in time come to replace the racemic one.

Give It Some ThoughtWhat are the requirements on the four groups attached to a carbon atom in order that it be a chiral center?


24.6 | introduction to BiochemistryThe functional groups discussed in Section 24.4 generate a vast array of molecules with very specific chemical reactivities. Nowhere is this specificity more apparent than in biochemistry—the chemistry of living organisms.

Before we discuss specific biochemical molecules, we can make some general observations. Many biologically important molecules are quite large, because organ-isms build biomolecules from smaller, simpler substances readily available in the bio-sphere. The synthesis of large molecules requires energy because most of the reactions are endothermic. The ultimate source of this energy is the Sun. Animals have essen-tially no capacity for using solar energy directly, and so depend on plant photosynthesis to supply the bulk of their energy needs. (Section 23.3)

In addition to requiring large amounts of energy, living organisms are highly organized. In thermodynamic terms, this high degree of organization means that the entropy of living systems is much lower than that of the raw materials from which the systems formed. Thus, living systems must continuously work against the spontaneous tendency toward increased entropy. (Section 19.3)

In the “Chemistry and Life” essays that appear throughout this text, we have intro-duced some important biochemical applications of fundamental chemical ideas. The remainder of this chapter will serve as only a brief introduction to other aspects of bio-chemistry. Nevertheless, you will see some patterns emerging. Hydrogen bonding, (Section 11.2), for example, is critical to the function of many biochemical systems, and the geometry of molecules (Section 9.1) can govern their biological importance and activity. Many of the large molecules in living systems are polymers (Section 12.8) of much smaller molecules. These biopolymers can be classified into three broad categories: proteins, polysaccharides (carbohydrates), and nucleic acids. Lipids are another common class of molecules in living systems, but they are usually large molecules, not biopolymers.

24.7 | ProteinsProteins are macromolecules present in all living cells. About 50% of your body’s dry mass is protein. Some proteins are structural components in animal tissues; they are a key part of skin, nails, cartilage, and muscles. Other proteins catalyze reactions, trans-port oxygen, serve as hormones to regulate specific body processes, and perform other tasks. Whatever their function, all proteins are chemically similar, being composed of smaller molecules called amino acids.

Amino AcidsAn amino acid is a molecule containing an amine group, ¬ NH2, and a carboxylic acid group, ¬ COOH. The building blocks of all proteins are a@amino acids, where the a (alpha) indicates that the amino group is located on the carbon atom immediately ad-jacent to the carboxylic acid group. Thus, there is always one carbon atom between the amino group and the carboxylic acid group.

The general formula for an a@amino acid is represented by

C or

H

R

COOH C

H

R

COO−H2N

α carbon

One of about 20 different groups

H3N+

The doubly ionized form, called a zwitterion, usually predominates at near-neutral pH values. This form is a result of the transfer of a proton from the carboxylic acid group to the amine group. (Section 16.10: “The Amphiprotic Behavior of Amino Acids”)

Amino acids differ from one another in the nature of their R groups. Twenty-two amino acids have been identified in nature, and ▶ Figure 24.17 shows the 20 of these 22 that are found in humans. Our bodies can synthesize 11 of these 20 amino acids in

▲ Figure 24.16 (S)-Ibuprofen. For relieving pain and reducing inflammation, the ability of this enantiomer far outweighs that of the (R) isomer. In the (R) isomer, the positions of the H and CH3 group on the far-right carbon are switched.

CH2

CH3

CH

CH3

CH3

COOH

CH

seCtION 24.7 Proteins 1069

sufficient amounts for our needs. The other 9 must be ingested and are called essential amino acids because they are necessary components of our diet.

The a@carbon atom of the amino acids, which is the carbon between the amino and carboxylate groups, has four different groups attached to it. The amino acids are thus chiral (except for glycine, which has two hydrogens attached to the central carbon). For historical reasons, the two enantiomeric forms of amino acids are often dis-tinguished by the labels d (from the Latin dexter, “right”) and l (from the Latin laevus, “left”). Nearly all the chiral amino acids found in living organisms have the

Tryptophan(Trp; W)

CH2

N

C COO−

H

H

H3N+

Phenylalanine(Phe; F)

CH2

C COO−

H

H3N+

Glutamic acid(Glu; E)

CH2

C COO−

H

H3N+

CH2

CO O−

Aspartic acid(Asp; D)

CH2

C

CO O−

COO−

H

H3N+

Histidine(His; H)

CH2

C COO−

H

NH

HN

H3N+

+

Lysine(Lys; K)

CH2

C

CH2

CH2

CH2

NH3

COO−

H

H3N+

+

Arginine(Arg; R)

CH2

C

CH2

CH2

NH

NH2

C

COO−

H

H3N+

NH2+

Glutamine(Gln; Q)

CH2

C COO−

H

H3N+

CH2

CO NH2

Asparagine(Asn; N)

CH2

C

CO NH2

COO−

H

H3N+

Tyrosine(Tyr; Y)

CH2

OH

C COO−

H

H3N+

Cysteine(Cys; C)

CH2

SH

C COO−

H

H3N+

CH2

OH

Serine(Ser; S)

C COO−

H

H3N+

Threonine(Thr; T)

HC

CH3

C COO−

OH

H

H3N+

H2C

C COO−

H

H2N

CH2

CH2

Proline(Pro; P)

+C COO−

H

CH2

CH2

S

CH2

Methionine(Met; M)

+H3NC COO−

H

CHCH3

CH2

CH3

Isoleucine(Ile; I)

+H3NC COO−

CH2+

H

H3N

CH3

CHCH3

Leucine(Leu; L)

C COO−

H

CHCH3

CH3

Valine(Val; V)

H3N+

CH3

C COO−

HAlanine(Ala; A)

H3N+

H

C COO−

H

H3N

Glycine(Gly; G)

+

Nonpolar amino acids

Polar amino acids Aromatic amino acids

Basic amino acids Acidic amino acids and their amide derivatives

▲ Figure 24.17 The 20 amino acids found in the human body. The blue shading shows the different R groups for each amino acid. The acids are shown in the zwitterionic form in which they exist in water at near-neutral pH values. The amino acid names shown in bold are the nine essential ones.

GO FiGureWhich group of amino acids has a net positive charge at pH 7?


l configuration at the chiral center. The principal exceptions are the proteins that make up the cell walls of bacteria, which can contain considerable quantities of the d isomers.

Polypeptides and ProteinsAmino acids are linked together into proteins by amide groups (Table 24.6):

O

R R

H

C N

[24.23]

Each amide group is called a peptide bond when it is formed by amino acids. A peptide bond is formed by a condensation reaction between the carboxyl group of one amino acid and the amino group of another amino acid. Alanine and glycine, for example, form the dipeptide glycylalanine:

O

CC

H

H CH3

C

H

O

CC

H

H CH3

C

H

Glycine (Gly; G) Alanine (Ala; A)

Glycylalanine (Gly–Ala; GA)

H3N

H3N

O−

O−

+

O

C

H

H

H N

H2O

O

C

H

N

O− ++

+

+

The amino acid that furnishes the carboxyl group for peptide-bond formation is named first, with a -yl ending; then the amino acid furnishing the amino group is named. Using the abbreviations shown in Figure 24.18, glycylalanine can be abbreviated as either Gly-Ala or GA. In this notation, it is understood that the unreacted amino group is on the left and the unreacted carboxyl group on the right.

The artificial sweetener aspartame (◀ Figure 24.18) is the methyl ester of the dipeptide formed from the amino acids aspartic acid and phenylalanine.

sOLuTiONAnalyze We are given the name of a substance with peptide bonds and asked to write its structural formula.

Plan The name of this substance suggests that three amino acids—alanine, glycine, and serine—have been linked together, forming a tripeptide. Note that the ending -yl has been added to each amino acid except for the last one, serine. By convention, the sequence of amino acids in peptides and proteins is written from the nitrogen end to the carbon end: The first-named amino acid (alanine, in this case) has a free amino group and the last-named one (serine) has a free carboxyl group.

Solve We first combine the carboxyl group of alanine with the amino group of glycine to form a peptide bond and then the

saMPLe exerCise 24.7 drawing the structural Formula of a Tripeptide

Draw the structural formula for alanylglycylserine.

carboxyl group of glycine with the amino group of serine to form another peptide bond:

Amino group Carboxyl group

C C

OH

CH3 CH2OH

C C

OH

N

H H

C

H

N

H

AlaA

GlyG

SerS

H3N O−C

O+

We can abbreviate this tripeptide as either Ala-Gly-Ser or AGS.

Aspartic acid(Asp)

Phenylalanine(Phe)

H CH2CH2

CH3C

C

H2N

H

C

O

N C O

H

C

O

O

OH

▲ Figure 24.18 Sweet stuff. The artificial sweetener aspartame is the methyl ester of a dipeptide.

GO FiGureHow many chiral carbons are there in one molecule of aspartame?

seCtION 24.7 Proteins 1071

Polypeptides are formed when a large number of amino acids 17302 are linked together by peptide bonds. Proteins are linear (that is, unbranched) polypeptide mole-cules with molecular weights ranging from about 6000 to over 50 million amu. Because up to 22 different amino acids are linked together in proteins and because proteins con-sist of hundreds of amino acids, the number of possible arrangements of amino acids within proteins is virtually limitless.

Protein StructureThe sequence of amino acids along a protein chain is called its primary structure and gives the protein its unique identity. A change in even one amino acid can alter the biochemical characteristics of the protein. For example, sickle-cell anemia is a genetic disorder resulting from a single replacement in a protein chain in hemoglobin. The chain that is affected contains 146 amino acids. The substitution of an amino acid with a hydrocarbon side chain for one that has an acidic functional group in the side chain alters the solubility properties of the hemoglobin, and normal blood flow is impeded.

(Section 13.6, “Sickle-Cell Anemia”)Proteins in living organisms are not simply long, flexible chains with random

shapes. Rather, the chains self-assemble into structures based on the intermolecu-lar forces we learned about in Chapter 11. This self-assembling leads to a protein’s secondary structure, which refers to how segments of the protein chain are oriented in a regular pattern, as seen in Figure 24.19.

One of the most important and common secondary structure arrangements is the A1alpha2@helix. As the a@helix of Figure 24.19 shows, the helix is held in posi-tion by hydrogen bonds between amide H atoms and carbonyl O atoms. The pitch of the helix and its diameter must be such that (1) no bond angles are strained and (2) the N ¬ H and C “ O functional groups on adjacent turns are in proper position for hydrogen bonding. An arrangement of this kind is possible for some amino acids along the chain but not for others. Large protein molecules may contain segments of the chain that have the a@helical arrangement interspersed with sections in which the chain is in a random coil.

The other common secondary structure of proteins is the B (beta) sheet. Beta sheets are made of two or more strands of peptides that hydrogen-bond from an amide H in one strand to a carbonyl O in the other strand (Figure 24.19).

Give It Some ThoughtIf you heat a protein to break the intramolecular hydrogen bonds, will you maintain the a@helical or b@sheet structure?

Proteins are not active biologically unless they are in a particular shape in solution. The process by which the protein adopts its biologically active shape is called folding. The shape of a protein in its folded form—determined by all the bends, kinks, and sec-tions of rodlike a@helical, b@sheet, or flexible coil components—is called the tertiary structure.

Practice exercise 1How many nitrogen atoms are in the tripeptide Arg-Asp-Gly? (a) 3, (b) 4, (c) 5, (d) 6, (e) 7.

Practice exercise 2Name the dipeptide and give the two ways of writing its abbreviation.

O−

CH2

H3N C C

OH

HOCH2

C

COOH

C

OH

N

H

+

chemistry of the nonmetals · chemistry of the nonmetals everything we see in the chapter-opening...

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