metallurgy and the chemistry of...

Metallurgy and the Chemistry of Metals

Crystals of salt composed of sodium anion and a complex sodium cation. The molecular models show the complex of an alkali metal cation (Na1 or K1) with an organic compound called crown ether. The large green sphere is the alkali metal anion.

885

20.1 Occurrence of Metals

20.2 Metallurgical Processes

20.3 Band Theory of Electrical Conductivity

20.4 Periodic Trends in Metallic Properties

20.5 The Alkali Metals

20.6 The Alkaline Earth Metals

20.7 Aluminum

Chapter Outline

Media PlayerAluminum Production (20.7)Chapter Summary

ARISEnd of Chapter Problems

A Look Ahead• We fi rst survey the occurrence of ores containing various metals. (20.1)

• We then study the sequence of steps from the preparation of the ores to the production of metals. We focus mainly on the metallurgy of iron and the mak-ing of steel. We also examine several methods of metal purifi cation. (20.2)

• Next, we study the properties of solids and see how the band theory explains the difference between conductors (metals) and insulators. We learn the spe-cial properties of semiconductors. (20.3)

• We briefl y examine the periodic trends in metallic properties. (20.4)

• For alkali metals we discuss sodium and potassium and focus on their prep-arations, properties, compounds, and uses. (20.5)

• For alkaline earth metals we discuss magnesium and calcium and focus on their preparations, properties, compounds, and uses. (20.6)

• Finally, we study the preparation, properties, compounds and uses of a Group 3A metal—aluminum. (20.7).

Up to this point we have concentrated mainly on fundamental principles: theories of chemical bonding, intermolecular forces, rates and mechanisms

of chemical reactions, equilibrium, the laws of thermodynamics, and electrochem-istry. An understanding of these topics is necessary for the study of the properties of representative metallic elements and their compounds. The use and refi nement of metals date back to early human history. For example, archeologists have found evidence that in the fi rst millennium a.d. inhabitants of Sri Lanka used monsoon winds to run iron-smelting furnaces to produce high-carbon steel. Through the years, these furnaces could have been sources of steel for the legendary Damascus swords, known for their sharpness and durability. In this chapter we will study the methods for extracting, refi ning, and puri-fying metals and examine the properties of metals that belong to the representa-tive elements. We will emphasize (1) the occurrence and preparation of metals, (2) the physical and chemical properties of some of their compounds, and (3) their uses in modern society and their roles in biological systems.

Student Interactive Activities

886 Metallurgy and the Chemistry of Metals

20.1 Occurrence of MetalsMost metals come from minerals. A mineral is a naturally occurring substance with a range of chemical composition. A mineral deposit concentrated enough to allow economical recovery of a desired metal is known as ore. Table 20.1 lists the principal types of minerals, and Figure 20.1 shows a classifi cation of metals according to their minerals. The most abundant metals, which exist as minerals in Earth’s crust, are aluminum, iron, calcium, magnesium, sodium, potassium, titanium, and manganese (see p. 52). Seawater is a rich source of some metal ions, including Na1, Mg21, and Ca21. Fur-thermore, vast areas of the ocean fl oor are covered with manganese nodules, which are made up mostly of manganese, along with iron, nickel, copper, and cobalt in a chemically combined state (Figure 20.2).

20.2 Metallurgical ProcessesMetallurgy is the science and technology of separating metals from their ores and of compounding alloys. An alloy is a solid solution either of two or more metals, or of a metal or metals with one or more nonmetals. The three principal steps in the recovery of a metal from its ore are (1) prepara-tion of the ore, (2) production of the metal, and (3) purifi cation of the metal.

Preparation of the OreIn the preliminary treatment of an ore, the desired mineral is separated from waste materials—usually clay and silicate minerals—which are collectively called the gangue. One very useful method for carrying out such a separation is called fl otation. In this process, the ore is fi nely ground and added to water containing oil and deter-gent. The liquid mixture is then beaten or blown to form a froth. The oil preferentially

Type Minerals

Uncombined metals Ag, Au, Bi, Cu, Pd, PtCarbonates BaCO3 (witherite), CaCO3 (calcite, limestone), MgCO3

(magnesite), CaCO3 ? MgCO3 (dolomite), PbCO3 (cerussite), ZnCO3 (smithsonite)

Halides CaF2 (fl uorite), NaCl (halite), KCl (sylvite), Na3AlF6 (cryolite)Oxides Al2O3 ? 2H2O (bauxite), Al2O3 (corundum), Fe2O3 (hematite),

Fe3O4 (magnetite), Cu2O (cuprite), MnO2 (pyrolusite), SnO2 (cassiterite), TiO2 (rutile), ZnO (zincite)

Phosphates Ca3(PO4)2 (phosphate rock), Ca5(PO4)3OH (hydroxyapatite)Silicates Be3Al2Si6O18 (beryl), ZrSiO4 (zircon), NaAlSi3O8 (albite),

Mg3(Si4O10)(OH)2 (talc)Sulfi des Ag2S (argentite), CdS (greenockite), Cu2S (chalcocite), FeS2

(pyrite), HgS (cinnabar), PbS (galena), ZnS (sphalerite)Sulfates BaSO4 (barite), CaSO4 (anhydrite), PbSO4 (anglesite),

SrSO4 (celestite), MgSO4 ? 7H2O (epsomite)

TABLE 20.1 Principal Types of Minerals

wets the mineral particles, which are then carried to the top in the froth, while the gangue settles to the bottom. The froth is skimmed off, allowed to collapse, and dried to recover the mineral particles. Another physical separation process makes use of the magnetic properties of certain minerals. Ferromagnetic metals are strongly attracted to magnets. The mineral magnetite (Fe3O4), in particular, can be separated from the gangue by using a strong electromagnet. Cobalt is another ferromagnetic metal. Mercury forms amalgams with a number of metals. An amalgam is an alloy of mercury with another metal or metals. Mercury can therefore be used to extract metal from ore. Mercury dissolves the silver and gold in an ore to form a liquid amalgam, which is easily separated from the remaining ore. The gold or silver is recovered by distilling off mercury.

Production of MetalsBecause metals in their combined forms always have positive oxidation numbers, the production of a free metal is a reduction process. Preliminary operations may be necessary to convert the ore to a chemical state more suitable for reduction. For example, an ore may be roasted to drive off volatile impurities and at the same time to convert the carbonates and sulfi des to the corresponding oxides, which can be reduced more conveniently to yield the pure metals:

CaCO3(s) ¡ CaO(s) 1 CO2(g) 2PbS(s) 1 3O2(g) ¡ 2PbO(s) 1 2SO2(g)

This last equation points out the fact that the conversion of sulfi des to oxides is a major source of sulfur dioxide, a notorious air pollutant (p. 786).

Unreactive metals such as gold and silver can be leached from the ores using the cyanide ions (see Section 21.3).

Unreactive metals such as gold and silver can be leached from the ores using the cyanide ions (see Section 21.3).

20.2 Metallurgical Processes 887

Figure 20.1 Metals and their best-known minerals. Lithium is found in spodumene (LiAlSi2O6), and beryllium in beryl (see Table 20.1). The rest of the alkaline earth metals are found in minerals that are carbonates and sulfates. The minerals for Sc, Y, and La are the phosphates. Some metals have more than one type of important mineral. For example, in addition to the sulfi de, iron is found as the oxides hematite (Fe2O3) and magnetite (Fe3O4); and aluminum, in addition to the oxide, is found in beryl (Be3Al2Si6O18). Technetium (Tc) is a synthetic element.

Other compounds;see caption

Sulfides

Chlorides

Oxides

Uncombined

Li

Na

K

Rb

Cs

Ca

Sr

Ba

Sc

Y

La

Ti

Zr

Hf

V

Nb

Ta

Cr

Mo

W

Mn

Tc

Re

Fe

Ru

Os

Co

Rh

Ir

Ni

Pd

Pt

Cu

Ag

Au

Zn

Cd

Hg

Ga

In

Tl

Sn

Pb Bi

Al

Be

Mg

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

8 107 7B

9 8B

11 1B

12 2B

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

Figure 20.2 Manganese nodules on the ocean fl oor.


How a pure metal is obtained by reduction from its combined form depends on the standard reduction potential of the metal (see Table 19.1). Table 20.2 outlines the reduction processes for several metals. Most major metallurgical processes now in use involve pyrometallurgy, procedures carried out at high temperatures. The reduction in these procedures may be accomplished either chemically or electrolytically.

Chemical ReductionWe can use a more electropositive metal as a reducing agent to separate a less elec-tropositive metal from its compound at high temperatures:

V2O5(s) 1 5Ca(l) ¡ 2V(l) 1 5CaO(s) TiCl4(g) 1 2Mg(l) ¡ Ti(s) 1 2MgCl2(l) Cr2O3(s) 1 2Al(s) ¡ 2Cr(l) 1 Al2O3(s)

3Mn3O4(s) 1 8Al(s) ¡ 9Mn(l) 1 4Al2O3(s)

In some cases, even molecular hydrogen can be used as a reducing agent, as in the preparation of tungsten (used as fi laments in lightbulbs) from tungsten(VI) oxide:

WO3(s) 1 3H2(g) ¡ W(s) 1 3H2O(g)

Electrolytic ReductionElectrolytic reduction is suitable for very electropositive metals, such as sodium, mag-nesium, and aluminum. The process is usually carried out on the anhydrous molten oxide or halide of the metal:

2MO(l) ¡ 2M (at cathode) 1 O2 (at anode) 2MCl(l) ¡ 2M (at cathode) 1 Cl2 (at anode)

We will describe the specifi c procedures later in this chapter.

The Metallurgy of IronIron exists in Earth’s crust in many different minerals, such as iron pyrite (FeS2), sider-ite (FeCO3), hematite (Fe2O3), and magnetite (Fe3O4, often represented as FeO ? Fe2O3). Of these, hematite and magnetite are particularly suitable for the extraction of iron. The metallurgical processing of iron involves the chemical reduction of the minerals by carbon (in the form of coke) in a blast furnace (Figure 20.3). The concentrated iron ore, limestone (CaCO3), and coke are introduced into the furnace from the top. A blast of hot air is forced up the furnace from the bottom—hence the name blast furnace.

A more electropositive metal has a more negative standard reduction potential (see Table 19.1).

A more electropositive metal has a more negative standard reduction potential (see Table 19.1).

The extraction of iron from FeS2 leads to SO2 production and acid rain (see Section 17.6).

The extraction of iron from FeS2 leads to SO2 production and acid rain (see Section 17.6).

Metal Reduction Process

Lithium, sodium, magnesium, calcium Electrolytic reduction of the molten chloride Aluminum Electrolytic reduction of anhydrous oxide (in

molten cryolite) Chromium, manganese, titanium, Reduction of the metal oxide with a more electropositive vanadium, iron, zinc metal, or reduction with coke and carbon monoxide Mercury, silver, platinum, copper, gold These metals occur in the free (uncombined) state or

can be obtained by roasting their sulfi des

TABLE 20.2 Reduction Processes for Some Common Metals

Dec

reas

ing

activ

ity

of m

etal

s

The oxygen gas reacts with the carbon in the coke to form mostly carbon monoxide and some carbon dioxide. These reactions are highly exothermic, and as the hot CO and CO2 gases rise, they react with the iron oxides in different temperature zones, as shown in Figure 20.3. The key steps in the extraction of iron are

3Fe2O3(s) 1 CO(g) ¡ 2Fe3O4(s) 1 CO2(g) Fe3O4(s) 1 CO(g) ¡ 3FeO(s) 1 CO2(g)

FeO(s) 1 CO(g) ¡ Fe(l) 1 CO2(g)

The limestone decomposes in the furnace as follows:

CaCO3(s) ¡ CaO(s) 1 CO2(g)

The calcium oxide then reacts with the impurities in the iron, which are mostly sand (SiO2) and aluminum oxide (Al2O3):

CaO(s) 1 SiO2(s) ¡ CaSiO3(l) CaO(s) 1 Al2O3(s) ¡ Ca(AlO2)2(l)

The mixture of calcium silicate and calcium aluminate that remains molten at the furnace temperature is known as slag. By the time the ore works its way down to the bottom of the furnace, most of it has already been reduced to iron. The temperature of the lower part of the furnace is above the melting point of impure iron, and so the molten iron at the lower level can be run off to a receiver. The slag, because it is less dense, forms the top layer above the molten iron and can be run off at that level, as shown in Figure 20.3.

CaCO3 and other compounds that are used to form a molten mixture with the impurities in the ore for easy removal are called fl ux.

CaCO3 and other compounds that are used to form a molten mixture with the impurities in the ore for easy removal are called fl ux.


Figure 20.3 A blast furnace. Iron ore, limestone, and coke are introduced at the top of the furnace. Iron is obtained from the ore by reduction with carbon.

3Fe2O3 + CO 2Fe3O4 + CO2

CaCO3 CaO + CO2

C + CO2 2CO

FeO + CO Fe + CO2

Iron melts Molten slag forms

2C + O2 2CO

200°C

700°C

1500°C

1200°C

2000°C

Hot air blast

Molten ironSlag

CO, CO2

Hot

gas

es ri

se

Solid

cha

rge

desc

ends

Charge (ore, limestone, coke)

Fe3O4 + CO 3FeO + CO2


Iron extracted in this way contains many impurities and is called pig iron; it may contain up to 5 percent carbon and some silicon, phosphorus, manganese, and sulfur. Some of the impurities stem from the silicate and phosphate minerals, while carbon and sulfur come from coke. Pig iron is granular and brittle. It has a relatively low melting point (about 1180°C), so it can be cast in various forms; for this reason it is also called cast iron.

SteelmakingSteel manufacturing is one of the most important metal industries. In the United States, the annual consumption of steel is well above 100 million tons. Steel is an iron alloy that contains from 0.03 to 1.4 percent carbon plus various amounts of other elements. The wide range of useful mechanical properties associated with steel is primarily a function of chemical composition and heat treatment of a particular type of steel. Whereas the production of iron is basically a reduction process (converting iron oxides to metallic iron), the conversion of iron to steel is essentially an oxidation process in which the unwanted impurities are removed from the iron by reaction with oxygen gas. One of several methods used in steelmaking is the basic oxygen process. Because of its ease of operation and the relatively short time (about 20 minutes) required for each large-scale (hundreds of tons) conversion, the basic oxygen process is by far the most common means of producing steel today. Figure 20.4 shows the basic oxygen process. Molten iron from the blast furnace is poured into an upright cylindrical vessel. Pressurized oxygen gas is introduced via a water-cooled tube above the molten metal. Under these conditions, manganese, phosphorus, and silicon, as well as excess carbon, react with oxygen to form oxides. These oxides are then reacted with the appropriate fl uxes (for example, CaO or SiO2) to form slag. The type of fl ux chosen depends on the composition of the iron.

Figure 20.4 The basic oxygen process of steelmaking. The capacity of a typical vessel is 100 tons of cast iron.

Moltensteel

O2

CO2, SO2

Molten steel + slag Slag

Vertical position Horizontal position

CaO or SiO2

If the main impurities are silicon and phosphorus, a basic fl ux such as CaO is added to the iron:

SiO2(s) 1 CaO(s) ¡ CaSiO3(l) P4O10(l) 1 6CaO(s) ¡ 2Ca3(PO4)2(l)

On the other hand, if manganese is the main impurity, then an acidic fl ux such as SiO2 is needed to form the slag:

MnO(s) 1 SiO2(s) ¡ MnSiO3(l)

The molten steel is sampled at intervals. When the desired blend of carbon and other impurities has been reached, the vessel is rotated to a horizontal position so that the molten steel can be tapped off (Figure 20.5). The properties of steel depend not only on its chemical composition but also on the heat treatment. At high temperatures, iron and carbon in steel combine to form iron carbide, Fe3C, called cementite:

3Fe(s) 1 C(s) ∆ Fe3C(s)

The forward reaction is endothermic, so that the formation of cementite is favored at high temperatures. When steel containing cementite is cooled slowly, the preceding equilibrium shifts to the left, and the carbon separates as small particles of graphite, which give the steel a gray color. (Very slow decomposition of cementite also takes place at room temperature.) If the steel is cooled rapidly, equilibrium is not attained and the carbon remains largely in the form of cementite, Fe3C. Steel containing cemen-tite is light in color, and it is harder and more brittle than that containing graphite. Heating the steel to some appropriate temperature for a short time and then cool-ing it rapidly in order to give it the desired mechanical properties is known as “tem-pering.” In this way, the ratio of carbon present as graphite and as cementite can be varied within rather wide limits. Table 20.3 shows the composition, properties, and uses of various types of steel.


Figure 20.5 Steelmaking.


Purifi cation of MetalsMetals prepared by reduction usually need further treatment to remove impurities. The extent of purifi cation, of course, depends on how the metal will be used. Three com-mon purifi cation procedures are distillation, electrolysis, and zone refi ning.

DistillationMetals that have low boiling points, such as mercury, magnesium, and zinc, can be separated from other metals by fractional distillation. One well-known method of fractional distillation is the Mond† process for the purifi cation of nickel. Carbon mon-oxide gas is passed over the impure nickel metal at about 70°C to form the volatile tetracarbonylnickel (b.p. 43°C), a highly toxic substance, which is separated from the less volatile impurities by distillation:

Ni(s) 1 4CO(g) ¡ Ni(CO)4(g)

Pure metallic nickel is recovered from Ni(CO)4 by heating the gas at 200°C:

Ni(CO)4(g) ¡ Ni(s) 1 4CO(g)

The carbon monoxide that is released is recycled back into the process.

ElectrolysisElectrolysis is another important purifi cation technique. The copper metal obtained by roasting copper sulfi de usually contains impurities such as zinc, iron, silver, and gold. The more electropositive metals are removed by an electrolysis process in which the impure copper acts as the anode and pure copper acts as the cathode in a sulfuric acid solution containing Cu21 ions (Figure 20.6). The reactions are

Anode (oxidation): Cu(s) ¡ Cu21(aq) 1 2e2

Cathode (reduction): Cu21(aq) 1 2e2 ¡ Cu(s)

Reactive metals in the copper anode, such as iron and zinc, are also oxidized at the anode and enter the solution as Fe21 and Zn21 ions. They are not reduced at the

†Ludwig Mond (1839–1909). British chemist of German origin. Mond made many important contributions to industrial chemistry. His method for purifying nickel by converting it to the volatile Ni(CO)4 compound has been described as having given “wings” to the metal.

Composition (Percent by Mass)*

Type C Mn P S Si Ni Cr Others Uses

Plain 1.35 1.65 0.04 0.05 0.06 — — Cu (0.2–0.6) Sheet products, tools

High-strength 0.25 1.65 0.04 0.05 0.15–0.9 0.4–1.0 0.3–1.3 Cu (0.01–0.08) Construction, steam turbines

Stainless 0.03–1.2 1.0–10 0.04–0.06 0.03 1–3 1–22 4.0–27 — Kitchen utensils, razor blades

TABLE 20.3 Types of Steel

*A single number indicates the maximum amount of the substance present.

Figure 20.6 Electrolytic purifi cation of copper.

Purecoppercathode

Cu2+

SO42–

Impurecopperanode

Battery

cathode, however. The less electropositive metals, such as gold and silver, are not oxidized at the anode. Eventually, as the copper anode dissolves, these metals fall to the bottom of the cell. Thus, the net result of this electrolysis process is the transfer of copper from the anode to the cathode. Copper prepared this way has a purity greater than 99.5 percent (Figure 20.7).

Zone Refi ningAnother often-used method of obtaining extremely pure metals is zone refi ning. In this process, a metal rod containing a few impurities is drawn through an electrical heating coil that melts the metal (Figure 20.8). Most impurities dissolve in the molten metal. As the metal rod emerges from the heating coil, it cools and the pure metal crystallizes, leaving the impurities in the molten metal portion that is still in the heat-ing coil. (This is analogous to the freezing of seawater, in which the solid that separates

The metal impurities separated from the copper anode are valuable by-products, the sale of which often pays for the electricity needed to drive the electrolysis.

The metal impurities separated from the copper anode are valuable by-products, the sale of which often pays for the electricity needed to drive the electrolysis.


Figure 20.7 Copper cathodes used in the electrorefi ning process.

Heating coil Metal rod Figure 20.8 Zone-refi ning technique for purifying metals. Top to bottom: An impure metal rod is moved slowly through a heating coil. As the metal rod moves forward, the impurities dissolve in the molten portion of the metal while pure metal crystallizes out in front of the molten zone.


is mostly pure solvent—water. In zone refi ning, the liquid metal acts as the solvent and the impurities as the solutes.) When the molten zone carrying the impurities, now at increased concentration, reaches the end of the rod, it is allowed to cool and is then cut off. Repeating this procedure a number of times results in metal with a purity greater than 99.99 percent.

20.3 Band Theory of Electrical ConductivityIn Section 11.6 we saw that the ability of metals to conduct heat and electricity can be explained with molecular orbital theory. To gain a better understanding of the conductivity properties of metals we must also apply our knowledge of quantum mechanics. The model we will use to study metallic bonding is band theory, so called because it states that delocalized electrons move freely through “bands” formed by overlapping molecular orbitals. We will also apply band theory to certain elements that are semiconductors.

ConductorsMetals are characterized by high electrical conductivity. Consider magnesium, for example. The electron confi guration of Mg is [Ne]3s2, so each atom has two valence electrons in the 3s orbital. In a metallic crystal, the atoms are packed closely together, so the energy levels of each magnesium atom are affected by the immediate neighbors of the atom as a result of orbital overlaps. In Chapter 10 we saw that, in terms of molecular orbital theory, the interaction between two atomic orbitals leads to the formation of a bonding and an antibonding molecular orbital. Because the number of atoms in even a small piece of magnesium is enormously large (on the order of 1020 atoms), the number of molecular orbitals they form is also very large. These molecular orbitals are so closely spaced on the energy scale that they are more appro-priately described as a “band” (Figure 20.9). The closely spaced fi lled energy levels make up the valence band. The upper half of the energy levels corresponds to the empty, delocalized molecular orbitals formed by the overlap of the 3p orbitals. This set of closely spaced empty levels is called the conduction band. We can imagine a metallic crystal as an array of positive ions immersed in a sea of delocalized valence electrons (see Figure 11.30). The great cohesive force resulting from the delocalization is partly responsible for the strength noted in most metals. Because the valence band and the conduction band are adjacent to each other, the

Figure 20.9 Formation of conduction bands in magnesium. The electrons in the 1s, 2s, and 2p orbitals are localized on each Mg atom. However, the 3s and 3p orbitals overlap to form delocalized molecular orbitals. Electrons in these orbitals can travel throughout the metal, and this accounts for the electrical conductivity of the metal.

⎧⎩⎨

⎧⎩⎨

2p

2s

1s

3s

Conduction band3p

12 +

Mg

12 +

Mg

12 +

Mg

12 +

Mg

12 +

Mg

Valence band

Ene

rgy

amount of energy needed to promote a valence electron to the conduction band is negligible. There, the electron can travel freely through the metal, because the conduc-tion band is void of electrons. This freedom of movement accounts for the fact that metals are good conductors, that is, they are capable of conducting electric current. Why don’t substances like wood and glass conduct electricity as metals do? Figure 20.10 provides an answer to this question. Basically, the electrical conductivity of a solid depends on the spacing and the state of occupancy of the energy bands. In magnesium and other metals, the valence bands are adjacent to the conduction bands, and, there-fore, these metals readily act as conductors. In wood and glass, on the other hand, the gap between the valence band and the conduction band is considerably greater than that in a metal. Consequently, much more energy is needed to excite an electron into the conduction band. Lacking this energy, electrons cannot move freely. Therefore, glass and wood are insulators, ineffective conductors of electricity.

SemiconductorsA number of elements are semiconductors, that is, they normally are not conductors, but will conduct electricity at elevated temperatures or when combined with a small amount of certain other elements. The Group 4A elements silicon and germanium are especially suited for this purpose. The use of semiconductors in transistors and solar cells, to name two applications, has revolutionized the electronic industry in recent decades, leading to increased miniaturization of electronic equipment. The energy gap between the fi lled and empty bands of these solids is much smaller than that for insulators (see Figure 20.10). If the energy needed to excite electrons from the valence band into the conduction band is provided, the solid becomes a conductor. Note that this behavior is opposite that of the metals. A metal’s ability to conduct electricity decreases with increasing temperature, because the enhanced vibration of atoms at higher temperatures tends to disrupt the fl ow of electrons. The ability of a semiconductor to conduct electricity can also be enhanced by adding small amounts of certain impurities to the element, a process called doping. Let us consider what happens when a trace amount of boron or phosphorus is added to solid silicon. (Only about fi ve out of every million Si atoms are replaced by B or P atoms.) The structure of solid silicon is similar to that of diamond; each Si atom is covalently bonded to four other Si atoms. Phosphorus ([Ne]3s23p3) has one more valence electron than silicon ([Ne]3s23p2), so there is a valence electron left over after four of them are used to form covalent bonds with silicon (Figure 20.11). This extra electron can be removed from the phosphorus atom by applying a voltage across the solid. The free electron can move through the structure and function as a conduction electron. Impurities of this type are known as donor impurities, because they provide conduction electrons. Solids containing donor impurities are called n-type semicon-ductors, where n stands for negative (the charge of the “extra” electron).

20.3 Band Theory of Electrical Conductivity 895

Figure 20.10 Comparison of the energy gaps between valence band and conduction band in a metal, a semiconductor, and an insulator. In a metal, the energy gap is virtually nonexistent; in a semiconductor, the energy gap is small; and in an insulator, the energy gap is very large, thus making the promotion of an electron from the valence band to the conduction band diffi cult.

Conduction band

Valence band

MetalE

nerg

y

Ene

rgy

Semiconductor

Ene

rgy

Insulator

Conduction band

Valence band

Energy gap

Conduction band

Valence band

Energy gap


The opposite effect occurs if boron is added to silicon. A boron atom has three valence electrons (1s22s22p1). Thus, for every boron atom in the silicon crystal there is a single vacancy in a bonding orbital. It is possible to excite a valence electron from a nearby Si into this vacant orbital. A vacancy created at that Si atom can then be fi lled by an electron from a neighboring Si atom, and so on. In this manner, elec-trons can move through the crystal in one direction while the vacancies, or “positive holes,” move in the opposite direction, and the solid becomes an electrical conductor. Impurities that are electron defi cient are called acceptor impurities. Semiconductors that contain acceptor impurities are called p-type semiconductors, where p stands for positive. In both the p-type and n-type semiconductors, the energy gap between the valence band and the conduction band is effectively reduced, so that only a small amount of energy is needed to excite the electrons. Typically, the conductivity of a semiconduc-tor is increased by a factor of 100,000 or so by the presence of impurity atoms. The growth of the semiconductor industry since the early 1960s has been truly remarkable. Today semiconductors are essential components of nearly all electronic equipment, ranging from radios and television sets to pocket calculators and comput-ers. One of the main advantages of solid-state devices over vacuum-tube electronics is that the former can be made on a single “chip” of silicon no larger than the cross section of a pencil eraser. Consequently, much more equipment can be packed into a small volume—a point of particular importance in space travel, as well as in hand-held calculators and microprocessors (computers-on-a-chip).

20.4 Periodic Trends in Metallic PropertiesMetals are lustrous in appearance, solid at room temperature (with the exception of mercury), good conductors of heat and electricity, malleable (can be hammered fl at), and ductile (can be drawn into wire). Figure 20.12 shows the positions of the representative metals and the Group 2B metals in the periodic table. (The transition metals are discussed in Chapter 22.) As we saw in Figure 9.5, the electronegativity of elements increases from left to right across a period and from bottom to top in a group. The metallic character of metals increases in just the opposite directions, that is, from right to left across a period and from top to bottom in a group. Because metals generally have low electronegativities, they tend to form cations and almost always have positive oxidation numbers in their compounds. However, beryllium and magnesium in Group 2A and metals in Group 3A and beyond also form cova-lent compounds. In Sections 20.5, 20.6, and 20.7 we will study the chemistry of selected metals from Group 1A (the alkali metals), Group 2A (the alkaline earth metals), and Group 3A (aluminum).

Figure 20.11 (a) Silicon crystal doped with phosphorus. (b) Silicon crystal doped with boron. Note the formation of a negative center in (a) and of a positive center in (b).

(a) (b)

e–+

20.5 The Alkali MetalsAs a group, the alkali metals (the Group 1A elements) are the most electropositive (or the least electronegative) elements known. They exhibit many similar properties, some of which are listed in Table 20.4. From their electron confi gurations we expect the oxidation number of these elements in their compounds to be 11 because the cations would be isoelectronic with the noble gases. This is indeed the case. The alkali metals have low melting points and are soft enough to be sliced with a knife (see Figure 8.14). These metals all possess a body-centered crystal structure (see Figure 11.29) with low packing effi ciency. This accounts for their low densities among metals. In fact, lithium is the lightest metal known. Because of their great chemical reactivity, the alkali metals never occur naturally in elemental form; they are found combined with halide, sulfate, carbonate, and silicate ions. In this section we will describe the chemistry of two members of Group 1A—sodium and potassium.

*Refers to the cation M1, where M denotes an alkali metal atom.†The half-reaction M1(aq) 1 e2 ¡ M(s).

Li Na K Rb Cs

Valence electron confi guration 2s1 3s1 4s1 5s1 6s1

Density (g/cm3) 0.534 0.97 0.86 1.53 1.87Melting point (°C) 179 97.6 63 39 28Boiling point (°C) 1317 892 770 688 678Atomic radius (pm) 152 186 227 248 265Ionic radius (pm)* 78 98 133 148 165Ionization energy (kJ/mol) 520 496 419 403 375Electronegativity 1.0 0.9 0.8 0.8 0.7Standard reduction potential (V)† 23.05 22.71 22.93 22.93 22.92

TABLE 20.4 Properties of Alkali Metals

20.5 The Alkali Metals 897

Figure 20.12 Representative metals and Group 2B metals according to their positions in the periodic table.

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

8 107 7B

9 8B

11 1B

12 2B

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

H

Li

Na

K

Rb

Cs

Fr

Ca

Sr

Ba

Ra

Sc

Y

La

Ac

Ti

Zr

Hf

Rf

V

Nb

Ta

Db

Cr

Mo

W

Sg

Mn

Tc

Re

Bh

Fe

Ru

Os

Hs

Co

Rh

Ir

Mt

Ni

Pd

Pt

Cu

Ag

Au

Zn

Cd

Hg

Ga

In

Tl

Ge

Sn

Pb

As

Sb

Bi

Se

Te

Po

Br

I

At

Kr

Xe

Rn

Al Si P S Cl Ar

B C N O F Ne

He

Be

Mg

Ds Rg


The chemistry of lithium, rubidium, and cesium is less important; all isotopes of francium, the last member of the group, are radioactive. Sodium and potassium are about equally abundant in nature. They occur in sili-cate minerals such as albite (NaAlSi3O8) and orthoclase (KAlSi3O8). Over long peri-ods of time (on a geologic scale), silicate minerals are slowly decomposed by wind and rain, and their sodium and potassium ions are converted to more soluble com-pounds. Eventually rain leaches these compounds out of the soil and carries them to the sea. Yet when we look at the composition of seawater, we fi nd that the concentra-tion ratio of sodium to potassium is about 28 to 1. The reason for this uneven distri-bution is that potassium is essential to plant growth, while sodium is not. Plants take up many of the potassium ions along the way, while sodium ions are free to move on to the sea. Other minerals that contain sodium or potassium are halite (NaCl), shown in Figure 20.13, Chile saltpeter (NaNO3), and sylvite (KCl). Sodium chloride is also obtained from rock salt (see p. 373). Metallic sodium is most conveniently obtained from molten sodium chloride by electrolysis in the Downs cell (see Section 19.8). The melting point of sodium chloride is rather high (801°C), and much energy is needed to keep large amounts of the sub-stance molten. Adding a suitable substance, such as CaCl2, lowers the melting point to about 600°C—a more convenient temperature for the electrolysis process. Metallic potassium cannot be easily prepared by the electrolysis of molten KCl because it is too soluble in the molten KCl to fl oat to the top of the cell for collection. Moreover, it vaporizes readily at the operating temperatures, creating hazardous con-ditions. Instead, it is usually obtained by the distillation of molten KCl in the presence of sodium vapor at 892°C. The reaction that takes place at this temperature is

Na(g) 1 KCl(l) ∆ NaCl(l) 1 K(g)

This reaction may seem strange given that potassium is a stronger reducing agent than sodium (see Table 20.4). However, potassium has a lower boiling point (770°C) than sodium (892°C), so it is more volatile at 892°C and distills off more easily. Accord-ing to Le Châtelier’s principle, constant removal of potassium vapor shifts the above equilibrium from left to right, assuring recovery of metallic potassium. Sodium and potassium are both extremely reactive, but potassium is the more reactive of the two. Both react with water to form the corresponding hydroxides. In a limited supply of oxygen, sodium burns to form sodium oxide (Na2O). However, in the presence of excess oxygen, sodium forms the pale-yellow peroxide:

2Na(s) 1 O2(g) ¡ Na2O2(s)

Sodium peroxide reacts with water to give an alkaline solution and hydrogen peroxide:

Na2O2(s) 1 2H2O(l) ¡ 2NaOH(aq) 1 H2O2(aq)

Like sodium, potassium forms the peroxide. In addition, potassium also forms the superoxide when it burns in air:

K(s) 1 O2(g) ¡ KO2(s)

When potassium superoxide reacts with water, oxygen gas is evolved:

2KO2(s) 1 2H2O(l) ¡ 2KOH(aq) 1 O2(g) 1 H2O2(aq)

This reaction is utilized in breathing equipment (Figure 20.14). Exhaled air contains both moisture and carbon dioxide. The moisture reacts with KO2 in the apparatus to

Remember that Ca21 is harder to reduce than Na1.Remember that Ca21 is harder to reduce than Na1.

Note that this is a chemical rather than electrolytic reduction.Note that this is a chemical rather than electrolytic reduction.

Figure 20.13 Halite (NaCl).

generate oxygen gas as shown in the preceding equation. Furthermore, KO2 also reacts with exhaled CO2, which produces more oxygen gas:

4KO2(s) 1 2CO2(g) ¡ 2K2CO3(s) 1 3O2(g)

Thus, a person using the apparatus can continue to breathe oxygen without being exposed to toxic fumes outside. Sodium and potassium metals dissolve in liquid ammonia to produce a beautiful blue solution:

Na ¡NH3 Na1 1 e2

K ¡NH3 K1 1 e2

Both the cation and the electron exist in the solvated form; the solvated electrons are responsible for the characteristic blue color of such solutions. Metal-ammonia solu-tions are powerful reducing agents (because they contain free electrons); they are useful in synthesizing both organic and inorganic compounds. It was discovered that the hitherto unknown alkali metal anions, M2, are also formed in such solutions. This means that an ammonia solution of an alkali metal contains ion pairs such as Na1Na2 and K1K2! (Keep in mind that in each case the metal cation exists as a complex ion with crown ether, an organic compound with a high affi nity for cations.) In fact, these “salts” are so stable that they can be isolated in crystalline form (see p. 884). This fi nding is of considerable theoretical interest, for it shows clearly that the alkali metals can have an oxidation number of 21, although 21 is not found in ordinary compounds. Sodium and potassium are essential elements of living matter. Sodium ions and potassium ions are present in intracellular and extracellular fl uids, and they are essen-tial for osmotic balance and enzyme functions. We now describe the preparations and uses of several of the important compounds of sodium and potassium.

Sodium ChlorideThe source, properties, and uses of sodium chloride were discussed in Chapter 9 (see p. 373).

20.5 The Alkali Metals 899

Figure 20.14 Self-contained breathing apparatus.


Sodium CarbonateSodium carbonate (called soda ash) is used in all kinds of industrial processes, includ-ing water treatment and the manufacture of soaps, detergents, medicines, and food additives. Today about half of all Na2CO3 produced is used in the glass industry (in soda-lime glass; see Section 11.7). Sodium carbonate ranks eleventh among the chem-icals produced in the United States (11 million tons in 2008). For many years Na2CO3 was produced by the Solvay† process, in which ammonia is fi rst dissolved in a satu-rated solution of sodium chloride. Bubbling carbon dioxide into the solution results in the precipitation of sodium bicarbonate as follows:

NH3(aq) 1 NaCl(aq) 1 H2CO3(aq) ¡ NaHCO3(s) 1 NH4Cl(aq)

Sodium bicarbonate is then separated from the solution and heated to give sodium carbonate:

2NaHCO3(s) ¡ Na2CO3(s) 1 CO2(g) 1 H2O(g)

However, the rising cost of ammonia and the pollution problem resulting from by-products have prompted chemists to look for other sources of sodium carbonate. One is the mineral trona [Na5(CO3)2(HCO3) ? 2H2O], large deposits of which have been found in Wyoming. When trona is crushed and heated, it decomposes as follows:

2Na5(CO3)2(HCO3) ? 2H2O(s) ¡ 5Na2CO3(s) 1 CO2(g) 1 3H2O(g)

The sodium carbonate obtained this way is dissolved in water, the solution is fi ltered to remove the insoluble impurities, and the sodium carbonate is crystallized as Na2CO3 ? 10H2O. Finally, the hydrate is heated to give pure, anhydrous sodium carbonate.

Sodium Hydroxide and Potassium HydroxideThe properties of sodium hydroxide and potassium hydroxide are very similar. These hydroxides are prepared by the electrolysis of aqueous NaCl and KCl solutions (see Section 19.8); both hydroxides are strong bases and very soluble in water. Sodium hydroxide is used in the manufacture of soap and many organic and inorganic com-pounds. Potassium hydroxide is used as an electrolyte in some storage batteries, and aqueous potassium hydroxide is used to remove carbon dioxide and sulfur dioxide from air.

Sodium Nitrate and Potassium NitrateLarge deposits of sodium nitrate (chile saltpeter) are found in Chile. It decomposes with the evolution of oxygen at about 500°C:

2NaNO3(s) ¡ 2NaNO2(s) 1 O2(g)

Potassium nitrate (saltpeter) is prepared beginning with the “reaction”

KCl(aq) 1 NaNO3(aq) ¡ KNO3(aq) 1 NaCl(aq)

The last plant using the Solvay process in the United States closed in 1986.The last plant using the Solvay process in the United States closed in 1986.

†Ernest Solvay (1838–1922). Belgian chemist. Solvay’s main contribution to industrial chemistry was the development of the process for the production of sodium carbonate that now bears his name.

This process is carried out just below 100°C. Because KNO3 is the least soluble salt at room temperature, it is separated from the solution by fractional crystallization. Like NaNO3, KNO3 decomposes when heated. Gunpowder consists of potassium nitrate, wood charcoal, and sulfur in the approx-imate proportions of 6:1:1 by mass. When gunpowder is heated, the reaction is

2KNO3(s) 1 S(l) 1 3C(s) ¡ K2S(s) 1 N2(g) 1 3CO2(g)

The sudden formation of hot expanding gases causes an explosion.

20.6 The Alkaline Earth MetalsThe alkaline earth metals are somewhat less electropositive and less reactive than the alkali metals. Except for the fi rst member of the family, beryllium, which resem-bles aluminum (a Group 3A metal) in some respects, the alkaline earth metals have similar chemical properties. Because their M21 ions attain the stable electron con-fi guration of the preceding noble gas, the oxidation number of alkaline earth metals in the combined form is almost always 12. Table 20.5 lists some common proper-ties of these metals. Radium is not included in the table because all radium isotopes are radioactive and it is diffi cult and expensive to study the chemistry of this Group 2A element.

MagnesiumMagnesium (see Figure 8.15) is the sixth most plentiful element in Earth’s crust (about 2.5 percent by mass). Among the principal magnesium ores are brucite, Mg(OH)2; dolomite, CaCO3 ? MgCO3 (Figure 20.15); and epsomite, MgSO4 ? 7H2O. Seawater is a good source of magnesium; there are about 1.3 g of magnesium in each kilogram of seawater. As is the case with most alkali and alkaline earth metals, metallic magnesium is obtained by electrolysis, in this case from its molten chloride, MgCl2 (obtained from seawater, see p. 158).

Figure 20.15 Dolomite (CaCO3 ? MgCO3).

*Refers to the cation M21, where M denotes an alkali earth metal atom.†The half-reaction is M21(aq) 1 2e2 ¡ M(s).

Be Mg Ca Sr Ba

Valence electron confi guration 2s2 3s2 4s2 5s2 6s2

Density (g/cm3) 1.86 1.74 1.55 2.6 3.5Melting point (8C) 1280 650 838 770 714Boiling point (8C) 2770 1107 1484 1380 1640Atomic radius (pm) 112 160 197 215 222Ionic radius (pm)* 34 78 106 127 143First and second ionization 899 738 590 548 502energies (kJ/mol) 1757 1450 1145 1058 958Electronegativity 1.5 1.2 1.0 1.0 0.9Standard reduction potential (V)† 21.85 22.37 22.87 22.89 22.90

TABLE 20.5 Properties of Alkaline Earth Metals

20.6 The Alkaline Earth Metals 901


The chemistry of magnesium is intermediate between that of beryllium and the heavier Group 2A elements. Magnesium does not react with cold water but does react slowly with steam:

Mg(s) 1 H2O(g) ¡ MgO(s) 1 H2(g)

It burns brilliantly in air to produce magnesium oxide and magnesium nitride (see Figure 4.9):

2Mg(s) 1 O2(g) ¡ 2MgO(s)3Mg(s) 1 N2(g) ¡ Mg3N2(s)

This property makes magnesium (in the form of thin ribbons or fi bers) useful in fl ash photography and fl ares. Magnesium oxide reacts very slowly with water to form magnesium hydroxide, a white solid suspension called milk of magnesia (see p. 746), which is used to treat acid indigestion:

MgO(s) 1 H2O(l) ¡ Mg(OH)2(s)

Magnesium is a typical alkaline earth metal in that its hydroxide is a strong base. [The only alkaline earth hydroxide that is not a strong base is Be(OH)2, which is amphoteric.] The major uses of magnesium are in lightweight structural alloys, for cathodic protection (see Section 19.7), in organic synthesis, and in batteries. Magnesium is essential to plant and animal life, and Mg21 ions are not toxic. It is estimated that the average adult ingests about 0.3 g of magnesium ions daily. Magnesium plays several important biological roles. It is present in intracellular and extracellular fl uids. Magne-sium ions are essential for the proper functioning of a number of enzymes. Magnesium is also present in the green plant pigment chlorophyll, which plays an important part in photosynthesis.

CalciumEarth’s crust contains about 3.4 percent calcium (see Figure 8.15) by mass. Calcium occurs in limestone, calcite, chalk, and marble as CaCO3; in dolomite as CaCO3 ? MgCO3 (Figure 20.15); in gypsum as CaSO4 ? 2H2O; and in fl uorite as CaF2 (Figure 20.16). Metallic calcium is best prepared by the electrolysis of molten calcium chloride (CaCl2). As we read down Group 2A from beryllium to barium, we note an increase in metallic properties. Unlike beryllium and magnesium, calcium (like strontium and barium) reacts with cold water to yield the corresponding hydroxide, although the rate of reaction is much slower than those involving the alkali metals (see Figure 4.14):

Ca(s) 1 2H2O(l) ¡ Ca(OH)2(aq) 1 H2(g)

Calcium hydroxide [Ca(OH)2] is commonly known as slaked lime or hydrated lime. Lime (CaO), which is also referred to as quicklime, is one of the oldest materials known to mankind. Quicklime is produced by the thermal decomposition of calcium carbonate (see Section 18.3):

CaCO3(s) ¡ CaO(s) 1 CO2(g)Figure 20.16 Fluorite (CaF2).

while slaked lime is produced by the reaction between quicklime and water:

CaO(s) 1 H2O(l) ¡ Ca(OH)2(aq)

Quicklime is used in metallurgy (see Section 20.2) and the removal of SO2 when fossil fuel is burned (see p. 786). Slaked lime is used in water treatment. For many years, farmers have used lime to lower the acidity of soil for their crops (a process called liming). Nowadays lime is also applied to lakes affected by acid rain (see Section 17.6). Metallic calcium has rather limited uses. It serves mainly as an alloying agent for metals like aluminum and copper and in the preparation of beryllium metal from its compounds. It is also used as a dehydrating agent for organic solvents. Calcium is an essential element in living matter. It is the major component of bones and teeth; the calcium ion is present in a complex phosphate salt, hydroxyapa-tite, Ca5(PO4)3OH. A characteristic function of Ca21 ions in living systems is the activation of a variety of metabolic processes. Calcium plays a vital role in heart action, blood clotting, muscle contraction, and nerve impulse transmission.

20.7 AluminumAluminum (see Figure 8.16) is the most abundant metal and the third most plentiful element in Earth’s crust (7.5 percent by mass). The elemental form does not occur in nature; its principal ore is bauxite (Al2O3 ? 2H2O). Other minerals containing alumi-num are orthoclase (KAlSi3O8), beryl (Be3Al2Si6O18), cryolite (Na3AlF6), and corun-dum (Al2O3) (Figure 20.17). Aluminum is usually prepared from bauxite, which is frequently contaminated with silica (SiO2), iron oxides, and titanium(IV) oxide. The ore is fi rst heated in sodium hydroxide solution to convert the silica into soluble silicates:

SiO2(s) 1 2OH2(aq) ¡ SiO322(aq) 1 H2O(l)

At the same time, aluminum oxide is converted to the aluminate ion (AlO22):

Al2O3(s) 1 2OH2(aq) ¡ 2AlO22(aq) 1 H2O(l)

Iron oxide and titanium oxide are unaffected by this treatment and are fi ltered off. Next, the solution is treated with acid to precipitate the insoluble aluminum hydroxide:

AlO22(aq) 1 H3O

1(aq) ¡ Al(OH)3(s)

After fi ltration, the aluminum hydroxide is heated to obtain aluminum oxide:

2Al(OH)3(s) ¡ Al2O3(s) 1 3H2O(g)

Anhydrous aluminum oxide, or corundum, is reduced to aluminum by the Hall† process. Figure 20.18 shows a Hall electrolytic cell, which contains a series of car-bon anodes. The cathode is also made of carbon and constitutes the lining inside the

20.7 Aluminum 903

Figure 20.17 Corundum (Al2O3).

Carbon anodes

Al2O3 inmolten cryolite

Moltenaluminum

Carboncathode

Figure 20.18 Electrolytic production of aluminum based on the Hall process.

†Charles Martin Hall (1863–1914). American inventor. While Hall was an undergraduate at Oberlin Col-lege, he became interested in fi nding an inexpensive way to extract aluminum. Shortly after graduation, when he was only 22 years old, Hall succeeded in obtaining aluminum from aluminum oxide in a backyard woodshed. Amazingly, the same discovery was made at almost the same moment in France, by Paul Héroult, another 22-year-old inventor working in a similar makeshift laboratory.


cell. The key to the Hall process is the use of cryolite, or Na3AlF6 (m.p. 1000°C), as the solvent for aluminum oxide (m.p. 2045°C). The mixture is electrolyzed to produce aluminum and oxygen gas:

Anode (oxidation): 3[2O22 ¡ O2(g) 1 4e2] Cathode (reduction): 4[Al31 1 3e2 ¡ Al(l)]

Overall: 2Al2O3 ¡ 4Al(l) 1 3O2(g)

Oxygen gas reacts with the carbon anodes (at elevated temperatures) to form carbon monoxide, which escapes as a gas. The liquid aluminum metal (m.p. 660.2°C) sinks to the bottom of the vessel, from which it can be drained from time to time during the procedure. Aluminum is one of the most versatile metals known. It has a low density (2.7 g/cm3) and high tensile strength (that is, it can be stretched or drawn out). Aluminum is malleable, it can be rolled into thin foils, and it is an excellent electrical conductor. Its conductivity is about 65 percent that of copper. However, because aluminum is cheaper and lighter than copper, it is widely used in high-voltage transmission lines. Although aluminum’s chief use is in aircraft construction, the pure metal itself is too soft and weak to withstand much strain. Its mechanical properties are greatly improved by alloying it with small amounts of metals such as copper, magnesium, and manganese, as well as silicon. Aluminum is not used by living systems and is generally considered to be nontoxic. As we read across the periodic table from left to right in a given period, we note a gradual decrease in metallic properties. Thus, although aluminum is considered an active metal, it does not react with water as do sodium and calcium. Aluminum reacts with hydrochloric acid and with strong bases as follows:

2Al(s) 1 6HCl(aq) ¡ 2AlCl3(aq) 1 3H2(g) 2Al(s) 1 2NaOH(aq) 1 2H2O(l) ¡ 2NaAlO2(aq) 1 3H2(g)

Aluminum readily forms the oxide Al2O3 when exposed to air:

4Al(s) 1 3O2(g) ¡ 2Al2O3(s)

A tenacious fi lm of this oxide protects metallic aluminum from further corrosion and accounts for some of the unexpected inertness of aluminum. Aluminum oxide has a very large exothermic enthalpy of formation (DH8f 5 21670 kJ/mol). This property makes aluminum suitable for use in solid propellants for rockets such as those used for some space shuttles. When a mixture of aluminum and ammonium perchlorate (NH4ClO4) is ignited, aluminum is oxidized to Al2O3, and the heat liberated in the reaction causes the gases that are formed to expand with great force. This action lifts the rocket. The great affi nity of aluminum for oxygen is illustrated nicely by the reaction of aluminum powder with a variety of metal oxides, particularly the transition metal oxides, to produce the corresponding metals. A typical reaction is

2Al(s) 1 Fe2O3(s) ¡ Al2O3(s) 1 2Fe(l) ¢H° 5 2822.8 kJ/mol

which can result in temperatures approaching 3000°C. This reaction, which is used in the welding of steel and iron, is called the thermite reaction (Figure 20.19). Aluminum chloride exists as a dimer:

AlAlD

Cl

GCl

DCl

qr

ClD

ClG

ClD

Molten cryolite provides a good conducting medium for electrolysis.Molten cryolite provides a good conducting medium for electrolysis.

Figure 20.19 The temperature of a thermite reaction can reach 30008C.

Media PlayerAluminum Production

Each of the bridging chlorine atoms forms a normal covalent bond and a coordinate covalent bond (indicated by S) with two aluminum atoms. Each aluminum atom is assumed to be sp3-hybridized, so the vacant sp3 hybrid orbital can accept a lone pair from the chlorine atom (Figure 20.20). Aluminum chloride undergoes hydrolysis as follows:

AlCl3(s) 1 3H2O(l) ¡ Al(OH)3(s) 1 3HCl(aq)

Aluminum hydroxide, like Be(OH)2, is amphoteric:

Al(OH)3(s) 1 3H1(aq) ¡ Al31(aq) 1 3H2O(l) Al(OH)3(s) 1 OH2(aq) ¡ Al(OH)4

2(aq)

In contrast to the boron hydrides, which are a well-defi ned series of compounds, aluminum hydride is a polymer in which each aluminum atom is surrounded octahe-drally by bridging hydrogen atoms (Figure 20.21). When an aqueous mixture of aluminum sulfate and potassium sulfate is evapo-rated slowly, crystals of KAl(SO4)2 ? 12H2O are formed. Similar crystals can be formed by substituting Na1 or NH4

1 for K1, and Cr31 or Fe31 for Al31. These com-pounds are called alums, and they have the general formula

M1M31 (SO4)2 ? 12H2O M1: K1, Na1, NH41

M31: Al31, Cr31, Fe31

Alums are examples of double salts, that is, salts that contain two different cations.

In 2002, chemists prepared the fi rst member of aluminum hydride (Al2H6), which possesses bridging H atoms like diborane, B2H6.

In 2002, chemists prepared the fi rst member of aluminum hydride (Al2H6), which possesses bridging H atoms like diborane, B2H6.

20.7 Aluminum 905

Figure 20.21 Structure of aluminum hydride. Note that this compound is a polymer. Each Al atom is surrounded octahedrally by six bridging H atoms.

Groundstate

Promotionof electron

sp3-Hybridizedstate

3s

3s

sp 3 orbitals

3p

3p

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

Figure 20.20 The sp3 hybridization of an Al atom in Al2Cl6. Each Al atom has one vacant sp3 hybrid orbital that can accept a lone pair from the bridging Cl atom.

1. Depending on their reactivities, metals exist in nature in either the free or combined state.

2. Recovering a metal from its ore is a three-stage process. First, the ore must be prepared. The metal is then separated, usually by a reduction process, and fi nally, it is purifi ed.

3. The methods commonly used for purifying metals are distillation, electrolysis, and zone refi ning.

4. Metallic bonds can be thought of as the force between positive ions immersed in a sea of electrons. In terms of band theory, the atomic orbitals merge to form energy bands. A substance is a conductor when electrons can be readily promoted to the conduction band, where they are free to move through the substance.

5. In insulators, the energy gap between the valence band and the conduction band is so large that electrons can-

Summary of Facts and Concepts

not be promoted into the conduction band. In semicon-ductors, electrons can cross the energy gap at higher temperatures, and therefore conductivity increases with increasing temperature as more electrons are able to reach the conduction band.

6. n-Type semiconductors contain donor impurities and extra electrons. p-Type semiconductors contain accep-tor impurities and “positive holes.”

7. The alkali metals are the most reactive of all the metal-lic elements. They have an oxidation number of 11 in their compounds. Under special conditions, some of them also form uninegative ions.

8. The alkaline earth metals are somewhat less reactive than the alkali metals. They almost always have an oxidation number of 12 in their compounds. The

Media PlayerChapter Summary

906

C H E M I S T R Y

in ActionRecycling Aluminum

A luminum beverage cans were virtually unknown in 1960; yet by the early 1970s over 1.3 billion pounds of aluminum

had been used for these containers. The reasons for aluminum’s

popularity in the beverage industry are that it is nontoxic, odor-less, tasteless, and lightweight. Furthermore, it is thermally con-ducting, so the fl uid inside the container can be chilled rapidly.

Left: Collecting aluminum cans for recycling.Right: Melting and purifying recycled aluminum.

Acceptor impurity, p. 896Alloy, p. 886Amalgam, p. 887Band theory, p. 894

Key Words

Conductor, p. 895Donor impurity, p. 895Ferromagnetic, p. 887Insulator, p. 895

Metallurgy, p. 886Mineral, p. 886n-Type semiconductor, p. 895Ore, p. 886

p-Type semiconductor, p. 896Pyrometallurgy, p. 888Semiconductors, p. 895

Electronic Homework Problems

The following problems are available at www.aris.mhhe.com if assigned by your instructor as electronic homework.

ARIS Problems: 20.11, 20.12, 20.16, 20.27, 20.30, 20.34, 20.39, 20.44, 20.45, 20.46, 20.47, 20.48, 20.54, 20.57, 20.58, 20.63, 20.65, 20.66, 20.69, 20.71, 20.74, 20.76.

907

The tremendous increase in the demand for aluminum does have a defi nite drawback, however. More than 3 billion pounds of the metal cans and foils are discarded in the United States annually. They litter the countryside and clog landfi lls. The best solution to this environmental problem, and the way to prevent the rapid depletion of a finite re-source, is recycling. What are the economic benefi ts of recycling aluminum? Let us compare the energy consumed in the production of alu-minum from bauxite with that consumed when aluminum is recycled. The overall reaction for the Hall process can be repre-sented as

Al2O3 (in molten cryolite) 1 3C(s) ¡ 2Al(l) 1 3CO(g)

for which DH° 5 1340 kJ/mol and DS° 5 586 J/K ? mol. At 1000°C, which is the temperature of the process, the standard free-energy change for the reaction is given by

¢G° 5 ¢H° 2 T¢S°

5 1340 kJ/mol 2 (1273 K)a 586 JK ? mol

ba 1 kJ1000 J

b 5 594 kJ/mol

Equation (19.3) states that DG° 5 2nFE°; therefore, the amount of electrical energy needed to produce 1 mole of Al from bauxite is 594 kJ/2, or 297 kJ. Recycling aluminum requires only enough energy to heat the metal to its melting point (660°C) plus the heat of fusion

(10.7 kJ/mol). The heat change where 1 mole of aluminum is heated from 25°C to 660°C is

heat input 5 ms¢t 5 (27.0 g)(0.900 J/g ? °C)(660 2 25)°C 5 15.4 kJ

where m is the molar mass, s is the specifi c heat of Al, and Dt is the temperature change. Thus, the total energy needed to recycle 1 mole of Al is given by

total energy 5 15.4 kJ 1 10.7 kJ 5 26.1 kJ

To compare the energy requirements of the two methods we write

energy needed to recycle 1 mol Al

energy needed to produce 1 mol Al by electrolysis

526.1 kJ297 kJ

3 100%

5 8.8%

Thus, by recycling aluminum cans we can save about 91 percent of the energy required to extract the metal from bauxite. Recy-cling most of the aluminum cans thrown away each year saves 20 billion kilowatt-hours of electricity—about 1 percent of the electric power used in the United States annually. (Watt is the unit for power, 1 watt 5 1 joule per second.)

properties of the alkaline earth elements become in-creasingly metallic from top to bottom in their periodic group.

9. Aluminum does not react with water due to the forma-tion of a protective oxide; its hydroxide is amphoteric.


Questions and Problems

Occurrence of MetalsReview Questions

20.1 Defi ne mineral, ore, and metallurgy.20.2 List three metals that are usually found in an uncom-

bined state in nature and three metals that are always found in a combined state in nature.

20.3 Write chemical formulas for the following minerals: (a) calcite, (b) dolomite, (c) fl uorite, (d) halite, (e) corundum, (f) magnetite, (g) beryl, (h) galena, (i) epsomite, ( j) anhydrite.

20.4 Name the following minerals: (a) MgCO3, (b) Na3AlF6, (c) Al2O3, (d) Ag2S, (e) HgS, (f ) ZnS, (g) SrSO4, (h) PbCO3, (i) MnO2, ( j) TiO2.

Metallurgical ProcessesReview Questions

20.5 Describe the main steps involved in the preparation of an ore.

20.6 What does roasting mean in metallurgy? Why is roast-ing a major source of air pollution and acid rain?

20.7 Describe with examples the chemical and electrolytic reduction processes used in the production of metals.

20.8 Describe the main steps used to purify metals.20.9 Describe the extraction of iron in a blast furnace.20.10 Briefl y discuss the steelmaking process.

Problems

20.11 In the Mond process for the purifi cation of nickel, CO is passed over metallic nickel to give Ni(CO)4:

Ni(s) 1 4CO(g) ∆ Ni(CO)4(g)

Given that the standard free energies of formation of CO(g) and Ni(CO)4(g) are 2137.3 kJ/mol and 2587.4 kJ/mol, respectively, calculate the equilib-rium constant of the reaction at 80°C. (Assume DG°f to be independent of temperature.)

20.12 Copper is purifi ed by electrolysis (see Figure 20.6). A 5.00-kg anode is used in a cell where the current is 37.8 A. How long (in hours) must the current run to dis solve this anode and electroplate it onto the cathode?

20.13 Consider the electrolytic procedure for purifying copper described in Figure 20.6. Suppose that a sam-ple of copper contains the following impurities: Fe, Ag, Zn, Au, Co, Pt, and Pb. Which of the metals will be oxidized and dissolved in solution and which will be unaffected and simply form the sludge that accumulates at the bottom of the cell?

20.14 How would you obtain zinc from sphalerite (ZnS)?

20.15 Starting with rutile (TiO2), explain how you would obtain pure titanium metal. (Hint: First convert TiO2 to TiCl4. Next, reduce TiCl4 with Mg. Look up physi-cal properties of TiCl4, Mg, and MgCl2 in a chemistry handbook.)

20.16 A certain mine produces 2.0 3 108 kg of copper from chalcopyrite (CuFeS2) each year. The ore contains only 0.80 percent Cu by mass. (a) If the density of the ore is 2.8 g/cm3, calculate the volume (in cm3) of ore removed each year. (b) Calculate the mass (in kg) of SO2 produced by roasting (assume chalcopyrite to be the only source of sulfur).

20.17 Which of the following compounds would require electrolysis to yield the free metals? Ag2S, CaCl2, NaCl, Fe2O3, Al2O3, TiCl4.

20.18 Although iron is only about two-thirds as abundant as aluminum in Earth’s crust, mass for mass it costs only about one-quarter as much to produce. Why?

Band Theory of Electrical ConductivityReview Questions

20.19 Defi ne the following terms: conductor, insulator, semiconducting elements, donor impurities, ac-ceptor impurities, n-type semiconductors, p-type semiconductors.

20.20 Briefl y discuss the nature of bonding in metals, insu-lators, and semiconducting elements.

20.21 Describe the general characteristics of n-type and p-type semiconductors.

20.22 State whether silicon would form n-type or p-type semiconductors with the following elements: Ga, Sb, Al, As.

Alkali MetalsReview Questions

20.23 How is sodium prepared commercially?20.24 Why is potassium usually not prepared electrolyti-

cally from one of its salts?20.25 Describe the uses of the following compounds: NaCl,

Na2CO3, NaOH, KOH, KO2.20.26 Under what conditions do sodium and potassium

form Na2 and K2 ions?

Problems

20.27 Complete and balance the following equations:(a) K(s) 1 H2O(l) ¡(b) NaH(s) 1 H2O(l) ¡(c) Na(s) 1 O2(g) ¡(d) K(s) 1 O2(g) ¡

Questions and Problems 909

Problems

20.43 Before Hall invented his electrolytic process, alumi-num was produced by the reduction of its chloride with an active metal. Which metals would you use for the production of aluminum in that way?

20.44 With the Hall process, how many hours will it take to deposit 664 g of Al at a current of 32.6 A?

20.45 Aluminum forms the complex ions AlCl42 and AlF6

32. Describe the shapes of these ions. AlCl6

32 does not form. Why? (Hint: Consider the relative sizes of Al31, F2, and Cl2 ions.)

20.46 The overall reaction for the electrolytic production of aluminum by means of the Hall process may be represented as

Al2O3 (s) 1 3C (s) ¡ 2Al(l) 1 3CO(g)

At 1000°C, the standard free-energy change for this process is 594 kJ/mol. (a) Calculate the mini-mum voltage required to produce 1 mole of alumi-num at this temperature. (b) If the actual voltage applied is exactly three times the ideal value, cal-culate the energy required to produce 1.00 kg of the metal.

20.47 In basic solution, aluminum metal is a strong reduc-ing agent and is oxidized to AlO2

2. Give balanced equations for the reaction of Al in basic solution with the following: (a) NaNO3, to give ammonia; (b) water, to give hydrogen; (c) Na2SnO3, to give metallic tin.

20.48 Write a balanced equation for the thermal decompo-sition of aluminum nitrate to form aluminum oxide, nitrogen dioxide, and oxygen gas.

20.49 Describe some of the properties of aluminum that make it one of the most versatile metals known.

20.50 The pressure of gaseous Al2Cl6 increases more rap-idly with temperature than predicted by the ideal gas equation even though Al2Cl6 behaves like an ideal gas. Explain.

20.51 Starting with aluminum, describe with balanced equa-tions how you would prepare (a) Al2Cl6, (b) Al2O3, (c) Al2(SO4)3, (d) NH4Al(SO4)2 ? 12H2O.

20.52 Explain the change in bonding when Al2Cl6 dissoci-ates to form AlCl3 in the gas phase.

Additional Problems20.53 In steelmaking, nonmetallic impurities such as P, S,

and Si are removed as the corresponding oxides. The inside of the furnace is usually lined with CaCO3 and MgCO3, which decompose at high temperatures to yield CaO and MgO. How do CaO and MgO help in the removal of the nonmetallic oxides?

20.54 When 1.164 g of a certain metal sulfi de was roasted in air, 0.972 g of the metal oxide was formed. If the

20.28 Write a balanced equation for each of the following reactions: (a) sodium reacts with water; (b) an aque-ous solution of NaOH reacts with CO2; (c) solid Na2CO3 reacts with a HCl solution; (d) solid NaHCO3 reacts with a HCl solution; (e) solid NaHCO3 is heated; (f) solid Na2CO3 is heated.

20.29 Sodium hydride (NaH) can be used as a drying agent for many organic solvents. Explain how it works.

20.30 Calculate the volume of CO2 at 10.0°C and 746 mmHg pressure obtained by treating 25.0 g of Na2CO3 with an excess of hydrochloric acid.

Alkaline Earth MetalsReview Questions

20.31 List the common ores of magnesium and calcium.20.32 How are the metals magnesium and calcium obtained

commercially?

Problems

20.33 From the thermodynamic data in Appendix 3, calculate the DH° values for the following decompositions:(a) MgCO3(s) ¡ MgO(s) 1 CO2(g)(b) CaCO3(s) ¡ CaO(s) 1 CO2(g)

Which of the two compounds is more easily decom-posed by heat?

20.34 Starting with magnesium and concentrated nitric acid, describe how you would prepare magnesium oxide. [Hint: First convert Mg to Mg(NO3)2. Next, MgO can be obtained by heating Mg(NO3)2.]

20.35 Describe two ways of preparing magnesium chloride.20.36 The second ionization energy of magnesium is only

about twice as great as the fi rst, but the third ioniza-tion energy is 10 times as great. Why does it take so much more energy to remove the third electron?

20.37 List the sulfates of the Group 2A metals in order of increasing solubility in water. Explain the trend. (Hint: You need to consult a chemistry handbook.)

20.38 Helium contains the same number of electrons in its outer shell as do the alkaline earth metals. Explain why helium is inert whereas the Group 2A metals are not.

20.39 When exposed to air, calcium fi rst forms calcium oxide, which is then converted to calcium hydroxide, and fi nally to calcium carbonate. Write a balanced equation for each step.

20.40 Write chemical formulas for (a) quicklime, (b) slaked lime, (c) limewater.

AluminumReview Questions

20.41 Describe the Hall process for preparing aluminum.20.42 What action renders aluminum inert?


oxidation number of the metal is 12, calculate the molar mass of the metal.

20.55 An early view of metallic bonding assumed that bond-ing in metals consisted of localized, shared electron-pair bonds between metal atoms. What evidence would help you to argue against this viewpoint?

20.56 Referring to Figure 20.6, would you expect H2O and H1 to be reduced at the cathode and H2O oxidized at the anode?

20.57 A 0.450-g sample of steel contains manganese as an impurity. The sample is dissolved in acidic solution and the manganese is oxidized to the permanganate ion MnO4

2. The MnO42 ion is reduced to Mn21 by

reacting with 50.0 mL of 0.0800 M FeSO4 solution. The excess Fe21 ions are then oxidized to Fe31 by 22.4 mL of 0.0100 M K2Cr2O7. Calculate the percent by mass of manganese in the sample.

20.58 Given that DG°f (Fe2O3) 5 2741.0 kJ/mol and that DG°f(Al2O3) 5 21576.4 kJ/mol, calculate DG° for the following reactions at 25°C:(a) 2Fe2O3(s) ¡ 4Fe(s) 1 3O2(g)(b) 2Al2O3(s) ¡ 4Al(s) 1 3O2(g)

20.59 Use compounds of aluminum as an example to explain what is meant by amphoterism.

20.60 When an inert atmosphere is needed for a metallurgi-cal process, nitrogen is frequently used. However, in the reduction of TiCl4 by magnesium, helium is used. Explain why nitrogen is not suitable for this process.

20.61 It has been shown that Na2 species form in the vapor phase. Describe the formation of the “disodium mole-cule” in terms of a molecular orbital energy-level diagram. Would you expect the alkaline earth metals to exhibit a similar property?

20.62 Explain each of the following statements: (a) An aqueous solution of AlCl3 is acidic. (b) Al(OH)3 is soluble in NaOH solutions but not in NH3 solution.

20.63 Write balanced equations for the following reac-tions: (a) the heating of aluminum carbonate; (b) the reaction between AlCl3 and K; (c) the reaction between solutions of Na2CO3 and Ca(OH)2.

20.64 Write a balanced equation for the reaction between calcium oxide and dilute HCl solution.

20.65 What is wrong with the following procedure for obtaining magnesium?

MgCO3 ¡ MgO(s) 1 CO2(g) MgO(s) 1 CO(g) ¡ Mg(s) 1 CO2(g)

20.66 Explain why most metals have a fl ickering appearance.

20.67 Predict the chemical properties of francium, the last member of Group 1A.

20.68 Describe a medicinal or health-related application for each of the following compounds: NaF, Li2CO3, Mg(OH)2, CaCO3, BaSO4, Al(OH)2NaCO3. (You would need to do a Web search for some of these compounds.)

20.69 The following are two reaction schemes involving magnesium. Scheme I: When magnesium burns in oxygen, a white solid (A) is formed. A dissolves in 1 M HCl to give a colorless solution (B). Upon ad-dition of Na2CO3 to B, a white precipitate is formed (C). On heating, C decomposes to D and a color-less gas is generated (E). When E is passed through limewater [an aqueous suspension of Ca(OH)2], a white precipitate appears (F). Scheme II: Magne-sium reacts with 1 M H2SO4 to produce a colorless solution (G). Treating G with an excess of NaOH produces a white precipitate (H). H dissolves in 1 M HNO3 to form a colorless solution. When the solution is slowly evaporated, a white solid (I) appears. On heating I, a brown gas is given off. Identify A–I and write equations representing the reactions involved.

20.70 Lithium and magnesium exhibit a diagonal relation-ship in some chemical properties. How does lithium resemble magnesium in its reaction with oxygen and nitrogen? Consult a handbook of chemistry and com-pare the solubilities of carbonates, fl uorides, and phosphates of these metals.

20.71 To prevent the formation of oxides, peroxides and superoxides, alkali metals are sometimes stored in an inert atmosphere. Which of the following gases should not be used for lithium? Why? Ne, Ar, N2, Kr.

20.72 Which of the following metals is not found in the free state in nature: Ag, Cu, Zn, Au, Pt?

Special Problems 911

Special Problems

20.73 After heating, a metal surface (such as that of a cooking pan or skillet) develops a color pattern like an oil slick on water. Explain.

20.74 Chemical tests of four metals A, B, C, and D show the following results.(a) Only B and C react with 0.5 M HCl to give H2 gas.(b) When B is added to a solution containing the

ions of the other metals, metallic A, C, and D are formed.

(c) A reacts with 6 M HNO3 but D does not. Arrange the metals in the increasing order as re-

ducing agents. Suggest four metals that fi t these descriptions.

20.75 The electrical conductance of copper metal decreases with temperature, but that of a CuSO4 solution increases with temperature. Explain.

20.76 As stated in the chapter, potassium superoxide (KO2) is a useful source of oxygen employed in breathing equipment. Calculate the pressure at which oxygen gas stored at 20°C would have the same density as the oxygen gas provided by KO2. The density of KO2 at 20°C is 2.15 g/cm3.

20.77 A sample of 10.00 g of sodium reacts with oxygen to form 13.83 g of sodium oxide (Na2O) and sodium peroxide (Na2O2). Calculate the percent composition of the mixture.

Nonmetallic Elements and Their Compounds

The nose cone of the space shuttle is made of graphite and silicon carbide and can withstand the tremendous heat generated when the vehicle enters Earth’s atmosphere. The models show graphite and silicon carbide, which has a diamond structure.

913


ARISEnd of Chapter Problems

A Look Ahead• This chapter starts by examining the general properties of the nonmetals. (21.1)

• We see that hydrogen does not have a unique position in the periodic table. We learn the preparation of hydrogen and study several different types of compounds containing hydrogen. We also discuss the hydrogenation reaction and the role hydrogen plays in energy production. (21.2)

• Next, we consider the inorganic world of carbon in terms of carbides, cyanides, and carbon monoxide and carbon dioxide. (21.3)

• Nitrogen is the most abundant element in the atmosphere. Its major com-pounds are ammonia, hydrazine, and several oxides. Nitric acid, a strong oxidizing agent, is a major industrial chemical. Phosphorus is the other important element in Group 5A. It is a major component of teeth and bones and in genetic materials like deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Phosphorus compounds include hydride and oxides. Phosphoric acid has many commercial applications. (21.4)

• Oxygen is the most abundant element in Earth’s crust. It forms compounds with most other elements as oxides, peroxides, and superoxides. Its allotropic form, ozone, is a strong oxidizing agent. Sulfur, the second member of Group 6A, also forms many compounds with metals and nonmetals. Sulfuric acid is the most important industrial chemical in the world. (21.5)

• The halogens are the most electronegative and most reactive of the nonmetals. We study their preparations, properties, reactions, and applications of their compounds. (21.6)

Of the 117 elements known, only 25 are nonmetallic elements. Unlike the metals, the chemistry of these elements is diverse. Despite their relatively

small number, most of the essential elements in biological systems are nonmetals (H, C, N, P, O, S, Cl, and I). This group of nonmetallic elements also includes the most unreactive of the elements—the noble gases. The unique properties of hydrogen set it aside from the rest of the elements in the periodic table. A whole branch of chemistry—organic chemistry—is based on carbon compounds. In this chapter, we continue our survey of the elements by concentrating on the nonmetals. The emphasis will again be on important chemical properties and on the roles of nonmetals and their compounds in industrial, chemical, and bio-logical processes.

21.1 General Properties of Nonmetals

21.2 Hydrogen

21.3 Carbon

21.4 Nitrogen and Phosphorus

21.5 Oxygen and Sulfur

21.6 The Halogens

Chapter Outline


914 Nonmetallic Elements and Their Compounds

21.1 General Properties of NonmetalsProperties of nonmetals are more varied than those of metals. A number of nonmetals are gases in the elemental state: hydrogen, oxygen, nitrogen, fl uorine, chlorine, and the noble gases. Only one, bromine, is a liquid. All the remaining nonmetals are solids at room temperature. Unlike metals, nonmetallic elements are poor conductors of heat and electricity; they exhibit both positive and negative oxidation numbers. A small group of elements, called metalloids, have properties characteristic of both metals and nonmetals. The metalloids boron, silicon, germanium, and arsenic are semiconducting elements (see Section 20.3). Nonmetals are more electronegative than metals. The electronegativity of ele-ments increases from left to right across any period and from bottom to top in any group in the periodic table (see Figure 9.5). With the exception of hydrogen, the non-metals are concentrated in the upper right-hand corner of the periodic table (Figure 21.1). Compounds formed by a combination of metals and nonmetals tend to be ionic, hav-ing a metallic cation and a nonmetallic anion. In this chapter, we will discuss the chemistry of a number of common and impor-tant nonmetallic elements: hydrogen; carbon (Group 4A); nitrogen and phosphorus (Group 5A); oxygen and sulfur (Group 6A); and fl uorine, chlorine, bromine, and iodine (Group 7A).

21.2 HydrogenHydrogen is the simplest element known—its most common atomic form contains only one proton and one electron. The atomic form of hydrogen exists only at very high temperatures, however. Normally, elemental hydrogen is a diatomic molecule, the product of an exothermic reaction between H atoms:

H(g) 1 H(g) ¡ H2(g) ¢H° 5 2436.4 kJ/mol

Molecular hydrogen is a colorless, odorless, and nonpoisonous gas. At 1 atm, liquid hydrogen has a boiling point of 2252.9°C (20.3 K).

Recall that there is no totally suitable position for hydrogen in the periodic table.

Recall that there is no totally suitable position for hydrogen in the periodic table.

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

8 107 7B

9 8B

11 1B

12 2B

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

1H

3Li

11Na

19K

37Rb

55Cs

87Fr

20Ca

38Sr

56Ba

88Ra

21Sc

39Y

57La

89Ac

22Ti

40Zr

72Hf

104Rf

23V

41Nb

73Ta

105Db

24Cr

42Mo

74W

106Sg

25Mn

43Tc

75Re

107Bh

26Fe

44Ru

76Os

108Hs

27Co

45Rh

77Ir

109Mt

110Ds

111Rg

112 113 114 115 116 (117) 118

28Ni

46Pd

78Pt

29Cu

47Ag

79Au

30Zn

48Cd

80Hg

31Ga

49In

81Tl

32Ge

50Sn

82Pb

33As

51Sb

83Bi

34Se

52Te

84Po

35Br

53I

85At

36Kr

54Xe

86Rn

13Al

14Si

15P

16S

17Cl

18Ar

5B

6C

7N

8O

9F

10Ne

2He

4Be

12Mg

Figure 21.1 Representative nonmetallic elements (in blue) and metalloids (gray).

Hydrogen is the most abundant element in the universe, accounting for about 70 percent of the universe’s total mass. It is the tenth most abundant element in Earth’s crust, where it is found in combination with other elements. Unlike Jupiter and Saturn, Earth does not have a strong enough gravitational pull to retain the lightweight H2 molecules, so hydrogen is not found in our atmosphere. The ground-state electron confi guration of H is 1s1. It resembles the alkali metals in that it can be oxidized to the H1 ion, which exists in aqueous solutions in the hydrated form. On the other hand, hydrogen resembles the halogens in that it forms the uninegative hydride ion (H2), which is isoelectronic with helium (1s2). Hydrogen is found in a large number of covalent compounds. It also has the unique capacity for hydrogen-bond formation (see Section 11.2). Hydrogen gas plays an important role in industrial processes. About 95 percent of the hydrogen produced is used captively; that is, it is produced at or near the plant where it is used for industrial processes, such as the synthesis of ammonia. The large-scale industrial preparation is the reaction between propane (from natural gas and also as a product of oil refi neries) and steam in the presence of a catalyst at 900°C:

C3H8(g) 1 3H2O(g) ¡ 3CO(g) 1 7H2(g)

In another process, steam is passed over a bed of red-hot coke:

C(s) 1 H2O(g) ¡ CO(g) 1 H2(g)

The mixture of carbon monoxide and hydrogen gas produced in this reaction is com-monly known as water gas. Because both CO and H2 burn in air, water gas was used as a fuel for many years. But because CO is poisonous, water gas has been replaced by natural gases, such as methane and propane. Small quantities of hydrogen gas can be prepared conveniently in the laboratory by reacting zinc with dilute hydrochloric acid (Figure 21.2):

Zn(s) 1 2HCl(aq) ¡ ZnCl2(aq) 1 H2(g)

Hydrogen gas can also be produced by the reaction between an alkali metal or an alkaline earth metal (Ca or Ba) and water (see Section 4.4), but these reactions are

Hydrogen typically has 11 oxidation state in its compounds, but in ionic hydrides it has a 21 oxidation state.

Hydrogen typically has 11 oxidation state in its compounds, but in ionic hydrides it has a 21 oxidation state.

Figure 21.2 Apparatus for the laboratory preparation of hydrogen gas. The gas is collected over water, as is also the case of oxygen gas (see Figure 5.12).

Zn

HCl

H2 gas

Water

21.2 Hydrogen 915


too violent to be suitable for the laboratory preparation of hydrogen gas. Very pure hydrogen gas can be obtained by the electrolysis of water, but this method consumes too much energy to be practical on a large scale.

Binary HydridesBinary hydrides are compounds containing hydrogen and another element, either a metal or a nonmetal. Depending on structure and properties, these hydrides are broadly divided into three types: (1) ionic hydrides, (2) covalent hydrides, and (3) interstitial hydrides.

Ionic HydridesIonic hydrides are formed when molecular hydrogen combines directly with any alkali metal or with the alkaline earth metals Ca, Sr, or Ba:

2Li(s) 1 H2(g) ¡ 2LiH(s) Ca(s) 1 H2(g) ¡ CaH2(s)

All ionic hydrides are solids that have the high melting points characteristic of ionic compounds. The anion in these compounds is the hydride ion, H2, which is a very strong Brønsted base. It readily accepts a proton from a proton donor such as water:

H2(aq) 1 H2O(l) ¡ OH2(aq) 1 H2(g)

Due to their high reactivity with water, ionic hydrides are frequently used to remove traces of water from organic solvents.

Covalent HydridesIn covalent hydrides, the hydrogen atom is covalently bonded to the atom of another element. There are two types of covalent hydrides—those containing discrete molec-ular units, such as CH4 and NH3, and those having complex polymeric structures, such as (BeH2)x and (AlH3)x, where x is a very large number. Figure 21.3 shows the binary ionic and covalent hydrides of the representative elements. The physical and chemical properties of these compounds change from ionic to covalent across a given period. Consider, for example, the hydrides of the second-period elements: LiH, BeH2, B2H6, CH4, NH3, H2O, and HF. LiH is an ionic com-pound with a high melting point (680°C). The structure of BeH2 (in the solid state) is polymeric; it is a covalent compound. The molecules B2H6 and CH4 are nonpolar. In contrast, NH3, H2O, and HF are all polar molecules in which the hydrogen atom is the positive end of the polar bond. Of this group of hydrides (NH3, H2O, and HF), only HF is acidic in water. As we move down any group in Figure 21.3, the compounds change from cova-lent to ionic. In Group 2A, for example, BeH2 and MgH2 are covalent, but CaH2, SrH2, and BaH2 are ionic.

Interstitial HydridesMolecular hydrogen forms a number of hydrides with transition metals. In some of these compounds, the ratio of hydrogen atoms to metal atoms is not a constant. Such compounds are called interstitial hydrides. For example, depending on conditions, the formula for titanium hydride can vary between TiH1.8 and TiH2. Many of the interstitial hydrides have metallic properties such as electrical con-ductivity. Yet it is known that hydrogen is defi nitely bonded to the metal in these compounds, although the exact nature of the bonding is often not clear.

This is an example of the diagonal relationship between Be and Al (see p. 344).

This is an example of the diagonal relationship between Be and Al (see p. 344).

Interstitial compounds are sometimes called nonstoichiometric compounds. Note that they do not obey the law of defi nite proportions (see Section 2.1).

Interstitial compounds are sometimes called nonstoichiometric compounds. Note that they do not obey the law of defi nite proportions (see Section 2.1).

21.2 Hydrogen 917

Molecular hydrogen interacts in a unique way with palladium (Pd). Hydrogen gas is readily adsorbed onto the surface of the palladium metal, where it dissociates into atomic hydrogen. The H atoms then “dissolve” into the metal. On heating and under the pressure of H2 gas on one side of the metal, these atoms diffuse through the metal and recombine to form molecular hydrogen, which emerges as the gas from the other side. Because no other gas behaves in this way with palladium, this process has been used to separate hydrogen gas from other gases on a small scale.

Isotopes of HydrogenHydrogen has three isotopes: 1

1H (hydrogen), 12H (deuterium, symbol D), and 1

3H (tritium, symbol T). The natural abundances of the stable hydrogen isotopes are hydrogen, 99.985 percent; and deuterium, 0.015 percent. Tritium is a radioactive isotope with a half-life of about 12.5 years. Table 21.1 compares some of the common properties of H2O with those of D2O. Deuterium oxide, or heavy water as it is commonly called, is used in some nuclear reactors as a coolant and a moderator of nuclear reactions (see Chapter 23). D2O can be separated from H2O by fractional distillation because H2O boils at a lower

The 11H isotope is also called protium.

Hydrogen is the only element whose isotopes are given different names.

The 11H isotope is also called protium.

Hydrogen is the only element whose isotopes are given different names.

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

8 107 7B

9 8B

11 1B

12 2B

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

Discrete molecular units

Polymeric structure; covalent compound

Ionic compoundLiH

NaH

KH

RbH

CsH

CaH2

SrH2

BaH2

GaH3

InH3

TlH3

GeH4

SnH4

PbH4

AsH3

SbH3

BiH3

H2Se

H2Te

HBr

HI

AlH3 SiH4 PH3 H2S HCl

B2H6 CH4 NH3 H2O HFBeH2

MgH2

Figure 21.3 Binary hydrides of the representative elements. In cases in which hydrogen forms more than one compound with the same element, only the formula of the simplest hydride is shown. The properties of many of the transition metal hydrides are not well characterized.

Property H2O D2O

Molar mass (g/mol) 18.02 20.03Melting point (°C) 0 3.8Boiling point (°C) 100 101.4Density at 4°C (g/cm3) 1.000 1.108

TABLE 21.1 Properties of H2O and D2O


temperature, as Table 21.1 shows. Another technique for its separation is electrolysis of water. Because H2 gas is formed about eight times as fast as D2 during electrolysis, the water remaining in the electrolytic cell becomes progressively enriched with D2O. Interestingly, the Dead Sea, which for thousands of years has entrapped water that has no outlet other than through evaporation, has a higher [D2O]/[H2O] ratio than water found elsewhere. Although D2O chemically resembles H2O in most respects, it is a toxic substance. The reason is that deuterium is heavier than hydrogen; thus, its compounds often react more slowly than those of the lighter isotope. Regular drinking of D2O instead of H2O could prove fatal because of the slower rate of transfer of D1 compared with that of H1 in the acid-base reactions involved in enzyme catalysis. This kinetic isotope effect is also manifest in acid ionization constants. For example, the ionization constant of acetic acid

CH3COOH(aq) ∆ CH3COO2(aq) 1 H1(aq) Ka 5 1.8 3 1025

is about three times as large as that of deuterated acetic acid:

CH3COOD(aq) ∆ CH3COO2(aq) 1 D1(aq) Ka 5 6 3 1026

HydrogenationHydrogenation is the addition of hydrogen to compounds containing multiple bonds, especially CP C and CqC bonds. A simple hydrogenation reaction is the conversion of ethylene to ethane:

HO CO CO H

HA

AH

HA

AH

ethylene ethane

H2 ! 88nCP CD

H

GH

HG

HD

This reaction is quite slow under normal conditions, but the rate can be greatly increased by the presence of a catalyst such as nickel or platinum. As in the Haber synthesis of ammonia (see Section 13.6), the main function of the catalyst is to weaken the H—H bond and facilitate the reaction. Hydrogenation is an important process in the food industry. Vegetable oils have considerable nutritional value, but some oils must be hydrogenated before we can use them because of their unsavory fl avor and their inappropriate molecular structures (that is, there are too many CP C bonds present). Upon exposure to air, these polyunsatu-rated molecules (that is, molecules with many CP C bonds) undergo oxidation to yield unpleasant-tasting products (oil that has oxidized is said to be rancid). In the hydroge-nation process, a small amount of nickel (about 0.1 percent by mass) is added to the oil and the mixture is exposed to hydrogen gas at high temperature and pressure. Afterward, the nickel is removed by fi ltration. Hydrogenation reduces the number of double bonds in the molecule but does not completely eliminate them. If all the double bonds are eliminated, the oil becomes hard and brittle. Under controlled conditions, suitable cooking oils and margarine may be prepared by hydrogenation from vegetable oils extracted from cottonseed, corn, and soybeans.

The Hydrogen EconomyThe world’s fossil fuel reserves are being depleted at an alarmingly fast rate. Faced with this dilemma, scientists have made intensive efforts in recent years to develop a

Platinum catalyst on alumina (Al2O3) used in hydrogenation.

method of obtaining hydrogen gas as an alternative energy source. Hydrogen gas could replace gasoline to power automobiles (after considerable modifi cation of the engine, of course) or be used with oxygen gas in fuel cells to generate electricity (see p. 860). One major advantage of using hydrogen gas in these ways is that the reactions are essentially free of pollutants; the end product formed in a hydrogen-powered engine or in a fuel cell would be water, just as in the burning of hydrogen gas in air:

2H2(g) 1 O2(g) ¡ 2H2O(l)

Of course, success of a hydrogen economy would depend on how cheaply we could produce hydrogen gas and how easily we could store it. Although electrolysis of water consumes too much energy for large-scale appli-cation, if scientists can devise a more practical method of “splitting” water mole-cules, we could obtain vast amounts of hydrogen from seawater. One approach that is currently in the early stages of development would use solar energy. In this scheme, a catalyst (a complex molecule containing one or more transition metal atoms, such as ruthenium) absorbs a photon from solar radiation and becomes ener-getically excited. In its excited state, the catalyst is capable of reducing water to molecular hydrogen.

The total volume of ocean water is about 1 3 1021 L. Thus, the ocean contains an almost inexhaustible supply of hydrogen.

The total volume of ocean water is about 1 3 1021 L. Thus, the ocean contains an almost inexhaustible supply of hydrogen.

919

C H E M I S T R Y

in ActionMetallic Hydrogen

Scientists have long been interested in how nonmetallic sub-stances, including hydrogen, behave under exceedingly

high pressure. It was predicted that when atoms or molecules are compressed, their bonding electrons might be delocalized, producing a metallic state. In 1996, physicists at the Lawrence Livermore Laboratory used a 60-foot-long gun to generate a shock compression onto a thin (0.5-mm) layer of liquid hydro-gen. For an instant, at pressures between 0.9 and 1.4 million atm, they were able to measure the electrical conductivity of the hydrogen sample and found that it was comparable to that of cesium metal at 2000 K. (The temperature of the hydrogen sam-ple rose as a result of compression, although it remained in the molecular form.) As the pressure fell rapidly, the metallic state of hydrogen disappeared. The Livermore experiment suggested that metallic hydro-gen, if it can be kept in a stable state, may act as a room- temperature superconductor. The fact that hydrogen becomes metallic at pressures lower than previously thought possible also has provided new insight into planetary science. For many years scientists were puzzled by Jupiter’s strong magnetic fi eld, which is 20 times greater than that of Earth. A planet’s magnetic fi eld results from the convection motion of electrically conduc-tive fl uid in its interior. (For example, Earth’s magnetic fi eld is due to the heat-driven motion of liquid iron within its core.)

Jupiter is composed of an outer layer of nonmetallic molecular hydrogen that continuously transforms hydrogen within the core to metallic fl uid hydrogen. It is now believed that this me-tallic layer is much closer to the surface (because the pressure needed to convert molecular hydrogen to metallic hydrogen is not as high as previously thought), which would account for Jupiter’s unusually strong magnetic fi eld.

Interior composition of Jupiter.

Insulating molecular hydrogen

Metallic molecular hydrogen

Metallic atomic hydrogen

Rock core


Some of the interstitial hydrides we have discussed would make suitable storage compounds for hydrogen. The reactions that form these hydrides are usually revers-ible, so hydrogen gas can be obtained simply by reducing the pressure of the hydro-gen gas above the metal. The advantages of using interstitial hydrides are as follows: (1) many metals have a high capacity to take up hydrogen gas—sometimes up to three times as many hydrogen atoms as there are metal atoms; and (2) because these hydrides are solids, they can be stored and transported more easily than gases or liquids. The Chemistry in Action essay on p. 919 describes what happens to hydrogen under pressure.

21.3 CarbonAlthough it constitutes only about 0.09 percent by mass of Earth’s crust, carbon is an essential element of living matter. It is found free in the form of diamond and graph-ite (see Figure 8.17), and it is also a component of natural gas, petroleum, and coal. (Coal is a natural dark-brown to black solid used as a fuel; it is formed from fossil-ized plants and consists of amorphous carbon with various organic and some inorganic compounds.) Carbon combines with oxygen to form carbon dioxide in the atmosphere and occurs as carbonate in limestone and chalk. Diamond and graphite are allotropes of carbon. Figure 21.4 shows the phase diagram of carbon. Although graphite is the stable form of carbon at 1 atm and 25°C, owners of diamond jewelry need not be alarmed, for the rate of the spontaneous process

C(diamond) ¡ C(graphite) ¢G° 5 22.87 kJ/mol

is extremely slow. Millions of years may pass before a diamond turns to graphite. Synthetic diamond can be prepared from graphite by applying very high pressures and temperatures. Figure 21.5 shows a synthetic diamond and its starting material, graphite. Synthetic diamonds generally lack the optical properties of natural diamonds. They are useful, however, as abrasives and in cutting concrete and many other hard substances, including metals and alloys. The uses of graphite are described on p. 484. Carbon has the unique ability to form long chains (consisting of more than 50 C atoms) and stable rings with fi ve or six members. This phenomenon is called catena-tion, the linking of like atoms. Carbon’s versatility is responsible for the millions of organic compounds (made up of carbon and hydrogen and other elements such as oxygen, nitrogen, and halogens) found on Earth. The chemistry of organic compounds is discussed in Chapter 24.

The carbon cycle is discussed on p. 781.The carbon cycle is discussed on p. 781.

The structures of diamond and graphite are shown in Figure 11.28.The structures of diamond and graphite are shown in Figure 11.28.

Figure 21.4 Phase diagram of carbon. Note that under atmospheric conditions, graphite is the stable form of carbon.

2 × 104

P (a

tm)

Diamond

GraphiteVapor

Liquid

3300

t (°C)

Carbides and CyanidesCarbon combines with metals to form ionic compounds called carbides, such as CaC2 and Be2C, in which carbon is in the form of C2

22 or C42 ions. These ions are strong Brønsted bases and react with water as follows:

C222(aq) 1 2H2O(l) ¡ 2OH2(aq) 1 C2H2(g)

C42(aq) 1 4H2O(l) ¡ 4OH2(aq) 1 CH4(g)

Carbon also forms a covalent compound with silicon. Silicon carbide, SiC, is called carborundum and is prepared as follows:

SiO2(s) 1 3C(s) ¡ SiC(s) 1 2CO(g)

Carborundum is also formed by heating silicon with carbon at 1500°C. Carborundum is almost as hard as diamond and it has the diamond structure; each carbon atom is tetrahedrally bonded to four Si atoms, and vice versa. It is used mainly for cutting, grinding, and polishing metals and glasses. Another important class of carbon compounds, the cyanides, contain the anion group :CqN:2. Cyanide ions are extremely toxic because they bind almost irre-versibly to the Fe(III) ion in cytochrome oxidase, a key enzyme in metabolic processes. Hydrogen cyanide, which has the aroma of bitter almonds, is even more dangerous because of its volatility (b.p. 26°C). A few tenths of 1 percent by volume of HCN in air can cause death within minutes. Hydrogen cyanide can be prepared by treating sodium cyanide or potassium cyanide with acid:

NaCN(s) 1 HCl(aq) ¡ NaCl(aq) 1 HCN(aq)

Because HCN (in solution, called hydrocyanic acid) is a very weak acid (Ka 5 4.9 3 10210), most of the HCN produced in this reaction is in the nonion-ized form and leaves the solution as hydrogen cyanide gas. For this reason, acids should never be mixed with metal cyanides in the laboratory without proper ventilation.

HCN is the gas used in gas execution chambers.HCN is the gas used in gas execution chambers.

Figure 21.5 A synthetic diamond and the starting material—graphite.

21.3 Carbon 921


Cyanide ions are used to extract gold and silver. Although these metals are usu-ally found in the uncombined state in nature, in other metal ores they may be present in relatively small concentrations and are more diffi cult to extract. In a typical process, the crushed ore is treated with an aqueous cyanide solution in the presence of air to dissolve the gold by forming the soluble complex ion [Au(CN)2]

2 :

4Au(s) 1 8CN2(aq) 1 O2(g) 1 2H2O(l) ¡ 4[Au(CN)2]2(aq) 1 4OH2(aq)

The complex ion [Au(CN)2]2 (along with some cation, such as Na1) is separated from

other insoluble materials by fi ltration and treated with an electropositive metal such as zinc to recover the gold:

Zn(s) 1 2[Au(CN)2]2(aq) ¡ [Zn(CN)4]

22(aq) 1 2Au(s)

Figure 21.6 shows an aerial view of cyanide ponds used for the extraction of gold.

Oxides of CarbonOf the several oxides of carbon, the most important are carbon monoxide, CO, and carbon dioxide, CO2. Carbon monoxide is a colorless, odorless gas formed by the incomplete combustion of carbon or carbon-containing compounds:

2C(s) 1 O2(g) ¡ 2CO(g)

Carbon monoxide is used in metallurgical process for extracting nickel (see p. 892), in organic synthesis, and in the production of hydrocarbon fuels with hydrogen. Indus-trially, it is prepared by passing steam over heated coke. Carbon monoxide burns readily in oxygen to form carbon dioxide:

2CO(g) 1 O2(g) ¡ 2CO2(g) ¢H° 5 2566 kJ/mol

Carbon monoxide is not an acidic oxide (it differs from carbon dioxide in that regard), and it is only slightly soluble in water. Carbon dioxide is a colorless and odorless gas. Unlike carbon monoxide, CO2 is nontoxic. It is an acidic oxide (see p. 695). Carbon dioxide is used in beverages, in fi re extinguishers, and in the manufacture of baking soda, NaHCO3, and soda ash, Na2CO3. Solid carbon dioxide, called dry ice, is used as a refrigerant (see Figure 11.42).

The role of CO as an indoor air pollutant is discussed on p. 793.The role of CO as an indoor air pollutant is discussed on p. 793.

Carbon dioxide is the primary greenhouse gas (see p. 781).Carbon dioxide is the primary greenhouse gas (see p. 781).

Figure 21.6 Cyanide ponds for extracting gold from metal ore.

923

C H E M I S T R Y

in ActionSynthetic Gas from Coal

T he very existence of our technological society depends on an abundant supply of energy. Although the United States

has only 5 percent of the world’s population, we consume about 20 percent of the world’s energy! At present, the two major sources of energy are nuclear fi ssion and fossil fuels (discussed in Chapters 23 and 24, respectively). Coal, oil (which is also known as petroleum), and natural gas (mostly methane) are col-lectively called fossil fuels because they are the end result of the decomposition of plants and animals over tens or hundreds of millions of years. Oil and natural gas are cleaner-burning and more effi cient fuels than coal, so they are preferred for most purposes. However, supplies of oil and natural gas are being depleted at an alarming rate, and research is under way to make coal a more versatile source of energy. Coal consists of many high-molar-mass carbon compounds that also contain oxygen, hydrogen, and small amounts of nitro-gen and sulfur. Coal constitutes about 90 percent of the world’s fossil fuel reserves. For centuries coal has been used as a fuel both in homes and in industry. However, underground coal mining is expensive and dangerous, and strip mining (that is, mining in an open pit after removal of the overlaying earth and rock) is tremendously harmful to the environment. Another problem, this one associated with the burning of coal, is the

formation of sulfur dioxide (SO2) from the sulfur-containing compounds. This process leads to the formation of “acid rain,” discussed on p. 786. One of the most promising methods for making coal a more effi cient and cleaner fuel involves the conversion of coal to a gaseous form, called syngas for “synthetic gas.” This pro-cess is called coal gasifi cation. In the presence of very hot steam and air, coal decomposes and reacts according to the following simplifi ed scheme:

C(s) 1 H2O(g) ¡ CO(g) 1 H2(g) C(s) 1 2H2(g) ¡ CH4(g)

The main component of syngas is methane. In addition, the fi rst reaction yields hydrogen and carbon monoxide gases and other useful by-products. Under suitable conditions, CO and H2 com-bine to form methanol:

CO(g) 1 2H2(g) ¡ CH3OH(l)

Methanol has many uses, for example, as a solvent and a start-ing material for plastics. Syngas is easier than coal to store and transport. What’s more, it is not a major source of air pollution because sulfur is removed in the gasifi cation process.

Underground coal mining.


21.4 Nitrogen and Phosphorus

NitrogenAbout 78 percent of air by volume is nitrogen. The most important mineral sources of nitrogen are saltpeter (KNO3) and Chile saltpeter (NaNO3). Nitrogen is an essential element of life; it is a component of proteins and nucleic acids. Molecular nitrogen is obtained by fractional distillation of air (the boiling points of liquid nitrogen and liquid oxygen are 2196°C and 2183°C, respectively). In the laboratory, very pure nitrogen gas can be prepared by the thermal decomposition of ammonium nitrite:

NH4NO2(s) ¡ 2H2O(g) 1 N2(g)

The N2 molecule contains a triple bond and is very stable with respect to disso-ciation into atomic species. However, nitrogen forms a large number of com-pounds with hydrogen and oxygen in which the oxidation number of nitrogen varies from 23 to 15 (Table 21.2). Most nitrogen compounds are covalent; however, when heated with certain metals, nitrogen forms ionic nitrides containing the N32 ion:

6Li(s) 1 N2(g) ¡ 2Li3N(s)

The nitrogen cycle is discussed on p. 770.The nitrogen cycle is discussed on p. 770.

Molecular nitrogen will boil off before molecular oxygen does during the fractional distillation of liquid air.

Molecular nitrogen will boil off before molecular oxygen does during the fractional distillation of liquid air.

OxidationNumber Compound Formula Structure

23 Ammonia NH3 AH

HO NO HO

22 Hydrazine N2H4 AH

HO NO NO HOAH

O

21 Hydroxylamine NH2OH HO NO OO HO OQAH

0 Nitrogen* (dinitrogen) N2 SNqNS

11 Nitrous oxide (dinitrogen monoxide) N2O SNqNO OSOQ

12 Nitric oxide (nitrogen monoxide) NO SNP OOQN

13 Nitrous acid HNO2 OQ O OQOP NO OO H

14 Nitrogen dioxide NO2 OQ OQN

SOO NP O

15 Nitric acid HNO3 OQSS

OQ

Q

OP NO OO HAO

TABLE 21.2 Common Compounds of Nitrogen

*We list the element here as a reference.

The nitride ion is a strong Brønsted base and reacts with water to produce ammonia and hydroxide ions:

N32(aq) 1 3H2O(l) ¡ NH3(g) 1 3OH2(aq)

AmmoniaAmmonia is one of the best-known nitrogen compounds. It is prepared industrially from nitrogen and hydrogen by the Haber process (see Section 13.6 and p. 646). It can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:

NH4Cl(aq) 1 NaOH(aq) ¡ NaCl(aq) 1 H2O(l) 1 NH3(g)

Ammonia is a colorless gas (b.p. 233.4°C) with an irritating odor. About three-quarters of the ammonia produced annually in the United States (about 18 million tons in 2005) is used in fertilizers. Liquid ammonia, like water, undergoes autoionization:

2NH3(l) ∆ NH14 1 NH2

2

or simply

NH3(l) ∆ H1 1 NH22

where NH22 is called the amide ion. Note that both H1 and NH2

2 are solvated with the NH3 molecules. (Here is an example of ion-dipole interaction.) At 250°C, the ion product [H1][NH2

2] is about 1 3 10233, considerably smaller than 1 3 10214 for water at 25°C. Nevertheless, liquid ammonia is a suitable solvent for many electrolytes, especially when a more basic medium is required or if the solutes react with water. The ability of liquid ammonia to dissolve alkali metals was discussed in Section 20.5.

HydrazineAnother important hydride of nitrogen is hydrazine:

NO ND

H

GH

HG

HD

OO

Each N atom is sp3-hybridized. Hydrazine is a colorless liquid that smells like ammo-nia. It melts at 2°C and boils at 114°C. Hydrazine is a base that can be protonated to give the N2H5

1 and N2H621 ions. As

a reducing agent, it can reduce Fe31 to Fe21, MnO42 to Mn21, and I2 to I2. Its reaction

with oxygen is highly exothermic:

N2H4(l) 1 O2(g) ¡ N2(g) 1 2H2O(l) DH° 5 2666.6 kJ/mol

Hydrazine and its derivative methylhydrazine, N2H3(CH3), together with the oxidizer dinitrogen tetroxide (N2O4), are used as rocket fuels. Hydrazine also plays a role in polymer synthesis and in the manufacture of pesticides.

Oxides and Oxoacids of NitrogenThere are many nitrogen oxides, but the three particularly important ones are: nitrous oxide, nitric oxide, and nitrogen dioxide.

The amide ion is a strong Brønsted base and does not exist in water.The amide ion is a strong Brønsted base and does not exist in water.

21.4 Nitrogen and Phosphorus 925

N2H4


Nitrous oxide, N2O, is a colorless gas with a pleasing odor and sweet taste. It is prepared by heating ammonium nitrate to about 270°C:

NH4NO3(s) ¡ N2O(g) 1 2H2O(g)

Nitrous oxide resembles molecular oxygen in that it supports combustion. It does so because it decomposes when heated to form molecular nitrogen and molecular oxygen:

2N2O(g) ¡ 2N2(g) 1 O2(g)

It is chiefl y used as an anesthetic in dental procedures and other minor surgery. Nitrous oxide is also called “laughing gas” because a person inhaling the gas becomes some-what giddy. No satisfactory explanation has yet been proposed for this unusual phys-iological response. Nitrous oxide is also used as the propellant in cans of whipped cream due to its high solubility in the whipped cream mixture. Nitric oxide, NO, is a colorless gas. The reaction of N2 and O2 in the atmosphere

N2(g) 1 O2(g) ∆ 2NO(g) ¢G° 5 173.4 kJ/mol

is a form of nitrogen fi xation (see p. 771). The equilibrium constant for the above reac-tion is very small at room temperature: KP is only 4.0 3 10231 at 25°C, so very little NO will form at that temperature. However, the equilibrium constant increases rapidly with temperature, for example, in a running auto engine. An appreciable amount of nitric oxide is formed in the atmosphere by the action of lightning. In the laboratory, the gas can be prepared by the reduction of dilute nitric acid with copper:

3Cu(s) 1 8HNO3(aq) ¡ 3Cu1NO322(aq) 1 4H2O(l) 1 2NO(g)

The nitric oxide molecule is paramagnetic, containing one unpaired electron. It can be represented by the following resonance structures:

mn !"OQPQNP O PQOQNP O

As we noted in Chapter 9, this molecule does not obey the octet rule. The properties of nitric oxide are discussed on p. 393. Unlike nitrous oxide and nitric oxide, nitrogen dioxide is a highly toxic yellow-brown gas with a choking odor. In the laboratory nitrogen dioxide is prepared by the action of concentrated nitric acid on copper (Figure 21.7):

Cu(s) 1 4HNO3(aq) ¡ Cu(NO3)2(aq) 1 2H2O(l) 1 2NO2(g)

Nitrogen dioxide is paramagnetic. It has a strong tendency to dimerize to dinitrogen tetroxide, which is a diamagnetic molecule:

2NO2 ∆ N2O4

This reaction occurs in both the gas phase and the liquid phase. Nitrogen dioxide is an acidic oxide; it reacts rapidly with cold water to form both nitrous acid, HNO2, and nitric acid:

2NO2(g) 1 H2O(l) ¡ HNO2(aq) 1 HNO3(aq)

This is a disproportionation reaction (see p. 144) in which the oxidation number of nitrogen changes from 14 (in NO2) to 13 (in HNO2) and 15 (in HNO3). Note that this reaction is quite different from that between CO2 and H2O, in which only one acid (carbonic acid) is formed.

According to Le Châtelier’s principle, the forward endothermic reaction is favored by heating.

According to Le Châtelier’s principle, the forward endothermic reaction is favored by heating.

The role of NO2 in smog formation is discussed on p. 789.The role of NO2 in smog formation is discussed on p. 789.

Neither N2O nor NO reacts with water.Neither N2O nor NO reacts with water.

Figure 21.7 The production of NO2 gas when copper reacts with concentrated nitric acid.

Nitric acid is one of the most important inorganic acids. It is a liquid (b.p. 82.6°C), but it does not exist as a pure liquid because it decomposes spontaneously to some extent as follows:

4HNO3(l) ¡ 4NO2(g) 1 2H2O(l) 1 O2(g)

The major industrial method of producing nitric acid is the Ostwald process, discussed in Section 13.6. The concentrated nitric acid used in the laboratory is 68 percent HNO3 by mass (density 1.42 g/cm3), which corresponds to 15.7 M. Nitric acid is a powerful oxidizing agent. The oxidation number of N in HNO3 is 15. The most common reduction products of nitric acid are NO2 (oxidation num-ber of N 5 14), NO (oxidation number of N 5 12), and NH4

1 (oxidation number of N 5 23). Nitric acid can oxidize metals both below and above hydrogen in the activity series (see Figure 4.16). For example, copper is oxidized by concentrated nitric acid, as discussed earlier. In the presence of a strong reducing agent, such as zinc metal, nitric acid can be reduced all the way to the ammonium ion:

4Zn(s) 1 10H1(aq) 1 NO32(aq) ¡ 4Zn21(aq) 1 NH4

1(aq) 1 3H2O(l)

Concentrated nitric acid does not oxidize gold. However, when the acid is added to concentrated hydrochloric acid in a 1:3 ratio by volume (one part HNO3 to three parts HCl), the resulting solution, called aqua regia, can oxidize gold, as follows:

Au(s) 1 3HNO3(aq) 1 4HCl(aq) ¡ HAuCl4(aq) 1 3H2O(l) 1 3NO2(g)

The oxidation of Au is promoted by the complexing ability of the Cl2 ion (to form the AuCl4

2 ion). Concentrated nitric acid also oxidizes a number of nonmetals to their correspond-ing oxoacids:

P4(s) 1 20HNO3(aq) ¡ 4H3PO4(aq) 1 20NO2(g) 1 4H2O(l) S(s) 1 6HNO3(aq) ¡ H2SO4(aq) 1 6NO2(g) 1 2H2O(l)

Nitric acid is used in the manufacture of fertilizers, dyes, drugs, and explosives. The Chemistry in Action essay on p. 931 describes a nitrogen-containing fertilizer that can be highly explosive.

PhosphorusLike nitrogen, phosphorus is a member of the Group 5A family; in some respects the chemistry of phosphorus resembles that of nitrogen. Phosphorus occurs most com-monly in nature as phosphate rocks, which are mostly calcium phosphate, Ca3(PO4)2, and fl uoroapatite, Ca5(PO4)3F (Figure 21.8). Elemental phosphorus can be obtained by heating calcium phosphate with coke and silica sand:

2Ca3(PO4)2(s) 1 10C(s) 1 6SiO2(s) ¡ 6CaSiO3(s) 1 10CO(g) 1 P4(s)

There are several allotropic forms of phosphorus, but only white phosphorus and red phosphorus (see Figure 8.18) are of importance. White phosphorus consists of discrete tetrahedral P4 molecules (Figure 21.9). A solid (m.p. 44.2°C), white phosphorus is insol-uble in water but quite soluble in carbon disulfi de (CS2) and in organic solvents such as chloroform (CHCl3). White phosphorus is a highly toxic substance. It bursts into fl ames spontaneously when exposed to air; hence it is used in incendiary bombs and grenades:

P4(s) 1 5O2(g) ¡ P4O10(s)

On standing, a concentrated nitric acid solution turns slightly yellow as a result of NO2 formation.

On standing, a concentrated nitric acid solution turns slightly yellow as a result of NO2 formation.



The high reactivity of white phosphorus is attributed to structural strain: The PO P bonds are compressed in the tetrahedral P4 molecule. White phosphorus was once used in matches, but because of its toxicity it has been replaced by tetraphosphorus trisulfi de, P4S3. When heated in the absence of air, white phosphorus is slowly converted to red phosphorus at about 300°C:

nP4(white phosphorus) ¡ (P42n(red phosphorus)

Red phosphorus has a polymeric structure (see Figure 21.9) and is more stable and less volatile than white phosphorus.

Hydride of PhosphorusThe most important hydride of phosphorus is phosphine, PH3, a colorless, very poi-sonous gas formed by heating white phosphorus in concentrated sodium hydroxide:

P4(s) 1 3NaOH(aq) 1 3H2O(l) ¡ 3NaH2PO2(aq) 1 PH3(g)

Figure 21.8 Phosphate mining.

Figure 21.9 The structures of white and red phosphorus. Red phosphorus is believed to have a chain structure, as shown.

Red phosphorus

White phosphorus

Phosphine is moderately soluble in water and more soluble in carbon disulfi de and organic solvents. Its aqueous solution is neutral, unlike that of ammonia. In liquid ammonia, phosphine dissolves to give NH4

1PH22. Phosphine is a strong

reducing agent; it reduces many metal salts to the corresponding metal. The gas burns in air:

PH3(g) 1 2O2(g) ¡ H3PO4(s)

Halides of PhosphorusPhosphorus forms binary compounds with halogens: the trihalides, PX3, and the pen-tahalides, PX5, where X denotes a halogen atom. In contrast, nitrogen can form only trihalides (NX3). Unlike nitrogen, phosphorus has a 3d subshell, which can be used for valence-shell expansion. We can explain the bonding in PCl5 by assuming that phos-phorus undergoes sp3d hybridization of its 3s, 3p, and 3d orbitals (see Example 10.4). The fi ve sp3d hybrid orbitals also account for the trigonal bipyramidal geometry of the PCl5 molecule (see Table 10.4). Phosphorus trichloride is prepared by heating white phosphorus in chlorine:

P4(l) 1 6Cl2(g) ¡ 4PCl3(g)

A colorless liquid (b.p. 76°C), PCl3 is hydrolyzed according to the equation:

PCl3(l) 1 3H2O(l) ¡ H3PO3(aq) 1 3HCl(g)

In the presence of an excess of chlorine gas, PCl3 is converted to phosphorus penta-chloride, which is a light-yellow solid:

PCl3(l) 1 Cl2(g) ¡ PCl5(s)

X-ray studies have shown that solid phosphorus pentachloride exists as [PCl41][PCl6

2], in which the PCl4

1 ion has a tetrahedral geometry and the PCl62 ion has an octahedral

geometry. In the gas phase, PCl5 (which has trigonal bipyramidal geometry) is in equilibrium with PCl3 and Cl2:

PCl5(g) ∆ PCl3(g) 1 Cl2(g)

Phosphorus pentachloride reacts with water as follows:

PCl5(s) 1 4H2O(l) ¡ H3PO4(aq) 1 5HCl(aq)

Oxides and Oxoacids of PhosphorusThe two important oxides of phosphorus are tetraphosphorus hexaoxide, P4O6, and tetraphosphorus decaoxide, P4O10 (Figure 21.10). The oxides are obtained by burning white phosphorus in limited and excess amounts of oxygen gas, respectively:

P4(s) 1 3O2(s) ¡ P4O6(s) P4(s) 1 5O2(g) ¡ P4O10(s)

Both oxides are acidic; that is, they are converted to acids in water. The compound P4O10 is a white fl occulent powder (m.p. 420°C) that has a great affi nity for water:

P4O10(s) 1 6H2O(l) ¡ 4H3PO4(aq)

For this reason, it is often used for drying gases and for removing water from solvents.



There are many oxoacids containing phosphorus. Some examples are phospho-rous acid, H3PO3; phosphoric acid, H3PO4; hypophosphorous acid, H3PO2; and tri-phosphoric acid, H5P3O10 (Figure 21.11). Phosphoric acid, also called orthophosphoric acid, is a weak triprotic acid (see p. 685). It is prepared industrially by the reaction of calcium phosphate with sulfuric acid:

Ca3(PO4)2(s) 1 3H2SO4(aq) ¡ 2H3PO4(aq) 1 3CaSO4(s)

In the pure form phosphoric acid is a colorless solid (m.p. 42.2°C). The phos-phoric acid we use in the laboratory is usually an 82 percent H3PO4 solution (by mass). Phosphoric acid and phosphates have many commercial applications in detergents, fertilizers, flame retardants, and toothpastes, and as buffers in car-bonated beverages. Like nitrogen, phosphorus is an element that is essential to life. It constitutes only about 1 percent by mass of the human body, but it is a very important 1 per-cent. About 23 percent of the human skeleton is mineral matter. The phosphorus content of this mineral matter, calcium phosphate, Ca3(PO4)2, is 20 percent. Our teeth are basically Ca3(PO4)2 and Ca5(PO4)3OH. Phosphates are also important components of the genetic materials deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

Phosphoric acid is the most important phosphorus-containing oxoacid.Phosphoric acid is the most important phosphorus-containing oxoacid.

Figure 21.10 The structures of P4O6 and P4O10. Note the tetrahedral arrangement of the P atoms in P4O10.

Phosphorus

Oxygen

P4O10P4O6

Figure 21.11 Structures of some common phosphorus-containing oxoacids.

H P

Triphosphoric acid (H5P3O 10)

Hypophosphorous acid (H3PO2)

Phosphoric acid (H3PO4)

AO

B¼S

OO H

O

Oþ³H

O HP

Phosphorous acid (H3PO3)

AO

B

¼S

OOO þ³

O

Oþ³

H

OH PA

OB¼S

O OO Hþ³

O

Oþ³H

A

¼SO

OH PA

OB

¼S

O OO Hþ³

O

Oþ³

HA

¼SO

OPA

B

¼S

Oþ³

O

HA

¼SO

OPA

B

¼S

Oþ³

O

HA

¼SO

O OH

931

C H E M I S T R Y

in ActionAmmonium Nitrate}The Explosive Fertilizer

Ammonium nitrate is the most important fertilizer in the world (see p. 108). It ranked fi fteenth among the industrial chemicals

produced in the United States in 2005 (8 million tons). Unfortu-nately, it is also a powerful explosive. In 1947 an explosion oc-curred aboard a ship being loaded with the fertilizer in Texas. The fertilizer was in paper bags and apparently blew up after sailors tried to stop a fi re in the ship’s hold by closing a hatch, thereby creating the compression and heat necessary for an explosion. More than six hundred people died as a result of the accident. More recent disasters involving ammonium nitrate took place at the World Trade Center in New York City in 1993 and at the Alfred P. Murrah Federal Building in Oklahoma City in 1995. A strong oxidizer, ammonium nitrate is stable at room tem-perature. At 250°C, it begins to decompose as follows:

NH4NO3(g) ¡ N2O(g) 1 2H2O(g)

At 300°C, different gaseous products and more heat are produced:

2NH4NO3(g) ¡ 2N2(g) 1 4H2O(g) 1 O2(g)

About 1.46 kJ of heat are generated per gram of the compound decomposed. When it is combined with a combustible mate-rial, such as fuel oil, the energy released increases almost

threefold. Ammonium nitrate can also be mixed with charcoal, fl our, sugar, sulfur, rosin, and paraffi n to form an explosive. Intense heat from the explosion causes the gases to expand rapidly, generating shock waves that destroy most objects in their path. Federal law regulates the sale of explosive-grade ammo-nium nitrate, which is used for 95 percent of all commercial blasting in road construction and mining. However, the wide availability of large quantities of ammonium nitrate and other substances that enhance its explosive power make it possible for anyone who is so inclined to construct a bomb. The bomb that destroyed the federal building in Oklahoma City is estimated to have contained 4000 pounds of ammonium nitrate and fuel oil, which was set off by another small explosive device. How can the use of ammonium nitrate by terrorists be pre-vented? The most logical approach is to desensitize or neutral-ize the compound’s ability to act as an explosive, but to date no satisfactory way has been found to do so without diminishing its value as a fertilizer. A more passive method is to add to the fertilizer an agent known as a taggant, which would allow law enforcement to trace the source of an ammonium nitrate explo-sive. A number of European countries now forbid the sale of ammonium nitrate without taggants, although the U.S. Congress has yet to pass such a law.

The Alfred P. Murrah building after a deadly explosion caused by an ammo-nium nitrate bomb. A bag of ammonium nitrate fertilizer, which is labeled as an explosive.


21.5 Oxygen and Sulfur

OxygenOxygen is by far the most abundant element in Earth’s crust, constituting about 46 per-cent of its mass. In addition, the atmosphere contains about 21 percent molecular oxygen by volume (23 percent by mass). Like nitrogen, oxygen in the free state is a diatomic molecule (O2). In the laboratory, oxygen gas can be obtained by heating potassium chlorate (see Figure 5.15):

2KClO3(s) ¡ 2KCl(s) 1 3O2(g)

The reaction is usually catalyzed by manganese(IV) dioxide, MnO2. Pure oxygen gas can be prepared by electrolyzing water (p. 866). Industrially, oxygen gas is prepared by the fractional distillation of liquefi ed air (p. 924). Oxygen gas is colorless and odorless. Oxygen is a building block of practically all biomolecules, accounting for about a fourth of the atoms in living matter. Molecular oxygen is the essential oxidant in the metabolic breakdown of food molecules. Without it, a human being cannot survive for more than a few minutes.

Properties of Diatomic OxygenAlthough oxygen has two allotropes, O2 and O3, when we speak of molecular oxygen, we normally mean O2. Ozone, O3, is less stable than O2. The O2 molecule is para-magnetic because it contains two unpaired electrons (see Section 10.7). A strong oxidizing agent, molecular oxygen is one of the most widely used industrial chemicals. Its main uses are in the steel industry (see Section 20.2) and in sewage treatment. Oxygen is also used as a bleaching agent for pulp and paper, in medicine to ease breathing diffi culties, in oxyacetylene torches, and as an oxidizing agent in many inorganic and organic reactions.

Oxides, Peroxides, and SuperoxidesOxygen forms three types of oxides: the normal oxide (or simply the oxide), which contains the O22 ion; the peroxide, which contains the O2

22 ion; and the superoxide, which contains the O2

2 ion:

2" 2"

superoxideoxide peroxide

OSOSQ OSOSQ OOSQ "OSOSQ OT

SQ

The ions are all strong Brønsted bases and react with water as follows:

Oxide: O22(aq) 1 H2O(l) ¡ 2OH2(aq) Peroxide: 2O2

22(aq) 1 2H2O(l) ¡ O2(g) 1 4OH2(aq)Superoxide: 4O2

2(aq) 1 2H2O(l) ¡ 3O2(g) 1 4OH2(aq)

Note that the reaction of O22 with water is a hydrolysis reaction, but those involving O2

22 and O22 are redox processes.

The nature of bonding in oxides changes across any period in the periodic table (see Figure 15.8). Oxides of elements on the left side of the periodic table, such as those of the alkali metals and alkaline earth metals, are generally ionic solids with high melting points. Oxides of the metalloids and of the metallic elements toward the middle of the periodic table are also solids, but they have much less ionic character. Oxides of nonmetals are covalent compounds that generally exist as liquids or gases at room temperature. The acidic character of the oxides increases from left to right.

The oxygen cycle is discussed on p. 771.The oxygen cycle is discussed on p. 771.

Consider the oxides of the third-period elements (see Table 8.4):

acidicbasic amphoteric

Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7

The basicity of the oxides increases as we move down a particular group. MgO does not react with water but reacts with acid as follows:

MgO(s) 1 2H1(aq) ¡ Mg21(aq) 1 H2O(l)

On the other hand, BaO, which is more basic, undergoes hydrolysis to yield the cor-responding hydroxide:

BaO(s) 1 H2O(l) ¡ Ba(OH)2(aq)

The best-known peroxide is hydrogen peroxide (H2O2). It is a colorless, syrupy liquid (m.p. 20.9°C), prepared in the laboratory by the action of cold dilute sulfuric acid on barium peroxide octahydrate:

BaO2 ? 8H2O(s) 1 H2SO4(aq) ¡ BaSO4(s) 1 H2O2(aq) 1 8H2O(l)

The structure of hydrogen peroxide is shown in Figure 21.12. Using the VSEPR method we see that the HO O and OO O bonds are bent around each oxygen atom in a con-fi guration similar to the structure of water. The lone-pair–bonding-pair repulsion is greater in H2O2 than in H2O, so that the HOO angle is only 97° (compared with 104.5° for HOH in H2O). Hydrogen peroxide is a polar molecule (m 5 2.16 D). Hydrogen peroxide readily decomposes when heated or exposed to sunlight or even in the presence of dust particles or certain metals, including iron and copper:

2H2O2(l) ¡ 2H2O(l) 1 O2(g) ¢H° 5 2196.4 kJ/mol

Note that this is a disproportionation reaction. The oxidation number of oxygen changes from 21 to 22 and 0. Hydrogen peroxide is miscible with water in all proportions due to its ability to hydrogen-bond with water. Dilute hydrogen peroxide solutions (3 percent by mass), available in drugstores, are used as mild antiseptics; more concentrated H2O2 solutions are employed as bleaching agents for textiles, fur, and hair. The high heat of decom-position of hydrogen peroxide also makes it a suitable component in rocket fuel. Hydrogen peroxide is a strong oxidizing agent; it can oxidize Fe21 ions to Fe31 ions in an acidic solution:

H2O2(aq) 1 2Fe21(aq) 1 2H1(aq) ¡ 2Fe31(aq) 1 2H2O(l)

It also oxidizes SO322 ions to SO4

22 ions:

H2O2(aq) 1 SO322(aq) ¡ SO4

22(aq) 1 H2O(l)

In addition, hydrogen peroxide can act as a reducing agent toward substances that are stronger oxidizing agents than itself. For example, hydrogen peroxide reduces silver oxide to metallic silver:

H2O2(aq) 1 Ag2O(s) ¡ 2Ag(s) 1 H2O(l) 1 O2(g)

and permanganate, MnO42, to manganese(II) in an acidic solution:

5H2O2 (aq) 1 2MnO42(aq) 1 6H1(aq) ¡ 2Mn21(aq) 1 5O2(g) 1 8H2O(l)

21.5 Oxygen and Sulfur 933

Figure 21.12 The structure of H2O2.

86°97°


If we want to determine hydrogen peroxide concentration, this reaction can be carried out as a redox titration, using a standard permanganate solution. There are relatively few known superoxides, that is, compounds containing the O2

2 ion. In general, only the most reactive alkali metals (K, Rb, and Cs) form superoxides. We should take note of the fact that both the peroxide ion and the superoxide ion are by-products of metabolism. Because these ions are highly reactive, they can infl ict great damage on living cells. Fortunately, our bodies are equipped with the enzymes catalase, peroxidase, and superoxide dismutase which convert these toxic substances to water and molecular oxygen.

OzoneOzone is a rather toxic, light-blue gas (b.p. 2111.3°C). Its pungent odor is noticeable around sources of signifi cant electrical discharges (such as a subway train). Ozone can be prepared from molecular oxygen, either photochemically or by subjecting O2 to an electrical discharge (Figure 21.13):

3O2(g) ¡ 2O3(g) ¢G° 5 326.8 kJ/mol

Because the standard free energy of formation of ozone is a large positive quantity [DG°f 5 (326.8/2) kJ/mol or 163.4 kJ/mol, ozone is less stable than molecular oxygen. The ozone molecule has a bent structure in which the bond angle is 116.5°:

ODOM

O S

SSSS

O

SS

SSS

Omn

OJOG

O

Ozone is mainly used to purify drinking water, to deodorize air and sewage gases, and to bleach waxes, oils, and textiles. Ozone is a very powerful oxidizing agent—its oxidizing power is exceeded only by that of molecular fl uorine (see Table 19.1). For example, ozone can oxidize sulfi des of many metals to the corresponding sulfates:

4O3(g) 1 PbS(s) ¡ PbSO4(s) 1 4O2(g)

Ozone oxidizes all the common metals except gold and platinum. In fact, a convenient test for ozone is based on its action on mercury. When exposed to ozone, mercury

High-voltage source

Metal foil on outer tubeOuter tube

O2

Inner tube

O3 plus some unreacted O2Figure 21.13 The preparation of O3 from O2 by electrical discharge. The outside of the outer tube and the inside of the inner tube are coated with metal foils that are connected to a high-voltage source. (The metal foil on the inside of the inner tube is not shown.) During the electrical discharge, O2 gas is passed through the tube. The O3 gas formed exits from the upper right-hand tube, along with some unreacted O2 gas.

Liquid ozone.

loses its metallic luster and sticks to glass tubing (instead of fl owing freely through it). This behavior is attributed to the change in surface tension caused by the forma-tion of mercury(II) oxide:

O3(g) 1 3Hg(l) ¡ 3HgO(s)

The benefi cial effect of ozone in the stratosphere and its undesirable action in smog formation were discussed in Chapter 17.

SulfurAlthough sulfur is not a very abundant element (it constitutes only about 0.06 percent of Earth’s crust by mass), it is readily available because it occurs commonly in nature in the elemental form. The largest known reserves of sulfur are found in sedimentary deposits. In addition, sulfur occurs widely in gypsum (CaSO4 ? 2H2O) and various sulfi de minerals such as pyrite (FeS2) (Figure 21.14). Sulfur is also present in natural gas as H2S, SO2, and other sulfur-containing compounds. Sulfur is extracted from underground deposits by the Frasch† process, shown in Figure 21.15. In this process, superheated water (liquid water heated to about 160°C under high pressure to prevent it from boiling) is pumped down the outermost pipe to melt the sulfur. Next, compressed air is forced down the innermost pipe. Liquid sulfur mixed with air forms an emulsion that is less dense than water and therefore rises to the surface as it is forced up the middle pipe. Sulfur produced in this manner, which amounts to about 10 million tons per year, has a purity of about 99.5 percent.


Figure 21.14 Pyrite (FeS2), commonly called “fool’s gold” because of its gold luster.

Figure 21.15 The Frasch process. Three concentric pipes are inserted into a hole drilled down to the sulfur deposit. Superheated water is forced down the outer pipe into the sulfur, causing it to melt. Molten sulfur is then forced up the middle pipe by compressed air.

Molten sulfur

Compressed air

Superheated water

Sulfur

†Herman Frasch (1851–1914). German chemical engineer. Besides inventing the process for obtaining pure sulfur, Frasch developed methods for refi ning petroleum.


There are several allotropic forms of sulfur, the most important being the rhom-bic and monoclinic forms. Rhombic sulfur is thermodynamically the most stable form; it has a puckered S8 ring structure:

S

SS

S S

HE

S

SSS

S

EH HS

SSS

S

S

ES

S

S

SS

SS

S

It is a yellow, tasteless, and odorless solid (m.p. 112°C) (see Figure 8.19) that is insoluble in water but soluble in carbon disulfi de. When heated, it is slowly converted to monoclinic sulfur (m.p. 119°C), which also consists of the S8 units. When liquid sulfur is heated above 150°C, the rings begin to break up, and the entangling of the sulfur chains results in a sharp increase in the liquid’s viscosity. Further heating tends to rupture the chains, and the viscosity decreases. Like nitrogen, sulfur shows a wide variety of oxidation numbers in its compounds (Table 21.3). The best-known hydrogen compound of sulfur is hydrogen sulfi de, which is prepared by the action of an acid on a sulfi de; for example,

FeS(s) 1 H2SO4(aq) ¡ FeSO4(aq) 1 H2S(g)

OxidationNumber Compound Formula Structure

22 Hydrogen sulfi de H2S S S

HD

SG

H

0 Sulfur* S8 S SSSS

HES

S

SS

SSS

EH HSS

SSS

S

SSS

SSS

E

11 Disulfur dichloride S2Cl2

SCl

ClS

EEOQOQOQ

OQSO S

12 Sulfur dichloride SCl2 S S

S SSSClCl SS

DSG

14 Sulfur dioxide SO2

SS

OS

O

OS

S

EE EE

16 Sulfur trioxide SO3 S S

S SSSO SS O

DSG

OB

TABLE 21.3 Common Compounds of Sulfur

*We list the element here as a reference.

S8

Today, hydrogen sulfi de used in qualitative analysis (see Section 16.11) is prepared by the hydrolysis of thioacetamide:

thioacetamide

! 2H2O ! H! 888n CH3O C ! H2S ! NH!

acetic acid

JS

GNH2

CH3O C J

O

G OO H

4

Hydrogen sulfi de is a colorless gas (b.p. 260.2°C) that smells like rotten eggs. (The odor of rotten eggs actually does come from hydrogen sulfi de, which is formed by the bacterial decomposition of sulfur-containing proteins.) Hydrogen sulfi de is a highly toxic substance that, like hydrogen cyanide, attacks respiratory enzymes. It is a very weak diprotic acid (see Table 15.5). In basic solution, H2S is a reducing agent. For example, it is oxidized by permanganate to elemental sulfur:

3H2S (aq) 1 2MnO42(aq) ¡ 3S(s) 1 2MnO2(s) 1 2H2O(l) 1 2OH2(aq)

Oxides of SulfurSulfur has two important oxides: sulfur dioxide, SO2; and sulfur trioxide, SO3. Sulfur dioxide is formed when sulfur burns in air:

S(s) 1 O2(g) ¡ SO2(g)

In the laboratory, it can be prepared by the action of an acid on a sulfi te; for example,

2HCl(aq) 1 Na2SO3(aq) ¡ 2NaCl(aq) 1 H2O(l) 1 SO2(g)

or by the action of concentrated sulfuric acid on copper:

Cu(s) 1 2H2SO4(aq) ¡ CuSO4(aq) 1 2H2O(l) 1 SO2(g)

Sulfur dioxide (b.p. 210°C) is a pungent, colorless gas that is quite toxic. As an acidic oxide, it reacts with water as follows:

SO2(g) 1 H2O(l) ∆ H1(aq) 1 HSO32(aq)

Sulfur dioxide is slowly oxidized to sulfur trioxide, but the reaction rate can be greatly enhanced by a platinum or vanadium oxide catalyst (see Section 13.6):

2SO2(g) 1 O2(g) ¡ 2SO3(g)

Sulfur trioxide dissolves in water to form sulfuric acid:

SO3(g) 1 H2O(l) ¡ H2SO4(aq)

The contributing role of sulfur dioxide to acid rain is discussed on p. 786.

Sulfuric AcidSulfuric acid is the world’s most important industrial chemical. It is prepared industri-ally by fi rst burning sulfur in air:

S(s) 1 O2(g) ¡ SO2(g)

Next is the key step of converting sulfur dioxide to sulfur trioxide:

2SO2(g) 1 O2(g) ¡ 2SO3(g)

There is no evidence for the formation of sulfurous acid, H2SO3, in water.There is no evidence for the formation of sulfurous acid, H2SO3, in water.

Approximately 50 million tons of sulfuric acid are produced annually in the United States.

Approximately 50 million tons of sulfuric acid are produced annually in the United States.



Vanadium(V) oxide (V2O5) is the catalyst used for the second step. Because the sul-fur dioxide and oxygen molecules react in contact with the surface of solid V2O5, the process is referred to as the contact process. Although sulfur trioxide reacts with water to produce sulfuric acid, it forms a mist of fi ne droplets of H2SO4 with water vapor that is hard to condense. Instead, sulfur trioxide is fi rst dissolved in 98 percent sulfuric acid to form oleum (H2S2O7):

SO3(g) 1 H2SO4(aq) ¡ H2S2O7(aq)

On treatment with water, concentrated sulfuric acid can be generated:

H2S2O7(aq) 1 H2O(l) ¡ 2H2SO4(aq)

Sulfuric acid is a diprotic acid (see Table 15.5). It is a colorless, viscous liquid (m.p. 10.4°C). The concentrated sulfuric acid we use in the laboratory is 98 percent H2SO4 by mass (density: 1.84 g/cm3), which corresponds to a concentration of 18 M. The oxidizing strength of sulfuric acid depends on its temperature and concentration. A cold, dilute sulfuric acid solution reacts with metals above hydrogen in the activity series (see Figure 4.15), thereby liberating molecular hydrogen in a displacement reaction:

Mg(s) 1 H2SO4(aq) ¡ MgSO4(aq) 1 H2(g)

This is a typical reaction of an active metal with an acid. The strength of sulfuric acid as an oxidizing agent is greatly enhanced when it is both hot and concentrated. In such a solution, the oxidizing agent is actually the sulfate ion rather than the hydrated proton, H1(aq). Thus, copper reacts with concentrated sulfuric acid as follows:

Cu(s) 1 2H2SO4(aq) ¡ CuSO4(aq) 1 SO2(g) 1 2H2O(l)

Depending on the nature of the reducing agents, the sulfate ion may be further reduced to elemental sulfur or the sulfi de ion. For example, reduction of H2SO4 by HI yields H2S and I2:

8HI(aq) 1 H2SO4(aq) ¡ H2S(aq) 1 4I2(s) 1 4H2O(l)

Concentrated sulfuric acid oxidizes nonmetals. For example, it oxidizes carbon to carbon dioxide and sulfur to sulfur dioxide:

C(s) 1 2H2SO4(aq) ¡ CO2(g) 1 2SO2(g) 1 2H2O(l) S(s) 1 2H2SO4(aq) ¡ 3SO2(g) 1 2H2O(l)

Other Compounds of SulfurCarbon disulfi de, a colorless, fl ammable liquid (b.p. 46°C), is formed by heating carbon and sulfur to a high temperature:

C(s) 1 2S(l) ¡ CS2(l)

It is only slightly soluble in water. Carbon disulfi de is a good solvent for sulfur, phosphorus, iodine, and nonpolar substances such as waxes and rubber. Another interesting compound of sulfur is sulfur hexafl uoride, SF6, which is pre-pared by heating sulfur in an atmosphere of fl uorine:

S(l) 1 3F2(g) ¡ SF6(g)

Sulfur hexafl uoride is a nontoxic, colorless gas (b.p. 263.8°C). It is the most inert of all sulfur compounds; it resists attack even by molten KOH. The structure and bonding

Vanadium oxide on alumina (Al2O3).

of SF6 were discussed in Chapters 9 and 10 and its critical phenomenon illustrated in Chapter 11 (see Figure 11.37).

21.6 The HalogensThe halogens—fl uorine, chlorine, bromine, and iodine—are reactive nonmetals (see Figure 8.20). Table 21.4 lists some of the properties of these elements. Although all halogens are highly reactive and toxic, the magnitude of reactivity and toxicity gen-erally decreases from fl uorine to iodine. The chemistry of fl uorine differs from that of the rest of the halogens in the following ways:

1. Fluorine is the most reactive of all the halogens. The difference in reactivity between fl uorine and chlorine is greater than that between chlorine and bromine. Table 21.4 shows that the FOF bond is considerably weaker than the ClOCl bond. The weak bond in F2 can be explained in terms of the lone pairs on the F atoms:

F—F SOQ OSQ

The small size of the F atoms (see Table 21.4) allows a close approach of the three lone pairs on each of the F atoms, resulting in a greater repulsion than that found in Cl2, which consists of larger atoms.

2. Hydrogen fl uoride, HF, has a high boiling point (19.5°C) as a result of strong intermolecular hydrogen bonding, whereas all other hydrogen halides have much lower boiling points (see Figure 11.6).

3. Hydrofl uoric acid is a weak acid, whereas all other hydrohalic acids (HCl, HBr, and HI) are strong acids.

4. Fluorine reacts with cold sodium hydroxide solution to produce oxygen difl uoride as follows:

2F2(g) 1 2NaOH(aq) ¡ 2NaF(aq) 1 H2O(l) 1 OF2(g)

Recall that the fi rst member of a group usually differs in properties from the rest of the members of the group (see p. 344).

Recall that the fi rst member of a group usually differs in properties from the rest of the members of the group (see p. 344).

Property F Cl Br I

Valence electron 2s22p5 3s23p5 4s24p5 5s25p5

confi gurationMelting point (°C) 2223 2102 27 114Boiling point (°C) 2187 235 59 183Appearance* Pale- Yellow- Red- Dark-violet vapor yellow green brown Dark metallic- gas gas liquid looking solidAtomic radius (pm) 72 99 114 133Ionic radius (pm)† 133 181 195 220Ionization energy (kJ/mol) 1680 1251 1139 1003Electronegativity 4.0 3.0 2.8 2.5Standard reduction 2.87 1.36 1.07 0.53potential (V)*Bond enthalpy (kJ/mol)* 150.6 242.7 192.5 151.0

TABLE 21.4 Properties of the Halogens

*These values and descriptions apply to the diatomic species X2, where X represents a halogen atom. The half-reaction is X2(g) 1 2e2 ¡ 2X2(aq).†Refers to the anion X2.

21.6 The Halogens 939


The same reaction with chlorine or bromine, on the other hand, produces a halide and a hypohalite:

X2(g) 1 2NaOH(aq) ¡ NaX(aq) 1 NaXO(aq) 1 H2O(l)

where X stands for Cl or Br. Iodine does not react under the same conditions.

5. Silver fl uoride, AgF, is soluble. All other silver halides (AgCl, AgBr, and AgI) are insoluble (see Table 4.2).

The element astatine also belongs to the Group 7A family. However, all isotopes of astatine are radioactive; its longest-lived isotope is astatine-210, which has a half-life of 8.3 h. Therefore, it is both diffi cult and expensive to study astatine in the laboratory. The halogens form a very large number of compounds. In the elemental state they form diatomic molecules, X2. In nature, however, because of their high reactivity, halogens are always found combined with other elements. Chlorine, bromine, and iodine occur as halides in seawater, and fl uorine occurs in the minerals fl uorite (CaF2) (see Figure 20.16) and cryolite (Na3AlF6).

Preparation and General Properties of the HalogensBecause fl uorine and chlorine are strong oxidizing agents, they must be prepared by electrolysis rather than by chemical oxidation of the fl uoride and chloride ions. Elec-trolysis does not work for aqueous solutions of fl uorides, however, because fl uorine is a stronger oxidizing agent than oxygen. From Table 19.1 we fi nd that

F2(g) 1 2e2 ¡ 2F2(aq) E° 5 2.87 V O2(g) 1 4H1(aq) 1 4e2 ¡ 2H2O(l) E° 5 1.23 V

If F2 were formed by the electrolysis of an aqueous fl uoride solution, it would imme-diately oxidize water to oxygen. For this reason, fl uorine is prepared by electrolyzing liquid hydrogen fl uoride containing potassium fl uoride to increase its conductivity, at about 70°C (Figure 21.16):

Anode (oxidation): 2F2 ¡ F2(g) 1 2e2

Cathode (reduction): 2H1 1 2e2 ¡ H2(g)

Overall reaction: 2HF(l) ¡ H2(g) 1 F2(g)

Carbon anode

H2 gas H2 gas

F2 gas

Steel cathode

Liquid HF

Diaphragm to prevent mixing of H2 and F2 gases

Figure 21.16 Electrolytic cell for the preparation of fl uorine gas. Note that because H2 and F2 form an explosive mixture, these gases must be separated from each other.

Chlorine gas, Cl2, is prepared industrially by the electrolysis of molten NaCl (see Section 19.8) or by the chlor-alkali process, the electrolysis of a concentrated aque-ous NaCl solution (called brine). (Chlor denotes chlorine and alkali denotes an alkali metal, such as sodium.) Two of the common cells used in the chlor-alkali process are the mercury cell and the diaphragm cell. In both cells the overall reaction is

2NaCl(aq) 1 2H2O(l) electrolysis

¬¬¬¡ 2NaOH(aq) 1 H2(g) 1 Cl2(g)

As you can see, this reaction yields two useful by-products, NaOH and H2. The cells are designed to separate the molecular chlorine from the sodium hydroxide solution and the molecular hydrogen to prevent side reactions such as

2NaOH(aq) 1 Cl2(g) ¡ NaOCl(aq) 1 NaCl(aq) 1 H2O(l) H2(g) 1 Cl2(g) ¡ 2HCl(g)

These reactions consume the desired products and can be dangerous because a mixture of H2 and Cl2 is explosive. Figure 21.17 shows the mercury cell used in the chlor-alkali process. The cathode is a liquid mercury pool at the bottom of the cell, and the anode is made of either graphite or titanium coated with platinum. Brine is continuously passed through the cell as shown in the diagram. The electrode reactions are

Anode (oxidation): 2Cl2(aq) ¡ Cl2(g) 1 2e2

Cathode (reduction): 2Na1(aq) 1 2e2 ¡Hg(l)

2Na/Hg

Overall reaction: 2NaCl(aq) ¡ 2Na/Hg 1 Cl2(g)

where Na/Hg denotes the formation of sodium amalgam. The chlorine gas generated this way is very pure. The sodium amalgam does not react with the brine solution but decomposes as follows when treated with pure water outside the cell:

2Na/Hg 1 2H2O(l) ¡ 2NaOH(aq) 1 H2(g) 1 2Hg(l)

the by-products are sodium hydroxide and hydrogen gas. Although the mercury is cycled back into the cell for reuse, some of it is always discharged with waste solu-tions into the environment, resulting in mercury pollution. This is a major drawback of the mercury cell. Figure 21.18 shows the industrial manufacture of chlorine gas. The half-cell reactions in a diaphragm cell are shown in Figure 21.19. The asbes-tos diaphragm is permeable to the ions but not to the hydrogen and chlorine gases and so prevents the gases from mixing. During electrolysis a positive pressure is applied on the anode side of the compartment to prevent the migration of the OH2 ions from the cathode compartment. Periodically, fresh brine solution is added to the cell and the sodium hydroxide solution is run off as shown. The diaphragm cell presents no


Brine

Graphite anode

Brine

Cl2

Hg plus Na/Hg

Hg cathode

Figure 21.17 Mercury cell used in the chlor-alkali process. The cathode contains mercury. The sodium-mercury amalgam is treated with water outside the cell to produce sodium hydroxide and hydrogen gas.


pollution problems. Its main disadvantage is that the sodium hydroxide solution is contaminated with unreacted sodium chloride. The preparation of molecular bromine and iodine from seawater by oxidation with chlorine was discussed in Section 4.4. In the laboratory, chlorine, bromine, and iodine can be prepared by heating the alkali halides (NaCl, KBr, or KI) in concentrated sulfuric acid in the presence of manganese(IV) oxide. A representative reaction is

MnO2(s) 1 2H2SO4(aq) 1 2NaCl(aq) ¡MnSO4(aq) 1 Na2SO4(aq) 1 2H2O(l) 1 Cl2(g)

Compounds of the HalogensMost of the halides can be classifi ed into two categories. The fl uorides and chlorides of many metallic elements, especially those belonging to the alkali metal and alkaline earth metal (except beryllium) families, are ionic compounds. Most of the halides of nonmetals such as sulfur and phosphorus are covalent compounds. As Figure 4.10 shows, the oxidation numbers of the halogens can vary from 21 to 17. The only exception is fl uorine. Because it is the most electronegative element, fl uorine can have only two oxidation numbers, 0 (as in F2) and 21, in its compounds.

The Hydrogen HalidesThe hydrogen halides, an important class of halogen compounds, can be formed by the direct combination of the elements:

H2(g) 1 X2(g) ∆ 2HX(g)

From Table 19.1 we see that the oxidizing strength decreases from Cl2 to Br2 to I2.From Table 19.1 we see that the oxidizing strength decreases from Cl2 to Br2 to I2.

Figure 21.18 The industrial manufacture of chlorine gas.

e–e–

Battery

Oxidation2Cl–(aq ) Cl2(g) + 2e–

Reduction2H2O(l) + 2e– H2(g) + 2OH–(aq )

Anode CathodeAsbestos

diaphragm

NaOH solution

Brine

Figure 21.19 Diaphragm cell used in the chlor-alkali process.

where X denotes a halogen atom. These reactions (especially the ones involving F2 and Cl2) can occur with explosive violence. Industrially, hydrogen chloride is pro-duced as a by-product in the manufacture of chlorinated hydrocarbons:

C2H6(g) 1 Cl2(g) ¡ C2H5Cl(g) 1 HCl(g)

In the laboratory, hydrogen fl uoride and hydrogen chloride can be prepared by react-ing the metal halides with concentrated sulfuric acid:

CaF2(s) 1 H2SO4(aq) ¡ 2HF(g) 1 CaSO4(s) 2NaCl(s) 1 H2SO4(aq) ¡ 2HCl(g) 1 Na2SO4(aq)

Hydrogen bromide and hydrogen iodide cannot be prepared this way because they are oxidized to elemental bromine and iodine. For example, the reaction between NaBr and H2SO4 is

2NaBr(s) 1 2H2SO4(aq) ¡ Br2(l) 1 SO2(g) 1 Na2SO4(aq) 1 2H2O(l)

Instead, hydrogen bromide is prepared by fi rst reacting bromine with phosphorus to form phosphorus tribromide:

P4(s) 1 6Br2(l) ¡ 4PBr3(l)

Next, PBr3 is treated with water to yield HBr:

PBr3(l) 1 3H2O(l) ¡ 3HBr(g) 1 H3PO3(aq)

Hydrogen iodide can be prepared in a similar manner. The high reactivity of HF is demonstrated by the fact that it attacks silica and silicates:

6HF(aq) 1 SiO2(s) ¡ H2SiF6(aq) 1 2H2O(l)

This property makes HF suitable for etching glass and is the reason that hydro-gen fl uoride must be kept in plastic or inert metal (for example, Pt) containers. Hydrogen fl uoride is used in the manufacture of Freons (see Chapter 17); for example,

CCl4(l) 1 HF(g) ¡ CFCl3(g) 1 HCl(g) CFCl3(g) 1 HF(g) ¡ CF2Cl2(g) 1 HCl(g)

It is also important in the production of aluminum (see Section 20.7). Hydrogen chloride is used in the preparation of hydrochloric acid, inorganic chlorides, and in various metallurgical processes. Hydrogen bromide and hydrogen iodide do not have any major industrial uses. Aqueous solutions of hydrogen halides are acidic. The strength of the acids increases as follows:

HF ! HCl , HBr , HI

Oxoacids of the HalogensThe halogens also form a series of oxoacids with the following general formulas:

HXO HXO2 HXO3 HXO4

hypohalous halous halic perhalic acid acid acid acid



Chlorous acid, HClO2, is the only known halous acid. All the halogens except fl uorine form halic and perhalic acids. The Lewis structures of the chlorine oxo-acids are

HSOSClSO OQ OQ SOSQ HSOSClSO OQ O OSQHSOSClSOQOQO SOSQ SOSQ

HSOSClSO OQ OSQO

O SOSQ

hypochlorous acid

perchloric acid chlorous acid

chloric acid

For a given halogen, the acid strength decreases from perhalic acid to hypohalous acid; the explanation of this trend is discussed in Section 15.9. Table 21.5 lists some of the halogen compounds. Periodic acid, HIO4, does not appear because this compound cannot be isolated in the pure form. Instead the formula H5IO6 is often used to represent periodic acid.

Uses of the HalogensFluorineApplications of the halogens and their compounds are widespread in industry, health care, and other areas. One is fl uoridation, the practice of adding small quantities of fl uorides (about 1 ppm by mass) such as NaF to drinking water to reduce dental caries. One of the most important inorganic fl uorides is uranium hexafl uoride, UF6, which is essential to the gaseous diffusion process for separating isotopes of uranium (U-235 and U-238). Industrially, fl uorine is used to produce polytetrafl uoroethylene, a polymer better known as Tefl on:

O(CF2OCF2)On

where n is a large number. Tefl on is used in electrical insulators, high-temperature plastics, cooking utensils, and so on.

ChlorineChlorine plays an important biological role in the human body, where the chloride ion is the principal anion in intracellular and extracellular fl uids. Chlorine is widely used as an industrial bleaching agent for paper and textiles. Ordinary household laundry bleach contains the active ingredient sodium hypochlorite (about 5 percent

*The number in parentheses indicates the oxidation number of the halogen.

Compound F Cl Br I

Hydrogen halide HF (21) HCl (21) HBr (21) HI (21)Oxides OF2 (21) Cl2O (11) Br2O (11) I2O5 (15) ClO2 (14) BrO2 (14) Cl2O7 (17)Oxoacids HFO (21) HClO (11) HBrO (11) HIO (11) HClO2 (13) HClO3 (15) HBrO3 (15) HIO3 (15) HClO4 (17) HBrO4 (17) H5IO6 (17)

TABLE 21.5 Common Compounds of Halogens*

by mass), which is prepared by reacting chlorine gas with a cold solution of sodium hydroxide:

Cl2(g) 1 2NaOH(aq) ¡ NaCl(aq) 1 NaClO(aq) 1 H2O(l)

Chlorine is also used to purify water and disinfect swimming pools. When chlorine dissolves in water, it undergoes the following reaction:

Cl2(g) 1 H2O(l) ¡ HCl(aq) 1 HClO(aq)

It is thought that the ClO2 ions destroy bacteria by oxidizing life-sustaining com-pounds within them. Chlorinated methanes, such as carbon tetrachloride and chloroform, are useful organic solvents. Large quantities of chlorine are used to produce insecticides, such as DDT. However, in view of the damage they infl ict on the environment, the use of many of these compounds is either totally banned or greatly restricted in the United States. Chlorine is also used to produce polymers such as poly(vinyl chloride).

BromineSo far as we know, bromine compounds occur naturally only in some marine organ-isms. Seawater is about 1 3 1023 M Br2; therefore, it is the main source of bromine. Bromine is used to prepare ethylene dibromide (BrCH2CH2Br), which is used as an insecticide and as a scavenger for lead (that is, to combine with lead) in gasoline to keep lead deposits from clogging engines. Studies have shown that ethylene dibromide is a very potent carcinogen. Bromine combines directly with silver to form silver bromide (AgBr), which is used in photographic fi lms.

IodineIodine is not used as widely as the other halogens. A 50 percent (by mass) alcohol solution of iodine, known as tincture of iodine, is used medicinally as an antiseptic. Iodine is an essential constituent of the thyroid hormone thyroxine:

HOO

H A

ANH2

JO

GOH

O CH2O CO CO OO

IG

ID

IG

ID

Iodine defi ciency in the diet may result in enlargement of the thyroid gland (known as goiter). Iodized table salt sold in the United States usually contains 0.01 percent KI or NaI, which is more than suffi cient to satisfy the 1 mg of iodine per week required for the formation of thyroxine in the human body. A compound of iodine that deserves mention is silver iodide, AgI. It is a pale-yellow solid that darkens when exposed to light. In this respect it is similar to silver bromide. Silver iodide is sometimes used in cloud seeding, a process for inducing rainfall on a small scale (Figure 21.20). The advantage of using silver iodide is that enormous numbers of nuclei (that is, small particles of silver iodide on which ice crystals can form) become available. About 1015 nuclei are produced from 1 g of AgI by vaporizing an acetone solution of silver iodide in a hot fl ame. The nuclei are then dispersed into the clouds from an airplane.



Figure 21.20 Cloud seeding using AgI particles.

1. Hydrogen atoms contain one proton and one electron. They are the simplest atoms. Hydrogen combines with many metals and nonmetals to form hydrides; some hy-drides are ionic and some are covalent.

2. There are three isotopes of hydrogen: 11H, 1

2H (deuterium), and 1

3H (tritium). Heavy water contains deuterium. 3. The important inorganic compounds of carbon are the

carbides; the cyanides, most of which are extremely toxic; carbon monoxide, also toxic and a major air pollutant; the carbonates and bicarbonates; and carbon dioxide, an end product of metabolism and a compo-nent of the global carbon cycle.

4. Elemental nitrogen, N2, contains a triple bond and is very stable. Compounds in which nitrogen has oxida-tion numbers from 23 to 15 are formed between nitro-gen and hydrogen and/or oxygen atoms. Ammonia, NH3, is widely used in fertilizers.

5. White phosphorus, P4, is highly toxic, very reactive, and fl ammable; the polymeric red phosphorus, (P4)n, is more stable. Phosphorus forms oxides and halides with oxidation numbers of 13 and 15 and several oxoacids.


The phosphates are the most important phosphorus compounds.

6. Elemental oxygen, O2, is paramagnetic and contains two unpaired electrons. Oxygen forms ozone (O3), oxides (O22), peroxides (O2

22), and superoxides (O22). The most

abundant element in Earth’s crust, oxygen is essential for life on Earth.

7. Sulfur is taken from Earth’s crust by the Frasch process as a molten liquid. Sulfur exists in a number of allo-tropic forms and has a variety of oxidation numbers in its compounds.

8. Sulfuric acid is the cornerstone of the chemical indus-try. It is produced from sulfur via sulfur dioxide and sulfur trioxide by means of the contact process.

9. The halogens are toxic and reactive elements that are found only in compounds with other elements. Fluorine and chlorine are strong oxidizing agents and are pre-pared by electrolysis.

10. The reactivity, toxicity, and oxidizing ability of the halogens decrease from fl uorine to iodine. The halogens all form binary acids (HX) and a series of oxoacids.


Carbide, p. 921Catenation, p. 920

Key Words

Chlor-alkali process, p. 941

Cyanide, p. 921Hydrogenation, p. 918

Electronic



ARIS Problems: 21.17, 21.26, 21.27, 21.28, 21.31, 21.32, 21.39, 21.40, 21.41, 21.42, 21.49, 21.50, 21.51,

21.56, 21.68, 21.70, 21.76, 21.78, 21.84, 21.90, 21.92, 21.94, 21.97.


General Properties of NonmetalsReview Questions

21.1 Without referring to Figure 21.1, state whether each of the following elements are metals, metalloids, or nonmetals: (a) Cs, (b) Ge, (c) I, (d) Kr, (e) W, (f) Ga, (g) Te, (h) Bi.

21.2 List two chemical and two physical properties that distinguish a metal from a nonmetal.

21.3 Make a list of physical and chemical properties of chlorine (Cl2) and magnesium. Comment on their differences with reference to the fact that one is a metal and the other is a nonmetal.

21.4 Carbon is usually classifi ed as a nonmetal. However, the graphite used in “lead” pencils conducts electric-ity. Look at a pencil and list two nonmetallic proper-ties of graphite.

HydrogenReview Questions

21.5 Explain why hydrogen has a unique position in the periodic table.

21.6 Describe two laboratory and two industrial prepara-tions for hydrogen.

21.7 Hydrogen exhibits three types of bonding in its compounds. Describe each type of bonding with an example.

21.8 What are interstitial hydrides?21.9 Give the name of (a) an ionic hydride and (b) a cova-

lent hydride. In each case describe the preparation and give the structure of the compound.

21.10 Describe what is meant by the “hydrogen economy.”

Problems

21.11 Elements number 17 and 20 form compounds with hydrogen. Write the formulas for these two compounds and compare their chemical behavior in water.

21.12 Give an example of hydrogen as (a) an oxidizing agent and (b) a reducing agent.

21.13 Compare the physical and chemical properties of the hydrides of each of the following elements: Na, Ca, C, N, O, Cl.

21.14 Suggest a physical method that would enable you to separate hydrogen gas from neon gas.

21.15 Write a balanced equation to show the reaction between CaH2 and H2O. How many grams of CaH2 are needed to produce 26.4 L of H2 gas at 20°C and 746 mmHg?

21.16 How many kilograms of water must be processed to obtain 2.0 L of D2 at 25°C and 0.90 atm pressure? Assume that deuterium abundance is 0.015 percent and that recovery is 80 percent.

21.17 Predict the outcome of the following reactions:(a) CuO(s) 1 H2(g) ¡(b) Na2O(s) 1 H2(g) ¡

21.18 Starting with H2, describe how you would prepare (a) HCl, (b) NH3, (c) LiOH.

CarbonReview Questions

21.19 Give an example of a carbide and a cyanide.21.20 How are cyanide ions used in metallurgy?21.21 Briefl y discuss the preparation and properties of car-

bon monoxide and carbon dioxide.21.22 What is coal?21.23 Explain what is meant by coal gasifi cation.21.24 Describe two chemical differences between CO

and CO2.

Problems

21.25 Describe the reaction between CO2 and OH2 in terms of a Lewis acid-base reaction such as that shown on p. 700.



21.26 Draw a Lewis structure for the C222 ion.

21.27 Balance the following equations:(a) Be2C(s) 1 H2O(l) ¡(b) CaC2(s) 1 H2O(l) ¡

21.28 Unlike CaCO3, Na2CO3 does not readily yield CO2 when heated. On the other hand, NaHCO3 undergoes thermal decomposition to produce CO2 and Na2CO3. (a) Write a balanced equation for the reaction. (b) How would you test for the CO2 evolved? [Hint: Treat the gas with limewater, an aqueous solution of Ca(OH)2.]

21.29 Two solutions are labeled A and B. Solution A con-tains Na2CO3 and solution B contains NaHCO3. De-scribe how you would distinguish between the two solutions if you were provided with a MgCl2 solu-tion. (Hint: You need to know the solubilities of MgCO3 and MgHCO3.)

21.30 Magnesium chloride is dissolved in a solution contain-ing sodium bicarbonate. On heating, a white precipi-tate is formed. Explain what causes the precipitation.

21.31 A few drops of concentrated ammonia solution added to a calcium bicarbonate solution cause a white precipitate to form. Write a balanced equation for the reaction.

21.32 Sodium hydroxide is hygroscopic—that is, it absorbs moisture when exposed to the atmosphere. A student placed a pellet of NaOH on a watch glass. A few days later, she noticed that the pellet was covered with a white solid. What is the identity of this solid? (Hint: Air contains CO2.)

21.33 A piece of red-hot magnesium ribbon will continue to burn in an atmosphere of CO2 even though CO2 does not support combustion. Explain.

21.34 Is carbon monoxide isoelectronic with nitrogen (N2)?

Nitrogen and PhosphorusReview Questions

21.35 Describe a laboratory and an industrial preparation of nitrogen gas.

21.36 What is meant by nitrogen fi xation? Describe a pro-cess for fi xation of nitrogen on an industrial scale.

21.37 Describe an industrial preparation of phosphorus.21.38 Why is the P4 molecule unstable?

Problems

21.39 Nitrogen can be obtained by (a) passing ammonia over red-hot copper(II) oxide and (b) heating ammo-nium dichromate [one of the products is Cr(III) ox-ide]. Write a balanced equation for each preparation.

21.40 Write balanced equations for the preparation of sodium nitrite by (a) heating sodium nitrate and (b) heating sodium nitrate with carbon.

21.41 Sodium amide (NaNH2) reacts with water to produce sodium hydroxide and ammonia. Describe this reac-tion as a Brønsted acid-base reaction.

21.42 Write a balanced equation for the formation of urea, (NH2)2CO, from carbon dioxide and ammonia. Should the reaction be run at a high or low pressure to maximize the yield?

21.43 Some farmers feel that lightning helps produce a bet-ter crop. What is the scientifi c basis for this belief?

21.44 At 620 K the vapor density of ammonium chloride relative to hydrogen (H2) under the same conditions of temperature and pressure is 14.5, although, according to its formula mass, it should have a vapor density of 26.8. How would you account for this discrepancy?

21.45 Explain, giving one example in each case, why ni-trous acid can act both as a reducing agent and as an oxidizing agent.

21.46 Explain why nitric acid can be reduced but not oxidized.

21.47 Write a balanced equation for each of the follow-ing processes: (a) On heating, ammonium nitrate produces nitrous oxide. (b) On heating, potassium nitrate produces potassium nitrite and oxygen gas. (c) On heating, lead nitrate produces lead(II) oxide, nitrogen dioxide (NO2), and oxygen gas.

21.48 Explain why, under normal conditions, the reaction of zinc with nitric acid does not produce hydrogen.

21.49 Potassium nitrite can be produced by heating a mixture of potassium nitrate and carbon. Write a balanced equation for this reaction. Calculate the theoretical yield of KNO2 produced by heating 57.0 g of KNO3 with an excess of carbon.

21.50 Predict the geometry of nitrous oxide, N2O, by the VSEPR method and draw resonance structures for the molecule. (Hint: The atoms are arranged as NNO.)

21.51 Consider the reaction

N2(g) 1 O2(g) ∆ 2NO(g)

Given that the DG° for the reaction at 298 K is 173.4 kJ/mol, calculate (a) the standard free energy of formation of NO, (b) KP for the reaction, and (c) Kc for the reaction.

21.52 From the data in Appendix 3, calculate DH° for the synthesis of NO (which is the fi rst step in the manu-facture of nitric acid) at 25°C:

4NH3(g) 1 5O2(g) ¡ 4NO(g) 1 6H2O(l)

21.53 Explain why two N atoms can form a double bond or a triple bond, whereas two P atoms normally can form only a single bond.

21.54 When 1.645 g of white phosphorus are dissolved in 75.5 g of CS2, the solution boils at 46.709°C, whereas pure CS2 boils at 46.300°C. The molal boiling-point

elevation constant for CS2 is 2.34°C/m. Calculate the molar mass of white phosphorus and give the mo-lecular formula.

21.55 Starting with elemental phosphorus, P4, show how you would prepare phosphoric acid.

21.56 Dinitrogen pentoxide is a product of the reaction be-tween P4O10 and HNO3. Write a balanced equation for this reaction. Calculate the theoretical yield of N2O5 if 79.4 g of P4O10 are reacted with an excess of HNO3. (Hint: One of the products is HPO3.)

21.57 Explain why (a) NH3 is more basic than PH3, (b) NH3 has a higher boiling point than PH3, (c) PCl5 exists but NCl5 does not, (d) N2 is more inert than P4.

21.58 What is the hybridization of phosphorus in the phos-phonium ion, PH4

1?

Oxygen and SulfurReview Questions

21.59 Describe one industrial and one laboratory prepara-tion of O2.

21.60 Give an account of the various kinds of oxides that exist and illustrate each type by two examples.

21.61 Hydrogen peroxide can be prepared by treating bar-ium peroxide with sulfuric acid. Write a balanced equation for this reaction.

21.62 Describe the Frasch process for obtaining sulfur.21.63 Describe the contact process for the production of

sulfuric acid.21.64 How is hydrogen sulfi de generated in the laboratory?

Problems

21.65 Draw molecular orbital energy level diagrams for O2, O2

2, and O222.

21.66 One of the steps involved in the depletion of ozone in the stratosphere by nitric oxide may be represented as

NO(g) 1 O3(g) ¡ NO2(g) 1 O2(g)

From the data in Appendix 3 calculate DG°, KP, and Kc for the reaction at 25°C.

21.67 Hydrogen peroxide is unstable and decomposes readily:

2H2O2(aq) ¡ 2H2O(l) 1 O2(g)

This reaction is accelerated by light, heat, or a cata-lyst. (a) Explain why hydrogen peroxide sold in drugstores comes in dark bottles. (b) The concentra-tions of aqueous hydrogen peroxide solutions are normally expressed as percent by mass. In the de-composition of hydrogen peroxide, how many liters of oxygen gas can be produced at STP from 15.0 g of a 7.50 percent hydrogen peroxide solution?

21.68 What are the oxidation numbers of O and F in HFO?

21.69 Oxygen forms double bonds in O2, but sulfur forms single bonds in S8. Explain.

21.70 In 2004, about 48 million tons of sulfuric acid were produced in the United States. Calculate the amount of sulfur (in grams and moles) used to produce that amount of sulfuric acid.

21.71 Sulfuric acid is a dehydrating agent. Write balanced equations for the reactions between sulfuric acid and the following substances: (a) HCOOH, (b) H3PO4, (c) HNO3, (d) HClO3. (Hint: Sulfuric acid is not de-composed by the dehydrating action.)

21.72 Calculate the amount of CaCO3 (in grams) that would be required to react with 50.6 g of SO2 emit-ted by a power plant.

21.73 SF6 exists but OF6 does not. Explain.21.74 Explain why SCl6, SBr6, and SI6 cannot be prepared.21.75 Compare the physical and chemical properties of

H2O and H2S.21.76 The bad smell of water containing hydrogen sulfi de

can be removed by the action of chlorine. The reac-tion is

H2S(aq) 1 Cl2(aq) ¡ 2HCl(aq) 1 S(s)

If the hydrogen sulfi de content of contaminated water is 22 ppm by mass, calculate the amount of Cl2 (in grams) required to remove all the H2S from 2.0 3 102 gallons of water. (1 gallon 5 3.785 L.)

21.77 Describe two reactions in which sulfuric acid acts as an oxidizing agent.

21.78 Concentrated sulfuric acid reacts with sodium iodide to produce molecular iodine, hydrogen sulfi de, and sodium hydrogen sulfate. Write a balanced equation for the reaction.

The HalogensReview Questions

21.79 Describe an industrial method for preparing each of the halogens.

21.80 Name the major uses of the halogens.

Problems

21.81 Metal chlorides can be prepared in a number of ways: (a) direct combination of metal and molecular chlo-rine, (b) reaction between metal and hydrochloric acid, (c) acid-base neutralization, (d) metal carbonate treated with hydrochloric acid, (e) precipitation reac-tion. Give an example for each type of preparation.

21.82 Sulfuric acid is a weaker acid than hydrochloric acid. Yet hydrogen chloride is evolved when concentrated sulfuric acid is added to sodium chloride. Explain.

21.83 Show that chlorine, bromine, and iodine are very much alike by giving an account of their behavior (a) with hydrogen, (b) in producing silver salts,



(c) as oxidizing agents, and (d) with sodium hydrox-ide. (e) In what respects is fl uorine not a typical halogen element?

21.84 A 375-gallon tank is fi lled with water containing 167 g of bromine in the form of Br2 ions. How many liters of Cl2 gas at 1.00 atm and 20°C will be required to oxidize all the bromide to molecular bromine?

21.85 Draw structures for (a) (HF)2 and (b) HF22.

21.86 Hydrogen fl uoride can be prepared by the action of sulfuric acid on sodium fl uoride. Explain why hydro-gen bromide cannot be prepared by the action of the same acid on sodium bromide.

21.87 Aqueous copper(II) sulfate solution is blue. When aqueous potassium fl uoride is added to the CuSO4 so-lution, a green precipitate is formed. If aqueous potas-sium chloride is added instead, a bright-green solution is formed. Explain what happens in each case.

21.88 What volume of bromine (Br2) vapor measured at 100°C and 700 mmHg pressure would be obtained if 2.00 L of dry chlorine (Cl2), measured at 15°C and 760 mmHg, were absorbed by a potassium bromide solution?

21.89 Use the VSEPR method to predict the geometries of the following species: (a) I3

2, (b) SiCl4, (c) PF5, (d) SF4.21.90 Iodine pentoxide, I2O5, is sometimes used to remove

carbon monoxide from the air by forming carbon di-oxide and iodine. Write a balanced equation for this reaction and identify species that are oxidized and reduced.

Additional Problems21.91 Write a balanced equation for each of the following

reactions: (a) Heating phosphorous acid yields phos-phoric acid and phosphine (PH3). (b) Lithium car-bide reacts with hydrochloric acid to give lithium chloride and methane. (c) Bubbling HI gas through an aqueous solution of HNO2 yields molecular iodine and nitric oxide. (d) Hydrogen sulfi de is oxidized by chlorine to give HCl and SCl2.

21.92 (a) Which of the following compounds has the great-est ionic character? PCl5, SiCl4, CCl4, BCl3 (b) Which of the following ions has the smallest ionic radius?

F2, C42, N32, O22 (c) Which of the following atoms has the highest ionization energy? F, Cl, Br, I (d) Which of the following oxides is most acidic? H2O, SiO2, CO2

21.93 Both N2O and O2 support combustion. Suggest one physical and one chemical test to distinguish be-tween the two gases.

21.94 What is the change in oxidation number for the fol-lowing reaction?

3O2 ¡ 2O3

21.95 Describe the bonding in the C222ion in terms of the

molecular orbital theory.21.96 Starting with deuterium oxide (D2O), describe how

you would prepare (a) NaOD, (b) DCl, (c) ND3, (d) C2D2, (e) CD4, (f) D2SO4.

21.97 Solid PCl5 exists as [PCl41][PCl6

2]. Draw Lewis structures for these ions. Describe the hybridization state of the P atoms.

21.98 Consider the Frasch process. (a) How is it possible to heat water well above 100°C without turning it into steam? (b) Why is water sent down the outermost pipe? (c) Why would excavating a mine and digging for sulfur be a dangerous procedure for obtaining the element?

21.99 Predict the physical and chemical properties of asta-tine, a radioactive element and the last member of Group 7A.

21.100 Lubricants used in watches usually consist of long-chain hydrocarbons. Oxidation by air forms solid polymers that eventually destroy the effectiveness of the lubricants. It is believed that one of the initial steps in the oxidation is removal of a hydrogen atom (hydrogen abstraction). By replacing the hydrogen atoms at reactive sites with deuterium atoms, it is possible to substantially slow down the overall oxi-dation rate. Why? (Hint: Consider the kinetic isotope effect.)

21.101 How are lightbulbs frosted? (Hint: Consider the ac-tion of hydrofl uoric acid on glass, which is made of silicon dioxide.)

Special Problems

21.102 Life evolves to adapt to its environment. In this respect, explain why life most frequently needs oxygen for survival, rather than the more abundant nitrogen.

21.103 As mentioned in Chapter 3, ammonium nitrate is the most important nitrogen-containing fertilizer in the world. Given only air and water as starting materials and any equipment and catalyst at your disposal,

describe how you would prepare ammonium nitrate. State conditions under which you can increase the yield in each step.

21.104 As we saw in Section 20.2, the reduction of iron ox-ides is accomplished by using carbon monoxide as a reducing agent. Starting with coke in a blast furnace, the following equilibrium plays a key role in the ex-traction of iron:

C(s) 1 CO2(g) ∆ 2CO(g)

Use the data in Appendix 3 to calculate the equilib-rium constant at 25°C and 1000°C. Assume DH° and DS° to be independent of temperature.

21.105 Assuming ideal behavior, calculate the density of gaseous HF at its normal boiling point (19.5°C). The experimentally measured density under the same conditions is 3.10 g/L. Account for the discrepancy between your calculated value and the experimental result.

21.106 A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500 mL of solution. Calculate the pH of the solution at 25°C.

Special Problems 951

Transition Metals Chemistry and Coordination Compounds

A solution showing the green color of chlorophyll. The red color is the fl uorescence of the molecule when irradiated with blue light. The model shows the chlorophyll molecule. The green sphere is the Mg21 ion.

953

AnimationChirality (22.4)Absorption of Color (22.5)


ARISExample Practice ProblemsEnd of Chapter Problems

A Look Ahead• We fi rst survey the general properties of transition metals, focusing on their

electron confi gurations and oxidation states. (22.1)

• Next, we study the chemistry of two representative transition metals—iron and copper. (22.2)

• We then consider the general characteristics of coordination compounds in terms of the nature of ligands and also cover the nomenclature of these com-pounds. (22.3)

• We see that the structure of coordination compounds can give rise to geometric and/or optical isomers. We become acquainted with the use of a polarimeter in studying optical isomers. (22.4)

• Crystal fi eld theory can satisfactorily explain the origin of color in and mag-netic properties of octahedral, tetrahedral, and square-planar complexes. (22.5)

• We examine the reactivity of coordination compounds and learn that they can be classifi ed as labile or inert with regard to ligand exchange reactions. (22.6)

• This chapter concludes with a discussion of several applications of coordina-tion compounds. (22.7)

The series of elements in the periodic table in which the d and f subshells are gradually fi lled are called the transition elements. There are about 50 transi-

tion elements, and they have widely varying and fascinating properties. To pre-sent even one interesting feature of each transition element is beyond the scope of this book. We will therefore limit our discussion to the transition elements that have incompletely fi lled d subshells and to their most commonly encountered property—the tendency to form complex ions.

22.1 Properties of the Transition Metals

22.2 Chemistry of Iron and Copper

22.3 Coordination Compounds

22.4 Structure of Coordination Compounds

22.5 Bonding in Coordination Compounds: Crystal Field Theory

22.6 Reactions of Coordination Compounds

22.7 Applications of Coordination Compounds

Chapter Outline


954 Transition Metals Chemistry and Coordination Compounds

22.1 Properties of the Transition MetalsTransition metals typically have incompletely fi lled d subshells or readily give rise to ions with incompletely fi lled d subshells (Figure 22.1). (The Group 2B metals—Zn, Cd, and Hg—do not have this characteristic electron confi guration and so, although they are sometimes called transition metals, they really do not belong in this category.) This attribute is responsible for several notable properties, including distinctive color-ing, formation of paramagnetic compounds, catalytic activity, and especially a great tendency to form complex ions. In this chapter we focus on the fi rst-row elements from scandium to copper, the most common transition metals. Table 22.1 lists some of their properties. As we read across any period from left to right, atomic numbers increase, elec-trons are added to the outer shell, and the nuclear charge increases by the addition of protons. In the third-period elements—sodium to argon—the outer electrons weakly shield one another from the extra nuclear charge. Consequently, atomic radii decrease rapidly from sodium to argon, and the electronegativities and ionization energies increase steadily (see Figures 8.5, 8.11, and 9.5). For the transition metals, the trends are different. Looking at Table 22.1 we see that the nuclear charge, of course, increases from scandium to copper, but electrons are being added to the inner 3d subshell. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than outer-shell elec-trons can shield one another, so the atomic radii decrease less rapidly. For the same reason, electronegativities and ionization energies increase only slightly from scan-dium across to copper compared with the increases from sodium to argon. Although the transition metals are less electropositive (or more electronegative) than the alkali and alkaline earth metals, their standard reduction potentials suggest that all of them except copper should react with strong acids such as hydrochloric acid to produce hydrogen gas. However, most transition metals are inert toward acids or react slowly with them because of a protective layer of oxide. A case in point is chromium: Despite a rather negative standard reduction potential, it is quite inert

Figure 22.1 The transition metals (blue squares). Note that although the Group 2B elements (Zn, Cd, Hg) are described as transition metals by some chemists, neither the metals nor their ions possess incompletely fi lled d subshells.

1 1A

2 2A

3 3B

4 4B

5 5B

6 6B

8 107 7B

9 8B

11 1B

12 2B

13 3A

14 4A

15 5A

16 6A

17 7A

18 8A

1H

3Li

11Na

19K

37Rb

55Cs

87Fr

20Ca

38Sr

56Ba

88Ra

21Sc

39Y

57La

89Ac

22Ti

40Zr

72Hf

104Rf

23V

41Nb

73Ta

105Db

24Cr

42Mo

74W

106Sg

25Mn

43Tc

75Re

107Bh

26Fe

44Ru

76Os

108Hs

27Co

45Rh

77Ir

109Mt

28Ni

46Pd

78Pt

29Cu

47Ag

79Au

30Zn

48Cd

80Hg

31Ga

49In

81Tl

32Ge

50Sn

82Pb

33As

51Sb

83Bi

34Se

52Te

84Po

35Br

53I

85At

36Kr

54Xe

86Rn

13Al

14Si

15P

16S

17Cl

18Ar

5B

6C

7N

8O

9F

10Ne

2He

4Be

12Mg

110Ds

111Rg

112 113 114 115 116 (117) 118

chemically because of the formation on its surfaces of chromium(III) oxide, Cr2O3. Consequently, chromium is commonly used as a protective and noncorrosive plating on other metals. On automobile bumpers and trim, chromium plating serves a decora-tive as well as a functional purpose.

General Physical PropertiesMost of the transition metals have a close-packed structure (see Figure 11.29) in which each atom has a coordination number of 12. Furthermore, these elements have rela-tively small atomic radii. The combined effect of closest packing and small atomic size results in strong metallic bonds. Therefore, transition metals have higher densities, higher melting points and boiling points, and higher heats of fusion and vaporization than the Group 1A, 2A, and 2B metals (Table 22.2).

Sc Ti V Cr Mn Fe Co Ni Cu

Electron confi guration M 4s23d1 4s23d2 4s23d3 4s13d5 4s23d5 4s23d6 4s23d7 4s23d8 4s13d10

M21 — 3d2 3d3 3d4 3d5 3d6 3d7 3d8 3d9

M31 [Ar] 3d1 3d2 3d3 3d4 3d5 3d6 3d7 3d8

Electronegativity 1.3 1.5 1.6 1.6 1.5 1.8 1.9 1.9 1.9Ionization energy(kJ/mol) First 631 658 650 652 717 759 760 736 745 Second 1235 1309 1413 1591 1509 1561 1645 1751 1958 Third 2389 2650 2828 2986 3250 2956 3231 3393 3578Radius (pm) M 162 147 134 130 135 126 125 124 128 M21 — 90 88 85 91 82 82 78 72 M31 83 68 74 64 66 67 64 — —Standard reduction potential (V)* 2 2.08 2 1.63 2 1.2 2 0.74 2 1.18 2 0.44 2 0.28 2 0.25 0.34

*The half-reaction is M21(aq) 1 2e2 ¡ M(s) (except for Sc and Cr, where the ions are Sc31 and Cr31, respectively).

TABLE 22.1 Electron Confi gurations and Other Properties of the First-Row Transition Metals

1A 2A Transition Metals 2B

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

Atomicradius (pm) 227 197 162 147 134 130 135 126 125 124 128 138Meltingpoint (°C) 63.7 838 1539 1668 1900 1875 1245 1536 1495 1453 1083 419.5Boilingpoint (°C) 760 1440 2730 3260 3450 2665 2150 3000 2900 2730 2595 906Density(g/cm3) 0.86 1.54 3.0 4.51 6.1 7.19 7.43 7.86 8.9 8.9 8.96 7.14

TABLE 22.2 Physical Properties of Elements K to Zn

22.1 Properties of the Transition Metals 955


Electron Confi gurationsThe electron confi gurations of the fi rst-row transition metals were discussed in Sec-tion 7.9. Calcium has the electron confi guration [Ar]4s2. From scandium across to copper, electrons are added to the 3d orbitals. Thus, the outer electron confi guration of scandium is 4s23d1, that of titanium is 4s23d2, and so on. The two exceptions are chromium and copper, whose outer electron confi gurations are 4s13d5 and 4s13d10, respectively. These irregularities are the result of the extra stability associated with half-fi lled and completely fi lled 3d subshells. When the fi rst-row transition metals form cations, electrons are removed fi rst from the 4s orbitals and then from the 3d orbitals. (This is the opposite of the order in which orbitals are fi lled in atoms.) For example, the outer electron confi guration of Fe21 is 3d6, not 4s23d4.

Oxidation StatesTransition metals exhibit variable oxidation states in their compounds. Figure 22.2 shows the oxidation states from scandium to copper. Note that the common oxidation

Sc Ti V Cr Mn Fe Co Ni Cu

+7

+6 +6 +6

+5 +5 +5 +5

+4 +4 +4 +4 +4 +4

+3 +3 +3 +3 +3 +3 +3 +3 +3

+2 +2 +2 +2 +2 +2 +2 +2

+1

Figure 22.2 Oxidation states of the fi rst-row transition metals. The most stable oxidation numbers are shown in color. The zero oxidation state is encountered in some compounds, such as Ni(CO)4 and Fe(CO)5.

Review of ConceptsLocate the transition metal atoms and ions in the periodic table shown here. Atoms: (1) [Kr]5s24d5. (2) [Xe]6s24f 145d4. Ions: (3) [Ar]3d3 (a 14 ion). (4) [Xe]4f 145d8 (a 13 ion). (See Table 7.3)

22.2 Chemistry of Iron and Copper 957

states for each element include 12, 13, or both. The 13 oxidation states are more stable at the beginning of the series, whereas toward the end the 12 oxidation states are more stable. The reason for this trend can be understood by examining the ioniza-tion energy plots in Figure 22.3. In general, the ionization energies increase gradually from left to right. However, the third ionization energy (when an electron is removed from the 3d orbital) increases more rapidly than the fi rst and second ionization ener-gies. Because it takes more energy to remove the third electron from the metals near the end of the row than from those near the beginning, the metals near the end tend to form M21 ions rather than M31 ions. The highest oxidation state for a transition metal is 17, for manganese (4s23d5). For elements to the right of Mn (Fe to Cu), oxidation numbers are lower. Transition metals usually exhibit their highest oxidation states in compounds with very electronegative elements such as oxygen and fl uorine—for example, V2O5, CrO3, and Mn2O7.

22.2 Chemistry of Iron and CopperFigure 22.4 on p. 958 shows the fi rst-row transition metals. In this section we will briefl y survey the chemistry of two of these elements—iron and copper—paying par-ticular attention to their occurrence, preparation, uses, and important compounds.

IronAfter aluminum, iron is the most abundant metal in Earth’s crust (6.2 percent by mass). It is found in many ores; some of the important ones are hematite, Fe2O3; siderite, FeCO3; and magnetite, Fe3O4 (Figure 22.5). The preparation of iron in a blast furnace and steelmaking were discussed in Section 20.2. Pure iron is a gray metal and is not particularly hard. It is an essential element in living systems. Iron reacts with hydrochloric acid to give hydrogen gas:

Fe(s) 1 2H1(aq) ¡ Fe21(aq) 1 H2(g)

Concentrated sulfuric acid oxidizes the metal to Fe31, but concentrated nitric acid renders the metal “passive” by forming a thin layer of Fe3O4 over the surface. One of the best-known reactions of iron is rust formation (see Section 19.7). The two

Recall that oxides in which the metal has a high oxidation number are covalent and acidic, whereas those in which the metal has a low oxidation number are ionic and basic (see Section 15.11).

Recall that oxides in which the metal has a high oxidation number are covalent and acidic, whereas those in which the metal has a low oxidation number are ionic and basic (see Section 15.11).

4000

Element

Ioni

zatio

n en

ergy

(kJ/

mol

)

0Ti

3000

2000

1000

Third

FeMnCrSc V Co Ni Cu

Second

First

Figure 22.3 Variation of the fi rst, second, and third ionization energies for the fi rst-row transition metals.


oxidation states of iron are 12 and 13. Iron(II) compounds include FeO (black), FeSO4 ? 7H2O (green), FeCl2 (yellow), and FeS (black). In the presence of oxygen, Fe21 ions in solution are readily oxidized to Fe31 ions. Iron(III) oxide is reddish brown, and iron(III) chloride is brownish black.

CopperCopper, a rare element (6.8 3 102 3 percent of Earth’s crust by mass), is found in nature in the uncombined state as well as in ores such as chalcopyrite, CuFeS2 (Figure 22.6). The reddish-brown metal is obtained by roasting the ore to give Cu2S and then metallic copper:

2CuFeS2(s) 1 4O2(g) ¡ Cu2S(s) 1 2FeO(s) 1 3SO2(g) Cu2S(s) 1 O2(g) ¡ 2Cu(l) 1 SO2(g)

Impure copper can be purifi ed by electrolysis (see Section 20.2). After silver, which is too expensive for large-scale use, copper has the highest electrical conductivity. It is also a good thermal conductor. Copper is used in alloys, electrical cables, plumbing (pipes), and coins.

Figure 22.4 The fi rst-row transition metals.

Scandium (Sc) Titanium (Ti) Vanadium (V)

Chromium (Cr) Manganese (Mn) Iron (Fe)

Cobalt (Co) Nickel (Ni) Copper (Cu)

Figure 22.5 The iron ore magnetite, Fe3O4.

Figure 22.6 Chalcopyrite, CuFeS2.

Copper reacts only with hot concentrated sulfuric acid and nitric acid (see Figure 21.7). Its two important oxidation states are 11 and 12. The 11 state is less stable and disproportionates in solution:

2Cu1(aq) ¡ Cu(s) 1 Cu21(aq)

All compounds of Cu(I) are diamagnetic and colorless except for Cu2O, which is red. The Cu(II) compounds are all paramagnetic and colored. The hydrated Cu21 ion is blue. Some important Cu(II) compounds are CuO (black), CuSO4 ? 5H2O (blue), and CuS (black).

22.3 Coordination CompoundsTransition metals have a distinct tendency to form complex ions (see p. 749). A coor-dination compound typically consists of a complex ion and counter ion. [Note that some coordination compounds such as Fe(CO)5 do not contain complex ions.] Our understanding of the nature of coordination compounds stems from the classic work of Alfred Werner,† who prepared and characterized many coordination compounds. In 1893, at the age of 26, Werner proposed what is now commonly referred to as Werner’s coordination theory. Nineteenth-century chemists were puzzled by a certain class of reactions that seemed to violate valence theory. For example, the valences of the elements in cobalt(III) chloride and in ammonia seem to be completely satisfi ed, and yet these two substances react to form a stable compound having the formula CoCl3 ? 6NH3. To explain this behavior, Werner postulated that most elements exhibit two types of valence: primary valence and secondary valence. In modern terminology, primary valence corresponds to the oxidation number and secondary valence to the coordina-tion number of the element. In CoCl3 ? 6NH3, according to Werner, cobalt has a primary valence of 3 and a secondary valence of 6. Today we use the formula [Co(NH3)6]Cl3 to indicate that the ammonia molecules and the cobalt atom form a complex ion; the chloride ions are not part of the complex but are held to it by ionic forces. Most, but not all, of the metals in coordination compounds are transition metals. The molecules or ions that surround the metal in a complex ion are called ligands (Table 22.3). The interactions between a metal atom and the ligands can be thought of as Lewis acid-base reactions. As we saw in Section 15.12, a Lewis base is a sub-stance capable of donating one or more electron pairs. Every ligand has at least one unshared pair of valence electrons, as these examples show:

ONAH HGD

H

S SHDOG

H!ClS SCqOSSOQ

Therefore, ligands play the role of Lewis bases. On the other hand, a transition metal atom (in either its neutral or positively charged state) acts as a Lewis acid, accepting (and sharing) pairs of electrons from the Lewis bases. Thus, the metal-ligand bonds are usually coordinate covalent bonds (see Section 9.9).

Recall that a complex ion contains a central metal ion bonded to one or more ions or molecules (see Section 16.10).

Recall that a complex ion contains a central metal ion bonded to one or more ions or molecules (see Section 16.10).

Ligands act as Lewis bases by donating electrons to metals, which act as Lewis acids.

Ligands act as Lewis bases by donating electrons to metals, which act as Lewis acids.

†Alfred Werner (1866–1919). Swiss chemist. Werner started as an organic chemist but became interested in coordination chemistry. For his theory of coordination compounds, Werner was awarded the Nobel Prize in Chemistry in 1913.

22.3 Coordination Compounds 959


The atom in a ligand that is bound directly to the metal atom is known as the donor atom. For example, nitrogen is the donor atom in the [Cu(NH3)4]

21 complex ion. The coordination number in coordination compounds is defi ned as the number of donor atoms surrounding the central metal atom in a complex ion. For example, the coordination number of Ag1 in [Ag(NH3)2]

1 is 2, that of Cu21 in [Cu(NH3)4]21

is 4, and that of Fe31 in [Fe(CN)6]32 is 6. The most common coordination numbers

are 4 and 6, but coordination numbers such as 2 and 5 are also known. Depending on the number of donor atoms present, ligands are classifi ed as mono-dentate, bidentate, or polydentate (see Table 22.3). H2O and NH3 are monodentate ligands with only one donor atom each. One bidentate ligand is ethylenediamine (sometimes abbreviated “en”):

H2NO CH2O CH2O NH2O O

The two nitrogen atoms can coordinate with a metal atom, as shown in Figure 22.7.

In a crystal lattice, the coordination number of an atom (or ion) is defi ned as the number of atoms (or ions) surrounding the atom (or ion).

In a crystal lattice, the coordination number of an atom (or ion) is defi ned as the number of atoms (or ions) surrounding the atom (or ion).

Name Structure

Monodentate ligands

Ammonia AH

HO NO HO

Carbon monoxide SCqOS

Chloride ion SClS!OQ

Cyanide ion [SCqNS]!

Thiocyanate ion [SSO CqNS]!OQ

Water HO OO HOQ

Bidentate ligands

Ethylenediamine S

S

O O

SSS S

S

SS

S

H2NO CH2O CH2O NH2

2!

CO CJ

O

GO

OM

OD

Oxalate ion

Polydentate ligand

Ethylenediaminetetraacetateion (EDTA)

GNO CH2O CH2O N

4!

OS

SS

S

S

O

S

S

S

SS

SOSB

SOSB

CA

SOS

CA

SOS

OD

CG

CH2

CGO

G CH2

G JO

D D

CH2

D D

CH2

OM

Q Q

TABLE 22.3 Some Common Ligands

H2N

CH2

CH2

NH2

NH2

CH2

CH2

H2N

NH2

H2N

CH2

CH2

(a) (b)

Figure 22.7 (a) Structure of a metal-ethylenediamine complex cation, such as [Co(en)3]21. Each ethylenediamine molecule provides two N donor atoms and is therefore a bidentate ligand. (b) Simplifi ed structure of the same complex cation.

O

C

CH2CH2

CH2

CH2

C

O

Pb

CH2CH2O

C

C

O

(a) (b)

O

O N

N

O

O

Figure 22.8 (a) EDTA complex of lead. The complex bears a net charge of 2 2 because each O donor atom has one negative charge and the lead ion carries two positive charges. Only the lone pairs that participate in bonding are shown. Note the octahedral geometry around the Pb21 ion. (b) Molecular model of the Pb21–EDTA complex. The yellow sphere is the Pb21 ion.


Review of ConceptsWhat is the difference between these two compounds: CrCl3 ? 6H2O and [Cr(H2O)6]Cl3?

Bidentate and polydentate ligands are also called chelating agents because of their ability to hold the metal atom like a claw (from the Greek chele, meaning “claw”). One example is ethylenediaminetetraacetate ion (EDTA), a polydentate ligand used to treat metal poisoning (Figure 22.8). Six donor atoms enable EDTA to form a very stable complex ion with lead. In this form, it is removed from the blood and tissues and excreted from the body. EDTA is also used to clean up spills of radioactive metals.

Oxidation Numbers of Metals in Coordination CompoundsAnother important property of coordination compounds is the oxidation number of the central metal atom. The net charge of a complex ion is the sum of the charges on the central metal atom and its surrounding ligands. In the [PtCl6]

22 ion, for example, each chloride ion has an oxidation number of 2 1, so the oxidation number of Pt must


Naming Coordination CompoundsNow that we have discussed the various types of ligands and the oxidation numbers of metals, our next step is to learn what to call these coordination compounds. The rules for naming coordination compounds are as follows:

1. The cation is named before the anion, as in other ionic compounds. The rule holds regardless of whether the complex ion bears a net positive or a negative charge. For example, in K3[Fe(CN)6] and [Co(NH3)4Cl2]Cl compound, we name the K1 and [Co(NH3)4Cl2]

1 cations fi rst, respectively.

2. Within a complex ion the ligands are named fi rst, in alphabetical order, and the metal ion is named last.

3. The names of anionic ligands end with the letter o, whereas a neutral ligand is usually called by the name of the molecule. The exceptions are H2O (aqua), CO (carbonyl), and NH3 (ammine). Table 22.4 lists some common ligands.

4. When several ligands of a particular kind are present, we use the Greek prefi xes di-, tri-, tetra-, penta-, and hexa- to name them. Thus, the ligands in the cation [Co(NH3)4Cl2]

1 are “tetraamminedichloro.” (Note that prefi xes are ignored when alphabetizing ligands.) If the ligand itself contains a Greek prefi x, we use the prefi xes bis (2), tris (3), and tetrakis (4) to indicate the number of ligands present. For example, the ligand ethylenediamine already contains di; therefore, if two such ligands are present the name is bis(ethylenediamine).

5. The oxidation number of the metal is written in Roman numerals following the name of the metal. For example, the Roman numeral III is used to indicate the

EXAMPLE 22.1

Specify the oxidation number of the central metal atom in each of the following compounds: (a) [Ru(NH3)5(H2O)]Cl2, (b) [Cr(NH3)6](NO3)3, (c) [Fe(CO)5], and (d) K4[Fe(CN)6].

Strategy The oxidation number of the metal atom is equal to its charge. First we examine the anion or the cation that electrically balances the complex ion. This step gives us the net charge of the complex ion. Next, from the nature of the ligands (charged or neutral species) we can deduce the net charge of the metal and hence its oxidation number.

Solution (a) Both NH3 and H2O are neutral species. Because each chloride ion carries a 2 1 charge, and there are two Cl2 ions, the oxidation number of Ru must be 12.

(b) Each nitrate ion has a charge of 2 1; therefore, the cation must be [Cr(NH3)6]31.

NH3 is neutral, so the oxidation number of Cr is 13.(c) Because the CO species are neutral, the oxidation number of Fe is zero.(d) Each potassium ion has a charge of 11; therefore, the anion is [Fe(CN)6]

42 . Next, we know that each cyanide group bears a charge of 2 1, so Fe must have an oxidation number of 12.

Practice Exercise Write the oxidation numbers of the metals in the compound K[Au(OH)4].

Similar problems: 22.13, 22.14.Similar problems: 22.13, 22.14.

be 14. If the ligands do not bear net charges, the oxidation number of the metal is equal to the charge of the complex ion. Thus, in [Cu(NH3)4]

21 each NH3 is neutral, so the oxidation number of Cu is 12. Example 22.1 deals with oxidation numbers of metals in coordination compounds.

13 oxidation state of chromium in [Cr(NH3)4Cl2]1, which is called tetraammine-

dichlorochromium(III) ion.

6. If the complex is an anion, its name ends in -ate. For example, in K4[Fe(CN)6] the anion [Fe(CN)6]

42 is called hexacyanoferrate(II) ion. Note that the Roman numeral II indicates the oxidation state of iron. Table 22.5 gives the names of anions containing metal atoms.

Examples 22.2 and 22.3 deal with the nomenclature of coordination compounds.

Name of Ligand in Ligand Coordination Compound

Bromide, Br2 BromoChloride, Cl2 ChloroCyanide, CN2 CyanoHydroxide, OH2 HydroxoOxide, O22 OxoCarbonate, CO3

22 CarbonatoNitrite, NO2

2 NitroOxalate, C2O4

22 OxalatoAmmonia, NH3 AmmineCarbon monoxide, CO CarbonylWater, H2O AquaEthylenediamine EthylenediamineEthylenediaminetetraacetate Ethylenediaminetetraacetato

TABLE 22.4 Names of Common Ligands in Coordination Compounds

Name of Metal in AnionicMetal Complex

Aluminum AluminateChromium ChromateCobalt CobaltateCopper CuprateGold AurateIron FerrateLead PlumbateManganese ManganateMolybdenum MolybdateNickel NickelateSilver ArgentateTin StannateTungsten TungstateZinc Zincate

Names of Anions Containing Metal Atoms

TABLE 22.5


EXAMPLE 22.2

Write the systematic names of the following coordination compounds: (a) Ni(CO)4, (b) NaAuF4, (c) K3[Fe(CN)6], (d) [Cr(en)3]Cl3.

Strategy We follow the preceding procedure for naming coordination compounds and refer to Tables 22.4 and 22.5 for names of ligands and anions containing metal atoms.

Solution (a) The CO ligands are neutral species and therefore the Ni atom bears no net charge. The compound is called tetracarbonylnickel(0) , or more commonly, nickel tetracarbonyl .

(b) The sodium cation has a positive charge; therefore, the complex anion has a negative charge (AuF4

2 ). Each fl uoride ion has a negative charge so the oxidation number of gold must be 13 (to give a net negative charge). The compound is called sodium tetrafl uoroaurate(III) .

(c) The complex ion is the anion and it bears three negative charges because each potassium ion bears a 11 charge. Looking at [Fe(CN)6]

32 , we see that the oxidation number of Fe must be 13 because each cyanide ion bears a 2 1 charge (2 6 total). The compound is potassium hexacyanoferrate(III) . This compound is commonly called potassium ferricyanide .

(Continued)


22.4 Structure of Coordination CompoundsIn studying the geometry of coordination compounds, we often fi nd that there is more than one way to arrange ligands around the central atom. Compounds rearranged in this fashion have distinctly different physical and chemical properties. Figure 22.9 shows four different geometric arrangements for metal atoms with monodentate ligands. In these diagrams, we see that structure and coordination number of the metal atom relate to each other as follows:

Coordination number Structure 2 Linear 4 Tetrahedral or square planar 6 Octahedral

Stereoisomers are compounds that are made up of the same types and numbers of atoms bonded together in the same sequence but with different spatial arrangements. There are two types of stereoisomers: geometric isomers and optical isomers. Coor-dination compounds may exhibit one or both types of isomerism. Note, however, that many coordination compounds do not have stereoisomers.

(d) As we noted earlier, en is the abbreviation for the ligand ethylenediamine. Because there are three chloride ions each with a 2 1 charge, the cation is [Cr(en)3]

31. The en ligands are neutral so the oxidation number of Cr must be 13. Because there are three en groups present and the name of the ligand already contains di (rule 4), the compound is called tris(ethylenediamine)chromium(III) chloride .

Practice Exercise What is the systematic name of [Cr(H2O)4Cl2]Cl?

Similar problems: 22.15, 22.16.

EXAMPLE 22.3

Write the formulas for the following compounds: (a) pentaamminechlorocobalt(III) chloride, (b) dichlorobis(ethylenediamine)platinum(IV) nitrate, (c) sodium hexanitrocobaltate(III).

Strategy We follow the preceding procedure and refer to Tables 22.4 and 22.5 for names of ligands and anions containing metal atoms.

Solution (a) The complex cation contains fi ve NH3 groups, a Cl2 ion, and a Co ion having a 13 oxidation number. The net charge of the cation must be 12, [Co(NH3)5Cl]21. Two chloride anions are needed to balance the positive charges. Therefore, the formula of the compound is [Co(NH3)5Cl]Cl2 .

(b) There are two chloride ions (2 1 each), two en groups (neutral), and a Pt ion with an oxidation number of 14. The net charge on the cation must be 12, [Pt(en)2Cl2]

21. Two nitrate ions are needed to balance the 12 charge of the complex cation. Therefore, the formula of the compound is [Pt(en)2Cl2](NO3)2 .

(c) The complex anion contains six nitro groups (2 1 each) and a cobalt ion with an oxidation number of 13. The net charge on the complex anion must be 2 3, [Co(NO2)6]

32 . Three sodium cations are needed to balance the 2 3 charge of the complex anion. Therefore, the formula of the compound is Na3[Co(NO2)6] .

Practice Exercise Write the formula for the following compound: tris(ethylene-diamine)cobalt(III) sulfate.

Similar problems: 22.17, 22.18.Similar problems: 22.17, 22.18.

Geometric IsomersGeometric isomers are stereoisomers that cannot be interconverted without breaking a chemical bond. Geometric isomers usually come in pairs. We use the terms “cis” and “trans” to distinguish one geometric isomer of a compound from the other. Cis means that two particular atoms (or groups of atoms) are adjacent to each other, and trans means that the atoms (or groups of atoms) are on opposite sides in the structural for-mula. The cis and trans isomers of coordination compounds generally have quite dif-ferent colors, melting points, dipole moments, and chemical reactivities. Figure 22.10 shows the cis and trans isomers of diamminedichloroplatinum(II). Note that although the types of bonds are the same in both isomers (two Pt—N and two Pt—Cl bonds), the spatial arrangements are different. Another example is tetraamminedichloro-cobalt(III) ion, shown in Figure 22.11.

Optical IsomersOptical isomers are nonsuperimposable mirror images. (“Superimposable” means that if one structure is laid over the other, the positions of all the atoms will match.) Like geometric isomers, optical isomers come in pairs. However, the optical isomers of a compound have identical physical and chemical properties, such as melting point, boiling point, dipole moment, and chemical reactivity toward molecules that are not optical isomers themselves. Optical isomers differ from each other in their interactions with plane-polarized light, as we will see. The structural relationship between two optical isomers is analogous to the rela-tionship between your left and right hands. If you place your left hand in front of a mirror, the image you see will look like your right hand (Figure 22.12). We say that your left hand and right hand are mirror images of each other. However, they are nonsuperimposable, because when you place your left hand over your right hand (with both palms facing down), they do not match. Figure 22.13 shows the cis and trans isomers of dichlorobis(ethylenediamine)-cobalt(III) ion and their images. Careful examination reveals that the trans isomer and its mirror image are superimposable, but the cis isomer and its mirror

M LLL

L

L

L

M M

L L

L L

L

L

L

LM

L

L

Linear Tetrahedral Square planar Octahedral

Figure 22.9 Common geometries of complex ions. In each case, M is a metal and L is a monodentate ligand.

cis-tetraamminedichlorocobalt(III) chloride (left) and trans-tetraammine-dichlorocobalt(III) chloride (right).

H3N

(a)

Cl

ClH3N

Pt

Cl

(b)

NH3

ClH3N

Pt

Figure 22.10 The (a) cis and (b) trans isomers of diammine-dichloroplatinum(II). Note that the two Cl atoms are adjacent to each other in the cis isomer and diagonally across from each other in the trans isomer.

22.4 Structure of Coordination Compounds 965


image are not. Therefore, the cis isomer and its mirror image are optical isomers. Optical isomers are described as chiral (from the Greek word for “hand”) because, like your left and right hands, chiral molecules are nonsuperimposable. Isomers that are superimposable with their mirror images are said to be achiral. Chiral molecules play a vital role in enzyme reactions in biological systems. Many drug molecules are chiral. It is interesting to note that frequently only one of a pair of chiral isomers is biologically effective. Chiral molecules are said to be optically active because of their ability to rotate the plane of polarization of polarized light as it passes through them. Unlike ordinary light, which vibrates in all directions, plane-polarized light vibrates only in a single plane. We use a polarimeter to measure the rotation of polarized light by optical isomers (Figure 22.14). A beam of unpolarized light fi rst passes through a Polaroid sheet, called the polarizer, and then through a sample tube containing a solution of an optically active, chiral compound. As the polarized light passes through the sample tube, its plane of polarization is rotated either to the right or to the left. This rotation can be measured directly by turning the analyzer in the appropriate direction until minimal light transmission is achieved (Figure 22.15). If the plane of polarization is rotated to the right, the isomer is said to be dextrorotatory (d); it is levorotatory (l) if the rotation is to the left. The d and l isomers of a chiral substance, called enan-tiomers, always rotate the light by the same amount, but in opposite directions. Thus, in an equimolar mixture of two enantiomers, called a racemic mixture, the net rotation is zero.

Polaroid sheets are used to make Polaroid glasses.Polaroid sheets are used to make Polaroid glasses.

Cl

(a)

Cl

Co

Cl

Cl

Co

Mirror

Cl

(b)

Cl

Co

Cl

Cl

Co

Mirror

Figure 22.13 The (a) cis and (b) trans isomers of dichlorobis(ethylenediamine)cobalt(III) ion and their mirror images. If you could rotate the mirror image in (b) 90° clockwise about the vertical position and place the ion over the trans isomer, you would fi nd that the two are superimposable. No matter how you rotated the cis isomer and its mirror image in (a), however, you could not superimpose one on the other.

Mirror imageof left hand

Left hand

Mirror

Figure 22.12 A left hand and its mirror image, which looks the same as the right hand.

AnimationChirality

NH3

(a)

Cl

Cl NH3

Co

NH3

NH3

NH3

(b)

Cl

ClH3N

Co

NH3

NH3

NH3

(d)

H3N

NH3H3N

Co

Cl

Cl

NH3

(c)

Cl

NH3H3N

Co

NH3

Cl

Figure 22.11 The (a) cis and (b) trans isomers of tetraamminedichlorocobalt(III) ion, [Co(NH3)4Cl2]1. The structure shown in (c) can be generated by rotating that in (a), and the structure shown in (d) can be generated by rotating that in (b). The ion has only two geometric isomers, (a) [or (c)] and (b) [or (d)].

22.5 Bonding in Coordination Compounds: Crystal Field Theory

A satisfactory theory of bonding in coordination compounds must account for proper-ties such as color and magnetism, as well as stereochemistry and bond strength. No single theory as yet does all this for us. Rather, several different approaches have been applied to transition metal complexes. We will consider only one of them here—crystal fi eld theory—because it accounts for both the color and magnetic properties of many coordination compounds.

Figure 22.15 With one Polaroid sheet over a picture, light passes through. With a second sheet of Polaroid placed over the fi rst so that the axes of polarization of the sheets are perpendicular, little or no light passes through. If the axes of polarization of the two sheets were parallel, light would pass through.

22.5 Bonding in Coordination Compounds: Crystal Field Theory 967

Fixedpolarizer

Polarimeter tube

Optically active substance in solution

Plane of polarization

Degree scaleAnalyzer 0°

–90°

180°

+90°

Lightsource

+

–

Figure 22.14 Operation of a polarimeter. Initially, the tube is fi lled with an achiral compound. The analyzer is rotated so that its plane of polarization is perpendicular to that of the polarizer. Under this condition, no light reaches the observer. Next, a chiral compound is placed in the tube as shown. The plane of polarization of the polarized light is rotated as it travels through the tube so that some light reaches the observer. Rotating the analyzer (either to the left or to the right) until no light reaches the observer again enables the angle of optical rotation to be measured.


We will begin our discussion of crystal fi eld theory with the most straightforward case, namely, complex ions with octahedral geometry. Then we will see how it is applied to tetrahedral and square-planar complexes.

Crystal Field Splitting in Octahedral ComplexesCrystal fi eld theory explains the bonding in complex ions purely in terms of elec-trostatic forces. In a complex ion, two types of electrostatic interaction come into play. One is the attraction between the positive metal ion and the negatively charged ligand or the negatively charged end of a polar ligand. This is the force that binds the ligands to the metal. The second type of interaction is electrostatic repulsion between the lone pairs on the ligands and the electrons in the d orbitals of the metals. As we saw in Chapter 7, d orbitals have different orientations, but in the absence of external disturbance they all have the same energy. In an octahedral complex, a central metal atom is surrounded by six lone pairs of electrons (on the six ligands), so all fi ve d orbitals experience electrostatic repulsion. The magnitude of this repul-sion depends on the orientation of the d orbital that is involved. Take the dx22y2 orbital as an example. In Figure 22.16, we see that the lobes of this orbital point toward corners of the octahedron along the x and y axes, where the lone-pair electrons are positioned. Thus, an electron residing in this orbital would experience a greater repul-sion from the ligands than an electron would in, say, the dxy orbital. For this reason, the energy of the dx22y2 orbital is increased relative to the dxy, dyz, and dxz orbitals. The dz2 orbital’s energy is also greater, because its lobes are pointed at the ligands along the z-axis. As a result of these metal-ligand interactions, the fi ve d orbitals in an octahedral complex are split between two sets of energy levels: a higher level with two orbitals (dx22y2 and dz2) having the same energy and a lower level with three equal-energy orbitals (dxy, dyz, and dxz), as shown in Figure 22.17. The crystal fi eld splitting (D) is the energy difference between two sets of d orbitals in a metal atom when ligands are present. The magnitude of D depends on the metal and the nature of the ligands; it has a direct effect on the color and magnetic properties of complex ions.

The name “crystal fi eld” is associated with the theory used to explain the properties of solid, crystalline materials. The same theory is used to study coordination compounds.

The name “crystal fi eld” is associated with the theory used to explain the properties of solid, crystalline materials. The same theory is used to study coordination compounds.

dz2 dx2 – y2

dyz dxzdxy

x

z

y

Figure 22.16 The fi ve d orbitals in an octahedral environment. The metal atom (or ion) is at the center of the octahedron, and the six lone pairs on the donor atoms of the ligands are at the corners.

ColorIn Chapter 7 we learned that white light, such as sunlight, is a combination of all colors. A substance appears black if it absorbs all the visible light that strikes it. If it absorbs no visible light, it is white or colorless. An object appears green if it absorbs all light but refl ects the green component. An object also looks green if it refl ects all colors except red, the complementary color of green (Figure 22.18). What has been said of refl ected light also applies to transmitted light (that is, the light that passes through the medium, for example, a solution). Consider the hydrated cupric ion, [Cu(H2O)6]

21, which absorbs light in the orange region of the spectrum so that a solution of CuSO4 appears blue to us. Recall from Chapter 7 that when the energy of a photon is equal to the difference between the ground state and an excited state, absorption occurs as the photon strikes the atom (or ion or compound), and an electron is promoted to a higher level. This knowledge enables us to calculate the energy change involved in the electron transition. The energy of a photon, given by Equation (7.2), is

E 5 hv

where h represents Planck’s constant (6.63 3 102 34 J ? s) and v is the frequency of the radiation, which is 5.00 3 1014/s for a wavelength of 600 nm. Here E 5 D, so we have

¢ 5 hv 5 (6.63 3 10234 J ? s)(5.00 3 1014ys) 5 3.32 3 10219 J

(Note that this is the energy absorbed by one ion.) If the wavelength of the photon absorbed by an ion lies outside the visible region, then the transmitted light looks the same (to us) as the incident light—white—and the ion appears colorless. The best way to measure crystal fi eld splitting is to use spectroscopy to determine the wavelength at which light is absorbed. The [Ti(H2O)6]

31 ion provides a straightfor-ward example, because Ti31 has only one 3d electron (Figure 22.19). The [Ti(H2O)6]

31 ion absorbs light in the visible region of the spectrum (Figure 22.20). The wavelength corresponding to maximum absorption is 498 nm [Figure 22.19(b)]. This information enables us to calculate the crystal fi eld splitting as follows. We start by writing

¢ 5 hv (22.1)

Also

v 5cl

A d-to-d transition must occur for a transition metal complex to show color. Therefore, ions with d0 or d10 electron confi gurations are usually colorless.

A d-to-d transition must occur for a transition metal complex to show color. Therefore, ions with d0 or d10 electron confi gurations are usually colorless.

AnimationAbsorption of Color

Crystal field splitting

Ene

rgy

dx2 – y2 dz2

dxy dyz dxz

Figure 22.17 Crystal fi eld splitting between d orbitals in an octahedral complex.

560 nm

580 nm650 nm

490 nm430 nm

700 nm400 nm

Figure 22.18 A color wheel with appropriate wavelengths. A compound that absorbs in the green region will appear red, the complementary color of green.



where c is the speed of light and l is the wavelength. Therefore,

¢ 5hcl

5(6.63 3 10234 J ? s)(3.00 3 108 mys)

(498 nm)(1 3 1029 my1 nm) 5 3.99 3 10219 J

This is the energy required to excite one [Ti(H2O)6]31 ion. To express this energy

difference in the more convenient units of kilojoules per mole, we write

¢ 5 (3.99 3 10219 Jyion)(6.02 3 1023 ionsymol) 5 240,000 Jymol 5 240 kJymol

Aided by spectroscopic data for a number of complexes, all having the same metal ion but different ligands, chemists calculated the crystal splitting for each ligand and established a spectrochemical series, which is a list of ligands arranged in increasing order of their abilities to split the d orbital energy levels:

I2 , Br2 , Cl2 , OH2 , F2 , H2O , NH3 , en , CN2 , CO

Equation (7.3) shows that E 5 hcyl.Equation (7.3) shows that E 5 hcyl.

The order in the spectrochemical series is the same no matter which metal atom (or ion) is present.

The order in the spectrochemical series is the same no matter which metal atom (or ion) is present.

Figure 22.20 Colors of some of the fi rst-row transition metal ions in solution. From left to right: Ti31, Cr31, Mn21, Fe31, Co21, Ni21, Cu21. The Sc31 and V51 ions are colorless.

dx2 – y2 dz2

dxy dyz dxz

Photon of energy h" dx2 – y2 dz2

dxy dyz dxz

(a)

700Wavelength (nm)

Abs

orpt

ion

600500400

(b)

Figure 22.19 (a) The process of photon absorption and (b) a graph of the absorption spectrum of [Ti(H2O)6]31. The energy of the incoming photon is equal to the crystal fi eld splitting. The maximum absorption peak in the visible region occurs at 498 nm.

Magnetic PropertiesThe magnitude of the crystal fi eld splitting also determines the magnetic properties of a complex ion. The [Ti(H2O)6]

31 ion, having only one d electron, is always paramagnetic. However, for an ion with several d electrons, the situation is less clearcut. Consider, for example, the octahedral complexes [FeF6]

32 and [Fe(CN)6]32

(Figure 22.21). The electron confi guration of Fe31 is [Ar]3d5, and there are two possible ways to distribute the fi ve d electrons among the d orbitals. According to Hund’s rule (see Section 7.8), maximum stability is reached when the electrons are placed in fi ve separate orbitals with parallel spins. But this arrangement can be achieved only at a cost; two of the fi ve electrons must be promoted to the higher-energy dx22y2 and dz2 orbitals. No such energy investment is needed if all fi ve electrons enter the dxy, dyz, and dxz orbitals. According to Pauli’s exclusion principle (p. 302), there will be only one unpaired electron present in this case. Figure 22.22 shows the distribution of electrons among d orbitals that results in low- and high-spin complexes. The actual arrangement of the electrons is determined by the amount of stability gained by having maximum parallel spins versus the invest-ment in energy required to promote electrons to higher d orbitals. Because F2 is a weak-fi eld ligand, the fi ve d electrons enter fi ve separate d orbitals with parallel spins to create a high-spin complex (see Figure 22.21). On the other hand, the cyanide ion is a strong-fi eld ligand, so it is energetically preferable for all fi ve electrons to be in the lower orbitals and therefore a low-spin complex is formed. High-spin complexes are more paramagnetic than low-spin complexes. The actual number of unpaired electrons (or spins) in a complex ion can be found by magnetic measurements, and in general, experimental fi ndings support predictions based on crystal fi eld splitting. However, a distinction between low- and high-spin complexes can be made only if the metal ion contains more than three and fewer than eight d electrons, as shown in Figure 22.22.

The magnetic properties of a complex ion depend on the number of unpaired electrons present.

The magnetic properties of a complex ion depend on the number of unpaired electrons present.

Review of ConceptsThe Cr31 ion forms octahedral complexes with two neutral ligands X and Y. The color of CrX6

31 is blue while that of CrY631 is yellow. Which is a stronger

fi eld ligand?

Fe3+ ion

Ene

rgy

dx2 – y2 dz2

dxy dyz dxz

dx2 – y2 dz2

dxy dyz dxz[FeF6]3–

(high spin)[Fe(CN)6]3–

(low spin)

Figure 22.21 Energy-level diagrams for the Fe31 ion and for the [FeF6]32 and [Fe(CN)6] 32 complex ions.


These ligands are arranged in the order of increasing value of D. Co and CN2 are called strong-fi eld ligands, because they cause a large splitting of the d orbital energy levels. The halide ions and hydroxide ion are weak-fi eld ligands, because they split the d orbitals to a lesser extent.


dx2 – y2 dz2

dxy dyz dxz

dx2 – y2 dz2

dxy dyz dxz

High spin Low spin

d 4

d 5

d 6

d 7

Figure 22.22 Orbital diagrams for the high-spin and low-spin octahedral complexes corresponding to the electron confi gurations d4, d5, d6, and d7. No such distinctions can be made for d1, d2, d3, d8, d9, and d10.

EXAMPLE 22.4

Predict the number of unpaired spins in the [Cr(en)3]21 ion.

Strategy The magnetic properties of a complex ion depend on the strength of the ligands. Strong-fi eld ligands, which cause a high degree of splitting among the d orbital energy levels, result in low-spin complexes. Weak-fi eld ligands, which cause a small degree of splitting among the d orbital energy levels, result in high-spin complexes.

Solution The electron confi guration of Cr21 is [Ar]3d 4. Because en is a strong-fi eld ligand, we expect [Cr(en)3]

21 to be a low-spin complex. According to Figure 22.22, all four electrons will be placed in the lower-energy d orbitals (dxy, dyz, and dxz) and there will be a total of two unpaired spins.

Practice Exercise How many unpaired spins are in [Mn(H2O)6]21? (H2O is a weak-

fi eld ligand.)

Similar problem: 22.35.Similar problem: 22.35.

Crystal field splitting

Ene

rgy

dxy dyz dxz

dx2 – y2 dz2

Figure 22.23 Crystal fi eld splitting between d orbitals in a tetrahedral complex.

Ene

rgy

dx2 – y2

dxy

dxz

dz2

dyz

Figure 22.24 Energy-level diagram for a square-planar complex. Because there are more than two energy levels, we cannot defi ne crystal fi eld splitting as we can for octahedral and tetrahedral complexes.

Tetrahedral and Square-Planar ComplexesSo far we have concentrated on octahedral complexes. The splitting of the d orbital energy levels in two other types of complexes—tetrahedral and square-planar—can also be accounted for satisfactorily by the crystal fi eld theory. In fact, the splitting pattern for a tetrahedral ion is just the reverse of that for octahedral complexes. In this case, the dxy, dyz, and dxz orbitals are more closely directed at the ligands and therefore have more energy than the dx22y2 and dz2 orbitals (Figure 22.23). Most tetrahedral complexes are high-spin complexes. Presumably, the tetrahedral arrange-ment reduces the magnitude of metal-ligand interactions, resulting in a smaller D value compared to the octahedral case. This is a reasonable assumption because the number of ligands is smaller in a tetrahedral complex. As Figure 22.24 shows, the splitting pattern for square-planar complexes is the most complicated of the three cases. Clearly, the dx22y2 orbital possesses the highest energy (as in the octahedral case), and the dxy orbital the next highest. However, the relative placement of the dz2 and the dxz and dyz orbitals cannot be determined simply by inspection and must be calculated.

22.6 Reactions of Coordination CompoundsComplex ions undergo ligand exchange (or substitution) reactions in solution. The rates of these reactions vary widely, depending on the nature of the metal ion and the ligands.

22.6 Reactions of Coordination Compounds 973


In studying ligand exchange reactions, it is often useful to distinguish between the stability of a complex ion and its tendency to react, which we call kinetic lability. Stability in this context is a thermodynamic property, which is measured in terms of the species’ formation constant Kf (see p. 749). For example, we say that the complex ion tetracyanonickelate(II) is stable because it has a large formation constant (Kf < 1 3 1030)

Ni21 1 4CN2 ∆ [Ni(CN)4]22

By using cyanide ions labeled with the radioactive isotope carbon-14, chemists have shown that [Ni(CN)4]

22 undergoes ligand exchange very rapidly in solution. The following equilibrium is established almost as soon as the species are mixed:

[Ni(CN)4]22 1 4*CN2 ∆ [Ni(*CN)4]

22 1 4CN2

where the asterisk denotes a 14C atom. Complexes like the tetracyanonickelate(II) ion are termed labile complexes because they undergo rapid ligand exchange reactions. Thus, a thermodynamically stable species (that is, one that has a large formation constant) is not necessarily unreactive. (In Section 13.4 we saw that the smaller the activation energy, the larger the rate constant, and hence the greater the rate.) A complex that is thermodynamically unstable in acidic solution is [Co(NH3)6]

31. The equilibrium constant for the following reaction is about 1 3 1020:

[Co(NH3)6]31 1 6H1 1 6H2O ∆ [Co(H2O)6]

31 1 6NH14

When equilibrium is reached, the concentration of the [Co(NH3)6]31 ion is very low.

However, this reaction requires several days to complete because of the inertness of the [Co(NH3)6]

31 ion. This is an example of an inert complex, a complex ion that undergoes very slow exchange reactions (on the order of hours or even days). It shows that a thermodynamically unstable species is not necessarily chemically reac-tive. The rate of reaction is determined by the energy of activation, which is high in this case. Most complex ions containing Co31, Cr31, and Pt21 are kinetically inert. Because they exchange ligands very slowly, they are easy to study in solution. As a result, our knowledge of the bonding, structure, and isomerism of coordination compounds has come largely from studies of these compounds.

22.7 Applications of Coordination CompoundsCoordination compounds are found in living systems and have many uses in the home, in industry, and in medicine. We describe a few examples here and in the Chemistry in Action essay on p. 976.

MetallurgyThe extraction of silver and gold by the formation of cyanide complexes (p. 922) and the purifi cation of nickel (p. 892) by converting the metal to the gaseous compound Ni(CO)4 are typical examples of the use of coordination compounds in metallurgical processes.

At equilibrium, there is a distribution of *CN2 ions in the complex ion.At equilibrium, there is a distribution of *CN2 ions in the complex ion.

Therapeutic Chelating AgentsEarlier we mentioned that the chelating agent EDTA is used in the treatment of lead poisoning. Certain platinum-containing compounds can effectively inhibit the growth of cancerous cells. A specifi c case is discussed on p. 978.

Chemical AnalysisAlthough EDTA has a great affi nity for a large number of metal ions (especially 21 and 31 ions), other chelates are more selective in binding. For example, dimethylglyoxime,

H3CG

H3CD

CP NO OH ACP NO OH

forms an insoluble brick-red solid with Ni21 and an insoluble bright-yellow solid with Pd21. These characteristic colors are used in qualitative analysis to identify nickel and palladium. Further, the quantities of ions present can be determined by gravimetric analysis (see Section 4.6) as follows: To a solution containing Ni21 ions, say, we add an excess of dimethylglyoxime reagent, and a brick-red precipitate forms. The precipitate is then fi ltered, dried, and weighed. Knowing the formula of the complex (Figure 22.25), we can readily calculate the amount of nickel present in the original solution.

DetergentsThe cleansing action of soap in hard water is hampered by the reaction of the Ca21 ions in the water with the soap molecules to form insoluble salts or curds. In the late 1940s the detergent industry introduced a “builder,” usually sodium tripolyphosphate, to circumvent this problem. The tripolyphosphate ion is an effective chelating agent that forms stable, soluble complexes with Ca21 ions. Sodium tripolyphosphate revo-lutionized the detergent industry. However, because phosphates are plant nutrients, waste waters containing phosphates discharged into rivers and lakes cause algae to grow, resulting in oxygen depletion. Under these conditions most or all aquatic life eventually succumbs. This process is called eutrophication. Consequently, many states have banned phosphate detergents since the 1970s, and manufacturers have reformu-lated their products to eliminate phosphates.

An aqueous suspension of bis(dimethylglyoximato)nickel(II).

Z Z

CH3

C

ZZ

C

H

O

OO

N

N

Ni

N

C

H3C

H3C

CH3

C

OH

N

Figure 22.25 Structure of nickel dimethylglyoxime. Note that the overall structure is stabilized by hydrogen bonds.

5!

SOO PO OO PO OO PO OS

SOSB

ASOSQ

SOSB

ASOSQ

SOSB

ASOSQ

OQ OQ OQ OQ

Tripolyphosphate ion.

22.7 Applications of Coordination Compounds 975

976

C H E M I S T R Y

in Action

C oordination compounds play many important roles in ani-mals and plants. They are essential in the storage and trans-

port of oxygen, as electron transfer agents, as catalysts, and in photosynthesis. Here we focus on coordination compounds con-taining iron and magnesium. Because of its central function as an oxygen carrier for metabolic processes, hemoglobin is probably the most studied of all the proteins. The molecule contains four folded long chains called subunits. Hemoglobin carries oxygen in the blood from the lungs to the tissues, where it delivers the oxygen mole-cules to myoglobin. Myoglobin, which is made up of only one subunit, stores oxygen for metabolic processes in muscle. The porphine molecule forms an important part of the hemoglobin structure. Upon coordination to a metal, the H1 ions that are bonded to two of the four nitrogen atoms in por-phine are displaced. Complexes derived from porphine are

Coordination Compounds in Living SystemsThe heme group in hemoglobin. The Fe21 ion is coordinated with the nitro-gen atoms of the heme group. The ligand below the porphyrin is the his-tidine group, which is attached to the protein. The sixth ligand is a water molecule.N

Fe

N N

NO

H H

N

HN

Protein

Simplifi ed structures of the porphine molecule and the Fe21-porphyrin com-plex. The dashed lines represent coordinate covalent bonds.

N

N

N

NFe

N

N

N

N

Fe2"-porphyrin

Porphine

HH

N

Fe

N N

N

O O

N

HN

(a)

N

Fe

N N

N

N

HN

(b)

N

Fe

N N

N

N

HN

(c)

O

O

O

O

Three possible ways for molecular oxygen to bind to the heme group in hemoglobin. The structure shown in (a) would have a coordination number of 7, which is considered unlikely for Fe(II) complexes. Although the end-on arrangement in (b) seems the most reasonable, evidence points to structure in (c) as the correct one. The structure shown in (c) is the most plausible.

Key Equation

D 5 hn (22.1) Calculation of crystal-fi eld splitting

1. Transition metals usually have incompletely fi lled d subshells and a pronounced tendency to form com-plexes. Compounds that contain complex ions are called coordination compounds.


2. The fi rst-row transition metals (scandium to copper) are the most common of all the transition metals; their chemistry is characteristic, in many ways, of the entire group.


977

called porphyrins, and the iron-porphyrin combination is called the heme group. The iron in the heme group has an oxidation number of 12; it is coordinated to the four nitrogen atoms in the porphine group and also to a nitrogen donor atom in a ligand that is attached to the protein. The sixth ligand is a water mole-cule, which binds to the Fe21 ion on the other side of the ring to complete the octahedral complex. This hemoglobin molecule is called deoxyhemoglobin and imparts a bluish tinge to venous blood. The water ligand can be replaced readily by molecular oxygen to form red oxyhemoglobin found in arterial blood. Each subunit contains a heme group, so each hemoglobin mole-cule can bind up to four O2 molecules. There are three possible structures for oxyhemoglobin. For a number of years, the exact arrangement of the oxygen mole-cule relative to the porphyrin group was not clear. Most experi-mental evidence suggests that the bond between O and Fe is bent relative to the heme group. The porphyrin group is a very effective chelating agent, and not surprisingly, we fi nd it in a number of biological sys-tems. The iron-heme complex is present in another class of pro-teins, called the cytochromes. The iron forms an octahedral complex in these proteins, but because both the histidine and the methionine groups are fi rmly bound to the metal ion, they can-not be displaced by oxygen or other ligands. Instead, the cyto-chromes act as electron carriers, which are essential to metabolic processes. In cytochromes, iron undergoes rapid reversible re-dox reactions:

Fe31 1 e2 ∆ Fe21

which are coupled to the oxidation of organic molecules such as the carbohydrates. The chlorophyll molecule, which is necessary for plant photosynthesis, also contains the porphyrin ring, but in this case the metal ion is Mg21 rather than Fe21.

The heme group in cytochrome c. The ligands above and below the porphyrin are the methionine group and histidine group of the protein, respectively.

NFe

N N

NS

H3C CH2

CH2

N

HN

Protein

N

HC

C

N

C

Mg

C

C

C

C

C

N

C

C

C

C

CH

N C

C

C

C

C

The porphyrin structure in chlorophyll. The dotted lines indicate the coordi-nate covalent bonds. The electron-delocalized portion of the molecule is shown in color.

3. Complex ions consist of a metal ion surrounded by li-gands. The donor atoms in the ligands each contribute an electron pair to the central metal ion in a complex.

4. Coordination compounds may display geometric and/or optical isomerism.

5. Crystal fi eld theory explains bonding in complexes in terms of electrostatic interactions. According to crystal fi eld theory, the d orbitals are split into two higher- energy and three lower-energy orbitals in an octahedral complex. The energy difference between these two sets of d orbitals is the crystal fi eld splitting.

6. Strong-fi eld ligands cause a large crystal fi eld splitting, and weak-fi eld ligands cause a small splitting. Electron spins tend to be parallel with weak-fi eld ligands and paired with strong-fi eld ligands, where a greater invest-ment of energy is required to promote electrons into the high-lying d orbitals.

7. Complex ions undergo ligand exchange reactions in solution.

8. Coordination compounds fi nd application in many dif-ferent areas, for example, as antidotes for metal poison-ing and in chemical analysis.

Chelating agent, p. 961Chiral, p. 966Coordination

compound, p. 959Coordination number, p. 960

Key Words

Crystal fi eld splitting (D), p. 968

Donor atom, p. 960Enantiomers, p. 966Geometric isomers, p. 965

Inert complex, p. 974Labile complex, p. 974Ligand, p. 959Optical isomers, p. 965Polarimeter, p. 966

Racemic mixture, p. 966Spectrochemical

series, p. 970Stereoisomers, p. 964



ARIS Problems: 22.11, 22.12, 22.13, 22.14, 22.15, 22.16, 22.17, 22.18, 22.23, 22.24, 22.35, 22.36,

22.37, 22.38, 22.46, 22.52, 22.55, 22.56, 22.65, 22.67, 22.68, 22.72, 22.74, 22.75.

978

C H E M I S T R Y

in Action

L uck often plays a role in major scientifi c breakthroughs, but it takes an alert and well-trained person to recognize the sig-

nifi cance of an accidental discovery and to take full advantage of it. Such was the case when, in 1964, the biophysicist Barnett Rosenberg and his research group at Michigan State University were studying the effect of an electric fi eld on the growth of bac-teria. They suspended a bacterial culture between two platinum

Cisplatin—The Anticancer Drugelectrodes and passed an electric current through it. To their sur-prise, they found that after an hour or so the bacteria cells ceased dividing. It did not take long for the group to determine that a platinum-containing substance extracted from the bacterial cul-ture inhibited cell division. Rosenberg reasoned that the platinum compound might be useful as an anticancer agent, because cancer involves uncon-trolled division of affected cells, so he set out to identify the substance. Given the presence of ammonia and chloride ions in solution during electrolysis, Rosenberg synthesized a number of platinum compounds containing ammonia and chlorine. The one that proved most effective at inhibiting cell division was cis- diamminedichloroplatinum(II) [Pt(NH3)2Cl2], also called cisplatin. The mechanism for the action of cisplatin is the chelation of DNA (deoxyribonucleic acid), the molecule that contains the genetic code. During cell division, the double-stranded DNA splits into two single strands, which must be accurately copied in order for the new cells to be identical to their parent cell. X-ray studies show that cisplatin binds to DNA by forming cross-links in which the two chlorides on cisplatin are replaced by nitrogen atoms in the adjacent guanine bases on the same strand of the DNA. (Guanine is one of the four bases in DNA. See Figure 25.17.) Consequently, the double-stranded structure assumes a bent confi guration at the binding site. Scientists be-lieve that this structural distortion is a key factor in inhibiting

Cisplatin, a bright yellow compound, is administered intravenously to cancer patients.


Properties of the Transition MetalsReview Questions

22.1 What distinguishes a transition metal from a repre-sentative metal?

22.2 Why is zinc not considered a transition metal?22.3 Explain why atomic radii decrease very gradually

from scandium to copper.22.4 Without referring to the text, write the ground-state

electron confi gurations of the fi rst-row transition metals. Explain any irregularities.

22.5 Write the electron confi gurations of the following ions: V51, Cr31, Mn21, Fe31, Cu21, Sc31, Ti41.

22.6 Why do transition metals have more oxidation states than other elements?

22.7 Give the highest oxidation states for scandium to copper.

22.8 Why does chromium seem to be less reactive than its standard reduction potential suggests?

Coordination CompoundsReview Questions

22.9 Defi ne the following terms: coordination compound, ligand, donor atom, coordination number, chelating agent.

22.10 Describe the interaction between a donor atom and a metal atom in terms of a Lewis acid-base reaction.

979

replication. The damaged cell is then destroyed by the body’s immune system. Because the binding of cisplatin to DNA requires both Cl atoms to be on the same side of the complex, the trans isomer of the compound is totally ineffective as an anticancer drug. Unfortunately, cisplatin can cause serious side effects, including severe kidney damage. Therefore, ongoing research efforts are directed toward fi nding related complexes that destroy cancer cells with less harm to healthy tissues.

cis-Pt(NH3)2Cl2

Cisplatin

Pt33°

NH3 N

H3

Cisplatin destroys the cancer cells’ ability to reproduce by changing the confi guration of their DNA. It binds to two sites on a strand of DNA, causing it to bend about 33º away from the rest of the strand. The structure of this DNA adduct was elucidated by Professor Stephen Lippard’s group at MIT.


Problems

22.11 Complete the following statements for the complex ion [Co(en)2(H2O)CN]21. (a) en is the abbreviation for . (b) The oxidation number of Co is . (c) The coordination number of Co is . (d) is a bidentate ligand.

22.12 Complete the following statements for the complex ion [Cr(C2O4)2(H2O)2]

2 . (a) The oxidation number of Cr is . (b) The coordination number of Cr is

. (c) is a bidentate ligand.22.13 Give the oxidation numbers of the metals in the follow-

ing species: (a) K3[Fe(CN)6], (b) K3[Cr(C2O4)3], (c) [Ni(CN)4]

22 .22.14 Give the oxidation numbers of the metals in the follow-

ing species: (a) Na2MoO4, (b) MgWO4, (c) Fe(CO)5.22.15 What are the systematic names for the following ions

and compounds?(a) [Co(NH3)4Cl2]

1 (c) [Co(en)2Br2]1

(b) Cr(NH3)3Cl3 (d) [Co(NH3)6]Cl3

22.16 What are the systematic names for the following ion and compounds?(a) [cis-Co(en)2Cl2]

1

(b) [Pt(NH3)5Cl]Cl3

(c) [Co(NH3)5Cl]Cl2

22.17 Write the formulas for each of the following ions and compounds: (a) tetrahydroxozincate(II), (b) penta-aquachlorochromium(III) chloride, (c) tetrabromo-cuprate(II), (d) ethylenediaminetetraacetatoferrate(II).

22.18 Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)- dichlorochromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammine-aquachlorocobalt(III) chloride.

Structure of Coordination CompoundsReview Questions

22.19 Defi ne the following terms: stereoisomers, geomet-ric isomers, optical isomers, plane-polarized light.

22.20 Specify which of the following structures can exhibit geometric isomerism: (a) linear, (b) square-planar, (c) tetrahedral, (d) octahedral.

22.21 What determines whether a molecule is chiral? How does a polarimeter measure the chirality of a molecule?

22.22 Explain the following terms: (a) enantiomers, (b) racemic mixtures.

Problems

22.23 The complex ion [Ni(CN)2Br2]22 has a square-planar

geometry. Draw the structures of the geometric isomers of this complex.

22.24 How many geometric isomers are in the following species? (a) [Co(NH3)2Cl4]

2 , (b) [Co(NH3)3Cl3]

22.25 Draw structures of all the geometric and optical isomers of each of the following cobalt complexes:(a) [Co(NH3)6]

31

(b) [Co(NH3)5Cl]21

(c) [Co(C2O4)3]32

22.26 Draw structures of all the geometric and optical isomers of each of the following cobalt complexes: (a) [Co(NH3)4Cl2]

1, (b) [Co(en)3]31.

Bonding in Coordination CompoundsReview Questions

22.27 Briefl y describe crystal fi eld theory.22.28 Defi ne the following terms: crystal fi eld splitting,

high-spin complex, low-spin complex, spectrochemi-cal series.

22.29 What is the origin of color in a coordination compound?

22.30 Compounds containing the Sc31 ion are colorless, whereas those containing the Ti31 ion are colored. Explain.

22.31 What factors determine whether a given complex will be diamagnetic or paramagnetic?

22.32 For the same type of ligands, explain why the crystal fi eld splitting for an octahedral complex is always greater than that for a tetrahedral complex.

Problems

22.33 The [Ni(CN)4]22 ion, which has a square-planar

geometry, is diamagnetic, whereas the [NiCl4]22 ion,

which has a tetrahedral geometry, is paramagnetic. Show the crystal fi eld splitting diagrams for those two complexes.

22.34 Transition metal complexes containing CN2 ligands are often yellow in color, whereas those containing H2O ligands are often green or blue. Explain.

22.35 Predict the number of unpaired electrons in the following complex ions: (a) [Cr(CN)6]

42 , (b) [Cr(H2O)6]

21.22.36 The absorption maximum for the complex ion

[Co(NH3)6]31 occurs at 470 nm. (a) Predict the color

of the complex and (b) calculate the crystal fi eld splitting in kJ/mol.

22.37 From each of the following pairs, choose the com-plex that absorbs light at a longer wavelength: (a) [Co(NH3)6]

21, [Co(H2O)6]21; (b) [FeF6]

32 , [Fe(CN)6]

32 ; (c) [Cu(NH3)4]21, [CuCl4]

22 .22.38 A solution made by dissolving 0.875 g of

Co(NH3)4Cl3 in 25.0 g of water freezes at 2 0.568C. Calculate the number of moles of ions produced when 1 mole of Co(NH3)4Cl3 is dissolved in water and sug-gest a structure for the complex ion present in this compound.

Reactions of Coordination CompoundsReview Questions

22.39 Defi ne the terms (a) labile complex, (b) inert complex.22.40 Explain why a thermodynamically stable species

may be chemically reactive and a thermodynami-cally unstable species may be unreactive.

Problems

22.41 Oxalic acid, H2C2O4, is sometimes used to clean rust stains from sinks and bathtubs. Explain the chemis-try underlying this cleaning action.

22.42 The [Fe(CN)6]32 complex is more labile than the

[Fe(CN)6]42 complex. Suggest an experiment that

would prove that [Fe(CN)6]32 is a labile complex.

22.43 Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fl uoride is added, a green precipitate is formed. When aqueous potassium chlo-ride is added instead, a bright-green solution is formed. Explain what is happening in these two cases.

22.44 When aqueous potassium cyanide is added to a solu-tion of copper(II) sulfate, a white precipitate, solu-ble in an excess of potassium cyanide, is formed. No precipitate is formed when hydrogen sulfi de is bub-bled through the solution at this point. Explain.

22.45 A concentrated aqueous copper(II) chloride solution is bright green in color. When diluted with water, the solution becomes light blue. Explain.

22.46 In a dilute nitric acid solution, Fe31 reacts with thio-cyanate ion (SCN2 ) to form a dark-red complex:

[Fe(H2O)6]31 1 SCN2 ∆

H2O 1 [Fe(H2O)5NCS]21

The equilibrium concentration of [Fe(H2O)5NCS]21

may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experi-ment, 1.0 mL of 0.20 M Fe(NO3)3 was mixed with 1.0 mL of 1.0 3 102 3 M KSCN and 8.0 mL of dilute HNO3. The color of the solution quantitatively indi-cated that the [Fe(H2O)5NCS]21 concentration was 7.3 3 102 5 M. Calculate the formation constant for [Fe(H2O)5NCS]21.

Additional Problems22.47 As we read across the fi rst-row transition metals

from left to right, the 12 oxidation state becomes more stable in comparison with the 13 state. Why is this so?

22.48 Which is a stronger oxidizing agent in aqueous solu-tion, Mn31 or Cr31? Explain your choice.

22.49 Carbon monoxide binds to Fe in hemoglobin some 200 times more strongly than oxygen. This is the reason why CO is a toxic substance. The metal-to-ligand sigma bond is formed by donating a lone

pair from the donor atom to an empty sp3d2 orbital on Fe. (a) On the basis of electronegativities, would you expect the C or O atom to form the bond to Fe? (b) Draw a diagram illustrating the overlap of the orbitals involved in the bonding.

22.50 What are the oxidation states of Fe and Ti in the ore ilmenite, FeTiO3? (Hint: Look up the ionization en-ergies of Fe and Ti in Table 22.1; the fourth ioniza-tion energy of Ti is 4180 kJ/mol.)

22.51 A student has prepared a cobalt complex that has one of the following three structures: [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl2, or [Co(NH3)4Cl2]Cl. Explain how the student would distinguish between these possi-bilities by an electrical conductance experiment. At the student’s disposal are three strong electrolytes—NaCl, MgCl2, and FeCl3—which may be used for comparison purposes.

22.52 Chemical analysis shows that hemoglobin contains 0.34 percent of Fe by mass. What is the minimum possible molar mass of hemoglobin? The actual mo-lar mass of hemoglobin is about 65,000 g. How do you account for the discrepancy between your mini-mum value and the actual value?

22.53 Explain the following facts: (a) Copper and iron have several oxidation states, whereas zinc has only one. (b) Copper and iron form colored ions, whereas zinc does not.

22.54 A student in 1895 prepared three coordination com-pounds containing chromium, with the following properties:

Cl2 Ions in Solution perFormula Color Formula Unit

(a) CrCl3 ? 6H2O Violet 3(b) CrCl3 ? 6H2O Light green 2(c) CrCl3 ? 6H2O Dark green 1

Write modern formulas for these compounds and suggest a method for confi rming the number of Cl2 ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates and Cr has a coordination number of 6 in all the compounds.)

22.55 The formation constant for the reaction Ag1 1 2NH3 ∆ [Ag(NH3)2]

1 is 1.5 3 107 and that for the reaction Ag1 1 2CN2 ∆ [Ag(CN)2]

2 is 1.0 3 1021 at 258C (see Table 16.3). Calculate the equilibrium constant and DG8 at 258C for the reaction

[Ag(NH3)2]1 1 2CN2 ∆ [Ag(CN)2]

2 1 2NH3



22.56 From the standard reduction potentials listed in Table 19.1 for Zn/Zn21 and Cu1/Cu21, calculate DG8 and the equilibrium constant for the reaction

Zn(s) 1 2Cu21(aq) ¡ Zn21(aq) 1 2Cu1(aq)

22.57 Using the standard reduction potentials listed in Table 19.1 and the Handbook of Chemistry and Physics, show that the following reaction is favor-able under standard-state conditions:

2Ag(s) 1 Pt21(aq) ¡ 2Ag1(aq) 1 Pt(s)

What is the equilibrium constant of this reaction at 258C?

22.58 The Co21-porphyrin complex is more stable than the Fe21-porphyrin complex. Why, then, is iron the metal ion in hemoglobin (and other heme-containing proteins)?

22.59 What are the differences between geometric isomers and optical isomers?

22.60 Oxyhemoglobin is bright red, whereas deoxyhemo-globin is purple. Show that the difference in color can be accounted for qualitatively on the basis of high-spin and low-spin complexes. (Hint: O2 is a strong-fi eld ligand; see Chemistry in Action essay on p. 976.)

22.61 Hydrated Mn21 ions are practically colorless (see Figure 22.20) even though they possess fi ve 3d elec-trons. Explain. (Hint: Electronic transitions in which there is a change in the number of unpaired electrons do not occur readily.)

22.62 Which of the following hydrated cations are color-less: Fe21(aq), Zn21(aq), Cu1(aq), Cu21(aq), V51(aq), Ca21(aq), Co21(aq), Sc31(aq), Pb21(aq)? Explain your choice.

22.63 Aqueous solutions of CoCl2 are generally either light pink or blue. Low concentrations and low tem-peratures favor the pink form while high concentra-tions and high temperatures favor the blue form. Adding hydrochloric acid to a pink solution of CoCl2 causes the solution to turn blue; the pink color is restored by the addition of HgCl2. Account for these observations.

22.64 Suggest a method that would allow you to distinguish between cis-Pt(NH3)2Cl2 and trans-Pt(NH3)2Cl2.

22.65 You are given two solutions containing FeCl2 and FeCl3 at the same concentration. One solution is light yellow and the other one is brown. Identify these solutions based only on color.

22.66 The label of a certain brand of mayonnaise lists EDTA as a food preservative. How does EDTA pre-vent the spoilage of mayonnaise?

22.67 The compound 1,1,1-trifl uoroacetylacetone (tfa) is a bidentate ligand:

OB

OB

CF3CCH2CCH3

It forms a tetrahedral complex with Be21 and a square planar complex with Cu21. Draw structures of these complex ions and identify the type of isomerism ex-hibited by these ions.

22.68 How many geometric isomers can the following square planar complex have?

a bG D

Pt D G d c

22.69 [Pt(NH3)2Cl2] is found to exist in two geometric iso-mers designated I and II, which react with oxalic acid as follows:

I 1 H2C2O4 ¡ [Pt(NH3)2C2O4] II 1 H2C2O4 ¡ [Pt(NH3)2(HC2O4)2]

Comment on the structures of I and II.22.70 The Kf for the complex ion formation between Pb21

and EDTA42

Pb21 1 EDTA42 ∆ Pb(EDTA)22

is 1.0 3 1018 at 258C. Calculate [Pb21] at equilib-rium in a solution containing 1.0 3 102 3 M Pb21 and 2.0 3 102 3 M EDTA42 .

22.71 Manganese forms three low-spin complex ions with the cyanide ion with the formulas [Mn(CN)6]

52 , [Mn(CN)6]

42 , and [Mn(CN)6]32 . For each complex

ion, determine the oxidation number of Mn and the number of unpaired d electrons present.

Special Problems

22.72 Commercial silver-plating operations frequently use a solution containing the complex Ag(CN)2

2 ion. Be-cause the formation constant (Kf) is quite large, this

procedure ensures that the free Ag1 concentration in solution is low for uniform electrodeposition. In one process, a chemist added 9.0 L of 5.0 M NaCN to

Answers to Practice Exercises

22.1 K: 11; Au: 13. 22.2 Tetraaquadichlorochromium(III) chloride. 22.3 [Co(en)3]2(SO4)3. 22.4 5.

Answers to Practice Exercises 983

90.0 L of 0.20 M AgNO3. Calculate the concentra-tion of free Ag1 ions at equilibrium. See Table 16.4 for Kf value.

22.73 Draw qualitative diagrams for the crystal-fi eld split-tings in (a) a linear complex ion ML2, (b) a trigonal-planar complex ion ML3, and (c) a trigonal-bipyramidal complex ion ML5.

22.74 (a) The free Cu(I) ion is unstable in solution and has a tendency to disproportionate:

2Cu1(aq2∆ Cu21(aq) 1 Cu(s)

Use the information in Table 19.1 (p. 846) to calculate the equilibrium constant for the reaction. (b) Based on your result in (a), explain why most Cu(I) compounds are insoluble.

22.75 Consider the following two ligand exchange reactions:

[Co(H2O)6]31 1 6NH3 ∆ [Co(NH3)6]

31 1 6H2O [Co(H2O)6]

31 1 3en ∆ [Co(en)3]31 1 6H2O

(a) Which of the reactions should have a larger DS8? (b) Given that the Co—N bond strength is approxi-mately the same in both complexes, which reaction will have a larger equilibrium constant? Explain your choices.

22.76 Copper is also known to exist in 13 oxidation state, which is believed to be involved in some biological electron transfer reactions. (a) Would you expect this oxidation state of copper to be stable? Explain. (b) Name the compound K3CuF6 and predict the geometry of the complex ion and its magnetic prop-erties. (c) Most of the known Cu(III) compounds have square planar geometry. Are these compounds diamagnetic or paramagnetic?

984

Dating Paintings with Prussian Blue

In 1995 a painting entitled Portrait of a Noblewoman was given to the Mead Art Museum in Amherst, Massachusetts, by an anonymous donor. The small bust-length portrait on wood

panel depicts an unidentifi ed adolescent girl of noble birth. The background, blue “wallpaper” with a gold fl eur-de-lys pattern above a wood-paneled wainscot, is characteristic of settings seen in full-length Renaissance portraits. The rich dress and French royal heraldic device sug-gest that the subject is a member of the royal court, if not the royal family itself. The painting was attributed to the school of the court painter, François Clouet (1522–1572). While in the donor’s possession, the painting was scratched by a cat. An art conservator who was asked to repair the damage became suspicious about the blue paint used to render the girl’s hat and the wallpaper. Subsequent analysis of microscopic samples of the paint revealed that the pigment was Prussian blue (ferric ferrocyanide, Fe4[Fe(CN)6]3), a coordination com-pound discovered by a German dyemaker between 1704 and 1707. Prior to the discovery of Prussian blue, there were three blue pigments available to paint-ers: azurite [Cu3(OH)2(CO3)2], smalt (a complex cobalt and arsenic compound), and ultramarine blue, which has the complex formula of CaNa7Al6Si6O24SO4. Prussian blue quickly came to be valued by painters for the intensity and transparency of its color, and it is commonly found in works painted after the early 1700s. In Prussian blue, the Fe21 ion is bonded to the carbon atom of the cyanide group in an octahedral arrangement, and each Fe31 ion is bonded through the nitrogen atom of the cyanide

C H E M I C A L

Mystery

C NFe2+ Fe3+

Portrait of a Noblewoman.

985

group with a similar octahedral symmetry. Thus, the cyanide group acts as a bidentate ligand as shown on p. 984. The blue color arises from the so-called intervalence charge transfer between the metal ions. If we designate these two sites of iron as FeA

21 and FeB31, where A and B denote different

sites defi ned by the ligands, we can represent the transfer of an electron from Fe21 to Fe31 as follows:

Fe21A , Fe31

B -¡hv Fe31A , Fe21

B

The right-hand side of the preceding equation has more energy than the left-hand side, and the result is an energy level and a light-absorbing scheme similar to that shown in Figure 22.19. It was Prussian blue’s unique color that drew the conservator’s suspicion and prompted an analysis that dated the painting at sometime after 1704. Additional analysis of the green pigment used for the jewels showed a mixture of Prussian blue and lemon yellow (zinc chro-mate, ZnCrO4), a pigment produced commercially beginning around 1850. Thus, Portrait of a Noblewoman is no longer ascribed to any sixteenth-century painter, but it is now used by the museum to teach art historians about fakes, forgeries, and mistaken attributions.

Chemical Clues1. Give the systematic name of Prussian blue. In what region of the visible spectrum does

the intervalence charge transfer absorb light?

2. How can you show that the color of Prussian blue arises from intervalence charge trans-fer and not from a transition within a single ion such as Fe(CN)6

32 or Fe(CN)642 ?

3. Write the formulas and give the systematic names for ferrous ferrocyanide and ferric fer-ricyanide. Will intervalence charge transfer occur in these two compounds?

4. Kinetically, the Fe(CN)642 ion is inert while the Fe(CN)6

32 ion is labile. Based on this knowledge, would you expect Prussian blue to be a poisonous cyanide compound? Explain.

5. When Prussian blue is added to a NaOH solution, an orange-brown precipitate forms. Iden-tify the precipitate.

6. How can the formation of Prussian blue be used to distinguish between Fe21 and Fe31 ions in solution?

Prussian blue can be prepared by mix-ing a FeCl3 solution with a K4Fe(CN)6 solution.

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