Download Chemistry of the Nonmetals

22
Chemistry of the Nonmetals
Everything we see in the chapter-opening photo is composed of nonmetals.
The water, of course, is H2O, and the sand is mostly SiO2. Although we cannot
see it, the air contains principally N2 and O2 with much lesser amounts of
other nonmetallic substances. The palm tree is also composed mostly of
nonmetallic elements.
In this chapter, we take a panoramic view of the descriptive chemistry of the nonmetallic elements, starting with hydrogen and progressing group by group across
the periodic table. We will consider how the elements occur in nature, how they
are isolated from their sources, and how they are used. We will emphasize hydrogen, oxygen, nitrogen, and carbon because these four nonmetals form many commercially important compounds and account for 99% of the atoms required by
living cells.
As you study this descriptive chemistry, it is important to look for trends rather
than trying to memorize all the facts presented. The periodic table is your most valuable
tool in this task.
22.1 | Periodic Trends and Chemical
Reactions
Recall that we can classify elements as metals, metalloids, and nonmetals.
(Section 7.6) Except for hydrogen, which is a special case, the nonmetals occupy
the upper right portion of the periodic table. This division of elements relates nicely
What’s
Ahead
22.1 Periodic Trends and Chemical Reactions We
begin with a review of periodic trends and types of chemical
reactions, which will help us focus on general patterns of behavior
as we examine each family in the periodic table.
22.2 Hydrogen The first nonmetal we consider, hydrogen, forms
compounds with most other nonmetals and with many metals.
▶ A tropical beach.
22.3 Group 8A: The Noble Gases Next, we consider the
noble gases, the elements of group 8A, which exhibit very limited
chemical reactivity.
22.4 Group 7A: The Halogens We then explore the most
electronegative elements: the halogens, group 7A.
22.5 Oxygen We next consider oxygen, the most abundant
element by mass in both Earth’s crust and the human body, and
the oxide and peroxide compounds it forms.
22.6 The Other Group 6A Elements: S, Se, Te, and Po We study the other members of group 6A (S, Se, Te, and Po), of
which sulfur is the most important.
22.9 Carbon We next focus on the inorganic compounds of
carbon.
22.7 Nitrogen We next consider nitrogen, a key component of
We then consider silicon, the element most abundant and
significant of the heavier members of group 4A.
our atmosphere. It forms compounds in which its oxidation number
ranges from - 3 to + 5, including such important compounds as
NH3 and HNO3.
22.8 The Other Group 5A Elements: P, As, Sb, and Bi Of the other members of group 5A (P, As, Sb, and Bi), we take a
closer look at phosphorus—the most commercially important one
and the only one that plays an important and beneficial role in
biological systems.
22.10 The Other Group 4A Elements: Si, Ge, Sn, and Pb 22.11 Boron Finally, we examine boron—the sole nonmetallic
element of group 3A.
954
chapter 22 Chemistry of the Nonmetals
Metals
Metalloids
to trends in the properties of the elements as summarized in
◀ Figure 22.1. Electronegativity, for example, increases as we move left
to right across a period and decreases as we move down a group. The
nonmetals thus have higher electronegativities than the metals. This
difference leads to the formation of ionic solids in reactions b
­ etween
(Sections 7.6, 8.2, and 8.4) In ­contrast,
metals and nonmetals.
compounds formed between two or more nonmetals are usually
(Sections 7.8 and 8.4)
molecular substances.
The chemistry exhibited by the first member of a nonmetal
group can differ from that of subsequent members in important
ways. Two differences are particularly notable: (1) The first mem (Section
ber is able to accommodate fewer bonded neighbors.
8.7) For example, nitrogen is able to bond to a maximum of three Cl
atoms, NCl3, whereas phosphorus can bond to five, PCl5. The small
size of nitrogen is largely responsible for this difference. (2) The first
member can more readily form p bonds. This trend is also due, in
part, to size because small atoms are able to approach each other
more closely. As a result, the overlap of p orbitals, which results in the formation of p
bonds, is more effective for the first element in each group (▼ Figure 22.2). More effec (Section 8.8) For
tive overlap means stronger p bonds, reflected in bond enthalpies.
example, the difference between the enthalpies of the C ¬ C bond and the C “ C bond
(Table 8.4); this large value reflects the “strength” of a carbon–
is about 270 kJ>mol
carbon p bond. On the other hand, the difference between Si ¬ Si and Si “ Si bonds is
only about 100 kJ/mol, significantly lower than that for carbon, reflecting much weaker
p bonding.
As we shall see, p bonds are particularly important in the chemistry of carbon,
nitrogen, and oxygen, each the first member in its group. The heavier elements in these
groups have a tendency to form only single bonds.
Nonmetals
Decreasing ionization energy
Increasing atomic radius
Decreasing electronegativity
Increasing metallic character
Increasing ionization energy
Decreasing atomic radius
Increasing electronegativity
Decreasing metallic character
▲ Figure 22.1 Trends in elemental properties.
Sample
Exercise 22.1 Identifying Elemental Properties
Of the elements Li, K, N, P, and Ne, which (a) is the most electronegative, (b) has the greatest
metallic character, (c) can bond to more than four atoms in a molecule, and (d) forms p bonds
most readily?
Solution
Analyze We are given a list of elements and asked to predict several properties that can be related
to periodic trends.
Plan We can use Figures 22.1 and 22.2 to guide us to the answers.
Solve
C C
Si
Si
(a) Electronegativity increases as we proceed toward the upper right portion of the periodic
table, excluding the noble gases. Thus, N is the most electronegative element of our
choices.
(b) Metallic character correlates inversely with electronegativity—the less electronegative an
element, the greater its metallic character. The element with the greatest metallic character
is therefore K, which is closest to the lower left corner of the periodic table.
Smaller nucleusto-nucleus distance,
more orbital overlap,
stronger π bond
Larger nucleusto-nucleus distance,
less orbital overlap,
weaker π bond
▲ Figure 22.2 P Bonds in period 2 and
period 3 elements.
(c) Nonmetals tend to form molecular compounds, so we can narrow our choice to the three
nonmetals on the list: N, P, and Ne. To form more than four bonds, an element must be able
to expand its valence shell to allow more than an octet of electrons around it. Valence-shell
expansion occurs for period 3 elements and below; N and Ne are both in period 2 and do
not undergo valence-shell expansion. Thus, the answer is P.
(d) Period 2 nonmetals form p bonds more readily than elements in period 3 and below. There
are no compounds known that contain covalent bonds to Ne. Thus, N is the element from
the list that forms p bonds most readily.
section 22.1 Periodic Trends and Chemical Reactions
955
Practice Exercise 1
Which description correctly describes a difference between the chemistry of oxygen and
sulfur?
(a) Oxygen is a nonmetal and sulfur is a metalloid. (b) Oxygen can form more than four
bonds whereas sulfur cannot. (c) Sulfur has a higher electronegativity than oxygen.
(d) Oxygen is better able to form p bonds than sulfur.
Practice Exercise 2
Of the elements Be, C, Cl, Sb, and Cs, which (a) has the lowest electronegativity, (b) has the
greatest nonmetallic character, (c) is most likely to participate in extensive p bonding,
(d) is most likely to be a metalloid?
The ready ability of period 2 elements to form p bonds is an important factor in
determining the elemental forms of these elements. Compare, for example, carbon and
silicon. Carbon has five major crystalline allotropes: diamond, graphite, buckminster (Sections 12.7 and 12.9) Diamond is a
fullerene, graphene, and carbon nanotubes.
covalent-network solid that has C ¬ C s bonds but no p bonds. Graphite, buckminsterfullerene, graphene, and carbon nanotubes have p bonds that result from the sideways
overlap of p orbitals. Elemental silicon, however, exists only as a diamond-like covalentnetwork solid with s bonds; it has no forms analogous to graphite, buckminsterfullerene, graphene, or carbon nanotubes, apparently because Si ¬ Si p bonds are too weak.
We likewise see significant differences in the dioxides of carbon and silicon as a
result of their relative abilities to form p bonds (▶ Figure 22.3). CO2 is a molecular
substance containing C “ O double bonds, whereas SiO2 is a covalent-network solid in
which four oxygen atoms are bonded to each silicon atom by single bonds, forming an
extended structure that has the empirical formula SiO2.
Fragment of extended SiO2 lattice;
Si forms only single bonds
Give It Some Thought
Nitrogen is found in nature as N21g2. Would you expect phosphorus to be found
in nature as P21g2? Explain.
Chemical Reactions
Because O2 and H2O are abundant in our environment, it is particularly important to
consider how these substances react with other compounds. About one-third of the reactions discussed in this chapter involve either O2 (oxidation or combustion reactions)
or H2O (especially proton-transfer reactions).
(Section 3.2), hydrogen-containing compounds proIn combustion reactions
duce H2O. Carbon-containing ones produce CO2 (unless the amount of O2 is insufficient, in which case CO or even C can form). Nitrogen-containing compounds tend
to form N2, although NO can form in special cases or in small amounts. A reaction
illustrating these points is:
4 CH3NH21g2 + 9 O21g2 ¡ 4 CO21g2 + 10 H2O1g2 + 2 N21g2[22.1]
The formation of H2O, CO2, and N2 reflects the high thermodynamic stability of these
substances, indicated by the large bond energies for the O ¬ H, C “ O, and N ‚ N
(Section 8.8)
bonds (463, 799, and 941 kJ/mol, respectively).
When dealing with proton-transfer reactions, remember that the weaker a
(Section 16.2) For example,
Brønsted–Lowry acid, the stronger its conjugate base.
H2, OH - , NH3, and CH4 are exceedingly weak proton donors that have no tendency to
act as acids in water. Thus, the species formed by removing one or more protons from
them are extremely strong bases. All of them react readily with water, removing protons
from H2O to form OH - . Two representative reactions are:
CH3- 1aq2 + H2O1l2 ¡ CH41g2 + OH - 1aq2[22.2]
N3 - 1aq2 + 3 H2O1l2 ¡ NH31aq2 + 3 OH - 1aq2[22.3]
CO2 ; C forms double bonds
▲ Figure 22.3 Comparison of the bonds in
SIO2 and CO2.
956
chapter 22 Chemistry of the Nonmetals
Sample
Exercise 22.2 Predicting the Products of Chemical Reactions
Predict the products formed in each of the following reactions, and write a balanced equation:
(a) CH3NHNH21g2 + O21g2 ¡
Solution
? (b) Mg3P21s2 + H2O1l2 ¡ ?
Analyze We are given the reactants for two chemical equations and asked to predict the products
and then balance the equations.
Plan We need to examine the reactants to see if we might recognize
a reaction type. In (a) the carbon compound is reacting with O2,
which suggests a combustion reaction. In (b) water reacts with an
ionic compound. The anion P3 - is a strong base and H2O is able to
act as an acid, so the reactants suggest an acid–base (proton-transfer)
reaction.
Solve
(a) Based on the elemental composition of the carbon compound, this combustion reaction should produce CO2, H2O,
and N2:
2+
3-
3-
(b) Mg3P2 is ionic, consisting of Mg and P ions. The P
ion, like N3 - , has a strong affinity for protons and reacts
with H2O to form OH - and PH3 (PH2 - , PH2- , and PH3 are
all exceedingly weak proton donors).
Mg1OH22 has low solubility in water and will precipitate.
Practice Exercise 1
When CaC2 reacts with water, what carbon-containing compound
forms?
2 CH3NHNH21g2 + 5 O21g2 ¡ 2 CO21g2 + 6 H2O1g2 + 2 N21g2
Mg3P21s2 + 6 H2O1l2 ¡ 2 PH31g2 + 3 Mg1OH221s2
Practice Exercise 2
Write a balanced equation for the reaction of solid sodium hydride
with water.
(a) CO, (b) CO2, (c) CH4, (d) C2H2, (e) H2CO3.
22.2 | Hydrogen
The English chemist Henry Cavendish (1731–1810) was the first to isolate hydrogen.
Because the element produces water when burned in air, the French chemist Antoine
(Figure 3.1) gave it the name hydrogen, which means “water producer”
Lavoisier
(Greek: hydro, water; gennao, to produce).
Hydrogen is the most abundant element in the universe. It is the nuclear
(Section 21.8)
fuel consumed by our Sun and other stars to produce energy.
Although about 75% of the known mass of the universe is hydrogen, it constitutes
only 0.87% of Earth’s mass. Most of the hydrogen on our planet is found associated
with oxygen. Water, which is 11% hydrogen by mass, is the most abundant hydrogen compound.
Isotopes of Hydrogen
The most common isotope of hydrogen, 11H, has a nucleus consisting of a single proton.
This isotope, sometimes referred to as protium,* makes up 99.9844% of naturally occurring hydrogen.
Two other isotopes are known: 21H, whose nucleus contains a proton and a neutron, and 31H, whose nucleus contains a proton and two neutrons. The 21H isotope,
deuterium, makes up 0.0156% of naturally occurring hydrogen. It is not radioactive
and is often given the symbol D in chemical formulas, as in D2O (deuterium oxide),
which is known as heavy water.
Because an atom of deuterium is about twice as massive as an atom of protium, the properties of deuterium-containing substances vary somewhat from those
*Giving unique names to isotopes is limited to hydrogen. Because of the proportionally large differences in
their masses, the isotopes of H show appreciably more differences in their properties than isotopes of heavier
elements.
section 22.2 Hydrogen
of the protium-containing analogs. For example, the normal melting and boiling
points of D2O are 3.81 °C and 101.42 °C, respectively, versus 0.00 °C and 100.00 °C
for H2O. Not surprisingly, the density of D2O at 25 °C 11.104 g>mL2 is greater than
that of H2O 10.997 g>mL2. Replacing protium with deuterium (a process called deuteration) can also have a profound effect on reaction rates, a phenomenon called a
kinetic-isotope effect. For example, heavy water can be obtained from the electrolysis
32 H2O1l2 ¡ 2 H21g2 + O21g24 of ordinary water because the small amount of naturally occurring D2O in the sample undergoes electrolysis more slowly than H2O and,
therefore, becomes concentrated during the reaction.
The third isotope, 31H, tritium, is radioactive, with a half-life of 12.3 yr:
3
1H
¡ 32He +
0
- 1e
t1>2 = 12.3 yr[22.4]
Because of its short half-life, only trace quantities of tritium exist naturally. The isotope
can be synthesized in nuclear reactors by neutron bombardment of lithium-6:
6
3Li
+ 10n ¡ 31H + 42He[22.5]
Deuterium and tritium are useful in studying reactions of compounds containing
hydrogen. A compound is “labeled” by replacing one or more ordinary hydrogen atoms
with deuterium or tritium at specific locations in a molecule. By comparing the locations
of the label atoms in reactants and products, the reaction mechanism can often be inferred.
When methyl alcohol 1CH3OH2 is placed in D2O, for example, the H atom of the O ¬ H
bond exchanges rapidly with the D atoms, forming CH3OD. The H atoms of the CH3 group
do not exchange. This experiment demonstrates the kinetic stability of C ¬ H bonds and
reveals the speed at which the O ¬ H bond in the molecule breaks and re-forms.
Properties of Hydrogen
Hydrogen is the only element that is not a member of any family in the periodic table.
Because of its ls1 electron configuration, it is generally placed above lithium in the table.
However, it is definitely not an alkali metal. It forms a positive ion much less readily than any alkali metal. The ionization energy of the hydrogen atom is 1312 kJ>mol,
whereas that of lithium is 520 kJ>mol.
Hydrogen is sometimes placed above the halogens in the periodic table because
the hydrogen atom can pick up one electron to form the hydride ion, H - , which has
the same electron configuration as helium. However, the electron affinity of hydrogen,
E = -73 kJ>mol, is not as large as that of any halogen. In general, hydrogen shows no
closer resemblance to the halogens than it does to the alkali metals.
Elemental hydrogen exists at room temperature as a colorless, odorless, tasteless gas
composed of diatomic molecules. We can call H2 dihydrogen, but it is more commonly
referred to as either molecular hydrogen or simply hydrogen. Because H2 is nonpolar
and has only two electrons, attractive forces between molecules are extremely weak. As a
result, its melting point 1 -259 °C2 and boiling point 1-253 °C2 are very low.
(Table 8.4)
The H ¬ H bond enthalpy 1436 kJ>mol2 is high for a single bond.
By comparison, the Cl ¬ Cl bond enthalpy is only 242 kJ>mol. Because H2 has a strong
bond, most reactions involving H2 are slow at room temperature. However, the molecule is readily activated by heat, irradiation, or catalysis. The activation generally produces hydrogen atoms, which are very reactive. Once H2 is activated, it reacts rapidly
and exothermically with a wide variety of substances.
Give It Some Thought
If H2 is activated to produce H+, what must the other product be?
Hydrogen forms strong covalent bonds with many other elements, including oxygen; the O ¬ H bond enthalpy is 463 kJ>mol. The formation of the strong O ¬ H bond
makes hydrogen an effective reducing agent for many metal oxides. When H2 is passed
over heated CuO, for example, copper is produced:
CuO1s2 + H21g2 ¡ Cu1s2 + H2O1g2[22.6]
957
958
chapter 22 Chemistry of the Nonmetals
When H2 is ignited in air, a vigorous reaction occurs, forming H2O. Air containing
as little as 4% H2 by volume is potentially explosive. Combustion of hydrogen–oxygen
mixtures is used in liquid-fuel rocket engines such as those of the Space Shuttle. The
hydrogen and oxygen are stored at low temperatures in liquid form.
Production of Hydrogen
When a small quantity of H2 is needed in the laboratory, it is usually obtained by
the reaction between an active metal such as zinc and a dilute strong acid such as
HCl or H2SO4:
Zn1s2 + 2 H+ 1aq2 ¡ Zn2 + 1aq2 + H21g2[22.7]
Large quantities of H2 are produced by reacting methane with steam at 1100 °C.
We can view this process as involving two reactions:
CH41g2 + H2O1g2 ¡ CO1g2 + 3 H21g2[22.8]
CO1g2 + H2O1g2 ¡ CO21g2 + H21g2[22.9]
Carbon heated with water to about 1000 °C is another source of H2:
C1s2 + H2O1g2 ¡ H21g2 + CO1g2[22.10]
This mixture, known as water gas, is used as an industrial fuel.
Electrolysis of water consumes too much energy and is consequently too costly
to be used commercially to produce H2. However, H2 is produced as a by-product in
the electrolysis of brine (NaCl) solutions in the course of commercial Cl2 and NaOH
manufacture:
electrolysis
2 NaCl1aq2 + 2 H2O1l2 ¡ H21g2 + Cl21g2 + 2 NaOH1aq2[22.11]
Give It Some Thought
What are the oxidation states of the H atoms in Equations 22.7–22.11?
Closer Look
The Hydrogen Economy
The reaction of hydrogen with oxygen is highly exothermic:
2 H21g2 + O21g2 ¡ 2 H2O1g2 ∆H = -483.6 kJ[22.12]
Because H2 has a low molar mass and a high enthalpy of combustion, it has a high energy density by mass. (That is, its combustion
produces high energy per gram.) Furthermore, the only product of the
reaction is water vapor, which means that hydrogen is environmentally cleaner than fossil fuels. Thus, the prospect of using hydrogen
widely as a fuel is attractive.
The term “hydrogen economy” is used to describe the concept of delivering and using hydrogen as a fuel in place of fossil fuels. In order to develop a hydrogen economy, it would be necessary
to generate elemental hydrogen on a large scale and arrange for its
transport and storage. These matters provide significant technical
challenges.
▶ Figure 22.4 illustrates various sources and uses of H2 fuel. The
generation of H2 through electrolysis of water is in principle the cleanest route, because this process—the reverse of Equation 22.11—produces
(Figure 1.7 and Section 20.9) However,
only hydrogen and oxygen.
the energy required to electrolyze water must come from somewhere. If
we burn fossil fuels to generate this energy, we have not advanced very
far toward a true hydrogen economy. If the energy for electrolysis came
instead from a hydroelectric or nuclear power plant, solar cells, or wind
generators, consumption of nonrenewable energy sources and undesired
production of CO2 could be avoided.
The storage of hydrogen is another technical obstacle that must
be overcome in developing a hydrogen economy. Although H2(g) has
a high energy density by mass, it has a low energy density by volume.
Thus, storing hydrogen as a gas requires a large volume compared
to the energy it delivers. There are also safety issues associated with
handling and storing the gas because its combustion can be explosive.
Storing hydrogen in the form of various hydride compounds such as
LiAlH4 is being investigated as a means of reducing the volume and
increasing the safety. One problem with this approach, however, is
that such compounds have high energy density by volume but low energy density by mass.
Related Exercises: 22.29, 22.30, 22.94
section 22.2 Hydrogen
Hydroelectric,
solar photovoltaics,
wind
ewable energies
Ren
Solar
thermal
Elec
tro
ly
s
Biomass
Electrical
nuclear
is
Natural
gas
Supply
Thermal
nuclear
Coal
H2
Fuel cell
engines
Demand
Processes,
syntheses
du
or
t
In
sp
str
y
n
Tra
Internal
combustion
engines
F u el c e lls
Turbines,
internal
combustion
engines
Thermal
Com- Residential
Other
mercial
B u i l di n g s
▲ Figure 22.4 The “hydrogen economy” would require hydrogen to be produced from various
sources and would use hydrogen in energy-related applications.
Uses of Hydrogen
Hydrogen is commercially important. About 5.0 * 1010 kg (50 million metric tons) is
produced annually across the world. About half of the H2 produced is used to syn (Section 15.2) Much of the remaining hythesize ammonia by the Haber process.
drogen is used to convert high-molecular-weight hydrocarbons from petroleum into
lower-molecular-weight hydrocarbons suitable for fuel (gasoline, diesel, and others) in
a process known as cracking. Hydrogen is also used to manufacture methanol via the
catalytic reaction of CO and H2 at high pressure and temperature:
CO1g2 + 2 H21g2 ¡ CH3OH1g2 [22.13]
Binary Hydrogen Compounds
Hydrogen reacts with other elements to form three types of compounds: (1) ionic hydrides, (2) metallic hydrides, and (3) molecular hydrides.
The ionic hydrides are formed by the alkali metals and by the heavier alkaline earths (Ca, Sr, and Ba). These active metals are much less electronegative than
hydrogen. Consequently, hydrogen acquires electrons from them to form hydride
ions 1H - 2:
Ca1s2 + H21g2 ¡ CaH21s2[22.14]
The hydride ion is very basic and reacts readily with compounds having even weakly
acidic protons to form H2:
H - 1aq2 + H2O1l2 ¡ H21g2 + OH - 1aq2[22.15]
Ionic hydrides can therefore be used as convenient (although expensive) sources of H2.
Calcium hydride 1CaH22 is used to inflate life rafts, weather balloons, and the like
where a simple, compact means of generating H2 is desired (Figure 22.5).
959
960
chapter 22 Chemistry of the Nonmetals
Go Figure
This reaction is exothermic. Is the beaker on the right warmer or colder than the beaker on the left?
CaH2
H2 gas
Color
change
indicates
presence
of OH−
H2O with
pH indicator
▲ Figure 22.5 The reaction of CaH2 with water.
The reaction between H - and H2O (Equation 22.15) is an acid–base reaction and
a redox reaction. The H - ion, therefore, is a good base and a good reducing agent. In
fact, hydrides are able to reduce O2 to OH - :
Go Figure
Which is the most thermodynamically
stable hydride? Which is the least
thermodynamically stable?
4A
5A
6A
7A
CH4(g) NH3(g) H2O(l) HF(g)
−50.8 −16.7 −237 −271
SiH4(g) PH3(g) H2S(g) HCl(g)
+56.9 +18.2 −33.0 −95.3
GeH4(g) AsH3(g) H2Se(g) HBr(g)
+117 +111
+71
−53.2
SbH3(g) H2Te(g) HI(g)
+187 +138 +1.30
▲ Figure 22.6 Standard free energies of
formation of molecular hydrides. All values
are kilojoules per mole of hydride.
2 NaH1s2 + O21g2 ¡ 2 NaOH1s2[22.16]
For this reason, hydrides are normally stored in an environment that is free of both
moisture and air.
Metallic hydrides are formed when hydrogen reacts with transition metals.
These compounds are so named because they retain their metallic properties. In
many metallic hydrides, the ratio of metal atoms to hydrogen atoms is not fixed or
in small whole numbers. The composition can vary within a range, depending on
reaction conditions. TiH2 can be produced, for example, but preparations usually
yield TiH1.8. These nonstoichiometric metallic hydrides are sometimes called interstitial hydrides. Because hydrogen atoms are small enough to fit between the sites
occupied by the metal atoms, many metal hydrides behave like interstitial alloys.
(Section 12.3)
The molecular hydrides, formed by nonmetals and metalloids, are either gases or liquids under standard conditions. The simple molecular hydrides are listed in ◀ Figure 22.6,
(Section 19.5) In each
together with their standard free energies of formation, ∆G°f .
family, the thermal stability 1measured as ∆G°f2 decreases as we move down the family.
(Recall that the more stable a compound is with respect to its elements under standard
conditions, the more negative ∆G°f is.)
22.3 | Group 8A: The Noble Gases
The elements of group 8A are chemically unreactive. Indeed, most of our references to
these elements have been in relation to their physical properties, as when we discussed
(Section 11.2) The relative inertness of these elements is due
intermolecular forces.
to the presence of a completed octet of valence-shell electrons (except He, which only
section 22.3 Group 8A: The Noble Gases
has a filled 1s shell). The stability of such an arrangement is reflected in the high ioniza (Section 7.4)
tion energies of the group 8A elements.
The group 8A elements are all gases at room temperature. They are components of
Earth’s atmosphere, except for radon, which exists only as a short-lived radioisotope.
(Section 21.9) Only argon is relatively abundant.
(Table 18.1)
Neon, argon, krypton, and xenon are used in lighting, display, and laser applications in which the atoms are excited electrically and electrons that are in a higher energy
(Section 6.2) Argon is used as a
state emit light as they fall to the ground state.
blanketing atmosphere in electric lightbulbs. The gas conducts heat away from the filament but does not react with it. Argon is also used as a protective atmosphere to prevent
oxidation in welding and certain high-temperature metallurgical processes.
Helium is in many ways the most important noble gas. Liquid helium is used as
a coolant to conduct experiments at very low temperatures. Helium boils at 4.2 K and
1 atm, the lowest boiling point of any substance. It is found in relatively high concentrations in many natural-gas wells from which it is isolated.
Noble-Gas Compounds
Because the noble gases are exceedingly stable, they react only under rigorous conditions. We expect the heavier ones to be most likely to form compounds because their
(Figure 7.9) A lower ionization energy suggests the
ionization energies are lower.
possibility of sharing an electron with another atom, leading to a chemical bond. In addition, because the group 8A elements (except helium) already contain eight electrons
in their valence shell, formation of covalent bonds will require an expanded valence
(Section 8.7)
shell. Valence-shell expansion occurs most readily with larger atoms.
The first noble-gas compound was reported in 1962. This discovery caused a sensation because it undercut the belief that the noble-gas elements were inert. The initial
study involved xenon in combination with fluorine, the element we would expect to be
most reactive in pulling electron density from another atom. Since that time chemists
have prepared several xenon compounds of fluorine and oxygen (▼ Table 22.1). The
fluorides XeF2, XeF4, and XeF6 are made by direct reaction of the elements. By varying
the ratio of reactants and altering reaction conditions, one of the three compounds can
be obtained. The oxygen-containing compounds are formed when the fluorides react
with water as, for example,
XeF61s2 + 3 H2O1l2 ¡ XeO31aq2 + 6 HF1aq2[22.17]
The other noble-gas elements form compounds much less readily than xenon. For
many years, only one binary krypton compound, KrF2, was known with certainty, and
Table 22.1 Properties of Xenon Compounds
XeF2
+2
129
∆H°f 1kJ , mol2 a
XeF4
+4
117
- 2181g2
XeF6
+6
49
- 2981g2
XeOF4
+6
-41 to - 28
+ 1461l2
XeO3
XeO2F2
Compound
Oxidation State of Xe
XeO4
Melting Point 1 °C 2
+6
-
b
+4021s2
+6
31
+1451s2
+8
c
a
-
At 25 °C, for the compound in the state indicated.
b
A solid; decomposes at 40 °C.
c
- 1091g2
A solid; decomposes at - 40 °C.
-
8A
2
He
10
Ne
18
Ar
36
Kr
54
Xe
86
Rn
961
962
chapter 22 Chemistry of the Nonmetals
it decomposes to its elements at -10 °C. Other compounds of krypton have been
isolated at very low temperatures (40 K).
Sample
Exercise 22.3 Predicting a Molecular Structure
Use the VSEPR model to predict the structure of XeF4.
Solution
Analyze We must predict the geometrical structure given only the
molecular formula.
F
Plan We must first write the Lewis structure for the molecule. We
F
then count the number of electron pairs (domains) around the Xe
atom and use that number and the number of bonds to predict the
geometry.
Solve There are 36 valence-shell electrons (8 from xenon and 7
from each fluorine). If we make four single Xe ¬ F bonds, each
fluorine has its octet satisfied. Xe then has 12 electrons in its
valence shell, so we expect an octahedral arrangement of six electron pairs. Two of these are nonbonded pairs. Because nonbonded
pairs require more volume than bonded pairs
(Section 9.2), it
is reasonable to expect these nonbonded pairs to be opposite each
other. The expected structure is square planar, as shown in
▶ Figure 22.7.
Comment The experimentally determined structure agrees with this
prediction.
Xe
F
F
▲ Figure 22.7 Xenon tetrafluoride.
Practice Exercise 1
Compounds containing the XeF3+ ion have been characterized.
Describe the electron-domain geometry and molecular geometry
of this ion.
(a) trigonal planar, trigonal planar; (b) tetrahedral, trigonal
pyramidal; (c) trigonal bipyramidal, T shaped; (d) tetrahedral,
tetrahedral; (e) octahedral, square planar.
Practice Exercise 2
Describe the electron-domain geometry and molecular geometry
of KrF2.
22.4 | Group 7A: The Halogens
7A
9
F
17
Cl
35
Br
53
I
85
At
The elements of group 7A, the halogens, have the outer-electron configuration ns2np5,
where n ranges from 2 through 6. The halogens have large negative electron affinities
(Section 7.5), and they most often achieve a noble-gas configuration by gaining an
electron, which results in a -1 oxidation state. Fluorine, being the most electronegative
element, exists in compounds only in the -1 state. The other halogens exhibit positive
oxidation states up to +7 in combination with more electronegative atoms such as O.
In the positive oxidation states, the halogens tend to be good oxidizing agents, readily
accepting electrons.
Chlorine, bromine, and iodine are found as the halides in seawater and in salt
deposits. Fluorine occurs in the minerals fluorspar 1CaF22, cryolite 1Na3AlF62, and fluorapatite 3Ca51PO423F4.* Only fluorspar is an important commercial source of fluorine.
All isotopes of astatine are radioactive. The longest-lived isotope is astatine-210,
which has a half-life of 8.1 h and decays mainly by electron capture. Because astatine is
so unstable, very little is known about its chemistry.
Properties and Production of the Halogens
Most properties of the halogens vary in a regular fashion as we go from fluorine to
iodine (▶ Table 22.2).
Under ordinary conditions the halogens exist as diatomic molecules. The molecules
(Section 11.2)
are held together in the solid and liquid states by dispersion forces.
Because I2 is the largest and most polarizable halogen molecule, the intermolecular forces
between I2 molecules are the strongest. Thus, I2 has the highest melting point and boiling point. At room temperature and 1 atm, I2 is a purple solid, Br2 is a red-brown liquid,
(Figure 7.27) Chlorine readily liquefies upon compression
and Cl2 and F2 are gases.
*Minerals are solid substances that occur in nature. They are usually known by their common names rather
than by their chemical names. What we know as rock is merely an aggregate of different minerals.
section 22.4 Group 7A: The Halogens
963
Table 22.2 Some Properties of the Halogens
Property
°
Atomic radius 1A2
Ionic radius, X - 1A° 2
F
Cl
Br
I
0.57
1.02
1.20
1.39
1.33
1.81
1.96
2.20
1681
1251
1140
1008
-328
-349
-325
- 295
Electronegativity
4.0
3.0
2.8
2.5
X ¬ X single-bond enthalpy 1kJ>mol2
155
242
193
151
2.87
1.36
1.07
0.54
First ionization energy 1kJ>mol2
Electron affinity 1kJ>mol2
Reduction potential (V):
1
2 X21aq2
+ e - ¡ X - 1aq2
at room temperature and is normally stored and handled in liquid form under pressure
in steel containers.
The comparatively low bond enthalpy of F2 1155 kJ>mol2 accounts in part for the
extreme reactivity of elemental fluorine. Because of its high reactivity, F2 is difficult to
work with. Certain metals, such as copper and nickel, can be used to contain F2 because
their surfaces form a protective coating of metal fluoride. Chlorine and the heavier halogens are also reactive, although less so than fluorine.
Because of their high electronegativities, the halogens tend to gain electrons from
other substances and thereby serve as oxidizing agents. The oxidizing ability of the
halogens, indicated by their standard reduction potentials, decreases going down the
group. As a result, a given halogen is able to oxidize the halide anions below it. For
example, Cl2 oxidizes Br - and I - but not F - , as seen in ▶ Figure 22.8.
Go Figure
Do Br2 and I2 appear to be more or
less soluble in CCl4 than in H2O?
Cl2(aq) added
NaF(aq)
NaBr(aq)
NaI(aq)
Sample
Exercise 22.4 Predicting Chemical Reactions among
the Halogens
Write the balanced equation for the reaction, if any, between (a) I - 1aq2 and Br21l2, (b) Cl - 1aq2
and I21s2.
Solution
Analyze We are asked to determine whether a reaction occurs when a particular halide and
halogen are combined.
Plan A given halogen is able to oxidize anions of the halogens below it in the periodic table.
Thus, in each pair the halogen having the smaller atomic number ends up as the halide ion. If the
halogen with the smaller atomic number is already the halide ion, there is no reaction. Thus, the
key to determining whether a reaction occurs is locating the elements in the periodic table.
Solve
(a) Br2 can oxidize (remove electrons from) the anions of the halogens below it in the periodic
table. Thus, it oxidizes I - :
2 I - 1aq2 + Br21aq2 ¡ I21s2 + 2 Br - 1aq2
(b) Cl - is the anion of a halogen above iodine in the periodic table. Thus, I2 cannot oxidize Cl - ;
there is no reaction.
Practice Exercise 1
Which is (are) able to oxidize Cl - ?
(a) F2
(b) F (c) Both Br2 and I2
(d) Both Br - and I Practice Exercise 2
Write the balanced chemical equation for the reaction between Br - 1aq2 and Cl21aq2.
CCl4 layer
No
reaction
I−
Br−
oxidized oxidized
to Br2
to I2
▲ Figure 22.8 Reaction of Cl2 with
aqueous solutions of NaF, NaBr, and NaI in
the presence of carbon tetrachloride. The top
liquid layer in each vial is water; the bottom
liquid layer is carbon tetrachloride. The
Cl21aq2, which has been added to each vial,
is colorless. The brown color in the carbon
tetrachloride layer indicates the presence of
Br2, whereas purple indicates the presence
of I2.
964
chapter 22 Chemistry of the Nonmetals
Notice in Table 22.2 that the standard reduction potential of F2 is exceptionally
high. As a result, fluorine gas readily oxidizes water:
F21aq2 + H2O1l2 ¡ 2 HF1aq2 +
1
2
E° = 1.80 V[22.18]
O21g2
Fluorine cannot be prepared by electrolytic oxidation of aqueous solutions of fluo (Section 20.9) In
ride salts because water is oxidized more readily than F - .
practice, the element is formed by electrolytic oxidation of a solution of KF in
anhydrous HF.
Chlorine is produced mainly by electrolysis of either molten or aqueous sodium
chloride. Both bromine and iodine are obtained commercially from brines containing
the halide ions; the reaction used is oxidation with Cl2.
Uses of the Halogens
Go Figure
What is the repeating unit in this
polymer?
▲ Figure 22.9 Structure of Teflon®, a
fluorocarbon polymer.
Fluorine is used to prepare fluorocarbons—very stable carbon–fluorine compounds used as refrigerants, lubricants, and plastics. Teflon® ( ◀ Figure 22.9) is a
polymeric fluorocarbon noted for its high thermal stability and lack of chemical
reactivity.
Chlorine is by far the most commercially important halogen. About 1 * 1010 kg
(10 million tons) of Cl2 is produced annually in the United States. In addition, hydrogen chloride production is about 4.0 * 109 kg (4.4 million tons) annually. About half
of this chlorine finds its way eventually into the manufacture of chlorine-containing
organic compounds, such as the vinyl chloride 1C2H3Cl2 used in making polyvinyl
(Section 12.8) Much of the remainder is used as a bleachchloride (PVC) plastics.
ing agent in the paper and textile industries.
When Cl2 dissolves in cold dilute base, it converts into Cl - and hypochlorite,
ClO :
-
-
Cl21aq2 + 2 OH - 1aq2 ∆ Cl 1aq2 + ClO 1aq2 + H2O1l2[22.19]
Sodium hypochlorite (NaClO) is the active ingredient in many liquid bleaches.
Chlorine is also used in water treatment to oxidize and thereby destroy bacteria.
(Section 18.4)
Give It Some Thought
What is the oxidation state of Cl in each Cl species in Equation 22.19?
A common use of iodine is as KI in table salt. Iodized salt provides the small
amount of iodine necessary in our diets; it is essential for the formation of thyroxin, a
hormone secreted by the thyroid gland. Lack of iodine in the diet results in an enlarged
thyroid gland, a condition called goiter.
The Hydrogen Halides
All the halogens form stable diatomic molecules with hydrogen. Aqueous solutions
of HCl, HBr, and HI are strong acids. The hydrogen halides can be formed by direct reaction of the elements. The most important means of preparing HF and HCl,
however, is by reacting a salt of the halide with a strong nonvolatile acid, as in the
reaction
∆
CaF21s2 + H2SO41l2 ¡ 2 HF1g2 + CaSO41s2[22.20]
Neither HBr nor HI can be prepared in this way, however, because H2SO4 oxidizes Br - and I - (▶ Figure 22.10). This difference in reactivity reflects the greater ease
of oxidation of Br - and I - relative to F - and Cl - . These undesirable oxidations are
avoided by using a nonvolatile acid, such as H3PO4, that is a weaker oxidizing agent
than H2SO4.
section 22.4 Group 7A: The Halogens
965
Go Figure
Are these reactions acid–base reactions or oxidation–reduction reactions?
H2SO4
NaI
NaBr
I2
formed
Br2
formed
▲ Figure 22.10 Reaction of H2SO4 with NaI and NaBr.
Sample
Exercise 22.5 Writing a Balanced Chemical Equation
Write a balanced equation for the formation of hydrogen bromide gas from the reaction of solid
sodium bromide with phosphoric acid.
Solution
Analyze We are asked to write a balanced equation for the reaction
between NaBr and H3PO4 to form HBr and another product.
Plan As in Equation 22.20, a metathesis reaction takes place.
(Section 4.2) Let’s assume that only one H in H3PO4 reacts.
(The actual number depends on the reaction conditions.) The
H2PO4- and Na+ will form NaH2PO4 as one product.
Solve The balanced equation is
NaBr1s2 + H3PO41l2 ¡ NaH2PO41s2 + HBr1g2
Practice Exercise 1
Which of the following are oxidized by H2SO4?
(a) Cl - , (b) Cl - and Br - , (c) Br - and I - , (d) Cl2, (e) Br2 and I2.
Practice Exercise 2
Write the balanced equation for the preparation of HI from NaI
and H3PO4.
The hydrogen halides form hydrohalic acid solutions when dissolved in water.
These solutions have the characteristic properties of acids, such as reactions with active
(Section 4.4) Hydrofluoric acid also reacts readily
metals to produce hydrogen gas.
with silica 1SiO22 and with silicates to form hexafluorosilicic acid 1H2SiF62:
SiO21s2 + 6 HF1aq2 ¡ H2SiF61aq2 + 2 H2O1l2[22.21]
Interhalogen Compounds
Because the halogens exist as diatomic molecules, diatomic molecules made up of two different halogen atoms exist. These compounds are the simplest examples of interhalogens,
compounds, such as ClF and IF5, formed between two halogen elements.
The vast majority of the higher interhalogen compounds have a central Cl, Br, or I
atom surrounded by fluorine atoms. The large size of the iodine atom allows the formation of IF3, IF5, and IF7, in which the oxidation state of I is +3, +5, and +7, respectively.
966
chapter 22 Chemistry of the Nonmetals
Table 22.3 The Stable Oxyacids of the Halogens
Formula of Acid
Oxidation State
of Halogen
Cl
Br
I
Acid Name
+1
HClO
HBrO
HIO
Hypohalous acid
+3
HClO2
—
—
Halous acid
+5
HClO3
HBrO3
HIO3
Halic acid
+7
HClO4
HBrO4
HIO4
Perhalic acid
With the smaller bromine and chlorine atoms, only compounds with 3 or 5 fluorines
form. The only higher interhalogen compounds that do not have outer F atoms are
ICl3 and ICl5; the large size of the I atom can accommodate 5 Cl atoms, whereas Br is
not large enough to allow even BrCl3 to form. All of the interhalogen compounds are
powerful oxidizing agents.
Oxyacids and Oxyanions
summarizes the formulas of the known oxyacids of the halogens and the
(Section 2.8) The acid strengths of the oxyacids increase with
way they are named*.
(Section 16.10) All the oxyincreasing oxidation state of the central halogen atom.
acids are strong oxidizing agents. The oxyanions, formed on removal of H+ from the
oxyacids, are generally more stable than the oxyacids. Hypochlorite salts are used as
bleaches and disinfectants because of the powerful oxidizing capabilities of the ClO ion. Chlorate salts are similarly very reactive. For example, potassium chlorate is used
to make matches and fireworks.
▲ Table 22.3
10 Al(s) + 6 NH4ClO4(s)
4 Al2O3(s) + 2 AlCl3(s)
+ 12 H2O(g) + 3 N2(g)
Give It Some Thought
Which do you expect to be the stronger oxidizing agent, NaBrO3 or NaClO3?
Perchloric acid and its salts are the most stable oxyacids and oxyanions. Dilute
solutions of perchloric acid are quite safe, and many perchlorate salts are stable except
when heated with organic materials. When heated, however, perchlorates can become
vigorous, even violent, oxidizers. Considerable caution should be exercised, therefore,
when handling these substances, and it is crucial to avoid contact between perchlorates
and readily oxidized material. The use of ammonium perchlorate 1NH4ClO42 as the
oxidizer in the solid booster rockets for the Space Shuttle demonstrates the oxidizing
power of perchlorates. The solid propellant contains a mixture of NH4ClO4 and powdered aluminum, the reducing agent. Each shuttle launch requires about 6 * 105 kg
(700 tons) of NH4ClO4 (◀ Figure 22.11).
22.5 | Oxygen
The large volume of gases
produced provides thrust
for the booster rockets
▲ Figure 22.11 Launch of the Space
Shuttle Columbia from the Kennedy Space
Center.
By the middle of the seventeenth century, scientists recognized that air contained a
component associated with burning and breathing. That component was not isolated until 1774, however, when English scientist Joseph Priestley discovered oxygen.
Lavoisier subsequently named the element oxygen, meaning “acid former.”
Oxygen is found in combination with other elements in a great variety of
compounds—water 1H2O2, silica 1SiO22, alumina 1Al2O32, and the iron oxides
*Fluorine forms one oxyacid, HOF. Because the electronegativity of fluorine is greater than that of oxygen,
we must consider fluorine to be in a - 1 oxidation state and oxygen to be in the 0 oxidation state in this
compound.
section 22.5 Oxygen
1Fe2O3, Fe3O42 are obvious examples. Indeed, oxygen is the most abundant element by
(Section 1.2) It is the oxidizing
mass both in Earth’s crust and in the human body.
agent for the metabolism of our foods and is crucial to human life.
Properties of Oxygen
Oxygen has two allotropes, O2 and O3. When we speak of molecular oxygen or simply oxygen, it is usually understood that we are speaking of dioxygen 1O22, the normal
form of the element; O3 is ozone.
At room temperature, dioxygen is a colorless, odorless gas. It condenses to a liquid
at -183 °C and freezes at -218 °C. It is only slightly soluble in water (0.04 g>L, or 0.001
M at 25 °C), but its presence in water is essential to marine life.
The electron configuration of the oxygen atom is 3He42s22p4. Thus, oxygen can
complete its octet of valence electrons either by picking up two electrons to form the
oxide ion 1O2 - 2 or by sharing two electrons. In its covalent compounds, it tends
to form either two single bonds, as in H2O, or a double bond, as in formaldehyde
1H2C “ O2. The O2 molecule contains a double bond. The bond in O2 is very strong
(bond enthalpy 495 kJ>mol). Oxygen also forms strong bonds with many other elements. Consequently, many oxygen-containing compounds are thermodynamically
more stable than O2. In the absence of a catalyst, however, most reactions of O2 have
high activation energies and thus require high temperatures to proceed at a suitable
rate. Once a sufficiently exothermic reaction begins, it may accelerate rapidly, producing a reaction of explosive violence.
Production of Oxygen
Nearly all commercial oxygen is obtained from air. The normal boiling point of O2 is
-183 °C, whereas that of N2, the other principal component of air, is -196 °C. Thus,
when air is liquefied and then allowed to warm, the N2 boils off, leaving liquid O2 contaminated mainly by small amounts of N2 and Ar.
In the laboratory, O2 can be obtained by heating either aqueous hydrogen peroxide
or solid potassium chlorate 1KClO32:
2 KClO31s2 ¡ 2 KCl1s2 + 3 O21g2[22.22]
Manganese dioxide 1MnO22 catalyzes both reactions.
Much of the O2 in the atmosphere is replenished through photosynthesis, in which
green plants use the energy of sunlight to generate O2 (along with glucose, C6H12O6)
from atmospheric CO2:
6 CO21g2 + 6 H2O1l2 ¡ C6H12O61aq2 + 6 O21g2
Uses of Oxygen
In industrial use, oxygen ranks behind only sulfuric acid 1H2SO42 and nitrogen 1N22.
About 3 * 1010 kg (30 million tons) of O2 is used annually in the United States. It is
shipped and stored either as a liquid or in steel containers as a compressed gas. About
70% of the O2 output, however, is generated where it is needed.
Oxygen is by far the most widely used oxidizing agent in industry. Over half of
the O2 produced is used in the steel industry, mainly to remove impurities from steel.
It is also used to bleach pulp and paper. (Oxidation of colored compounds often gives
colorless products.) Oxygen is used together with acetylene 1C2H22 in oxyacetylene
welding (▶ Figure 22.12). The reaction between C2H2 and O2 is highly exothermic,
producing temperatures in excess of 3000 °C.
2 C2H2(g) + 5 O2(g)
4 CO2(g) + 2 H2O(g);
∆H° = −2510 kJ
Ozone
Ozone is a pale blue, poisonous gas with a sharp, irritating odor. Many people can detect
as little as 0.01 ppm in air. Exposure to 0.1 to 1 ppm produces headaches, burning eyes,
and irritation to the respiratory passages.
▲ Figure 22.12 Welding with an
oxyacetylene torch.
967
968
chapter 22 Chemistry of the Nonmetals
The O3 molecule possesses p electrons that are delocalized over the three oxygen
(Section 8.6) The molecule dissociates readily, forming reactive oxygen atoms:
atoms.
O31g2 ¡ O21g2 + O1g2 ∆H° = 105 kJ[22.23]
Ozone is a stronger oxidizing agent than dioxygen. Ozone forms oxides with many
elements under conditions where O2 will not react; indeed, it oxidizes all the common
metals except gold and platinum.
Ozone can be prepared by passing electricity through dry O2 in a flow-through
apparatus. The electrical discharge causes the O2 bond to break, resulting in reactions
like those described in Section 18.1. During thunderstorms, ozone is generated (and
can be smelled, if you are too close) from lightning strikes:
electricity
3 O21g2 ¡ 2 O31g2 ∆H° = 285 kJ[22.24]
Ozone is sometimes used to treat drinking water. Like Cl2, ozone kills bacteria and
oxidizes organic compounds. The largest use of ozone, however, is in the preparation
of pharmaceuticals, synthetic lubricants, and other commercially useful organic compounds, where O3 is used to sever carbon–carbon double bonds.
Ozone is an important component of the upper atmosphere, where it screens out
ultraviolet radiation and so protects us from the effects of these high-energy rays. For this
(Section 18.2)
reason, depletion of stratospheric ozone is a major scientific concern.
In the lower atmosphere, ozone is considered an air pollutant and is a major constituent
(Section 18.2) Because of its oxidizing power, ozone damages living systems
of smog.
and structural materials, especially rubber.
Oxides
The electronegativity of oxygen is second only to that of fluorine. As a result, oxygen
has negative oxidation states in all compounds except OF2 and O2F2. The -2 oxidation
state is by far the most common. Compounds that contain oxygen in this oxidation
state are called oxides.
Nonmetals form covalent oxides, most of which are simple molecules with low
melting and boiling points. Both SiO2 and B2O3, however, have extended structures.
Most nonmetal oxides combine with water to give oxyacids. Sulfur dioxide 1SO22, for
example, dissolves in water to give sulfurous acid 1H2SO32:
SO21g2 + H2O1l2 ¡ H2SO31aq2[22.25]
This reaction and that of SO3 with H2O to form H2SO4 are largely responsible for acid
(Section 18.2) The analogous reaction of CO2 with H2O to form carbonic
rain.
acid 1H2CO32 causes the acidity of carbonated water.
Oxides that form acids when they react with water are called either acidic anhydrides
(anhydride means “without water”) or acidic oxides. A few nonmetal oxides, especially
ones with the nonmetal in a low oxidation state—such as N2O, NO, and CO—do not
react with water and are not acidic anhydrides.
Give It Some Thought
What acid is produced by the reaction of I2O5 with water?
Most metal oxides are ionic compounds. The ionic oxides that dissolve in water form
hydroxides and, consequently, are called either basic anhydrides or basic oxides. Barium
oxide, for example, reacts with water to form barium hydroxide (▶ Figure 22.13). These
kinds of reactions are due to the high basicity of the O2 - ion and its virtually complete
hydrolysis in water:
O2 - 1aq2 + H2O1l2 ¡ 2 OH - 1aq2[22.26]
Even those ionic oxides that are insoluble in water tend to dissolve in strong acids.
Iron(III) oxide, for example, dissolves in acids:
Fe2O31s2 + 6 H+ 1aq2 ¡ 2 Fe3 + 1aq2 + 3 H2O1l2[22.27]
section 22.5 Oxygen
969
Go Figure
Is this reaction a redox reaction?
H2O
with
indicator
Pink color
indicates
basic
solution
BaO(s)
+
H2O(l)
Ba(OH)2(aq)
▲ Figure 22.13 Reaction of a basic oxide with water.
This reaction is used to remove rust 1Fe2O3 # nH2O2 from iron or steel before a protective coat of zinc or tin is applied.
Oxides that can exhibit both acidic and basic characters are said to be amphoteric.
(Section 17.5) If a metal forms more than one oxide, the basic character of the
oxide decreases as the oxidation state of the metal increases (▼ Table 22.4).
4 KO2(s) + 2 H2O(l, from breath)
4 K+(aq) + 4 OH−(aq) + 3 O2(g)
2 OH−(aq) + CO2(g, from breath)
H2O(l) + CO32−(aq)
Table 22.4 Acid–Base Character of Chromium Oxides
Oxide
Oxidation State of Cr
Nature of Oxide
CrO
+2
Basic
Cr2O3
+3
Amphoteric
CrO3
+6
Acidic
Peroxides and Superoxides
Compounds containing O ¬ O bonds and oxygen in the -1 oxidation state are peroxides. Oxygen has an oxidation state of - 12 in O2- , which is called the superoxide ion. The
most active (easily oxidized) metals (K, Rb, and Cs) react with O2 to give superoxides
1KO2, RbO2, and CsO22. Their active neighbors in the periodic table (Na, Ca, Sr, and
Ba) react with O2, producing peroxides 1Na2O2, CaO2, SrO2, and BaO22. Less active
(Section 7.6)
metals and nonmetals produce normal oxides.
When superoxides dissolve in water, O2 is produced:
4 KO21s2 + 2 H2O1l2 ¡ 4 K+ 1aq2 + 4 OH - 1aq2 + 3 O21g2[22.28]
Because of this reaction, potassium superoxide is used as an oxygen source in masks
worn by rescue workers (▶ Figure 22.14). For proper breathing in toxic environments,
oxygen must be generated in the mask and exhaled carbon dioxide in the mask must be
eliminated. Moisture in the breath causes the KO2 to decompose to O2 and KOH, and
the KOH removes CO2 from the exhaled breath:
2 OH - 1aq2 + CO21g2 ¡ H2O1l2 + CO32 - 1aq2[22.29]
▲ Figure 22.14 A self-contained breathing
apparatus.
970
chapter 22 Chemistry of the Nonmetals
Go Figure
Does H2O2 have a dipole moment?
Hydrogen peroxide (◀ Figure 22.15) is the most familiar and commercially important peroxide. Pure hydrogen peroxide is a clear, syrupy liquid that melts at -0.4 °C.
Concentrated hydrogen peroxide is dangerously reactive because the decomposition to
water and oxygen is very exothermic:
2 H2O21l2 ¡ 2 H2O1l2 + O21g2 ∆H° = -196.1 kJ[22.30]
This is another example of a disproportionation reaction, in which an element is
simultaneously oxidized and reduced. The oxidation number of oxygen changes
from -1 to -2 and 0.
Hydrogen peroxide is marketed as a chemical reagent in aqueous solutions of up
to about 30% by mass. A solution containing about 3% H2O2 by mass is sold in drugstores and used as a mild antiseptic. Somewhat more concentrated solutions are used to
bleach fabrics.
The peroxide ion is a by-product of metabolism that results from the reduction of O2.
The body disposes of this reactive ion with enzymes such as peroxidase and
catalase.
111°
▲ Figure 22.15 Molecular structure of
hydrogen peroxide. The repulsive interaction
of the O ¬ H bonds with the lone-pairs of
electrons on each O atom restricts the free
rotation around the O ¬ O single bond.
6A
8
O
16
S
34
Se
52
Te
84
Po
22.6 | The Other Group 6A Elements:
S, Se, Te, and Po
The other group 6A elements are sulfur, selenium, tellurium, and polonium. In this section, we will survey the properties of the group as a whole and then examine the chemistry of sulfur, selenium, and tellurium. We will not discuss polonium, which has no
stable isotopes and is found only in minute quantities in radium-containing minerals.
General Characteristics of the Group 6A Elements
The group 6A elements possess the general outer-electron configuration ns2np4 with
n ranging from 2 to 6. Thus, these elements attain a noble-gas electron configuration
by adding two electrons, which results in a -2 oxidation state. Except for oxygen, the
group 6A elements are also commonly found in positive oxidation states up to +6, and
they can have expanded valence shells. Thus, we have such compounds as SF6, SeF6,
and TeF6 with the central atom in the +6 oxidation state.
▼ Table 22.5 summarizes some properties of the group 6A elements.
Occurrence and Production of S, Se, and Te
Sulfur, selenium, and tellurium can all be mined from the earth. Large underground
deposits are the principal source of elemental sulfur (◀ Figure 22.16). Sulfur also occurs widely as sulfide 1S2 - 2 and sulfate 1SO42 - 2 minerals. Its presence as a minor component of coal and petroleum poses a major problem. Combustion of these “unclean”
Table 22.5 Some Properties of the Group 6A Elements
▲ Figure 22.16 Massive amounts of sulfur
are extracted every year from the earth.
Property
°
Atomic radius 1A2
X2 - ionic radius 1A2
°
First ionization energy 1kJ>mol2
Electron affinity 1kJ>mol2
O
S
Se
Te
0.66
1.05
1.21
1.38
1.40
1.84
1.98
2.21
1314
1000
941
869
- 141
- 200
-195
-190
Electronegativity
3.5
2.5
2.4
2.1
X ¬ X single-bond enthalpy 1kJ>mol2
146*
266
172
126
1.23
0.14
- 0.40
- 0.72
Reduction potential to H2X in acidic
solution (V)
*Based on O ¬ O bond energy in H2O2.
section 22.6 The Other Group 6A Elements: S, Se, Te, and Po
971
(Section 18.2) Much effort has
fuels leads to serious pollution by sulfur oxides.
been directed at removing this sulfur, and these efforts have increased the availability
of sulfur.
Selenium and tellurium occur in rare minerals, such as Cu2Se, PbSe, Cu2Te, and
PbTe, and as minor constituents in sulfide ores of copper, iron, nickel, and lead.
Properties and Uses of Sulfur, Selenium,
and Tellurium
Elemental sulfur is yellow, tasteless, and nearly odorless. It is insoluble in water and
exists in several allotropic forms. The thermodynamically stable form at room temperature is rhombic sulfur, which consists of puckered S8 rings with each sulfur atom forming two bonds (Figure 7.26). Rhombic sulfur melts at 113 °C.
Most of the approximately 1 * 1010 kg (10 million tons) of sulfur produced in
the United States each year is used to manufacture sulfuric acid. Sulfur is also used to
vulcanize rubber, a process that toughens rubber by introducing cross-linking between
(Section 12.8)
polymer chains.
Selenium and tellurium do not form eight-membered rings in their elemental
(Section 7.8) The most stable allotropes of these elements are crystalline
forms.
substances containing helical chains of atoms (▶ Figure 22.17). In all the allotropes
each atom forms two bonds to its neighbors. Each atom is close to atoms in adjacent
chains, and it appears that some sharing of electron pairs between these atoms occurs.
The electrical conductivity of elemental selenium is low in the dark but increases
greatly upon exposure to light. This property is exploited in photoelectric cells and light
meters. Photocopiers also depend on the photoconductivity of selenium. Photocopy
machines contain a belt or drum coated with a film of selenium. This drum is electrostatically charged and then exposed to light reflected from the image being photocopied. The charge drains from the regions where the selenium film has been made
conductive by exposure to light. A black powder (the toner) sticks only to the areas
that remain charged. The photocopy is made when the toner is transferred to a sheet of
plain paper.
Se
Se
Se
Se
Se
Se
Se
Se
Se
Se
Se
Se
Se
Se
▲ Figure 22.17 Portion of helical chains
making up the structure of crystalline
selenium.
Sulfides
When an element is less electronegative than sulfur, sulfides that contain S2 - form.
Many metallic elements are found in the form of sulfide ores, such as PbS (galena) and
HgS (cinnabar). A series of related ores containing the disulfide ion, S22 - (analogous to
the peroxide ion), are known as pyrites. Iron pyrite, FeS2, occurs as golden yellow cubic
crystals (▶ Figure 22.18). Because it has been occasionally mistaken for gold by miners,
iron pyrite is often called fool’s gold.
One of the most important sulfides is hydrogen sulfide 1H2S2. This substance is
normally prepared by action of dilute acid on iron(II) sulfide:
FeS1s2 + 2 H+ 1aq2 ¡ H2S1aq2 + Fe2 + 1aq2[22.31]
One of hydrogen sulfide’s most readily recognized properties is its odor, which
is most frequently encountered as the offensive odor of rotten eggs. Hydrogen sulfide
is toxic but our noses can detect H2S in extremely low, nontoxic concentrations. A
sulfur-containing organic molecule, such as dimethyl sulfide, 1CH322S, which is similarly odoriferous and can be detected by smell at a level of one part per trillion, is added
to natural gas as a safety factor to give it a detectable odor.
Oxides, Oxyacids, and Oxyanions of Sulfur
Sulfur dioxide, formed when sulfur burns in air, has a choking odor and is poisonous.
The gas is particularly toxic to lower organisms, such as fungi, so it is used to sterilize
dried fruit and wine. At 1 atm and room temperature, SO2 dissolves in water to produce a 1.6 M solution. The SO2 solution is acidic, and we describe it as sulfurous acid
1H2SO32.
▲ Figure 22.18 Iron pyrite (FeS2, on the
right) with gold for comparison.
972
chapter 22 Chemistry of the Nonmetals
Salts of SO32 - (sulfites) and HSO3- (hydrogen sulfites or bisulfites) are well
known. Small quantities of Na2SO3 or NaHSO3 are used as food additives to prevent
bacterial spoilage. However, they are known to increase asthma symptoms in approximately 5% of asthmatics. Thus, all food products with sulfites must now carry a warning label disclosing their presence (◀ Figure 22.19).
Although combustion of sulfur in air produces mainly SO2, small amounts of SO3
are also formed. The reaction produces chiefly SO2 because the activation-energy barrier for oxidation to SO3 is very high unless the reaction is catalyzed. Interestingly,
the SO3 by-product is used industrially to make H2SO4, which is the ultimate product of the reaction between SO3 and water. In the manufacture of sulfuric acid, SO2 is
obtained by burning sulfur and then oxidized to SO3, using a catalyst such as V2O5 or
platinum. The SO3 is dissolved in H2SO4 because it does not dissolve quickly in water,
and then the H2S2O7 formed in this reaction, called pyrosulfuric acid, is added to water
to form H2SO4:
▲ Figure 22.19 Food label warning
of sulfites.
SO31g2 + H2SO41l2 ¡ H2S2O71l2[22.32]
H2S2O71l2 + H2O1l2 ¡ 2 H2SO41l2[22.33]
Give It Some Thought
What is the net reaction of Equations 22.32 and 22.33?
Commercial sulfuric acid is 98% H2SO4. It is a dense, colorless, oily liquid that
boils at 340 °C. It is a strong acid, a good dehydrating agent (▼ Figure 22.20), and a
moderately good oxidizing agent.
Year after year, the production of sulfuric acid is the largest of any chemical produced in the United States. About 4 * 1010 kg (40 million tons) is produced annually
in this country. Sulfuric acid is employed in some way in almost all manufacturing.
Sulfuric acid is a strong acid, but only the first hydrogen is completely ionized in
aqueous solution:
H2SO41aq2 ¡ H+ 1aq2 + HSO4- 1aq2[22.34]
HSO4- 1aq2 ∆ H+ 1aq2 + SO42 - 1aq2 Ka = 1.1 * 10 - 2[22.35]
Go Figure
In this reaction, what has happened to the H and O atoms in the sucrose?
H2SO4
Table sugar
(sucrose),
C12H22O11
Pure carbon, C
▲ Figure 22.20 Sulfuric acid dehydrates table sugar to produce elemental carbon.
section 22.7 Nitrogen
Consequently, sulfuric acid forms both sulfates 1SO42- salts2 and
bisulfates (or hydrogen sulfates, HSO4- salts). Bisulfate salts are
common components of the “dry acids” used for adjusting the
pH of swimming pools and hot tubs; they are also components of
many toilet bowl cleaners.
The term thio indicates substitution of sulfur for oxygen, and
the thiosulfate ion 1S2O32 - 2 is formed by boiling an alkaline solution of SO32 - with elemental sulfur:
8 SO32 - 1aq2 + S81s2 ¡ 8 S2O32 - 1aq2
973
Go Figure
What are the oxidation states of the sulfur atoms in the
S2O32 - ion?
2−
2−
[22.36]
The structures of the sulfate and thiosulfate ions are compared in
▶ Figure 22.21.
22.7 | Nitrogen
▲ Figure 22.21 Structures of the sulfate (left) and thiosulfate
(right) ions.
Nitrogen constitutes 78% by volume of Earth’s atmosphere, where it occurs as N2 molecules. Although nitrogen is a key element in living organisms, compounds of nitrogen
are not abundant in Earth’s crust. The major natural deposits of nitrogen compounds
are those of KNO3 (saltpeter) in India and NaNO3 (Chile saltpeter) in Chile and other
desert regions of South America.
Properties of Nitrogen
Nitrogen is a colorless, odorless, tasteless gas composed of N2 molecules. Its melting
point is -210 °C, and its normal boiling point is -196 °C.
The N2 molecule is very unreactive because of the strong triple bond between
nitrogen atoms (the N ‚ N bond enthalpy is 941 kJ>mol, nearly twice that for the bond
(Table 8.4) When substances burn in air, they normally react with O2 but
in O2.
not with N2.
The electron configuration of the nitrogen atom is 3He42s22p3. The element
exhibits all formal oxidation states from +5 to -3 (▶ Table 22.6). The +5, 0, and -3
oxidation states are the most common and generally the most stable of these. Because
nitrogen is more electronegative than all other elements except fluorine, oxygen, and
chlorine, it exhibits positive oxidation states only in combination with these three
elements.
Production and Uses of Nitrogen
Elemental nitrogen is obtained in commercial quantities by fractional distillation of
liquid air. About 4 * 1010 kg (40 million tons) of N2 is produced annually in the
United States.
Because of its low reactivity, large quantities of N2 are used as an inert gaseous
blanket to exclude O2 in food processing, manufacture of chemicals, metal fabrication,
and production of electronic devices. Liquid N2 is employed as a coolant to freeze foods
rapidly.
The largest use of N2 is in the manufacture of nitrogen-containing fertilizers, which
provide a source of fixed nitrogen. We have previously discussed nitrogen fixation in
the “Chemistry and Life” box in Section 14.7 and in the “Chemistry Put to Work” box
in Section 15.2. Our starting point in fixing nitrogen is the manufacture of ammonia
(Section 15.2) The ammonia can then be converted into a
via the Haber process.
variety of useful, simple nitrogen-containing species (Figure 22.22).
Hydrogen Compounds of Nitrogen
Ammonia is one of the most important compounds of nitrogen. It is a colorless, toxic
gas that has a characteristic irritating odor. As noted in previous discussions, the NH3
(Section 16.7)
molecule is basic 1Kb = 1.8 * 10 - 52.
Table 22.6 Oxidation States
of Nitrogen
Oxidation
State
Examples
+5
N2O5, HNO3, NO3-
+4
NO2, N2O4
+3
HNO2, NO2- , NF3
+2
NO
+1
N2O, H2N2O2, N2O22 - , HNF2
0
N2
-1
NH2OH, NH2F
-2
N2H4
-3
NH3, NH4+, NH2-
974
chapter 22 Chemistry of the Nonmetals
Go Figure
In which of these species is the oxidation number of nitrogen + 3?
N2H4
(hydrazine)
N2
(dinitrogen)
NH3
(ammonia)
NO
(nitric oxide)
NO2
(nitrogen dioxide)
NH4+
(ammonium salts)
NO2−
(nitrite salts)
HNO3
(nitric acid)
NO3−
(nitrate salts)
▲ Figure 22.22 Sequence of conversion of N2 into common nitrogen compounds.
Go Figure
Is the N ¬ N bond length in these
molecules shorter or longer than the
N ¬ N bond length in N2?
In the laboratory, NH3 can be prepared by the action of NaOH on an ammonium
salt. The NH4+ ion, which is the conjugate acid of NH3, transfers a proton to OH - . The
resultant NH3 is volatile and is driven from the solution by mild heating:
NH4Cl1aq2 + NaOH1aq2 ¡ NH31g2 + H2O1l2 + NaCl1aq2[22.37]
Commercial production of NH3 is achieved by the Haber process:
N21g2 + 3 H21g2 ¡ 2 NH31g2[22.38]
About 1 * 1010 kg (10 million tons) of ammonia is produced annually in the United
States. About 75% is used for fertilizer.
Hydrazine 1N2H42 is another important hydride of nitrogen. The hydrazine molecule contains an N ¬ N single bond (◀ Figure 22.23). Hydrazine is quite poisonous. It
can be prepared by the reaction of ammonia with hypochlorite ion 1OCl - 2 in aqueous
solution:
▲ Figure 22.23 Hydrazine (top, N2H4) and
methylhydrazine (bottom, CH3NHNH2).
2 NH31aq2 + OCl - 1aq2 ¡ N2H41aq2 + Cl - 1aq2 + H2O1l2[22.39]
The reaction involves several intermediates, including chloramine 1NH2Cl2. The poisonous NH2Cl bubbles out of solution when household ammonia and chlorine bleach
(which contains OCl - ) are mixed. This reaction is one reason for the frequently cited
warning not to mix bleach and household ammonia.
Pure hydrazine is a strong and versatile reducing agent. The major use of hydrazine and compounds related to it, such as methylhydrazine (Figure 22.23), is as
rocket fuel.
Sample
Exercise 22.6 Writing a Balanced Equation
Hydroxylamine 1NH2OH2 reduces copper(II) to the free metal in acid solutions. Write a
balanced equation for the reaction, assuming that N2 is the oxidation product.
Solution
Analyze We are asked to write a balanced oxidation–reduction equation in which NH2OH is
converted to N2 and Cu2 + is converted to Cu.
Plan Because this is a redox reaction, the equation can be balanced by the method of half-
reactions discussed in Section 20.2. Thus, we begin with two half-reactions, one involving the
NH2OH and N2 and the other involving Cu2 + and Cu.
Solve The unbalanced and incomplete
half-reactions are
Cu2 + 1aq2 ¡ Cu1s2
NH2OH1aq2 ¡ N21g2
Cu2 + 1aq2 + 2 e - ¡ Cu1s2
section 22.7 Nitrogen
Balancing these equations as
described in Section 20.2 gives
Adding these half-reactions gives
the balanced equation:
2 NH2OH1aq2 ¡ N21g2 + 2 H2O1l2 + 2 H+1aq2 + 2 e -
Cu2 + 1aq2 + 2 NH2OH1aq2 ¡ Cu1s2 + N21g2 + 2 H2O1l2 + 2 H+1aq2
Practice Exercise 1
In power plants, hydrazine is used to prevent corrosion of the metal parts of steam boilers by
the O2 dissolved in the water. The hydrazine reacts with O2 in water to give N2 and H2O.
Write a balanced equation for this reaction.
Practice Exercise 2
Methylhydrazine, N2H3CH31l2, is used with the oxidizer dinitrogen tetroxide, N2O41l2, to
power the steering rockets of the Space Shuttle orbiter. The reaction of these two substances
produces N2, CO2, and H2O. Write a balanced equation for this reaction.
Oxides and Oxyacids of Nitrogen
Nitrogen forms three common oxides: N2O (nitrous oxide), NO (nitric oxide), and
NO2 (nitrogen dioxide). It also forms two unstable oxides that we will not discuss,
N2O3 (dinitrogen trioxide) and N2O5 (dinitrogen pentoxide).
Nitrous oxide 1N2O2 is also known as laughing gas because a person becomes
giddy after inhaling a small amount. This colorless gas was the first substance used as
a general anesthetic. It is used as the compressed gas propellant in several aerosols and
foams, such as in whipped cream. It can be prepared in the laboratory by carefully heating ammonium nitrate to about 200 °C:
∆
NH4NO31s2 ¡ N2O1g2 + 2 H2O1g2[22.40]
Nitric oxide (NO) is also a colorless gas but, unlike N2O, it is slightly toxic. It can
be prepared in the laboratory by reduction of dilute nitric acid, using copper or iron as
a reducing agent:
3 Cu1s2 + 2 NO3- 1aq2 + 8 H+ 1aq2 ¡ 3 Cu2 + 1aq2 + 2 NO1g2 + 4 H2O1l2[22.41]
Nitric oxide is also produced by direct reaction of N2 and O2 at high temperatures. This
(Section 18.2) The
reaction is a significant source of nitrogen oxide air pollutants.
direct combination of N2 and O2 is not used for commercial production of NO, however, because the yield is low, the equilibrium constant Kp at 2400 K being only 0.05.
(Section 15.7, “Chemistry Put to Work: Controlling Nitric Oxide Emissions”)
The commercial route to NO (and hence to other oxygen-containing compounds
of nitrogen) is via the catalytic oxidation of NH3:
Pt catalyst
4 NH31g2 + 5 O21g2 ¡
4 NO1g2 + 6 H2O1g2[22.42]
850 °C
This reaction is the first step in the Ostwald process, by which NH3 is converted commercially into nitric acid 1HNO32.
When exposed to air, nitric oxide reacts readily with O2 (▼ Figure 22.24):
2 NO1g2 + O21g2 ¡ 2 NO21g2[22.43]
NO2(g)
NO(g)
O2 in air
▲ Figure 22.24 Formation of NO2(g) as NO(g) combines with O2(g) in the air.
975
976
chapter 22 Chemistry of the Nonmetals
When dissolved in water, NO2 forms nitric acid:
Go Figure
Which is the shortest NO bond in
these two molecules?
O
H
H
O
O
N
N
O
O
▲ Figure 22.25 Structures of nitric acid
(top) and nitrous acid (bottom).
3 NO21g2 + H2O1l2 ¡ 2 H+ 1aq2 + 2 NO3- 1aq2 + NO1g2[22.44]
Nitrogen is both oxidized and reduced in this reaction, which means it disproportionates. The NO can be converted back into NO2 by exposure to air (Equation 22.43) and
thereafter dissolved in water to prepare more HNO3.
NO is an important neurotransmitter in the human body. It causes the muscles
that line blood vessels to relax, thus allowing an increased passage of blood (see the
“Chemistry and Life”).
Nitrogen dioxide 1NO22 is a yellow-brown gas (Figure 22.24). Like NO, it is a major
(Section 18.2) It is poisonous and has a choking odor. As disconstituent of smog.
cussed in Section 15.1, NO2 and N2O4 exist in equilibrium:
2 NO21g2 ∆ N2O41g2 ∆H ° = -58 kJ[22.45]
The two common oxyacids of nitrogen are nitric acid 1HNO32 and nitrous acid
1HNO22 (◀ Figure 22.25). Nitric acid is a strong acid. It is also a powerful oxidizing
agent, as indicated by the standard reduction potential in the reaction
NO3- 1aq2 + 4 H+ 1aq2 + 3 e - ¡ NO1g2 + 2 H2O1l2 E° = +0.96 V[22.46]
Concentrated nitric acid attacks and oxidizes most metals except Au, Pt, Rh, and Ir.
About 7 * 109 kg (8 million tons) of nitric acid is produced annually in the United
States. Its largest use is in the manufacture of NH4NO3 for fertilizers. It is also used
in the production of plastics, drugs, and explosives. Among the explosives made from
nitric acid are nitroglycerin, trinitrotoluene (TNT), and nitrocellulose. The following
reaction occurs when nitroglycerin explodes:
4 C3H5N3O91l2 ¡ 6 N21g2 + 12 CO21g2 + 10 H2O1g2 + O21g2[22.47]
All the products of this reaction contain very strong bonds and are gases. As a result,
the reaction is very exothermic, and the volume of the products is far larger than the
volume occupied by the reactant. Thus, the expansion resulting from the heat gener (Section 8.8, “Explosives and Alfred
ated by the reaction produces the explosion.
Nobel”)
Nitrous acid is considerably less stable than HNO3 and tends to disproportionate into NO and HNO3. It is normally made by action of a strong acid, such as
H2SO4, on a cold solution of a nitrite salt, such as NaNO2. Nitrous acid is a weak acid
1Ka = 4.5 * 10 - 42.
Give It Some Thought
What are the oxidation numbers of the nitrogen atoms in
(a) nitric acid
(b) nitrous acid
Chemistry and Life
Nitroglycerin, Nitric Oxide,
and Heart Disease
During the 1870s, an interesting observation was made in Alfred
Nobel’s dynamite factories. Workers who suffered from heart disease
that caused chest pains when they exerted themselves found relief
from the pains during the workweek. It quickly became apparent that
nitroglycerin, present in the air of the factory, acted to enlarge blood
vessels. Thus, this powerfully explosive chemical became a standard treatment for angina pectoris, the chest pains accompanying
heart failure. It took more than 100 yrs to discover that nitroglycerin was converted in the vascular smooth muscle into NO, which
is the chemical agent actually causing dilation of the blood vessels.
In 1998, the Nobel Prize in Physiology or Medicine was awarded
to Robert F. Furchgott, Louis J. Ignarro, and Ferid Murad for their
discoveries of the detailed pathways by which NO acts in the cardiovascular system. It was a sensation that this simple, common air
pollutant could exert important functions in mammals, including
humans.
As useful as nitroglycerin is to this day in treating angina pectoris, it has a limitation in that prolonged administration results in
development of tolerance, or desensitization, of the vascular muscle
to further vasorelaxation by nitroglycerin. The bioactivation of nitroglycerin is the subject of active research in the hope that a means of
circumventing desensitization can be found.
977
section 22.8 The Other Group 5A Elements: P, As, Sb, and Bi
22.8 | The Other Group 5A Elements:
P, As, Sb, and Bi
Of the other group 5A elements—phosphorus, arsenic, antimony, and bismuth—
phosphorus has a central role in several aspects of biochemistry and environmental
chemistry.
5A
7
N
General Characteristics of the Group 5A Elements
15
P
The group 5A elements have the outer-shell electron configuration ns2np3, with n ranging from 2 to 6. A noble-gas configuration is achieved by adding three electrons to form
the -3 oxidation state. Ionic compounds containing X3 - ions are not common, however. More commonly, the group 5A element acquires an octet of electrons via covalent
bonding and oxidation numbers ranging from -3 to +5.
Because of its lower electronegativity, phosphorus is found more frequently in
positive oxidation states than is nitrogen. Furthermore, compounds in which phosphorus has the +5 oxidation state are not as strongly oxidizing as the corresponding
compounds of nitrogen. Compounds in which phosphorus has a -3 oxidation state are
much stronger reducing agents than are the corresponding nitrogen compounds.
Some properties of the group 5A elements are listed in ▼ Table 22.7. The general
pattern is similar to what we saw with other groups: Size and metallic character increase
as atomic number increases in the group.
The variation in properties among group 5A elements is more striking than that
seen in groups 6A and 7A. Nitrogen at the one extreme exists as a gaseous diatomic
molecule, clearly nonmetallic. At the other extreme, bismuth is a reddish white, metalliclooking substance that has most of the characteristics of a metal.
The values listed for X ¬ X single-bond enthalpies are not reliable because it is difficult to obtain such data from thermochemical experiments. However, there is no doubt
about the general trend: a low value for the N ¬ N single bond, an increase at phosphorus,
and then a gradual decline to arsenic and antimony. From observations of the elements in
the gas phase, it is possible to estimate the X ‚ X triple-bond enthalpies. Here, we see a
trend that is different from that for the X ¬ X single bond. Nitrogen forms a much stronger triple bond than the other elements, and there is a steady decline in the triple-bond
enthalpy down through the group. These data help us to appreciate why nitrogen alone
of the group 5A elements exists as a diatomic molecule in its stable state at 25 °C. All the
other elements exist in structural forms with single bonds between the atoms.
Occurrence, Isolation, and Properties of Phosphorus
Phosphorus occurs mainly in the form of phosphate minerals. The principal source of
phosphorus is phosphate rock, which contains phosphate principally as Ca31PO422.
The element is produced commercially by the reduction of calcium phosphate with
carbon in the presence of SiO2:
1500 °C
2 Ca31PO4221s2 + 6 SiO21s2 + 10 C1s2 ¡ P41g2 + 6 CaSiO31l2 + 10 CO1g2
[22.48]
Table 22.7 Properties of the Group 5A Elements
Property
N
P
As
Sb
Bi
0.71
1.07
1.19
1.39
1.48
1402
1012
947
834
703
Electron affinity 1kJ>mol2
X ‚ X triple-bond enthalpy 1kJ>mol2
°
Atomic radius 1A2
First ionization energy 1kJ>mol2
7 0
- 72
- 78
- 103
- 91
Electronegativity
3.0
2.1
2.0
1.9
1.9
X ¬ X single-bond enthalpy 1kJ>mol2*
163
200
150
120
—
941
490
380
295
192
*Approximate values only.
33
As
51
Sb
83
Bi
978
chapter 22 Chemistry of the Nonmetals
The phosphorus produced in this fashion is the allotrope known as white phosphorus.
This form distills from the reaction mixture as the reaction proceeds.
Phosphorus exists in several allotropic forms. White phosphorus consists of P4 tetrahedra (◀ Figure 22.26). The bond angles in this molecule, 60°, are unusually small,
so there is much strain in the bonding, which is consistent with the high reactivity of
white phosphorus. This allotrope bursts spontaneously into flames if exposed to air.
When heated in the absence of air to about 400 °C, white phosphorus is converted to a
more stable allotrope known as red phosphorus, which does not ignite on contact with
air. Red phosphorus is also considerably less poisonous than the white form. We will
denote elemental phosphorus as simply P(s).
Phosphorus Halides
Phosphorus forms a wide range of compounds with the halogens, the most important
of which are the trihalides and pentahalides. Phosphorus trichloride 1PCl32 is commercially the most significant of these compounds and is used to prepare a wide
variety of products, including soaps, detergents, plastics, and insecticides.
Phosphorus chlorides, bromides, and iodides can be made by direct oxidation
of elemental phosphorus with the elemental halogen. PCl3, for example, which is
a liquid at room temperature, is made by passing a stream of dry chlorine gas over
white or red phosphorus:
White phosphorus
Red phosphorus
▲ Figure 22.26 White and red phosphorus.
Despite the fact that both contain only
phosphorus atoms, these two forms of
phosphorus differ greatly in reactivity. The
white allotrope, which reacts violently with
oxygen, must be stored under water so that it
is not exposed to air. The much less reactive
red form does not need to be stored this way.
2 P1s2 + 3 Cl21g2 ¡ 2 PCl31l2[22.49]
If excess chlorine gas is present, an equilibrium is established between PCl3 and PCl5.
PCl31l2 + Cl21g2 ∆ PCl51s2[22.50]
The phosphorus halides react readily with water, and most fume in air because of
reaction with water vapor. In the presence of excess water, the products are the corresponding phosphorus oxyacid and hydrogen halide:
PBr31l2 + 3 H2O1l2 ¡ H3PO31aq2 + 3 HBr1aq2[22.51]
PCl51l2 + 4 H2O1l2 ¡ H3PO41aq2 + 5 HCl1aq2[22.52]
Give It Some Thought
Which oxyacid is produced when PF3 reacts with water?
Oxy Compounds of Phosphorus
Probably the most significant phosphorus compounds are those in which the element is combined with oxygen. Phosphorus(III) oxide 1P4O62 is obtained by allowing white phosphorus to oxidize in a limited supply of oxygen. When oxidation takes
place in the presence of excess oxygen, phosphorus(V) oxide 1P4O102 forms. This
compound is also readily formed by oxidation of P4O6. These two oxides represent
the two most common oxidation states for phosphorus, +3 and +5. The structural
relationship between P4O6 and P4O10 is shown in ▶ Figure 22.27. Notice the resemblance these molecules have to the P4 molecule (Figure 22.27); all three substances
have a P4 core.
Phosphorus(V) oxide is the anhydride of phosphoric acid 1H3PO42, a weak triprotic acid. In fact, P4O10 has a very high affinity for water and is consequently used as
a drying agent. Phosphorus(III) oxide is the anhydride of phosphorous acid 1H3PO32, a
weak diprotic acid (▶ Figure 22.28).*
*Note that the element phosphorus 1FOS # for # us2 has a -us suffix, whereas the first word in the name
phosphorous 1fos # FOR # us2 acid has an -ous suffix.
section 22.8 The Other Group 5A Elements: P, As, Sb, and Bi
One characteristic of phosphoric and phosphorous acids is their tendency to
(Section 12.8) For example, two
undergo condensation reactions when heated.
H3PO4 molecules are joined by the elimination of one H2O molecule to form H4P2O7:
O
HO
O
OH + HO
P
O
P
OH
OH
HO
OH
O
O
P
P
OH
OH + H2O
979
Go Figure
How do the electron domains about P in
P4O6 differ from those about P in P4O10?
O
P
OH
[22.53]
Phosphoric acid and its salts find their most important uses in detergents and fertilizers.
The phosphates in detergents are often in the form of sodium tripolyphosphate 1Na5P3O102.
The phosphate ions “soften” water by binding their oxygen groups to the metal
ions that contribute to the hardness of water. This keeps the metal ions from interfering with the action of the detergent. The phosphates also keep the pH above 7 and thus
prevent the detergent molecules from becoming protonated.
Most mined phosphate rock is converted to fertilizers. The Ca31PO422 in phosphate rock is insoluble 1Ksp = 2.0 * 10 - 292. It is converted to a soluble form for use in
fertilizers by treatment with sulfuric or phosphoric acid. The reaction with phosphoric
acid yields Ca1H2PO422:
These atoms are eliminated as H2O
Ca31PO4221s2 + 4 H3PO41aq2 ¡ 3 Ca2+ 1aq2 + 6 H2PO4- 1aq2[22.54]
Although the solubility of Ca1H2PO422 allows it to be assimilated by plants, it also
allows it to be washed from the soil and into bodies of water, thereby contributing to
(Section 18.4)
water pollution.
Phosphorus compounds are important in biological systems. The element occurs in
phosphate groups in RNA and DNA, the molecules responsible for the control of protein
biosynthesis and transmission of genetic information. It also occurs in adenosine triphosphate (ATP), which stores energy in biological cells and has the following structure:
▲ Figure 22.27 Structures of P4O6 (top) and
P4O10 (bottom).
NH2
N
HC
N
O
O
−O
P
O
O−
C
C
C
N
N
CH
This H not acidic
because P–H bond
is nonpolar
O
O
P
O
P
O−
O−
CH2
O
C H
H C
H C
C H
OH
OH
Adenosine
The P ¬ O ¬ P bond of the end phosphate group is broken by hydrolysis with
water, forming adenosine diphosphate (ADP):
O
O
O
−O
P
O−
O
P
O
P
O−
Adenosine + H2O
O
O−
ATP
O
O
HO
P
O
O−
P
O−
ADP
[22.55]
O
O
Adenosine +
−O
P
O−
OH
▲ Figure 22.28 Structures of H3PO4 (top)
and H3PO3 (bottom).
980
chapter 22 Chemistry of the Nonmetals
This reaction releases 33 kJ of energy under standard conditions, but in the living
cell, the Gibbs free energy change for the reaction is closer to -57 kJ>mol. The concentration of ATP inside a living cell is in the range of 1–10 mM, which means a typical
human metabolizes her or his body mass of ATP in one day! ATP is continually made
from ADP and continually converted back to ADP, releasing energy that can be harnessed by other cellular reactions.
Chemistry and Life
Arsenic in Drinking Water
Arsenic, in the form of its oxides, has been known as a poison for centuries. The current Environmental Protection Agency (EPA) standard for
arsenic in public water supplies is 10 ppb (equivalent to 10 mg>L). Most
regions of the United States tend to have low to moderate (2–10 ppb)
groundwater arsenic levels (▼ Figure 22.29). The western region
tends to have higher levels, coming mainly from natural geological
sources in the area. Estimates, for example, indicate that 35% of watersupply wells in Arizona have arsenic concentrations above 10 ppb.
The problem of arsenic in drinking water in the United States
is dwarfed by the problem in other parts of the world—especially in
Bangladesh, where the problem is tragic. Historically, surface water
Alaska
Hawaii
EXPLANATION
At least 25% of samples exceed 50 ppm As
At least 25% of samples exceed 10 ppm As
At least 25% of samples exceed 5 ppm As
At least 25% of samples exceed 3 ppm As
At least 25% of samples exceed 1 ppm As
Insufficient data
Puerto
Rico
▲ Figure 22.29 Geographic distribution of arsenic in groundwater.
sources in that country have been contaminated with microorganisms, causing significant health problems. During the
1970s, international agencies, headed by the United Nations
Children’s Fund (UNICEF), began investing millions of dollars of aid money in Bangladesh for wells to provide “clean”
drinking water. Unfortunately, no one tested the well water
for the presence of arsenic; the problem was not discovered
until the 1980s. The result has been the biggest outbreak of
mass poisoning in history. Up to half of the country’s estimated 10 million wells have arsenic concentrations above
50 ppb.
In water, the most common forms of arsenic are
the arsenate ion and its protonated hydrogen anions
(AsO43 - , HAsO42 - , and H2AsO4- ) and the arsenite ion and
its protonated forms 1AsO33 - , HAsO32 - , H2AsO3- , and
H3AsO32. These species are collectively referred to by the
oxidation number of the arsenic as arsenic(V) and arsenic(III),
respectively. Arsenic(V) is more prevalent in oxygen-rich
(aerobic) surface waters, whereas arsenic(III) is more likely to
occur in oxygen-poor (anaerobic) groundwaters.
One of the challenges in determining the health effects
of arsenic in drinking waters is the different chemistry of
arsenic(V) and arsenic(III), as well as the different concentrations required for physiological responses in different individuals. In Bangladesh, skin lesions were the first sign of
the arsenic problem. Statistical studies correlating arsenic
levels with the occurrence of disease indicate a lung and bladder cancer risk arising from even low levels of arsenic.
The current technologies for removing arsenic perform most effectively when treating arsenic in the form of
arsenic(V), so water treatment strategies require preoxidation of the drinking water. Once in the form of arsenic(V),
there are a number of possible removal strategies. For example, Fe3 + can be added to precipitate FeAsO4, which is then
removed by filtration.
22.9 | Carbon
Carbon constitutes only 0.027% of Earth’s crust. Although some carbon occurs in
elemental form as graphite and diamond, most is found in combined form. Over half
occurs in carbonate compounds, and carbon is also found in coal, petroleum, and natural gas. The importance of the element stems in large part from its occurrence in all
living organisms: Life as we know it is based on carbon compounds.
Elemental Forms of Carbon
We have seen that carbon exists in several allotropic crystalline forms: graphite, diamond, fullerenes, carbon nanotubes, and graphene. Fullerenes, nanotubes, and graphene
are discussed in Chapter 12; here we focus on graphite and diamond.
section 22.9 Carbon
Graphite is a soft, black, slippery solid that has a metallic luster and conducts electricity. It consists of parallel sheets of sp2@hybridized carbon atoms held together by
(Section 12.7) Diamond is a clear, hard solid in which the carbon
dispersion forces.
3
(Section 12.7) Diamond is denser
atoms form an sp @hybridized covalent network.
3
than graphite (d = 2.25 g>cm for graphite; d = 3.51 g>cm3 for diamond). At approximately 100,000 atm at 3000 °C, graphite converts to diamond. In fact, almost any
carbon-containing substance, if put under sufficiently high pressure, forms diamonds;
scientists at General Electric in the 1950s used peanut butter to make diamonds. About
3 * 104 kg of industrial-grade diamonds are synthesized each year, mainly for use in
cutting, grinding, and polishing tools.
Graphite has a well-defined crystalline structure, but it also exists in two common amorphous forms: carbon black and charcoal. Carbon black is formed when
hydrocarbons are heated in a very limited supply of oxygen, such as in this methane
reaction:
CH41g2 + O21g2 ¡ C1s2 + 2 H2O1g2[22.56]
Carbon black is used as a pigment in black inks; large amounts are also used in making
automobile tires.
Charcoal is formed when wood is heated strongly in the absence of air. Charcoal
has an open structure, giving it an enormous surface area per unit mass. “Activated
charcoal,” a pulverized form of charcoal whose surface is cleaned by heating with
steam, is widely used to adsorb molecules. It is used in filters to remove offensive odors
from air and colored or bad-tasting impurities from water.
Oxides of Carbon
Carbon forms two principal oxides: carbon monoxide (CO) and carbon dioxide 1CO22.
Carbon monoxide is formed when carbon or hydrocarbons are burned in a limited supply of oxygen:
2 C1s2 + O21g2 ¡ 2 CO1g2[22.57]
CO is a colorless, odorless, tasteless gas that is toxic because it binds to hemoglobin
in the blood and thus interferes with oxygen transport. Low-level poisoning results in
headaches and drowsiness; high-level poisoning can cause death.
Carbon monoxide is unusual in that it has a nonbonding pair of electrons on carbon: ∶C ‚ O∶. It is isoelectronic with N2, so you might expect CO to be equally unreactive. Moreover, both substances have high bond energies (1072 kJ>mol for C ‚ O
and 941 kJ>mol for N ‚ N). Because of the lower nuclear charge on carbon (compared
with either N or O), however, the carbon nonbonding pair is not held as strongly as
that on N or O. Consequently, CO is better able to function as a Lewis base than is
N2; for example, CO can coordinate its nonbonding pair to the iron of hemoglobin,
displacing O2, but N2 cannot. In addition, CO forms a variety of covalent compounds,
known as metal carbonyls, with transition metals. Ni1CO24, for example, is a volatile,
toxic compound formed by warming metallic nickel in the presence of CO. The formation of metal carbonyls is the first step in the transition-metal catalysis of a variety of
reactions of CO.
Carbon monoxide has several commercial uses. Because it burns readily, forming
CO2, it is employed as a fuel:
2 CO1g2 + O21g2 ¡ 2 CO21g2
∆H° = -566 kJ[22.58]
Carbon monoxide is an important reducing agent, widely used in metallurgical operations to reduce metal oxides, such as the iron oxides:
Fe3O41s2 + 4 CO1g2 ¡ 3 Fe1s2 + 4 CO21g2[22.59]
Carbon dioxide is produced when carbon-containing substances are burned in
excess oxygen, such as in this reaction involving ethanol:
C2H5OH1l2 + 3 O21g2 ¡ 2 CO21g2 + 3 H2O1g2[22.60]
981
982
chapter 22 Chemistry of the Nonmetals
Chemistry Put to Work
Carbon Fibers and Composites
The properties of graphite are anisotropic; that is, they differ in different directions through the solid. Along the carbon planes, graphite
possesses great strength because of the number and strength of the
carbon–carbon bonds in this direction. The bonds between planes are
relatively weak, however, making graphite weak in that direction.
Fibers of graphite can be prepared in which the carbon planes
are aligned to varying extents parallel to the fiber axis. These fibers are
lightweight (density of about 2 g>cm3) and chemically quite unreactive. The oriented fibers are made by first slowly pyrolyzing (decomposing by action of heat) organic fibers at about 150 to 300 °C. These
fibers are then heated to about 2500 °C to graphitize them (convert
amorphous carbon to graphite). Stretching the fiber during pyrolysis
helps orient the graphite planes parallel to the fiber axis. More amorphous carbon fibers are formed by pyrolysis of organic fibers at lower
temperatures (1200 to 400 °C). These amorphous materials, commonly called carbon fibers, are the type most often used in commercial
materials.
Composite materials that take advantage of the strength, stability, and low density of carbon fibers are widely used. Composites are combinations of two or more materials. These materials are
present as separate phases and are combined to form structures that
take advantage of certain desirable properties of each component. In
carbon composites, the graphite fibers are often woven into a fabric
that is embedded in a matrix that binds them into a solid structure.
Strong acid
CO2(g)
The fibers transmit loads evenly throughout the matrix. The finished
composite is thus stronger than any one of its components.
Carbon composite materials are used widely in a number of applications, including high-performance graphite sports equipment such
as tennis racquets, golf clubs, and bicycle wheels (▼ Figure 22.30).
Heat-resistant composites are required for many aerospace applications, where carbon composites now find wide use.
▲ Figure 22.30 Carbon composites in commercial products.
It is also produced when many carbonates are heated:
∆
CaCO31s2 ¡ CaO1s2 + CO21g2[22.61]
In the laboratory, CO2 can be produced by the action of acids on carbonates
(◀ Figure 22.31):
CaCO3
▲ Figure 22.31 CO2 formation from the
reaction between an acid and calcium
carbonate in rock.
CO32 - 1aq2 + 2 H+ 1aq2 ¡ CO21g2 + H2O1l2[22.62]
Carbon dioxide is a colorless, odorless gas. It is a minor component of Earth’s atmo (Section 18.2) Although it
sphere but a major contributor to the greenhouse effect.
is not toxic, high concentrations of CO2 increase respiration rate and can cause suffocation. It is readily liquefied by compression. When cooled at atmospheric pressure,
however, CO2 forms a solid rather than liquefying. The solid sublimes at atmospheric
pressure at -78 °C. This property makes solid CO2, known as dry ice, valuable as a
refrigerant. About half of the CO2 consumed annually is used for refrigeration. The
other major use of CO2 is in the production of carbonated beverages. Large quantities are also used to manufacture washing soda 1Na2CO3 # 10 H2O2, used to precipitate
metal ions that interfere with the cleansing action of soap, and baking soda 1NaHCO32.
Baking soda is so named because this reaction occurs during baking:
NaHCO31s2 + H+ 1aq2 ¡ Na+ 1aq2 + CO21g2 + H2O1l2[22.63]
The H+ 1aq2 is provided by vinegar, sour milk, or the hydrolysis of certain salts. The
bubbles of CO2 that form are trapped in the baking dough, causing it to rise.
Give It Some Thought
Yeast are living organisms that make bread rise in the absence of baking soda
and acid. What must the yeast be producing to make bread rise?
section 22.9 Carbon
Carbonic Acid and Carbonates
Carbon dioxide is moderately soluble in H2O at atmospheric pressure. The resulting
solution is moderately acidic because of the formation of carbonic acid 1H2CO32:
CO21aq2 + H2O1l2 ∆ H2CO31aq2[22.64]
Carbonic acid is a weak diprotic acid. Its acidic character causes carbonated beverages
to have a sharp, slightly acidic taste.
Although carbonic acid cannot be isolated, hydrogen carbonates (bicarbonates)
and carbonates can be obtained by neutralizing carbonic acid solutions. Partial neutralization produces HCO3- , and complete neutralization gives CO32 - . The HCO3- ion is
a stronger base than acid 1Kb = 2.3 * 10 - 8; Ka = 5.6 * 10 - 112. The carbonate ion is
much more strongly basic 1Kb = 1.8 * 10 - 42.
The principal carbonate minerals are calcite 1CaCO32, magnesite 1MgCO32, dolomite 3MgCa1CO3224, and siderite 1FeCO32. Calcite is the principal mineral in limestone and the main constituent of marble, chalk, pearls, coral reefs, and the shells of
marine animals such as clams and oysters. Although CaCO3 has low solubility in pure
water, it dissolves readily in acidic solutions with evolution of CO2:
CaCO31s2 + 2 H+ 1aq2 ∆ Ca2 + 1aq2 + H2O1l2 + CO21g2[22.65]
Because water containing CO2 is slightly acidic (Equation 22.64), CaCO3 dissolves
slowly in this medium:
CaCO31s2 + H2O1l2 + CO21g2 ¡ Ca2 + 1aq2 + 2 HCO3- 1aq2[22.66]
This reaction occurs when surface waters move underground through limestone
deposits. It is the principal way Ca2 + enters groundwater, producing “hard water.” If
the limestone deposit is deep enough underground, dissolution of the limestone produces a cave.
One of the most important reactions of CaCO3 is its decomposition into CaO and
CO2 at elevated temperatures (Equation 22.61). About 2 * 1010 kg (20 million tons) of
calcium oxide, known as lime or quicklime, is produced in the United States annually.
Because calcium oxide reacts with water to form Ca1OH22, it is an important commercial base. It is also important in making mortar, the mixture of sand, water, and CaO
used to bind bricks, blocks, or rocks together. Calcium oxide reacts with water and CO2
to form CaCO3, which binds the sand in the mortar:
CaO1s2 + H2O1l2 ¡ Ca2 + 1aq2 + 2 OH - 1aq2[22.67]
Ca2 + 1aq2 + 2 OH - 1aq2 + CO21aq2 ¡ CaCO31s2 + H2O1l2[22.68]
Carbides
The binary compounds of carbon with metals, metalloids, and certain nonmetals are
called carbides. The more active metals form ionic carbides, and the most common of
these contain the acetylide ion 1C22 - 2. This ion is isoelectronic with N2, and its Lewis
structure, 3:C ‚ C:42 - , has a carbon–carbon triple bond. The most important ionic
carbide is calcium carbide 1CaC22, produced by the reduction of CaO with carbon at
high temperature:
2 CaO1s2 + 5 C1s2 ¡ 2 CaC21s2 + CO21g2[22.69]
The carbide ion is a very strong base that reacts with water to form acetylene
1H ¬ C ‚ C ¬ H2:
CaC21s2 + 2 H2O1l2 ¡ Ca1OH221aq2 + C2H21g2[22.70]
Calcium carbide is therefore a convenient solid source of acetylene, which is used in
welding (Figure 22.13).
Interstitial carbides are formed by many transition metals. The carbon atoms
occupy open spaces (interstices) between the metal atoms in a manner analogous to the
(Section 22.2) This process generally makes the metal harder.
interstitial hydrides.
983
984
chapter 22 Chemistry of the Nonmetals
Tungsten carbide (WC), for example, is very hard and very heat-resistant and, thus,
used to make cutting tools.
Covalent carbides are formed by boron and silicon. Silicon carbide (SiC), known as
Carborundum®, is used as an abrasive and in cutting tools. Almost as hard as diamond,
SiC has a diamond-like structure with alternating Si and C atoms.
22.10 | The Other Group 4A Elements:
Si, Ge, Sn, and Pb
4A
6
C
14
Si
32
Ge
50
Sn
82
Pb
The trend from nonmetallic to metallic character as we go down a family is strikingly
evident in group 4A. Carbon is a nonmetal; silicon and germanium are metalloids; tin
and lead are metals. In this section, we consider a few general characteristics of group
4A and then look more thoroughly at silicon.
General Characteristics of the Group 4A Elements
The group 4A elements possess the outer-shell electron configuration ns2np2. The electronegativities of the elements are generally low (▼ Table 22.8); carbides that formally
contain C4 - ions are observed only in the case of a few compounds of carbon with very
active metals. Formation of 4+ ions by electron loss is not observed for any of these elements; the ionization energies are too high. The +4 oxidation state is common, however, and is found in the vast majority of the compounds of the group 4A elements. The
+2 oxidation state is found in the chemistry of germanium, tin, and lead, however, and
it is the principal oxidation state for lead. Carbon, except in highly unusual examples,
forms a maximum of four bonds. The other members of the family are able to form
(Section 8.7)
more than four bonds. Table 22.8 shows that the strength of a bond between two atoms of a given element decreases as we go down group 4A. Carbon–carbon bonds are quite strong.
Carbon, therefore, has a striking ability to form compounds in which carbon atoms
are bonded to one another in extended chains and rings, which accounts for the large
number of organic compounds that exist. Other elements can form chains and rings,
but these bonds are far less important in the chemistries of these other elements. The
Si ¬ Si bond strength 1226 kJ>mol2, for example, is much lower than the Si ¬ O bond
strength 1386 kJ>mol2. As a result, the chemistry of silicon is dominated by the formation of Si ¬ O bonds, and Si ¬ Si bonds play a minor role.
Occurrence and Preparation of Silicon
Silicon is the second most abundant element, after oxygen, in Earth’s crust. It occurs
in SiO2 and in an enormous variety of silicate minerals. The element is obtained by the
reduction of molten silicon dioxide with carbon at high temperature:
SiO21l2 + 2 C1s2 ¡ Si1l2 + 2 CO1g2[22.71]
Elemental silicon has a diamond-like structure. Crystalline silicon is a gray metallic-looking solid that melts at 1410 °C. The element is a semiconductor, as we saw in
Chapters 7 and 12, and is used to make solar cells and transistors for computer chips.
To be used as a semiconductor, it must be extremely pure, possessing less than 10 - 7%
Table 22.8 Some Properties of the Group 4A Elements
Property
C
Si
Ge
Sn
Pb
Atomic radius 1A° 2
0.76
1.11
1.20
1.39
1.46
1086
786
762
709
716
Electronegativity
2.5
1.8
1.8
1.8
1.9
X ¬ X single-bond enthalpy 1kJ>mol2
348
226
188
151
—
First ionization energy 1kJ>mol2
section 22.10 The Other Group 4A Elements: Si, Ge, Sn, and Pb
(1 ppb) impurities. One method of purification is to treat the element with Cl2 to form
SiCl4, a volatile liquid that is purified by fractional distillation and then converted back
to elemental silicon by reduction with H2:
SiCl41g2 + 2 H21g2 ¡ Si1s2 + 4 HCl1g2[22.72]
The process known as zone refining can further purify the element (▶ Figure 22.30). As
a heated coil is passed slowly along a silicon rod, a narrow band of the element is melted.
As the molten section is swept slowly along the length of the rod, the impurities concentrate in this section, following it to the end of the rod. The purified top portion of the rod
crystallizes as 99.999999999% pure silicon.
Go Figure
What limits the range of temperatures
you can use for zone refining of
silicon?
Silicates
Silicon dioxide and other compounds that contain silicon and oxygen make up over
90% of Earth’s crust. In silicates, a silicon atom is surrounded by four oxygens and
silicon is found in its most common oxidation state, +4. The orthosilicate ion, SiO44 - ,
is found in very few silicate minerals, but we can view it as the “building block” for
many mineral structures. As ▼ Figure 22.33 shows, adjacent tetrahedra are linked by
a common oxygen atom. Two tetrahedra joined in this way, called the disilicate ion,
contain two Si atoms and seven O atoms. Silicon and oxygen are in the +4 and -2
oxidation states, respectively, in all silicates, so the overall charge of any silicate ion
must be consistent with these oxidation states. For example, the charge on Si2O7 is
1221+42 + 1721-22 = -6; it is the Si2O76 - ion.
In most silicate minerals, silicate tetrahedra are linked together to form chains,
sheets, or three-dimensional structures. We can connect two vertices of each
tetrahedron to two other tetrahedra, for example, leading to an infinite chain with
an g O ¬ Si ¬ O ¬ Si g backbone as shown in Figure 22.33(b). Notice that each
silicon in this structure has two unshared (terminal) oxygens and two shared (bridging)
oxygens. The stoichiometry is then 2112 + 211>22 = 3 oxygens per silicon. Thus, the
formula unit for this chain is SiO32 - . The mineral enstatite 1MgSiO32 has this kind of
structure, consisting of rows of single-strand silicate chains with Mg2 + ions between
the strands to balance charge.
In Figure 22.33(c), each silicate tetrahedron is linked to three others, forming an
infinite sheet structure. Each silicon in this structure has one unshared oxygen and
three shared oxygens. The stoichiometry is then 1112 + 31½2 = 2½ oxygens per
silicon. The simplest formula of this sheet is Si2O52 - . The mineral talc, also known
as talcum powder, has the formula Mg31Si2O5221OH22 and is based on this sheet
Molten section
As heating coil
slowly moves down,
impurities concentrate in molten
section, leaving
behind ultrapure Si
Silicon rod
Inert atmosphere
▲ Figure 22.32 Zone-refining apparatus for
the production of ultrapure silicon.
Si2O52−
formula
unit
4−
O
O
Si
SiO32−
formula
unit
O
O
Silicate ion
Fragment of silicate chain
Fragment of silicate sheet
(a)
(b)
(c)
▲ Figure 22.33 Silicate chains and sheets.
985
986
chapter 22 Chemistry of the Nonmetals
▲ Figure 22.34 Serpentine asbestos.
structure. The Mg2 + and OH - ions lie between the silicate sheets. The slippery feel of
talcum powder is due to the silicate sheets sliding relative to one another.
Numerous minerals are based on silicates, and many are useful as clays, ceramics, and other materials. A few silicates have harmful effects on human health, the
best-known example being asbestos, a general term applied to a group of fibrous silicate minerals. The structure of these minerals is either chains of silicate tetrahedra
or sheets formed into rolls. The result is that the minerals have a fibrous character
(◀ Figure 22.34). Asbestos minerals were once widely used as thermal insulation,
especially in high-temperature applications, because of the great chemical stability
of the silicate structure. In addition, the fibers can be woven into asbestos cloth,
which was used for fireproof curtains and other applications. However, the fibrous
structure of asbestos minerals poses a health risk because the fibers readily penetrate
soft tissues, such as the lungs, where they can cause diseases, including cancer. The
use of asbestos as a common building material has therefore been discontinued.
When all four vertices of each SiO4 tetrahedron are linked to other tetrahedra,
the structure extends in three dimensions. This linking of the tetrahedra forms quartz
1SiO22. Because the structure is locked together in a three-dimensional array much like
(Section 12.7), quartz is harder than strand- or sheet-type silicates.
diamond
Sample
Exercise 22.7 Determining an Empirical Formula
The mineral chrysotile is a noncarcinogenic asbestos mineral that is based on the sheet structure shown in Figure 22.33(c). In addition to silicate tetrahedra, the mineral contains Mg2 + and
OH - ions. Analysis of the mineral shows that there are 1.5 Mg atoms per Si atom. What is the
empirical formula for chrysotile?
Solution
Analyze A mineral is described that has a sheet silicate structure with Mg2 + and OH - ions to balance
charge and 1.5 Mg for each 1 Si. We are asked to write the empirical formula for the mineral.
Plan As shown in Figure 22.33(c), the silicate sheet structure has the simplest formula Si2O52 - . We
first add Mg2 + to give the proper Mg:Si ratio. We then add OH - ions to obtain a neutral compound.
Solve The observation that the Mg:Si ratio equals 1.5 is consistent with three Mg2 + ions per
Si2O52 - unit. The addition of three Mg2 + ions would make Mg31Si2O524 + . In order to achieve
charge balance in the mineral, there must be four OH - per Si2O52 - . Thus, the formula of chrysotile
is Mg31Si2O521OH24. Since this is not reducible to a simpler formula, this is the empirical formula.
Practice Exercise 1
In the mineral beryl, six silicate tetrahedra are connected to form a
ring as shown here. The negative charge of this polyanion is balanced
by Be2 + and Al3 + cations. If elemental analysis gives a Be:Si ratio of
1:2 and an Al:Si ratio of 1:3 what is the empirical formula of beryl: (a)
Be2Al3Si6O19, (b) Be3Al2 (SiO4)6, (c) Be3Al2Si6O18, (d) BeAl2Si6O15?
Practice Exercise 2
The cyclosilicate ion consists of three silicate tetrahedra linked
together in a ring. The ion contains three Si atoms and nine O
atoms. What is the overall charge on the ion?
Beryl
Glass
Quartz melts at approximately 1600 °C, forming a tacky liquid. In the course of melting, many silicon–oxygen bonds are broken. When the liquid cools rapidly, silicon–
oxygen bonds are re-formed before the atoms are able to arrange themselves in a regular
fashion. An amorphous solid, known as quartz glass or silica glass, results. Many substances can be added to SiO2 to cause it to melt at a lower temperature. The common
glass used in windows and bottles, known as soda-lime glass, contains CaO and Na2O
section 22.11 Boron
987
in addition to SiO2 from sand. The CaO and Na2O are produced by heating two inexpensive chemicals, limestone 1CaCO32 and soda ash 1Na2CO32, which decompose at
high temperatures:
CaCO31s2 ¡ CaO1s2 + CO21g2[22.73]
Na2CO31s2 ¡ Na2O1s2 + CO21g2[22.74]
Other substances can be added to soda-lime glass to produce color or to change the
properties of the glass in various ways. The addition of CoO, for example, produces the
deep blue color of “cobalt glass.” Replacing Na2O with K2O results in a harder glass that
has a higher melting point. Replacing CaO with PbO results in a denser “lead crystal”
glass with a higher refractive index. Lead crystal is used for decorative glassware; the
higher refractive index gives this glass a particularly sparkling appearance. Addition
of nonmetal oxides, such as B2O3 and P4O10, which form network structures related
to the silicates, also changes the properties of the glass. Adding B2O3 creates a “borosilicate” glass with a higher melting point and a greater ability to withstand temperature changes. Such glasses, sold commercially under trade names such as Pyrex® and
Kimax®, are used where resistance to thermal shock is important, such as in laboratory
glassware or coffeemakers.
Silicones
Silicones consist of O ¬ Si ¬ O chains in which the two remaining bonding positions
on each silicon are occupied by organic groups such as CH3:
H3C
...
H 3C
CH3
Si
O
CH3
H3C
Si
O
CH3
...
Si
O
O
Depending on chain length and degree of cross-linking, silicones can be either
oils or rubber-like materials. Silicones are nontoxic and have good stability toward
heat, light, oxygen, and water. They are used commercially in a wide variety of products, including lubricants, car polishes, sealants, and gaskets. They are also used for
waterproofing fabrics. When applied to a fabric, the oxygen atoms form hydrogen
bonds with the molecules on the surface of the fabric. The hydrophobic (waterrepelling) organic groups of the silicone are then left pointing away from the surface
as a barrier.
Give It Some Thought
Distinguish among the substances silicon, silica, and silicone.
22.11 | Boron
Boron is the only group 3A element that can be considered nonmetallic and thus
is our final element in this chapter. The element has an extended network structure with a melting point 12300 °C2 that is intermediate between the melting points
of carbon 13550 °C2 and silicon 11410 °C2. The electron configuration of boron is
3He42s22p1.
In the family of compounds called boranes, the molecules contain only boron
and hydrogen. The simplest borane is BH3. This molecule contains only six valence
electrons and is therefore an exception to the octet rule. As a result, BH3 reacts with
itself to form diborane 1B2H62. This reaction can be viewed as a Lewis acid–base reaction in which one B ¬ H bonding pair of electrons in each BH3 molecule is donated
to the other. As a result, diborane is an unusual molecule in which hydrogen atoms
form a bridge between two B atoms (▶ Figure 22.35). Such hydrogens, called bridging
hydrogens, exhibit interesting chemical reactivity, which you may learn about in a more
advanced chemistry course.
3A
5
B
13
Al
31
Ga
49
In
81
Tl
▲ Figure 22.35 The structure of diborane
1B2H62.
988
chapter 22 Chemistry of the Nonmetals
Sharing hydrogen atoms between the two boron atoms compensates somewhat
for the deficiency in valence electrons around each boron. Nevertheless, diborane is
an extremely reactive molecule, spontaneously flammable in air in a highly exothermic
reaction:
B2H61g2 + 3 O21g2 ¡ B2O31s2 + 3 H2O1g2
∆H ° = -2030 kJ[22.75]
Boron and hydrogen form a series of anions called borane anions. Salts of the borohydride ion 1BH4- 2 are widely used as reducing agents. For example, sodium borohydride 1NaBH42 is a commonly used reducing agent for certain organic compounds.
Give It Some Thought
Recall that the hydride ion is H - . What is the oxidation state of boron in sodium
borohydride?
The only important oxide of boron is boric oxide 1B2O32. This substance is the
anhydride of boric acid, which we may write as H3BO3 or B1OH23. Boric acid is so weak
an acid 1Ka = 5.8 * 10 - 102 that solutions of H3BO3 are used as an eyewash. Upon
heating, boric acid loses water by a condensation reaction similar to that described for
phosphorus in Section 22.8:
4 H3BO31s2 ¡ H2B4O71s2 + 5 H2O1g2[22.76]
The diprotic acid H2B4O7 is tetraboric acid. The hydrated sodium salt Na2B4O7 #
10 H2O, called borax, occurs in dry lake deposits in California and can also be prepared from
other borate minerals. Solutions of borax are alkaline, and the substance is used in various
laundry and cleaning products.
Sample
Integrative Exercise Putting Concepts Together
The interhalogen compound BrF3 is a volatile, straw-colored liquid. The compound exhibits appreciable electrical conductivity because of autoionization (“solv” refers to BrF3 as the solvent):
2 BrF31l2 ∆ BrF2+1solv2 + BrF4- 1solv2
(a) What are the molecular structures of the BrF2+ and BrF4- ions?
(b) The electrical conductivity of BrF3 decreases with increasing temperature. Is the autoionization
process exothermic or endothermic?
(c) One chemical characteristic of BrF3 is that it acts as a Lewis acid toward fluoride ions. What do
we expect will happen when KBr is dissolved in BrF3?
Solution
(a) The BrF2+ ion has 7 + 2172 - 1 = 20 valence-shell electrons.
The Lewis structure for the ion is
F
Br
F
+
F
Because there are four electron domains around the central Br
atom, the resulting electron domain geometry is tetrahedral
(Section 9.2). Because bonding pairs of electrons occupy two
of these domains, the molecular geometry is bent:
+
−
F
+
F
F
Because there are six electron domains around the central Br atom
in this ion, the electron-domain geometry is octahedral. The two
nonbonding pairs of electrons are located opposite each other on
the octahedron, leading to a square-planar molecular geometry:
−
The BrF4- ion has 7 + 4172 + 1 = 36 electrons, leading to the
Lewis structure
Br
−
Chapter Summary and Key Terms
(b) The observation that conductivity decreases as temperature
increases indicates that there are fewer ions present in the solution at the higher temperature. Thus, increasing the temperature
causes the equilibrium to shift to the left. According to Le Châtelier’s principle, this shift indicates that the reaction is exothermic
as it proceeds from left to right.
(Section 15.7)
F− +
(c) A Lewis acid is an electron-pair acceptor.
(Section 16.11)
The fluoride ion has four valence-shell electron pairs and can act
as a Lewis base (an electron-pair donor). Thus, we can envision
the following reaction occurring:
F−
989
−
F
F
Br
F
+
BrF3
F
F
Br
F
F
BrF4−
Chapter Summary and Key Terms
The Periodic Table and Chemical Reactions (Section
22.1) The periodic table is useful for organizing and remembering
the descriptive chemistry of the elements. Among elements of a given
group, size increases with increasing atomic number, and electronegativity and ionization energy decrease. Nonmetallic character parallels
electronegativity, so the most nonmetallic elements are found in the
upper right portion of the periodic table.
Among the nonmetallic elements, the first member of each group
differs dramatically from the other members; it forms a maximum of
four bonds to other atoms and exhibits a much greater tendency to
form p bonds than the heavier elements in its group.
Because O2 and H2O are abundant in our world, we focus on
two important and general reaction types as we discuss the nonmetals: oxidation by O2 and proton-transfer reactions involving H2O or
aqueous solutions.
Hydrogen (Section 22.2) Hydrogen has three isotopes: protium
111H2, deuterium 121H2, and tritium 131H2. Hydrogen is not a member
of any particular periodic group, although it is usually placed above
lithium. The hydrogen atom can either lose an electron, forming H+,
or gain one, forming H - (the hydride ion). Because the H ¬ H bond
is relatively strong, H2 is fairly unreactive unless activated by heat or
a catalyst. Hydrogen forms a very strong bond to oxygen, so the reactions of H2 with oxygen-containing compounds usually lead to the
formation of H2O. Because the bonds in CO and CO2 are even stronger than the O ¬ H bond, the reaction of H2O with carbon or certain
organic compounds leads to the formation of H2. The H+1aq2 ion is
able to oxidize many metals, forming H21g2. The electrolysis of water
also forms H21g2.
The binary compounds of hydrogen are of three general types:
ionic hydrides (formed by active metals), metallic hydrides (formed by
transition metals), and molecular hydrides (formed by nonmetals). The
ionic hydrides contain the H - ion; because this ion is extremely basic,
ionic hydrides react with H2O to form H2 and OH - .
Group 8A: The Noble Gases and Group 7A: Halogens
(Sections 22.3 and 22.4) The noble gases (group 8A) exhibit a
very limited chemical reactivity because of the exceptional stability
of their electron configurations. The xenon fluorides and oxides and
KrF2 are the best-established compounds of the noble gases.
The halogens (group 7A) occur as diatomic molecules. All except
fluorine exhibit oxidation states varying from - 1 to + 7. Fluorine is
the most electronegative element, so it is restricted to the oxidation
states 0 and - 1. The oxidizing power of the element (the tendency
to form the - 1 oxidation state) decreases as we proceed down the
group.
The hydrogen halides are among the most useful compounds of
these elements; these gases dissolve in water to form the hydrohalic
acids, such as HCl(aq). Hydrofluoric acid reacts with silica . The
interhalogens are compounds formed between two different halogen
elements. Chlorine, bromine, and iodine form a series of oxyacids, in
which the halogen atom is in a positive oxidation state. These compounds and their associated oxyanions are strong oxidizing agents.
Oxygen and the Other Group 6A Elements (Sections
22.5 and 22.6) Oxygen has two allotropes, O2 and O3 (ozone).
Ozone is unstable compared to O2, and it is a stronger oxidizing agent
than O2. Most reactions of O2 lead to oxides, compounds in which
oxygen is in the -2 oxidation state. The soluble oxides of nonmetals generally produce acidic aqueous solutions; they are called acidic
anhydrides or acidic oxides. In contrast, soluble metal oxides produce
basic solutions and are called basic anhydrides or basic oxides. Many
metal oxides that are insoluble in water dissolve in acid, accompanied
by the formation of H2O. Peroxides contain O ¬ O bonds and oxygen
in the -1 oxidation state. Peroxides are unstable, decomposing to O2
and oxides. In such reactions peroxides are simultaneously oxidized
and reduced, a process called disproportionation. Superoxides contain
the O2- ion in which oxygen is in the - 12 oxidation state.
Sulfur is the most important of the other group 6A elements. It
has several allotropic forms; the most stable one at room temperature consists of S8 rings. Sulfur forms two oxides, SO2 and SO3, vand
both are important atmospheric pollutants. Sulfur trioxide is the
anhydride of sulfuric acid, the most important sulfur compound and
the most-produced industrial chemical. Sulfuric acid is a strong acid
and a good dehydrating agent. Sulfur forms several oxyanions as well,
including the SO32 - (sulfite), SO42 - (sulfate), and S2O32 - (thiosulfate) ions. Sulfur is found combined with many metals as a sulfide,
in which sulfur is in the -2 oxidation state. These compounds often
react with acids to form hydrogen sulfide 1H2S2, which smells like
rotten eggs.
Nitrogen and the Other Group 5A Elements (Sections
22.7 and 22.8) Nitrogen is found in the atmosphere as N2 mol-
ecules. Molecular nitrogen is chemically very stable because of the
strong N ‚ N bond. Molecular nitrogen can be converted into ammonia via the Haber process. Once the ammonia is made, it can be converted into a variety of different compounds that exhibit nitrogen
oxidation states ranging from - 3 to + 5. The most important industrial conversion of ammonia is the Ostwald process, in which ammonia
is oxidized to nitric acid 1HNO32.
Nitrogen has three important oxides: nitrous oxide 1N2O2, nitric
oxide (NO), and nitrogen dioxide 1NO22. Nitrous acid 1HNO22 is a
weak acid; its conjugate base is the nitrite ion 1NO2- 2. Another important nitrogen compound is hydrazine 1N2H42.
Phosphorus is the most important of the remaining group 5A
elements. It occurs in nature as phosphate minerals. Phosphorus has
990
chapter 22 Chemistry of the Nonmetals
several allotropes, including white phosphorus, which consists of
P4 tetrahedra. In reaction with the halogens, phosphorus forms trihalides PX3 and pentahalides PX5. These compounds undergo hydrolysis to produce an oxyacid of phosphorus and HX.
Phosphorus forms two oxides, P4O6 and P4O10. Their corresponding acids, phosphorous acid and phosphoric acid, undergo
condensation reactions when heated. Phosphorus compounds are
important in biochemistry and as fertilizers.
Carbon and the Other Group 4A Elements (Sections
22.9 and 22.10) The allotropes of carbon include diamond, graph-
ite, fullerenes, carbon nanotubes, and graphene. Amorphous forms
of graphite include charcoal and carbon black. Carbon forms two common oxides, CO and CO2. Aqueous solutions of CO2 produce the
weak diprotic acid carbonic acid 1H2CO32, which is the parent acid of
hydrogen carbonate and carbonate salts. Binary compounds of carbon
are called carbides. Carbides may be ionic, interstitial, or covalent. Calcium carbide 1CaC22 contains the strongly basic acetylide ion 1C22 - 2,
which reacts with water to form acetylene.
The other group 4A elements show great diversity in physical and
chemical properties. Silicon, the second most abundant element, is a semiconductor. It reacts with Cl2 to form SiCl4, a liquid at room temperature,
a reaction that is used to help purify silicon from its native minerals. Silicon forms strong Si ¬ O bonds and therefore occurs in a variety of silicate
minerals. Silica is SiO2; silicates consist of SiO4 tetrahedra, linked together
at their vertices to form chains, sheets, or three-dimensional structures.
The most common three-dimensional silicate is quartz 1SiO22. Glass is an
amorphous (noncrystalline) form of SiO2. Silicones contain O ¬ Si ¬ O
chains with organic groups bonded to the Si atoms. Like silicon, germanium is a metalloid; tin and lead are metallic.
Boron (Section 22.11) Boron is the only group 3A element that
is a nonmetal. It forms a variety of compounds with hydrogen called
boron hydrides, or boranes. Diborane 1B2H62 has an unusual structure
with two hydrogen atoms that bridge between the two boron atoms.
Boranes react with oxygen to form boric oxide 1B2O32, in which boron
is in the +3 oxidation state. Boric oxide is the anhydride of boric acid
1H3BO32. Boric acid readily undergoes condensation reactions.
Learning Outcomes After studying this chapter, you should be able to:
• Use periodic trends to explain the basic differences between the
• Know the sources of the common nonmetals, how they are obtained,
• Explain two ways in which the first element in a group differs from
• Understand how phosphoric and phosphorous acids undergo con-
• Be able to determine electron configurations, oxidation numbers,
• Explain how the bonding and structures of silicates relate to their
elements of a group or period. (Section 22.1)
subsequent elements in the group. (Section 22.1)
and molecular shapes of elements and compounds. (Sections
22.2–22.11)
and how they are used. (Sections 22.2–22.11)
densation reactions. (Section 22.8)
chemical formulas and properties. (Section 22.10)
Exercises
Visualizing Concepts
22.1 (a) One of these structures is a stable compound; the other is
not. Identify the stable compound, and explain why it is stable. Explain why the other compound is not stable. (b) What
is the geometry around the central atoms of the stable compound? [Section 22.1]
H
H
C
H
H
C
H
Si
H
H
Si
H
22.2(a) Identify the type of chemical reaction represented by the
following diagram. (b) Place appropriate charges on the species on both sides of the equation. (c) Write the chemical
equation for the reaction. [Section 22.1]
+
+
22.3 Which of the following species (there may be more than
one) is/are likely to have the structure shown here: (a)
XeF4, (b) BrF4+, (c) SiF4, (d) TeCl4, (e) HClO4? The colors
do not reflect atom identities.) [Sections 22.3, 22.4, 22.6,
and 22.10]
22.4You have two glass bottles, one containing oxygen and one
filled with nitrogen. How could you determine which one is
which? [Sections 22.5 and 22.7]
22.5 Write the molecular formula and Lewis structure for each of
the following oxides of nitrogen: [Section 22.7]
Exercises
22.6 Which property of the group 6A elements might be the one depicted in the graph shown here: (a) electronegativity, (b) first
ionization energy, (c) density, (d) X ¬ X single-bond enthalpy,
(e) electron affinity? Explain your answer. [Sections 22.5 and 22.6]
S
X
Te
Po
22.7 The atomic and ionic radii of the first three group 6A
elements are
Ionic
Atomic
radius
radius
(Å)
(Å)
O
0.66
S
Ar
n
where X and Y may be the same or different, and n may have
a value from +1 to - 2. (b) Which of the compounds is likely
to be the strongest Brønsted base? Explain. [Sections 22.1,
22.7, and 22.9]
Periodic Trends and Chemical Reactions
(Section 22.1)
1.40
22.12Identify each of the following elements as a metal, nonmetal,
or metalloid: (a) gallium, (b) molybdenum, (c) tellurium, (d)
arsenic, (e) xenon, (f) ruthenium.
S
2−
1.84
Se
Se2−
1.98
22.8Which property of the third-row nonmetallic elements
might be the one depicted below: (a) first ionization energy,
(b) atomic radius, (c) electronegativity, (d) melting point,
(e) X ¬ X single-bond enthalpy? Explain both your choice
and why the other choices would not be correct. [Sections
22.3, 22.4, 22.6, 22.8, and 22.10]
Cl
Y
O2−
(a) Explain why the atomic radius increases in moving downward
in the group. (b) Explain why the ionic radii are larger than the
atomic radii. (c) Which of the three anions would you expect to be
the strongest base in water? Explain. [Sections 22.5 and 22.6]
S
O
22.11 Identify each of the following elements as a metal, nonmetal,
or metalloid: (a) phosphorus, (b) strontium, (c) manganese,
(d) selenium, (e) sodium, (f) krypton.
1.05
1.20
P
O
22.10(a) Draw the Lewis structures for at least four species that
have the general formula
Se
Si
22.9 Which of the following compounds would you expect to be
the most generally reactive, and why? (Each corner in these
structures represents a CH2 group.) [Section 22.8]
O
O
991
22.13Consider the elements O, Ba, Co, Be, Br, and Se. From this
list, select the element that (a) is most electronegative, (b) exhibits a maximum oxidation state of + 7, (c) loses an electron
most readily, (d) forms p bonds most readily, (e) is a transition metal, (f) is a liquid at room temperature and pressure.
22.14Consider the elements Li, K, Cl, C, Ne, and Ar. From this list,
select the element that (a) is most electronegative, (b) has the
greatest metallic character, (c) most readily forms a positive
ion, (d) has the smallest atomic radius, (e) forms p bonds
most readily, (f) has multiple allotropes.
22.15 Explain the following observations: (a) The highest fluoride
compound formed by nitrogen is NF3, whereas phosphorus
readily forms PF5. (b) Although CO is a well-known compound, SiO does not exist under ordinary conditions. (c)
AsH3 is a stronger reducing agent than NH3.
22.16Explain the following observations: (a) HNO3 is a stronger
oxidizing agent than H3PO4. b) Silicon can form an ion with
six fluorine atoms, SiF62 - , whereas carbon is able to bond
to a maximum of four, CF4. (c) There are three compounds
formed by carbon and hydrogen that contain two carbon atoms each (C2H2, C2H4, and C2H6), whereas silicon forms only
one analogous compound 1Si2H62.
22.17 Complete and balance the following equations:
(a) NaOCH31s2 + H2O1l2 ¡
(b) CuO1s2 + HNO31aq2 ¡
∆
(c) WO31s2 + H21g2 ¡
(d) NH2OH1l2 + O21g2 ¡
(e) Al4C31s2 + H2O1l2 ¡
22.18Complete and balance the following equations:
(a) Mg3N21s2 + H2O1l2 ¡
(b) C3H7OH1l2 + O21g2 ¡
∆
(c) MnO21s2 + C1s2 ¡
(d) AlP1s2 + H2O1l2 ¡
(e) Na2S1s2 + HCl1aq2 ¡
992
chapter 22 Chemistry of the Nonmetals
Hydrogen, the Noble Gases, and the Halogens
(Sections 22.2, 22.3, and 22.4)
22.19 (a) Give the names and chemical symbols for the three isotopes
of hydrogen. (b) List the isotopes in order of decreasing natural
abundance. (c) Which hydrogen isotope is radioactive? (d) Write
the nuclear equation for the radioactive decay of this isotope.
22.20Are the physical properties of H2O different from D2O? Explain.
22.21 Give a reason why hydrogen might be placed along with the
group 1A elements of the periodic table.
22.22What does hydrogen have in common with the halogens?
Explain.
22.23 Write a balanced equation for the preparation of H2 using (a) Mg
and an acid, (b) carbon and steam, (c) methane and steam.
22.24List (a) three commercial means of producing H2, (b) three
industrial uses of H2.
22.25 Complete and balance the following equations:
(a) NaH1s2 + H2O1l2 ¡
(b) Fe1s2 + H2SO41aq2 ¡
(c) H21g2 + Br21g2 ¡
(d) Na1l2 + H21g2 ¡
(e) PbO1s2 + H21g2 ¡
22.26Write balanced equations for each of the following reactions
(some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen
gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium
hydride reacts with water to generate hydrogen gas.
22.27 Identify the following hydrides as ionic, metallic, or molecular: (a) BaH2, (b) H2Te, (c) TiH1.7.
22.28Identify the following hydrides as ionic, metallic, or molecular:
(a) B2H6, (b) RbH, (c) Th4H1.5.
22.29 Describe two characteristics of hydrogen that are favorable
for its use as a general energy source in vehicles.
22.30The H2 >O2 fuel cell converts elemental hydrogen and oxygen
into water, producing, theoretically, 1.23 V. What is the most
sustainable way to obtain hydrogen to run a large number of
fuel cells? Explain.
22.31Why does xenon form stable compounds with fluorine,
whereas argon does not?
22.32A friend tells you that the “neon” in neon signs is a compound
of neon and aluminum. Can your friend be correct? Explain.
22.33 Write the chemical formula for each of the following, and indicate the oxidation state of the halogen or noble-gas atom in each:
(a) calcium hypobromite, (b) bromic acid, (c) xenon trioxide, (d)
perchlorate ion, (e) iodous acid, (f) iodine pentafluoride.
22.34Write the chemical formula for each of the following compounds, and indicate the oxidation state of the halogen or
noble-gas atom in each: (a) chlorate ion, (b) hydroiodic acid,
(c) iodine trichloride, (d) sodium hypochlorite, (e) perchloric
acid, (f) xenon tetrafluoride.
22.35 Name the following compounds and assign oxidation states
to the halogens in them: (a) Fe1ClO323, (b) HClO2, (c) XeF6,
(d) BrF5, (e) XeOF4, (f) HIO3.
22.36Name the following compounds and assign oxidation states
to the halogens in them: (a) KClO3, (b) Ca1IO322, (c) AlCl3,
(d) HBrO3, (e) H5IO6, (f) XeF4.
22.37 Explain each of the following observations: (a) At room temperature I2 is a solid, Br2 is a liquid, and Cl2 and F2 are both gases.
(b) F2 cannot be prepared by electrolytic oxidation of aqueous F solutions. (c) The boiling point of HF is much higher than those
of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order F2 7 Cl2 7 Br2 7 I2.
22.38Explain the following observations: (a) For a given oxidation
state, the acid strength of the oxyacid in aqueous solution decreases in the order chlorine 7 bromine 7 iodine. (b) Hydrofluoric acid cannot be stored in glass bottles. (c) HI cannot
be prepared by treating NaI with sulfuric acid. (d) The interhalogen ICl3 is known, but BrCl3 is not.
Oxygen and the Other Group 6A Elements
(Sections 22.5 and 22.6)
22.39 Write balanced equations for each of the following reactions.
(a) When mercury(II) oxide is heated, it decomposes to form
O2 and mercury metal. (b) When copper(II) nitrate is heated
strongly, it decomposes to form copper(II) oxide, nitrogen
dioxide, and oxygen. (c) Lead(II) sulfide, PbS(s), reacts with
ozone to form PbSO41s2 and O21g2. (d) When heated in air,
ZnS(s) is converted to ZnO. (e) Potassium peroxide reacts
with CO21g2 to give potassium carbonate and O2. (f) Oxygen
is converted to ozone in the upper atmosphere.
22.40Complete and balance the following equations:
(a) CaO1s2 + H2O1l2 ¡
(b) Al2O31s2 + H+1aq2 ¡
(c) Na2O21s2 + H2O1l2 ¡
(d) N2O31g2 + H2O1l2 ¡
(e) KO21s2 + H2O1l2 ¡
(f) NO1g2 + O31g2 ¡
22.41 Predict whether each of the following oxides is acidic, basic,
amphoteric, or neutral: (a) NO2, (b) CO2, (c) Al2O3, (d) CaO.
22.42Select the more acidic member of each of the following pairs:
(a) Mn2O7 and MnO2, (b) SnO and SnO2, (c) SO2 and SO3,
(d) SiO2 and SO2, (e) Ga2O3 and In2O3, (f) SO2 and SeO2.
22.43 Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 6A
element in each: (a) selenous acid, (b) potassium hydrogen
sulfite, (c) hydrogen telluride, (d) carbon disulfide, (e) calcium sulfate, (f) cadmium sulfide, (g) zinc telluride.
22.44Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 6A element in each: (a) sulfur tetrachloride, (b) selenium trioxide,
(c) sodium thiosulfate, (d) hydrogen sulfide, (e) sulfuric acid,
(f) sulfur dioxide, (g) mercury telluride.
22.45 In aqueous solution, hydrogen sulfide reduces (a) Fe3 + to Fe2 + ,
(b) Br2 to Br - , (c) MnO4- to Mn2 + , (d) HNO3 to NO2. In all
cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.
22.46An aqueous solution of SO2 reduces (a) aqueous KMnO4 to
MnSO41aq2, (b) acidic aqueous K2Cr2O7 to aqueous Cr3 + , (c)
aqueous Hg21NO322 to mercury metal. Write balanced equations for these reactions.
22.47 Write the Lewis structure for each of the following species,
and indicate the structure of each: (a) SeO32 - ; (b) S2Cl2;
(c) chlorosulfonic acid, HSO3Cl (chlorine is bonded to sulfur).
22.48The SF5- ion is formed when SF41g2 reacts with fluoride salts
containing large cations, such as CsF(s). Draw the Lewis structures
for SF4 and SF5- , and predict the molecular structure of each.
Exercises
22.49 Write a balanced equation for each of the following reactions:
(a) Sulfur dioxide reacts with water. (b) Solid zinc sulfide reacts
with hydrochloric acid. (c) Elemental sulfur reacts with sulfite ion
to form thiosulfate. (d) Sulfur trioxide is dissolved in sulfuric acid.
22.50Write a balanced equation for each of the following reactions.
(You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based
on your study of this chapter.) (a) Hydrogen selenide can be
prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess
Cl2 from chlorine-bleached fabrics. The thiosulfate ion forms
SO42 - and elemental sulfur, while Cl2 is reduced to Cl - .
Nitrogen and the Other Group 5A Elements
(Sections 22.7 and 22.8)
22.51 Write the chemical formula for each of the following compounds,
and indicate the oxidation state of nitrogen in each: (a) sodium
nitrite, (b) ammonia, (c) nitrous oxide, (d) sodium cyanide,
(e) nitric acid, (f) nitrogen dioxide, (g) nitrogen, (h) boron nitride.
22.52Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each:
(a) nitric oxide, (b) hydrazine, (c) potassium cyanide, (d) sodium nitrite, (e) ammonium chloride, (f) lithium nitride.
22.53 Write the Lewis structure for each of the following species,
describe its geometry, and indicate the oxidation state of the
nitrogen: (a) HNO2, (b) N3- , (c) N2H5+, (d) NO3- .
22.54Write the Lewis structure for each of the following species,
describe its geometry, and indicate the oxidation state of the
nitrogen: (a) NH4+, (b) NO2- , (c) N2O, (d) NO2.
22.55 Complete and balance the following equations:
(a) Mg3N21s2 + H2O1l2 ¡
(b) NO1g2 + O21g2 ¡
(c) N2O51g2 + H2O1l2 ¡
(d) NH31aq2 + H+1aq2 ¡
(e) N2H41l2 + O21g2 ¡
Which ones of these are redox reactions?
22.56Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with
formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of
nitric oxide. (d) Hydrazine is burned in excess fluorine gas,
forming NF3. (e) Hydrazine reduces CrO42 - to Cr1OH24- in
base (hydrazine is oxidized to N2).
22.57 Write complete balanced half-reactions for (a) oxidation of
nitrous acid to nitrate ion in acidic solution, (b) oxidation of
N2 to N2O in acidic solution.
22.58Write complete balanced half-reactions for (a) reduction of
nitrate ion to NO in acidic solution, (b) oxidation of HNO2 to
NO2 in acidic solution.
22.59 Write a molecular formula for each compound, and indicate
the oxidation state of the group 5A element in each formula:
(a) phosphorous acid, (b) pyrophosphoric acid, (c) antimony
trichloride, (d) magnesium arsenide, (e) diphosphorus pentoxide, (f) sodium phosphate.
22.60Write a chemical formula for each compound or ion, and indicate the oxidation state of the group 5A element in each formula: (a) phosphate ion, (b) arsenous acid, (c) antimony(III)
sulfide, (d) calcium dihydrogen phosphate, (e) potassium
phosphide, (f) gallium arsenide.
993
22.61 Account for the following observations: (a) Phosphorus
forms a pentachloride, but nitrogen does not. (b) H3PO2 is a
monoprotic acid. (c) Phosphonium salts, such as PH4Cl, can
be formed under anhydrous conditions, but they cannot be
made in aqueous solution. (d) White phosphorus is more reactive than red phosphorus.
22.62Account for the following observations: (a) H3PO3 is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a
phosphate fertilizer. (d) Phosphorus does not exist at room
temperature as diatomic molecules, but nitrogen does. (e) Solutions of Na3PO4 are quite basic.
22.63 Write a balanced equation for each of the following reactions:
(a) preparation of white phosphorus from calcium phosphate,
(b) hydrolysis of PBr3, (c) reduction of PBr3 to P4 in the gas
phase, using H2.
22.64Write a balanced equation for each of the following reactions:
(a) hydrolysis of PCl5, (b) dehydration of phosphoric acid
(also called orthophosphoric acid) to form pyrophosphoric
acid, (c) reaction of P4O10 with water.
Carbon, the Other Group 4A Elements, and
Boron (Sections 22.9, 22.10, and 22.11)
22.65 Give the chemical formula for (a) hydrocyanic acid, (b) nickel
tetracarbonyl, (c) barium bicarbonate, (d) calcium acetylide,
(e) potassium carbonate.
22.66Give the chemical formula for (a) carbonic acid, (b) sodium
cyanide, (c) potassium hydrogen carbonate, (d) acetylene,
(e) iron pentacarbonyl.
22.67 Complete and balance the following equations:
∆
(a) ZnCO31s2 ¡
(b) BaC21s2 + H2O1l2 ¡
(c) C2H21g2 + O21g2 ¡
(d) CS21g2 + O21g2 ¡
(e) Ca1CN221s2 + HBr1aq2 ¡
22.68Complete and balance the following equations:
(a) CO21g2 + OH - 1aq2 ¡
(b) NaHCO31s2 + H+1aq2 ¡
∆
(c) CaO1s2 + C1s2 ¡
∆
(d) C1s2 + H2O1g2 ¡
(e) CuO1s2 + CO1g2 ¡
22.69Write a balanced equation for each of the following reactions: (a) Hydrogen cyanide is formed commercially by passing a mixture of methane, ammonia, and air over a catalyst at
800 °C. Water is a by-product of the reaction. (b) Baking soda
reacts with acids to produce carbon dioxide gas. (c) When
barium carbonate reacts in air with sulfur dioxide, barium
sulfate and carbon dioxide form.
22.70Write a balanced equation for each of the following reactions:
(a) Burning magnesium metal in a carbon dioxide atmosphere reduces the CO2 to carbon. (b) In photosynthesis, solar energy is used to produce glucose 1C6H12O62 and O2 from
carbon dioxide and water. (c) When carbonate salts dissolve
in water, they produce basic solutions.
22.71 Write the formulas for the following compounds, and indicate the oxidation state of the group 4A element or of boron
in each: (a) boric acid, (b) silicon tetrabromide, (c) lead(II)
chloride, (d) sodium tetraborate decahydrate (borax), (e) boric oxide, (f) germanium dioxide.
994
chapter 22 Chemistry of the Nonmetals
22.72Write the formulas for the following compounds, and indicate
the oxidation state of the group 4A element or of boron in each:
(a) silicon dioxide, (b) germanium tetrachloride, (c) sodium borohydride, (d) stannous chloride, (e) diborane, (f) boron trichloride.
22.73 Select the member of group 4A that best fits each description:
(a) has the lowest first ionization energy, (b) is found in oxidation states ranging from - 4 to +4, (c) is most abundant in
Earth’s crust.
22.74Select the member of group 4A that best fits each description:
(a) forms chains to the greatest extent, (b) forms the most basic oxide, (c) is a metalloid that can form 2 + ions.
22.75 (a) What is the characteristic geometry about silicon in all silicate minerals? (b) Metasilicic acid has the empirical formula
H2SiO3. Which of the structures shown in Figure 22.34 would
you expect metasilicic acid to have?
22.76Speculate as to why carbon forms carbonate rather than silicate analogs.
22.77 (a) Determine the number of calcium ions in the chemical formula of the mineral hardystonite, CaxZn1Si2O72. (b) Determine
the number of hydroxide ions in the chemical formula of the
mineral pyrophyllite , Al21Si2O5221OH2x.
22.78(a) Determine the number of sodium ions in the chemical formula of albite, NaxAlSi3O8. (b) Determine the number of hydroxide ions in the chemical formula of tremolite,
Ca2Mg51Si4O11221OH2x.
22.79 (a) How does the structure of diborane 1B2H62 differ from that
of ethane 1C2H62? (b) Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that
the hydrogen atoms in diborane are described as “hydridic”?
22.80Write a balanced equation for each of the following reactions: (a) Diborane reacts with water to form boric acid and
­molecular hydrogen. (b) Upon heating, boric acid undergoes
a condensation reaction to form tetraboric acid. (c) Boron
­oxide dissolves in water to give a solution of boric acid.
Additional Exercises
22.81 Indicate whether each of the following statements is true
or false (a) H21g2 and D21g2 are allotropic forms of hydrogen. (b) ClF3 is an interhalogen compound. (c) MgO(s)
is an acidic anhydride. (d) SO21g2 is an acidic anhydride.
(e) 2 H3PO41l2 S H4P2O71l2+ H2O1g2 is an example of a condensation reaction. (f) Tritium is an isotope of the element
hydrogen. (g) 2SO21g2+ O21g2 S 2SO31g2 is an example of a
disproportionation reaction.
22.82Although the ClO4- and IO4- ions have been known for a long
time, BrO4- was not synthesized until 1965. The ion was synthesized by oxidizing the bromate ion with xenon difluoride, producing xenon, hydrofluoric acid, and the perbromate ion. (a) Write
the balanced equation for this reaction. (b) What are the oxidation
states of Br in the Br-containing species in this reaction?
22.83Write a balanced equation for the reaction of each of the following
compounds with water: (a) SO21g2, (b) Cl2O71g2, (c) Na2O21s2,
(d) BaC21s2, (e) RbO21s2 (f) Mg3N21s2, (g) NaH(s).
22.84 What is the anhydride for each of the following acids:
(a) H2SO4, (b) HClO3, (c) HNO2, (d) H2CO3, (e) H3PO4?
22.85 Hydrogen peroxide is capable of oxidizing (a) hydrazine to
N2 and H2O, (b) SO2 to SO42 - , (c) NO2- to NO3- , (d) H2S1g2
to S(s), (e) Fe2 + to Fe3 + . Write a balanced net ionic equation
for each of these redox reactions.
22.86Explain why SO2 can be used as a reducing agent but SO3 cannot.
22.87 A sulfuric acid plant produces a considerable amount of heat.
This heat is used to generate electricity, which helps reduce
operating costs. The synthesis of H2SO4 consists of three main
chemical processes: (a) oxidation of S to SO2, (b) oxidation of
SO2 to SO3, (c) the dissolving of SO3 in H2SO4 and its reaction with water to form H2SO4. If the third process produces
130 kJ>mol, how much heat is produced in preparing a mole
of H2SO4 from a mole of S? How much heat is produced in
preparing 5000 pounds of H2SO4?
22.88 (a) What is the oxidation state of P in PO43 - and of N in NO3- ?
(b) Why doesn’t N form a stable NO43 - . ion analogous to P?
22.89 (a) The P4, P4O6 and P4O10 molecules have a common structural feature of four P atoms arranged in a tetrahedron
(Figures 22.27 and 22.28). Does this mean that the bonding
between the P atoms is the same in all these cases? Explain.
(b) Sodium trimetaphosphate 1Na3P3O92 and sodium tetrametaphosphate 1Na4P4O122 are used as water-softening agents.
They contain cyclic P3O93 - and P4O124 - ions, respectively.
Propose reasonable structures for these ions.
22.90Ultrapure germanium, like silicon, is used in semiconductors.
Germanium of “ordinary” purity is prepared by the high-temperature reduction of GeO2 with carbon. The Ge is converted
to GeCl4 by treatment with Cl2 and then purified by distillation;
GeCl4 is then hydrolyzed in water to GeO2 and reduced to the
elemental form with H2. The element is then zone refined. Write
a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from GeO2.
22.91 (a) Determine the charge of the aluminosilicate ion whose
composition is AlSi3O10. (b) Using Figure 22.33, propose a
reasonable description of the structure of this aluminosilicate.
Integrative Exercises
[22.92] (a) How many grams of H2 can be stored in 100.0 kg of the
alloy FeTi if the hydride FeTiH2 is formed? (b) What volume
does this quantity of H2 occupy at STP? (c) If this quantity of
hydrogen was combusted in air to produce liquid water, how
much energy could be produced?
[22.93] Using the thermochemical data in Table 22.1 and
Appendix C, calculate the average Xe ¬ F bond enthalpies in
XeF2, XeF4, and XeF6, respectively. What is the significance
of the trend in these quantities?
22.94 Hydrogen gas has a higher fuel value than natural gas on a
mass basis but not on a volume basis. Thus, hydrogen is not
competitive with natural gas as a fuel transported long distances through pipelines. Calculate the heats of combustion
of H2 and CH4 (the principal component of natural gas)
(a) per mole of each, (b) per gram of each, (c) per cubic meter
of each at STP. Assume H2O1l2 as a product.
22.95 Using ∆G°f for ozone from Appendix C, calculate the equilibrium constant for Equation 22.24 at 298.0 K, assuming no
electrical input.
Designing an Experiment
22.96 The solubility of Cl2 in 100 g of water at STP is 310 cm3. Assume
that this quantity of Cl2 is dissolved and equilibrated as follows:
Cl21aq2 + H2O ∆ Cl - 1aq2 + HClO1aq2 + H+1aq2
(a) If the equilibrium constant for this reaction is 4.7 * 10 - 4,
calculate the equilibrium concentration of HClO formed. (b)
What is the pH of the final solution?
[22.97] When ammonium perchlorate decomposes thermally, the
products of the reaction are N21g2, O21g2, H2O1g2, and HCl(g).
(a) Write a balanced equation for the reaction. (Hint: You might
find it easier to use fractional coefficients for the products.)
(b) Calculate the enthalpy change in the reaction per mole of
NH4ClO4. The standard enthalpy of formation of NH4ClO41s2
is - 295.8 kJ. (c) When NH4ClO41s2 is employed in solid-fuel
booster rockets, it is packed with powdered aluminum. Given
the high temperature needed for NH4ClO41s2 decomposition
and what the products of the reaction are, what role does the
aluminum play? (d) Calculate the volume of all the gases that
would be produced at STP, assuming complete reaction of one
pound of ammonium perchlorate.
22.98The dissolved oxygen present in any highly pressurized,
high-temperature steam boiler can be extremely corrosive
to its metal parts. Hydrazine, which is completely miscible
with water, can be added to remove oxygen by reacting with
it to form nitrogen and water. (a) Write the balanced equation for the reaction between gaseous hydrazine and oxygen.
(b) Calculate the enthalpy change accompanying this reaction.
(c) Oxygen in air dissolves in water to the extent of 9.1 ppm at
20 °C at sea level. How many grams of hydrazine are required
to react with all the oxygen in 3.0 * 104 L (the volume of a
small swimming pool) under these conditions?
22.99 One method proposed for removing SO2 from the flue gases of
power plants involves reaction with aqueous H2S. Elemental sulfur is the product. (a) Write a balanced chemical equation for the
reaction. (b) What volume of H2S at 27 °C and 760 torr would be
required to remove the SO2 formed by burning 2.0 tons of coal
containing 3.5% S by mass? (c) What mass of elemental sulfur is
produced? Assume that all reactions are 100% efficient.
22.100 The maximum allowable concentration of H2S1g2 in air is
20 mg per kilogram of air (20 ppm by mass). How many
grams of FeS would be required to react with hydrochloric
acid to produce this concentration at 1.00 atm and 25 °C in an
average room measuring 12 ft * 20 ft * 8 ft? (Under these
conditions, the average molar mass of air is 29.0 g>mol.)
22.101 The standard heats of formation of H2O1g2, H2S1g2, H2Se1g2,
and H2Te1g2 are - 241.8, - 20.17, + 29.7, and + 99.6 kJ>
mol, respectively. The enthalpies necessary to convert the
elements in their standard states to one mole of gaseous atoms are 248, 277, 227, and 197 kJ>mol of atoms for O, S, Se,
995
and Te, respectively. The enthalpy for dissociation of H2
is 436 kJ>mol. Calculate the average H ¬ O, H ¬ S, H ¬ Se,
and H ¬ Te bond enthalpies, and comment on their trend.
22.102 Manganese silicide has the empirical formula MnSi and melts
at 1280 °C. It is insoluble in water but does dissolve in aqueous HF. (a) What type of compound do you expect MnSi to
be: metallic, molecular, covalent-network, or ionic? (b) Write
a likely balanced chemical equation for the reaction of MnSi
with concentrated aqueous HF.
[22.103] Chemists tried for a long time to make molecular compounds
containing silicon–silicon double bonds; they finally succeeded
in 1981. The trick is having large, bulky R groups on the silicon
atoms to make R2Si “ SiR2 compounds. What experiments
could you do to prove that a new compound has a silicon–silicon
double bond rather than a silicon–silicon single bond?
22.104 Hydrazine has been employed as a reducing agent for metals.
Using standard reduction potentials, predict whether the following metals can be reduced to the metallic state by hydrazine under standard conditions in acidic solution: (a) Fe2 + ,
(b) Sn2 + , (c) Cu2 + , (d) Ag +, (e) Cr3 + , (f) Co3 + .
22.105 Both dimethylhydrazine, 1CH322NNH2, and methylhydrazine, CH3NHNH2, have been used as rocket fuels. When dinitrogen tetroxide 1N2O42 is used as the oxidizer, the products
are H2O, CO2, and N2. If the thrust of the rocket depends
on the volume of the products produced, which of the substituted hydrazines produces a greater thrust per gram total
mass of oxidizer plus fuel? (Assume that both fuels generate
the same temperature and that H2O1g2 is formed.)
22.106 Carbon forms an unusual unstable oxide of formula C3O2, called
carbon suboxide. Carbon suboxide is made by using P2O5 to
dehydrate the dicarboxylic acid called malonic acid, which has
the formula HOOC ¬ CH2 ¬ COOH. (a) Write a balanced
reaction for the production of carbon suboxide from malonic
acid. (b) How many grams of carbon suboxide could be made
from 20.00 g of malonic acid? (c) Suggest a Lewis structure for
C3O2. (Hint: The Lewis structure of malonic acid suggests which
atoms are connected to which.) (d) By using the information in
Table 8.5, predict the C ¬ C and C ¬ O bond lengths in C3O2.
(e) Sketch the Lewis structure of a product that could result by
the addition of 2 mol of H2 to 1 mol of C3O2.
22.107 Borazine, 1BH231NH23, is an analog of C6H6, benzene. It can
be prepared from the reaction of diborane with ammonia,
with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and
hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine.
(c) How many grams of borazine can be prepared from 2.00 L
of ammonia at STP, assuming diborane is in excess?
Designing an Experiment
You are given samples of five substances. At room temperature, three are colorless gases, one is
a colorless liquid, and one is a white solid. You are told that the substances are NF3, PF3, PCl3,
PF5, and PCl5. Let’s design experiments to determine which substance is which, using concepts
from this and earlier chapters.
(a) Assuming that you don’t have access to either the internet or to a handbook of chemistry
(as is the case during your exams!), design experiments that would allow you to identify the
substances. (b) How might you proceed differently if you had access to data from the internet?
(c) Which of the substances could undergo reaction to add more atoms around the central
atom? What types of reactions might you choose to test this hypothesis? (d) Based on what you
know about intermolecular forces, which of the substances is likely the solid?
23
Transition Metals and
Coordination Chemistry
The colors of our world are beautiful, but to a chemist they are also
informative—providing insights into the structure and bonding of matter.
Compounds of the transition metals constitute an important group of
colored substances. Some of them are used in pigments; others produce
the colors in glass and precious gems. The use of vivid green, yellow, and
blue colors in the paintings of impressionists like Monet, Cezanne, and van Gogh was
made possible by the development of synthetic pigments in the 1800s. Three such pigments used extensively by the impressionists were cobalt blue, CoAl2O4, chrome yellow,
PbCrO4, and emerald green, Cu4(CH3COO)2(AsO2)6. In each case, the presence of a
transition metal ion is directly responsible for the color of the pigment—Co2+ in cobalt
blue, Cr6+ in chrome yellow, and Cu2+ in emerald green.
The color of a given transition-metal compound depends upon not only the
transition-metal ion but also upon the identity and geometry of the ions and/or
molecules that surround it. To appreciate the importance of the transition-metal ion
surroundings, consider the colors of the minerals azurite, Cu3(CO3)2(OH)2, and malachite, Cu2(CO3)(OH)2 (Figure 23.1). Both minerals contain Cu2+ ions surrounded by
carbonate and hydroxide ions, but subtle differences in the local surroundings of the
Cu2+ ion lead to the contrasting colors of these two minerals. In this chapter, we will
learn why transition-metal compounds are colored and how changes in the identity
and environment of the transition-metal ion lead to changes in color.
What’s
Ahead
23.1 The Transition Metals We examine the physical
properties, electron configurations, oxidation states, and magnetic
properties of the transition metals.
23.2 Transition-Metal Complexes We introduce the
concepts of metal complexes and ligands and provide a brief
history of the development of coordination chemistry.
▶ WATER LILLIES AND THE JAPANESE
BRIDGE, was painted by Claude Monet in
1899.
23.3 Common Ligands in Coordination Chemistry We
examine some common geometries found in coordination
complexes and how the geometries relate to coordination numbers.
23.4 Nomenclature and Isomerism in Coordination
Chemistry We introduce the nomenclature used for
coordination compounds. We see that coordination compounds
exhibit isomerism, in which two compounds have the same
composition but different structures, and then look at two types:
structural isomers and stereoisomers.
23.5 Color and Magnetism in Coordination
Chemistry We discuss color and magnetism in coordination
compounds, emphasizing the visible portion of the
electromagnetic spectrum and the notion of complementary
colors. We then see that many transition-metal complexes are
paramagnetic because they contain unpaired electrons.
23.6 Crystal-Field Theory We explore how crystal-field
theory allows us to explain some of the interesting spectral and
magnetic properties of coordination compounds.
998
chapter 23 Transition Metals and Coordination Chemistry
Transition metals and their compounds are important for much more than their
color. For example, they are used as catalysts and as magnets, and they play an important role in biology.
In earlier chapters, we saw that metal ions can function as Lewis acids, forming
(Section 16.11)
covalent bonds with molecules and ions functioning as Lewis bases.
We have encountered many ions and compounds that result from such interactions,
such as 3Ag(NH3)24 + in
Section 17.5 and hemoglobin in
Section 13.6. In this
chapter, we focus on the rich and important chemistry associated with such complex
assemblies of metal ions surrounded by molecules and ions. Metal compounds of this
kind are called coordination compounds, and the branch of chemistry that focuses on
them is called coordination chemistry.
23.1 | The Transition Metals
▲ Figure 23.1 Crystals of blue azurite and
green malachite. These semiprecious stones
were ground up and used as pigments in
the Middle Ages and Renaissance, but they
were eventually replaced with blue and green
pigments that have superior chemical stability.
8B
3B
3
21
Sc
4B
4
22
Ti
5B
5
23
V
6B
6
24
Cr
7B
7
25
Mn
26
Fe
9
27
Co
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
71
Lu
72
Hf
73
Ta
74
W
75
Re
76
Os
77
Ir
8
The part of the periodic table in which the d orbitals are being filled as we move left to
right across a row is the home of the transition metals (◀ Figure 23.2).
(Section 6.8)
With some exceptions (e.g., platinum, gold), metallic elements are found in nature
as solid inorganic compounds called minerals. Notice from ▼ Table 23.1 that minerals
are identified by common names rather than chemical names.
Most transition metals in minerals have oxidation states
ranging from +1 to +4. To obtain a pure metal from its mineral, various chemical processes must be performed to reduce
the metal to the 0 oxidation state. Metallurgy is the science and technology of extracting metals from their natural sources and preparing them for practical use. It usually
involves several steps: (1) mining, that is, removing the relevant
ore (a mixture of minerals) from the ground, (2) concentrat1B
2B
10
11
12
ing the ore or otherwise preparing it for further treatment,
(3) reducing the ore to obtain the free metal, (4) purifying the metal,
30
29
28
Cu Zn
Ni
and (5) mixing it with other elements to modify its properties. This
last process produces an alloy, a metallic material composed of two
48
47
46
Pd Ag Cd
or more elements.
(Section 12.3)
78
Pt
79
Au
80
Hg
Physical Properties
Some physical properties of the period 4 (also known as “firstrow”) transition metals are listed in ▶ Table 23.2. The properties of
the heavier transition metals vary similarly across periods 5 and 6.
▲ Figure 23.2 The position of the transition
metals in the periodic table. They are the
B groups in periods 4, 5, and 6. The short
lived, radioactive transition metals from
period 7 are not shown.
Table 23.1 Principal Mineral Sources of Some Transition Metals
Metal
Mineral
Mineral Composition
Chromium
Chromite
FeCr2O4
Cobalt
Cobaltite
CoAsS
Copper
Chalcocite
Cu2S
Chalcopyrite
CuFeS2
Malachite
Cu2CO3(OH)2
Hematite
Fe2O3
Magnetite
Fe3O4
Iron
Manganese
Pyrolusite
MnO2
Mercury
Cinnabar
HgS
Molybdenum
Molybdenite
MoS2
Titanium
Rutile
TiO2
Ilmenite
FeTiO3
Sphalerite
ZnS
Zinc
999
section 23.1 The Transition Metals
Table 23.2 Properties of the Period 4 Transition Metals
Group
3B
Element:
Sc
Ground state electron configuration
1
3d 4s
4B
5B
Ti
2
2
3d 4s
6B
V
2
3
3d 4s
7B
Cr
2
5
8B
Mn
1
5
3d 4s
3d 4s
Fe
2
6
3d 4s
1B
Co
2
7
3d 4s
Ni
2
8
3d 4s
2B
Cu
2
10
Zn
1
10
3d 4s
3d 4s2
First ionization energy (kJ/mol)
631
658
650
653
717
759
758
737
745
906
Metallic radius (Å)
1.64
1.47
1.35
1.29
1.37
1.26
1.25
1.25
1.28
1.37
3.0
4.5
6.1
7.9
7.2
7.9
8.7
8.9
8.9
7.1
Melting point (°C)
1541
1660
1917
1857
1244
1537
1494
1455
1084
420
Crystal structure*
hcp
hcp
bcc
bcc
**
bcc
hcp
fcc
fcc
hcp
3
Density (g/cm )
*Abbreviations for crystal structures are hcp = hexagonal close packed, fcc = face centered cubic, bcc = body centered cubic.
(Section 12.3)
**Manganese has a more complex crystal structure.
▶ Figure 23.3 shows the atomic radius observed in close-packed metallic structures as
a function of group number.* The trends seen in the graph are a result of two competing
forces. On the one hand, increasing effective nuclear charge favors a decrease in radius as
we move left to right across each period.
(Section 7.2) On the other hand, the metallic
bonding strength increases until we reach the middle of each period and then decreases as we
fill antibonding orbitals.
(Section 12.4) As a general rule, a bond shortens as it becomes
stronger.
(Section 8.8) For groups 3B through 6B, these two effects
work cooperatively and the result is a marked decrease in radius. In ele1.9
ments to the right of group 6B, the two effects counteract each other,
reducing the decrease and eventually leading to an increase in radius.
Go Figure
Does the variation in radius of the
transition metals follow the same
trend as the effective nuclear charge
on moving from left to right across the
periodic table?
Period 4 (Sc–Zn)
Period 5 (Y– Cd)
Period 6 (Lu–Hg)
1.8
Which element has the largest bonding atomic radius: Sc, Fe, or Au?
Atomic radius (Å)
Give It Some Thought
1.7
1.6
In general, atomic radii increase as we move down in a given group
in the periodic table because of the increasing principal quantum num1.5
ber of the outer-shell electrons. ( Section 7.3) Note in Figure 23.2,
however, that once we move beyond the group 3B elements, the period
1.4
5 and period 6 transition elements in a given group have virtually the
same radii. In group 5B, for example, tantalum in period 6 has virtually
the same radius as niobium in period 5. This interesting and important
1.3
effect has its origin in the lanthanide series, elements 57 through 70. The
filling of 4f orbitals through the lanthanide elements
(Figure 6.31)
1.2
causes a steady increase in the effective nuclear charge, producing a
3B 4B
size decrease, called the lanthanide contraction, that just offsets the
increase we expect as we go from period 5 transition metals to period 6.
Thus, the period 5 and period 6 transition metals in each group have about the same radii
and similar chemical properties. For example, the chemical properties of the group 4B
metals zirconium (period 5) and hafnium (period 6) are remarkably similar. These two
metals always occur together in nature, and they are very difficult to separate.
Electron Configurations and Oxidation States
Transition metals owe their location in the periodic table to the filling of the d subshells,
as you saw in Figure 6.31. Many of the chemical and physical properties of transition
metals result from the unique characteristics of the d orbitals. For a given transitionmetal atom, the valence (n - 1)d orbitals are smaller than the corresponding valence
ns and np orbitals. In quantum mechanical terms, the (n - 1)d orbital wave functions
drop off more rapidly as we move away from the nucleus than do the ns and np orbital
*Note that the radii defined in this way, often referred to as metallic radii, differ somewhat from the bonding
atomic radii defined in Section 7.3.
5B
6B
7B
Group
8B
1B
2B
▲ Figure 23.3 Radii of transition metals as
a function of group number.
1000
chapter 23 Transition Metals and Coordination Chemistry
Go Figure
In which transition-metal ion of this group are the 3d orbitals completely filled?
▲ Figure 23.4 Aqueous solutions of transition metal ions. Left to right: Co2+ , Ni2+ , Cu2+ , and
Zn2+ . The counterion is nitrate in all cases.
wave functions. This characteristic feature of the d orbitals limits their interaction with
orbitals on neighboring atoms, but not so much that they are insensitive to surrounding atoms. As a result, electrons in these orbitals behave sometimes like valence electrons and sometimes like core electrons. The details depend on location in the periodic
table and the atom’s environment.
When transition metals are oxidized, they lose their outer s electrons before they
lose electrons from the d subshell.
(Section 7.4) The electron configuration of Fe is
3Ar43d 64s2, for example, whereas that of Fe2+ is 3Ar43d 6. Formation of Fe3+ requires
loss of one 3d electron, giving 3Ar43d 5. Most transition-metal ions contain partially
occupied d subshells, which are responsible in large part for three characteristics:
Go Figure
For which of the ions shown in this
figure are the 4s orbitals empty? For
which ions are the 3d orbitals empty?
1.Transition metals often have more than one stable oxidation state.
2.Many transition-metal compounds are colored, as shown in ▲ Figure 23.4.
3.Transition metals and their compounds often exhibit magnetic properties.
◀ Figure 23.5 shows the common nonzero oxidation states for the
period
4 transition metals. The +2 oxidation state, which is common for
Most frequently seen
most
transition
metals, is due to the loss of the two outer 4s electrons. This
Less common
oxidation state is found for all these elements except Sc, where the 3+ ion
3B 4B 5B 6B 7B
8B
1B 2B
with an [Ar] configuration is particularly stable.
Oxidation states above +2 are due to successive losses of 3d electrons.
+8
From Sc through Mn the maximum oxidation state increases from +3 to +7,
equaling in each case the total number of 4s plus 3d electrons in the atom.
+6
Thus, manganese has a maximum oxidation state of 2 + 5 = +7. As we
+4
move to the right beyond Mn in Figure 23.5, the maximum oxidation state
decreases. This decrease is due in part to the attraction of d orbital electrons
+2
to the nucleus, which increases faster than the attraction of the s orbital elec0
trons to the nucleus as we move left to right across the periodic table. In other
Sc Ti V Cr Mn Fe Co Ni Cu Zn
words, in each period the d electrons become more corelike as the atomic
▲ Figure 23.5 Nonzero oxidation states of
number increases. By the time we get to zinc, it is not possible to remove electrons from
the period 4 transition metals.
the 3d orbitals through chemical oxidation.
section 23.1 The Transition Metals
In the transition metals of periods 5 and 6, the increased size of the 4d and 5d
orbitals makes it possible to attain maximum oxidation states as high as +8, which is
achieved in RuO4 and OsO4. In general, the maximum oxidation states are found only
when the metals are combined with the most electronegative elements, especially O, F,
and in some cases Cl.
1001
Go Figure
Describe how the representation shown
for the paramagnetic material would
change if the material were placed in a
magnetic field.
Give It Some Thought
Why does Ti5+ not exist?
Magnetism
The spin an electron possesses gives the electron a magnetic moment, a property that causes the electron to behave like a tiny
magnet. In a diamagnetic solid, defined as one in which all the
electrons in the solid are paired, the spin-up and spin-down elecFerromagnetic; spins
Paramagnetic; spins
trons cancel one another.
(Section 9.8) Diamagnetic subaligned parallel to each
random; spins do align
stances are generally described as being nonmagnetic, but when a
other
if in a magnetic field
diamagnetic substance is placed in a magnetic field, the motions of
(a)
(b)
the electrons cause the substance to be very weakly repelled by the
magnet. In other words, these supposedly nonmagnetic substances
do show some very faint magnetic character in the presence of a
magnetic field.
A substance in which the atoms or ions have one or more
unpaired electrons is paramagnetic.
(Section 9.8) In a paramagnetic solid, the electrons on one atom or ion do not influence
the unpaired electrons on neighboring atoms or ions. As a result,
the magnetic moments on the atoms or ions are randomly oriented
and constantly changing direction, as shown in ▶ Figure 23.6(a).
When a paramagnetic substance is placed in a magnetic field, howAntiferromagnetic; spins
Ferrimagnetic; unequal spins
ever, the magnetic moments tend to align parallel to one another,
align in opposite directions
align in opposite directions but do
producing a net attractive interaction with the magnet. Thus,
and cancel each other
not completely cancel each other
unlike a diamagnetic substance, which is weakly repulsed by a
(c)
(d)
magnetic field, a paramagnetic substance is attracted to a magnetic
field.
▲ Figure 23.6 The relative orientation of
When you think of a magnet, you probably picture a simple
electron spins in various types of magnetic
iron magnet. Iron exhibits ferromagnetism, a form of magnetism much stronger than substances.
paramagnetism. Ferromagnetism arises when the unpaired electrons of the atoms or
ions in a solid are influenced by the orientations of the electrons in neighboring atoms
or ions. The most stable (lowest-energy) arrangement is when the spins of electrons on
adjacent atoms or ions are aligned in the same direction, as in Figure 23.6(b). When
a ferromagnetic solid is placed in a magnetic field, the electrons tend to align strongly
in a direction parallel to the magnetic field. The attraction to the magnetic field that
results may be as much as one million times stronger than that for a paramagnetic
substance.
When a ferromagnet is removed from an external magnetic field, the interactions between the electrons cause the ferromagnetic substance to maintain a magnetic
moment. We then refer to it as a permanent magnet (▶ Figure 23.7).
The only ferromagnetic transition metals are Fe, Co, and Ni, but many alloys also
exhibit ferromagnetism, which is in some cases stronger than the ferromagnetism of
the pure metals. Particularly powerful ferromagnetism is found in compounds containing both transition metals and lanthanide metals. Two of the most important examples
are SmCo5 and Nd2Fe14B.
Two additional types of magnetism involving ordered arrangements of unpaired
electrons are depicted in Figure 23.6. In materials that exhibit antiferromagnetism ▲ Figure 23.7 A permanent magnet.
[Figure 23.6(c)], the unpaired electrons on a given atom or ion align so that their Permanent magnets are made from
spins are oriented in the direction opposite the spin direction on neighboring atoms. ferromagnetic and ferrimagnetic materials.
1002
chapter 23 Transition Metals and Coordination Chemistry
This means that the spin-up and spin-down electrons cancel each other. Examples of
antiferromagnetic substances are chromium metal, FeMn alloys, and such transitionmetal oxides as Fe2O3, LaFeO3, and MnO.
A substance that exhibits ferrimagnetism [Figure 23.6(d)] has both ferromagnetic
and antiferromagnetic characteristics. Like an antiferromagnet, the unpaired electrons
align so that the spins in adjacent atoms or ions point in opposite directions. However,
unlike an antiferromagnet, the net magnetic moments of the spin-up electrons are not
fully canceled by the spin-down electrons. This can happen because the magnetic centers have different numbers of unpaired electrons (NiMnO3), because the number of
magnetic sites aligned in one direction is larger than the number aligned in the other
direction (Y3Fe5O12), or because both these conditions apply (Fe3O4). Because the magnetic moments do not cancel, the properties of ferrimagnetic materials are similar to
the properties of ferromagnetic materials.
Give It Some Thought
How do you think spin–spin interactions of unpaired electrons on adjacent atoms
in a substance are affected by the interatomic distance?
Ferromagnets, ferrimagnets, and antiferromagnets all become paramagnetic when
heated above a critical temperature. This happens when the thermal energy is sufficient
to overcome the forces determining the spin directions of the electrons. This temperature is called the Curie temperature, TC , for ferromagnets and ferrimagnets and the Néel
temperature, TN , for antiferromagnets.
23.2 | Transition-Metal Complexes
The transition metals occur in many interesting and important molecular forms. Species that are assemblies of a central transition-metal ion bonded to a group of surrounding molecules or ions, such as 3Ag(NH3)24 + and 3Fe(H2O)643+ , are called metal
complexes, or merely complexes.* If the complex carries a net charge, it is generally
called a complex ion.
(Section 17.5) Compounds that contain complexes are known
as coordination compounds.
The molecules or ions that bond to the metal ion in a complex are known as
ligands (from the Latin word ligare, “to bind”). There are two NH3 ligands bonded
to Ag + in the complex ion 3Ag(NH3)24 + , for instance, and six H2O ligands bonded to
Fe3+ in 3Fe(H2O)643+ . Each ligand functions as a Lewis base and so donates a pair of
electrons to form the ligand–metal bond.
(Section 16.11) Thus, every ligand has at
least one unshared pair of valence electrons. Four of the most frequently encountered
ligands,
H
O
H
H
N
H
Cl −
C
N−
H
illustrate that most ligands are either polar molecules or anions. In forming a complex,
the ligands are said to coordinate to the metal.
Give It Some Thought
Is the interaction between an ammonia ligand and a metal cation a Lewis acid–
base interaction? If so, which species acts as the Lewis acid?
*Most of the coordination compounds we examine in this chapter contain transition-metal ions, although
ions of other metals can also form complexes.
section 23.2 Transition-Metal Complexes
The Development of Coordination Chemistry:
Werner’s Theory
Because compounds of the transition metals are beautifully colored, the chemistry of
these elements fascinated chemists even before the periodic table was introduced. During the late 1700s through the 1800s, the many coordination compounds that were isolated and studied had properties that were puzzling in light of the bonding theories
prevailing at the time. ▼ Table 23.3, for example, lists a series of CoCl3 9NH3 compounds that have strikingly different colors. Note that the third and fourth species have
different colors even though the originally assigned formula was the same for both,
CoCl3 # 4 NH3.
The modern formulations of the compounds in Table 23.3 are based on various
lines of experimental evidence. For example, all four compounds are strong electrolytes
(Section 4.1) but yield different numbers of ions when dissolved in water.
Dissolving CoCl3 # 6 NH3 in water yields four ions per formula unit (3Co(NH3)643+
plus three Cl- ions), whereas CoCl3 # 5 NH3 yields only three ions per formula unit
(3Co(NH3)5Cl42+ and two Cl- ions). Furthermore, the reaction of the compounds with
excess aqueous silver nitrate leads to the precipitation of different amounts of AgCl(s).
When CoCl3 # 6 NH3 is treated with excess AgNO3(aq), 3 mol of AgCl(s) precipitate
per mole of complex, which means all three Cl- ions in the complex can react to form
AgCl(s). By contrast, when CoCl3 # 5 NH3 is treated with excess AgNO3(aq), only 2 mol
of AgCl(s) precipitate per mole of complex, telling us that one of the Cl- ions in the
complex does not react. These results are summarized in Table 23.3.
In 1893 the Swiss chemist Alfred Werner (1866–1919) proposed a theory that successfully explained the observations in Table 23.3. In a theory that became the basis for
understanding coordination chemistry, Werner proposed that any metal ion exhibits
both a primary valence and a secondary valence. The primary valence is the oxidation
state of the metal, which is +3 for the complexes in Table 23.3.
(Section 4.4) The
secondary valence is the number of atoms bonded to the metal ion, which is also called
the coordination number. For these cobalt complexes, Werner deduced a coordination number of 6 with the ligands in an octahedral arrangement around the Co3+ ion.
Werner’s theory provided a beautiful explanation for the results in Table 23.3. The
NH3 molecules are ligands bonded to the Co3+ ion (through the nitrogen atom as we will
see later); if there are fewer than six NH3 molecules, the remaining ligands are Cl- ions. The
central metal and the ligands bound to it constitute the coordination sphere of the complex.
In writing the chemical formula for a coordination compound, Werner suggested
using square brackets to signify the makeup of the coordination sphere in any given
compound. He therefore proposed that CoCl3 # 6 NH3 and CoCl3 # 5 NH3 are better
written as 3Co(NH3)64Cl3 and 3Co(NH3)5Cl4Cl2, respectively. He further proposed
that the chloride ions that are part of the coordination sphere are bound so tightly
that they do not dissociate when the complex is dissolved in water. Thus, dissolving
3Co(NH3)5Cl4Cl2 in water produces a 3Co(NH3)5Cl42+ ion and two Cl- ions.
Werner’s ideas also explained why there are two forms of CoCl3 # 4 NH3. Using
Werner’s postulates, we write the formula as 3Co(NH3)4Cl24Cl. As shown in Figure 23.8,
there are two ways to arrange the ligands in the 3Co(NH3)4Cl24 + complex, called the cis
and trans forms. In the cis form, the two chloride ligands occupy adjacent vertices of
the octahedral arrangement. In trans@3Co(NH3)4Cl24 + the two chlorides are opposite
Table 23.3 Properties of Some Ammonia Complexes of Cobalt(III)
Original
Formulation
CoCl3 # 6 NH3
CoCl3 # 5 NH3
CoCl3 # 4 NH3
CoCl3 # 4 NH3
Color
Ions per
“Free” Cl− Ions
Formula Unit per Formula Unit
Orange
4
3
Purple
3
2
Green
2
1
Violet
2
1
Modern Formulation
3Co(NH3)64Cl3
3Co(NH3)5Cl4Cl2
trans@3Co(NH3)4Cl24Cl
cis@3Co(NH3)4Cl24Cl
1003
1004
chapter 23 Transition Metals and Coordination Chemistry
Go Figure
Is there another way to arrange the chloride ions in the 3Co(NH3)4Cl24+ ion besides the two shown in this figure?
+
+
Two Cl on
opposite
sides of
metal ion
Two Cl on
same side of
metal ion
cis isomer
trans isomer
▲ Figure 23.8 Isomers of [Co(NH3)4Cl2]+. The cis isomer is violet, and the trans isomer
is green.
each other. It is this difference in positions of the Cl ligands that leads to two compounds, one violet and one green.
The insight Werner provided into the bonding in coordination compounds is even
more remarkable when we realize that his theory predated Lewis’s ideas of covalent
bonding by more than 20 years! Because of his tremendous contributions to coordination chemistry, Werner was awarded the 1913 Nobel Prize in Chemistry.
S a mp l e
Exercise 23.1 Identifying the Coordination Sphere
of a Complex
Palladium(II) tends to form complexes with coordination number 4. A compound has the
composition PdCl2 # 3 NH3. (a) Write the formula for this compound that best shows the coordination structure. (b) When an aqueous solution of the compound is treated with excess
AgNO3(aq), how many moles of AgCl(s) are formed per mole of PdCl2 # 3 NH3?
Solution
Analyze We are given the coordination number of Pd(II) and a chemical formula indicating that
the complex contains NH3 and Cl-. We are asked to determine (a) which ligands are attached to
Pd(II) in the compound and (b) how the compound behaves toward AgNO3 in aqueous solution.
Plan (a) Because of their charge, the Cl- ions can be either in the coordination sphere, where
they are bonded directly to the metal, or outside the coordination sphere, where they are bonded
ionically to the complex. The electrically neutral NH3 ligands must be in the coordination
sphere, if we assume four ligands bonded to the Pd(II) ion. (b) Any chlorides in the coordination
sphere do not precipitate as AgCl.
Solve
(a) By analogy to the ammonia complexes of cobalt(III) shown in Figure 23.7, we predict that
the three NH3 are ligands attached to the Pd(II) ion. The fourth ligand around Pd(II) is one
chloride ion. The second chloride ion is not a ligand; it serves only as a counterion (a noncoordinating ion that balances charge) in the compound. We conclude that the formula showing the structure best is 3Pd(NH3)3Cl4Cl.
section 23.2 Transition-Metal Complexes
(b) Because only the non-ligand Cl- can react, we expect to produce 1 mol of AgCl(s) per mole
of complex. The balanced equation is
3Pd(NH3)3Cl4Cl(aq) + AgNO3(aq) ¡ 3Pd(NH3)3Cl4NO3(aq) + AgCl(s)
This is a metathesis reaction
3Pd(NH3)3Cl4+ complex ion.
(Section 4.2) in which one of the cations is the
Practice Exercise 1
When the compound RhCl3 # 4 NH3 is dissolved in water and treated with excess AgNO3(aq)
one mole of AgCl(s) is formed for every mole of RhCl3 # 4 NH3. What is the correct way
to write the formula of this compound? (a) 3Rh(NH3)4Cl34, (b) 3Rh(NH3)4Cl24Cl,
(c) 3Rh(NH3)4Cl4Cl2, (d) 3Rh(NH3)44Cl3, (e) 3RhCl34(NH3)4.
Practice Exercise 2
Predict the number of ions produced per formula unit when the compound CoCl2 # 6 H2O
dissolves in water to form an aqueous solution.
The Metal–Ligand Bond
The bond between a ligand and a metal ion is a Lewis acid–base interaction.
(Section 16.11) Because the ligands have available pairs of electrons, they can function as Lewis bases (electron-pair donors). Metal ions (particularly transition-metal
ions) have empty valence orbitals, so they can act as Lewis acids (electron-pair acceptors). We can picture the bond between the metal ion and ligand as the result of their
sharing a pair of electrons initially on the ligand:
H
H
Ag+(aq) + 2 N
H
H(aq)
H
N Ag N
H
+
H
[23.1]
H (aq)
H
The formation of metal–ligand bonds can markedly alter the properties we observe
for the metal ion. A metal complex is a distinct chemical species that has physical and
chemical properties different from those of the metal ion and ligands from which it is
formed. As one example, ▼ Figure 23.9 shows the color change that occurs when aqueous
solutions of NCS- (colorless) and Fe3+ (yellow) are mixed, forming 3Fe(H2O)5NCS42+ .
Go Figure
Write a balanced chemical equation for the reaction depicted in this figure.
NH4NCS(aq)
solution
[Fe(H2O)6]3+(aq)
solution
Red [Fe(H2O)5NCS]2+
forms
▲ Figure 23.9 Reaction of Fe3+ (aq) and NCS− (aq).
1005
1006
chapter 23 Transition Metals and Coordination Chemistry
Complex formation can also significantly change other properties of metal ions,
such as their ease of oxidation or reduction. Silver ion, for example, is readily reduced
in water,
Ag + (aq) + e- ¡ Ag(s)
E° = +0.799 V[23.2]
but the 3Ag(CN)24 - ion is not so easily reduced because complexation by CN- ions
stabilizes silver in the +1 oxidation state:
3Ag(CN)24 - (aq) + e- ¡ Ag(s) + 2 CN- (aq) E° = -0.31 V[23.3]
Hydrated metal ions are complexes in which the ligand is water. Thus, Fe3+ (aq)
consists largely of 3Fe(H2O)643+ .
(Section 16.11) It is important to realize that
ligands can undergo reaction. For example, we saw in Figure 16.16 that a water molecule in 3Fe(H2O)643+ (aq) can be deprotonated to yield 3Fe(H2O)5OH42+ (aq) and
H+ (aq). The iron ion retains its oxidation state; the coordinated hydroxide ligand, with
a 1- charge, reduces the complex charge to 2+. Ligands can also be displaced from the
coordination sphere by other ligands, if the incoming ligands bind more strongly to the
metal ion than the original ones. For example, ligands such as NH3, NCS- , and CNcan replace H2O in the coordination sphere of metal ions.
Charges, Coordination Numbers, and Geometries
The charge of a complex is the sum of the charges on the metal and on the ligands. In
3Cu(NH3)44SO4 we can deduce the charge on the complex ion because we know that
the charge of the sulfate ion is 2-. Because the compound is electrically neutral, the
complex ion must have a 2+ charge, 3Cu(NH3)442+ . We can then use the charge of
the complex ion to deduce the oxidation number of copper. Because the NH3 ligands
are uncharged molecules, the oxidation number of copper must be +2:
+2 + 4(0) = +2
[Cu(NH3)4]2+
S a mp l e
Exercise 23.2 Determining the Oxidation Number of a Metal in a Complex
What is the oxidation number of the metal in 3Rh(NH3)5Cl4(NO3)2?
Solution
Analyze We are given the chemical formula of a coordination compound and asked to determine the oxidation number of its metal
atom.
Plan To determine the oxidation number of Rh, we need to figure out
what charges are contributed by the other groups. The overall charge
is zero, so the oxidation number of the metal must balance the charge
due to the rest of the compound.
Solve The NO3 group is the nitrate anion, which has a 1 - charge. The
NH3 ligands carry zero charge, and the Cl is a coordinated chloride
ion, which has a 1 - charge. The sum of all the charges must be zero:
Practice Exercise 1
In which of the following compounds does the transition-metal
have the highest oxidation number? (a) 3Co(NH3)4Cl2],
(b) K23PtCl64, (c) Rb33MoO3F34, (d) Na3Ag(CN)24,
(e) K43Mn(CN)64.
Practice Exercise 2
What is the charge of the complex formed by a platinum(II) metal
ion surrounded by two ammonia molecules and two bromide
ions?
+ 5(0) + (−1) + 2(−1) = 0
[Rh(NH3)5Cl](NO3)2
The oxidation number of rhodium, x, must therefore be +3.
Recall that the number of atoms directly bonded to the metal atom in a complex
is the coordination number of the complex. Thus, the silver ion in 3Ag(NH3)24 + has a
coordination number of 2, and the cobalt ion has a coordination number of 6 in all four
complexes in Table 23.3.
section 23.3 Common Ligands in Coordination Chemistry
1007
Some metal ions have only one observed coordination number. The coordination number of chromium(III) and cobalt(III) is invariably 6, for example, and that of
Go Figure
platinum(II) is always 4. For most metals, however, the coordination number is different for
In the drawings on the right-hand side,
different ligands. In these complexes, the most common coordination numbers are 4 and 6.
what does the solid wedge connecting
The coordination number of a metal ion is often influenced by the relative sizes of the atoms represent? What does the
metal ion and the ligands. As the ligand gets larger, fewer of them can coordinate to the metal dashed wedge connecting atoms
ion. Thus, iron(III) is able to coordinate to six fluorides in 3FeF643- but to only four chlorides represent?
in 3FeCl44 - . Ligands that transfer substantial negative charge to the metal
A tetrahedron has four
also produce reduced coordination numbers. For example, six ammonia
2+
triangular faces and
molecules can coordinate to nickel(II), forming 3Ni(NH3)64 , but only
four equivalent vertices.
2−
four cyanide ions can coordinate to this ion, forming 3Ni(CN)44 .
Cl
The most common coordination geometries for coordination comFe
plexes are shown in ▶ Figure 23.10. Complexes in which the coordinaCl
Cl
tion number is 4 have two geometries—tetrahedral and square planar.
Cl
The tetrahedral geometry is the more common of the two and is especially prevalent among nontransition metals. The square planar geomTetrahedral geometry
etry is characteristic of transition-metal ions with eight d electrons in
the valence shell, such as platinum(II) and gold(III). Complexes with a
The metal and all four
coordination number of 6 almost always have an octahedral geometry.
ligands lie in the same plane.
Even though the octahedron can be drawn as a square with one ligand
2−
Cl
above and another below the plane, all six vertices are equivalent.
Cl
23.3 | Common Ligands in
Coordination Chemistry
Pt
Cl
Cl
Square planar geometry
An octahedron has eight
The ligand atom that binds to the central metal ion in a coordination
triangular faces and six
complex is called the donor atom of the ligand. Ligands having only one
equivalent vertices.
3−
donor atom are called monodentate ligands (from the Latin, meaning
F
F
“one-toothed”). These ligands are able to occupy only one site in a coordinaF Fe
F
tion sphere. Ligands having two donor atoms are bidentate ligands (“twoF
toothed”), and those having three or more donor atoms are polydentate
F
ligands (“many-toothed”). In both bidentate and polydentate species, the
multiple donor atoms can simultaneously bond to the metal ion, thereby
Octahedral geometry
occupying two or more sites in a coordination sphere. Table 23.4 gives ex▲ Figure 23.10 Common geometries of
amples of all three types of ligands.
coordination complexes. In complexes having
Because they appear to grasp the metal between two or more donor atoms, coordination number 4, the geometry is
bidentate and polydentate ligands are also known as chelating agents (pronounced typically either tetrahedral or square planar. In
“KEE-lay-ting”; from the Greek chele, “claw”).
complexes having coordination number 6, the
One common chelating agent is the bidentate ligand ethylenediamine, denoted en: geometry is nearly always octahedral.
CH2
H2N
CH2
NH2
in which each donor nitrogen atom has one nonbonding electron pair. These donor
atoms are sufficiently far apart to allow both of them to bond to the metal ion in adjacent positions. The 3Co(en)343 + complex ion, which contains three ethylenediamine
ligands in the octahedral coordination sphere of cobalt(III), is shown in Figure 23.11.
Notice that in the image on the right the en is written in a shorthand notation as two
nitrogen atoms connected by an arc.
The ethylenediaminetetraacetate ion, 3EDTA44- , is an important polydentate ligand
that has six donor atoms. It can wrap around a metal ion using all six donor atoms, as shown
in Figure 23.12, although it sometimes binds to a metal using only five of its donor atoms.
Give It Some Thought
Both H2O and ethylenediamine (en) have two nonbonding pairs of electrons.
Why cannot water act as a bidentate ligand?
1008
chapter 23 Transition Metals and Coordination Chemistry
Table 23.4 Some Common Ligands
Ligand Type
Monodentate
Examples
H2O
F
−
Cl
−
Water
NH3 Ammonia
N ]−
[C
Fluoride ion
Chloride ion
[S
C
or
N
]−
[O
Thiocyanate ion
O
H2C
CH2
N
N
N
Bipyridine
(bipy or bpy)
Ethylenediamine (en)
Polydentate
H2N
CH2
CH2
O
CH2
NH2
NH
N
O
P
O
O
Diethylenetriamine
O
O
Nitrite ion
O
O
O
2−
O
O
C
O
Carbonate ion
5−
O
P
2−
C
Oxalate ion
Ortho-phenanthroline
(o-phen)
O
H2C
O
C
NH2
N
Hydroxide ion
]−
or
Bidentate
H2N
H ]−
[O
Cyanide ion
P
O
O
Triphosphate ion
4−
O
O
O
C
CH2
O
C
CH2
N
CH2
CH2
CH2
C
O
CH2
C
O
N
O
O
4−
Ethylenediaminetetraacetate ion (EDTA
)
3+
3+
H2
C
H2C
H2C
−
H2 H2N
N
Co
N
H2 H N
2
CH2
NH2
=
NH2
C
H2
CH2
[Co(en)3]3+
▲ Figure 23.11 The 3 Co(en)3 4 3+ ion. The ligand is ethylenediamine.
▲ Figure 23.12 The complex ion
3 Co(EDTA) 4 −. The ligand is the polydentate
ethylenediaminetetraacetate ion, whose full
representation is given in Table 23.2. This
representation shows how the two N and four
O donor atoms coordinate to cobalt.
In general, the complexes formed by chelating ligands (that is, bidentate and polydentate ligands) are more stable than the complexes formed by related monodentate
ligands. The equilibrium formation constants for 3Ni(NH3)642+ and 3Ni(en)342+ illustrate this observation:
3Ni(H2O)642+ (aq) + 6 NH3(aq) ∆ 3Ni(NH3)642+ (aq) + 6 H2O(l)
Kf = 1.2 * 109[23.4]
3Ni(H2O)642+ (aq) + 3 en(aq) ∆ 3Ni(en)342+ (aq) + 6 H2O(l)
Kf = 6.8 * 1017[23.5]
section 23.3 Common Ligands in Coordination Chemistry
Although the donor atom is nitrogen in both instances, 3Ni(en)342+ has a formation
constant that is more than 108 times larger than that of 3Ni(NH3)642+ . This trend of
generally larger formation constants for bidentate and polydentate ligands, known as
the chelate effect, is examined in the “A Closer Look” essay on page 1010.
Chelating agents are often used to prevent one or more of the customary reactions
of a metal ion without removing the ion from solution. For example, a metal ion that
interferes with a chemical analysis can often be complexed and its interference thereby
removed. In a sense, the chelating agent hides the metal ion. For this reason, scientists
sometimes refer to these ligands as sequestering agents.
Phosphate ligands, such as sodium tripolyphosphate, Na53OPO2OPO2OPO34, are
used to sequester Ca2+ and Mg2+ ions in hard water so that these ions cannot interfere
with the action of soap or detergents.
Chelating agents are used in many prepared foods, such as salad dressings and frozen desserts, to complex trace metal ions that catalyze decomposition reactions. Chelating agents are used in medicine to remove toxic heavy metal ions that have been
ingested, such as Hg2+ , Pb2+ , and Cd2+ . One method of treating lead poisoning, for
example, is to administer Na2Ca(EDTA). The EDTA chelates the lead, allowing it to be
removed from the body via urine.
Give It Some Thought
Cobalt(III) has a coordination number of 6 in all its complexes.
Is the carbonate ion a monodentate or bidentate ligand in the
3Co(NH3)4(CO3)4 + ion?
Metals and Chelates in Living Systems
1009
Go Figure
What is the coordination number
of the metal ion in heme b? In
chlorophyll a?
NH
N
N
HN
Porphine
Ten of the 29 elements known to be necessary for human life are
transition metals.
(Section 2.7, “Elements Required by Living
Organisms”) These ten elements—V, Cr, Mn, Fe, Co, Ni, Cu, Zn,
Mo, and Cd—form complexes with a variety of groups present in
N
N
biological systems.
2+
Fe
Although our bodies require only small quantities of metals,
N
N
deficiencies can lead to serious illness. A deficiency of manganese,
for example, can lead to convulsive disorders. Some epilepsy patients
have been helped by the addition of manganese to their diets.
Among the most important chelating agents in nature are those
OH
O
OH
derived from the porphine molecule (▶ Figure 23.13). This molecule
O
can coordinate to a metal via its four nitrogen donor atoms. Once
Heme b
porphine bonds to a metal ion, the two H atoms on the nitrogens are
displaced to form complexes called porphyrins. Two important porCH2
phyrins are hemes, in which the metal ion is Fe(II), and chlorophylls,
CH3
with a Mg(II) central ion.
H3C
CH3
Figure 23.14 shows a schematic structure of myoglobin, a proN
N
2
+
tein that contains one heme group. Myoglobin is a globular protein,
Mg
one that folds into a compact, roughly spherical shape. Myoglobin
N
N
CH3
is found in the cells of skeletal muscle, particularly in seals, whales,
H3C
and porpoises. It stores oxygen in cells, one molecule of O2 per myoCH3
CH3
CH3
globin, until it is needed for metabolic activities. Hemoglobin, the
O
O
OCH3
protein that transports oxygen in human blood, is made up of four H3C
O
O
2
heme-containing subunits, each of which is very similar to myogloChlorophyll a
bin. One hemoglobin can bind up to four O2 molecules.
In both myoglobin and hemoglobin, the iron is coordinated to the four nitrogen ▲ Figure 23.13 Porphine and two
atoms of a porphyrin and to a nitrogen atom from the protein chain (Figure 23.15). In porphyrins, heme b and chlorophyll a. Fe(II)
hemoglobin, the sixth position around the iron is occupied either by O2 (in oxyhemo- and Mg(II) ions replace the two blue H atoms
globin, the bright red form) or by water (in deoxyhemoglobin, the purplish red form). in porphine and bond with all four nitrogens
in heme b and chlorophyll a, respectively.
(The oxy form is the one shown in Figure 23.15.)
1010
chapter 23 Transition Metals and Coordination Chemistry
Heme
Go Figure
▲ Figure 23.14 Myoglobin. This ribbon diagram does not show most of the atoms.
What is the coordination number of iron
in the heme unit shown here? What is
the identity of the donor atoms?
O
O
Heme
N
HN
Protein (globin)
▲ Figure 23.15 Coordination sphere of the
hemes in oxymyoglobin and oxyhemoglobin.
Carbon monoxide is poisonous because the equilibrium binding constant of
human hemoglobin for CO is about 210 times greater than that for O2. As a result, a
relatively small quantity of CO can inactivate a substantial fraction of the hemoglobin in the blood by displacing the O2 molecule from the heme-containing subunit. For
example, a person breathing air that contains only 0.1% CO takes in enough CO after a
few hours to convert up to 60% of the hemoglobin (Hb) into COHb, thereby reducing
the blood’s normal oxygen-carrying capacity by 60%.
Under normal conditions, a nonsmoker breathing unpolluted air has about
0.3 to 0.5% COHb in her or his blood. This amount arises mainly from the production
of small quantities of CO in the course of normal body chemistry and from the small
amount of CO present in clean air. Exposure to higher concentrations of CO causes
the COHb level to increase, which in turn leaves fewer Hb sites to which O2 can bind.
If the level of COHb becomes too high, oxygen transport is effectively shut down and
death occurs. Because CO is colorless and odorless, CO poisoning occurs with very little warning. Improperly ventilated combustion devices, such as kerosene lanterns and
stoves, thus pose a potential health hazard.
A Closer Look
Entropy and the Chelate Effect
We learned in Section 19.5 that chemical processes are favored by positive
entropy changes and by negative enthalpy changes. The special stability
associated with the formation of chelates, called the chelate effect, can be
explained by comparing the entropy changes that occur with monodentate ligands with the entropy changes that occur with polydentate ligands.
We begin with the reaction in which two H2O ligands of the
square-planar Cu(II) complex 3Cu(H2O)442+ are replaced by monodentate NH3 ligands at 27 °C:
3Cu(H2O)442+(aq) + 2 NH3(aq) ∆
3Cu(H2O)2(NH3)242+(aq) + 2 H2O(l)
∆H° = - 46 kJ; ∆S° = -8.4 J>K; ∆G° = -43 kJ
The thermodynamic data tell us about the relative abilities of
H2O and NH3 to serve as ligands in this reaction. In general, NH3
binds more tightly to metal ions than does H2O, so this substitution
reaction is exothermic (∆H 6 0). The stronger bonding of the NH3
ligands also causes the 3Cu(H2O)2(NH3)242+ ion to be more rigid,
which is probably the reason ∆S° is slightly negative.
We can use Equation 19.20, ∆G° = - RT ln K, to calculate the equilibrium constant of the reaction at 27 °C. The result,
K = 3.1 * 107, tells us that the equilibrium lies far to the right, favoring replacement of H2O by NH3. For this equilibrium, therefore,
the enthalpy change, ∆H ° = - 46 kJ, is large enough and negative
enough to overcome the entropy change, ∆S° = - 8.4 J>K.
Now let’s use a single bidentate ethylenediamine (en) ligand in
our substitution reaction:
3Cu(H2O)442+(aq) + en(aq) ∆
∆H° = -54 kJ;
3Cu(H2O)2(en)42+(aq) + 2 H2O(l)
∆S° = +23 J>K;
∆G° = - 61 kJ
The en ligand binds slightly more strongly to the Cu2+ ion than two NH3
ligands, so the enthalpy change here ( -54 kJ) is slightly more negative
than for 3Cu(H2O)2(NH3)242+( - 46 kJ). There is a big difference in the
section 23.3 Common Ligands in Coordination Chemistry
entropy change, however: ∆S° is -8.4 J>K for the NH3 reaction but
+23 J>K for the en reaction. We can explain the positive ∆S° value using
concepts discussed in Section 19.3. Because a single en ligand occupies
two coordination sites, two molecules of H2O are released when one en
ligand bonds. Thus, there are three product molecules in the reaction but
only two reactant molecules. The greater number of product molecules
leads to the positive entropy change for the equilibrium.
The slightly more negative value of ∆H ° for the en reaction
( -54 kJ versus - 46 kJ) coupled with the positive entropy change leads
to a much more negative value of ∆G° (–61 kJ for en, -43 kJ for NH3)
and thus a larger equilibrium constant: K = 4.2 * 1010.
We can combine our two equations using Hess’s law
(Section 5.6) to calculate the enthalpy, entropy, and free-energy changes
that occur for en to replace ammonia as ligands on Cu(II):
3Cu(H2O)2(NH3)242+(aq) + en(aq) ∆
3Cu(H2O)2(en)42+(aq) + 2 NH3(aq)
∆H ° = ( -54 kJ) - ( -46 kJ) = -8 kJ
∆S° = ( +23 J>K) - (-8.4 J>K) = + 31 J>K
∆G° = (-61 kJ) - (-43 kJ) = -18 kJ
Notice that at 27 °C, the entropic contribution ( -T∆S°) to the
free-energy change, ∆G° = ∆H ° - T∆S° (Equation 19.12), is negative and greater in magnitude than the enthalpic contribution (∆H °).
The equilibrium constant for the NH3 9en reaction, 1.4 * 103, shows
that the replacement of NH3 by en is thermodynamically favorable.
The chelate effect is important in biochemistry and molecular biology. The additional thermodynamic stabilization provided
by entropy effects helps stabilize biological metal–chelate complexes, such as porphyrins, and can allow changes in the oxidation
state of the metal ion while retaining the structural integrity of the
complex.
Related Exercises: 23.31, 23.32, 23.96, 23.98
The chlorophylls, which are porphyrins that contain Mg(II) (Figure 23.13),
are the key components in the conversion of solar energy into forms that can
be used by living organisms. This process, called photosynthesis, occurs in the
leaves of green plants:
Go Figure
Which peak in this curve corresponds
to the lowest-energy transition by an
electron in a chlorophyll molecule?
6 CO2(g) + 6 H2O(l) ¡ C6H12O6(aq) + 6 O2(g) [23.6]
The formation of 1 mol of glucose, C6H12O6, requires the absorption of 48 mol
of photons from sunlight or other sources of light. Chlorophyll-containing pigments in the leaves of plants absorb the photons. Figure 23.13 shows that the
chlorophyll molecule has a series of alternating, or conjugated, double bonds
in the ring surrounding the metal ion. This system of conjugated double bonds
makes it possible for chlorophyll to absorb light strongly in the visible region of
the spectrum. As ▶ Figure 23.16 shows, chlorophyll is green because it absorbs
red light (maximum absorption at 655 nm) and blue light (maximum absorption
at 430 nm) and transmits green light.
Photosynthesis is nature’s solar energy–conversion machine, and thus all
living systems on Earth depend on photosynthesis for continued existence.
Give It Some Thought
What property of the porphine ligand makes it possible for chlorophyll to
play a role in plant photosynthesis?
Light absorption
1011
400
500
600
Wavelength (nm)
700
▲ Figure 23.16 The absorption of sunlight
by chlorophyll.
Chemistry and Life
The Battle for Iron
In Living Systems
Because living systems have difficulty assimilating enough iron to
satisfy their nutritional needs, iron-deficiency anemia is a common
problem in humans. Chlorosis, an iron deficiency in plants that makes
leaves turn yellow, is also commonplace.
Living systems have difficulty assimilating iron because most
iron compounds found in nature are not very soluble in water. Microorganisms have adapted to this problem by secreting an iron-binding
compound, called a siderophore, that forms an extremely stable water-soluble complex with iron(III). One such complex is ferrichrome
(Figure 23.17). The iron-binding strength of a siderophore is so great
that it can extract iron from iron oxides.
When ferrichrome enters a living cell, the iron it carries is removed through an enzyme-catalyzed reaction that reduces the
strongly bonding iron(III) to iron(II), which is only weakly complexed
by the siderophore (Figure 23.18 ). Microorganisms thus acquire
iron by excreting a siderophore into their immediate environment
and then taking the resulting iron complex into the cell.
In humans, iron is assimilated from food in the intestine. A protein called transferrin binds iron and transports it across the intestinal wall to distribute it to other tissues in the body. The normal adult
body contains about 4 g of iron. At any one time, about 3 g of this iron
is in the blood, mostly in the form of hemoglobin. Most of the remainder is carried by transferrin.
A bacterium that infects the blood requires a source of iron if it is
to grow and reproduce. The bacterium excretes a siderophore into the
1012
chapter 23 Transition Metals and Coordination Chemistry
H
H
O
H
H
C
N
C
CH2
CH2
C
H CH3
C
N
N
N
CH2 CH2
O
O
O
C
H
C
H CH3
O
C
C
O
N
N
CH2 CH2
C
O
C
H
N
O
Fe3+
−
e
Ferrichrome
N
CH3
CH2 C
C
H
H
H
C
N
Fe3+
Fe3+
Fe2+
CH2 C
O
O
Siderophore
H
CH2
C
Cell wall
O
Fe3+
H
H
▲ Figure 23.18 The iron-transport system of a bacterial cell.
O
H
▲ Figure 23.17 Ferrichrome.
blood to compete with transferrin for iron. The equilibrium constants
for forming the iron complex are about the same for transferrin and
siderophores. The more iron available to the bacterium, the more rapidly it can reproduce and thus the more harm it can do.
Several years ago, New Zealand clinics regularly gave iron supplements to infants soon after birth. However, the incidence of certain
bacterial infections was eight times higher in treated than in untreated
infants. Presumably, the presence of more iron in the blood than absolutely necessary makes it easier for bacteria to obtain the iron needed
for growth and reproduction.
In the United States, it is common medical practice to supplement
infant formula with iron sometime during the first year of life. However, iron supplements are not necessary for infants who breast-feed
because breast milk contains two specialized proteins, lactoferrin and
transferrin, which provide sufficient iron while denying its availability
to bacteria. Even for infants fed with infant formulas, supplementing
with iron during the first several months of life may be ill-advised.
For bacteria to continue to multiply in the blood, they must
synthesize new supplies of siderophores. Synthesis of siderophores
in bacteria slows, however, as the temperature is increased above the
normal body temperature of 37 °C and stops completely at 40 °C. This
suggests that fever in the presence of an invading microbe is a mechanism used by the body to deprive bacteria of iron.
Related Exercise: 23.74
23.4 | Nomenclature and Isomerism
in Coordination Chemistry
When complexes were first discovered, they were named after the chemist who originally
prepared them. A few of these names persist, as, for example, with the dark red substance
NH43Cr(NH3)2(NCS)44, which is still known as Reinecke’s salt. Once the structures of
complexes were more fully understood, it became possible to name them in a more systematic manner. Let’s use two substances to illustrate how coordination compounds are named:
[Co(NH3)5Cl]Cl2
Cation
Anion
Pentaamminechlorocobalt(III)
Chloride
5 NH3
ligands
Na2[MoOCl4]
Cl−
ligand
Cobalt in
+3 oxidation
state
Cation
Anion
Sodium
Tetrachlorooxomolybdate(IV)
4 Cl−
ligands
Oxide, O2−
ligand
Molybdenum
in +4 oxidation
state
section 23.4 Nomenclature and Isomerism in Coordination Chemistry
1.In naming complexes that are salts, the name of the cation is given before the
name of the anion. Thus, in 3Co(NH3)5Cl4Cl2 we name the 3Co(NH3)5Cl42+ cation and then the Cl- .
2.In naming complex ions or molecules, the ligands are named before the metal.
Ligands are listed in alphabetical order, regardless of their charges. Prefixes that give
the number of ligands are not considered part of the ligand name in determining
alphabetical order. Thus, the 3Co(NH3)5Cl42+ ion is pentaamminechlorocobalt(III).
(Be careful to note, however, that the metal is written first in the chemical formula.)
3.The names of anionic ligands end in the letter o, but electrically neutral ligands ordinarily bear the name of the molecules (▲ Table 23.5). Special names
are used for H2O (aqua), NH3 (ammine), and CO (carbonyl). For example,
3Fe(CN)2(NH3)2(H2O)24 + is the diamminediaquadicyanoiron(III) ion.
4.Greek prefixes (di-, tri-, tetra-, penta-, hexa-) are used to indicate the number of each
kind of ligand when more than one is present. If the ligand contains a Greek prefix
(for example, ethylenediamine) or is polydentate, the alternate prefixes bis-, tris-,
tetrakis-, pentakis-, and hexakis- are used and the ligand name is placed in parentheses. For example, the name for 3Co(en)34Br3 is tris(ethylenediamine)-cobalt(III)
bromide.
5.If the complex is an anion, its name ends in -ate. The compound K43Fe(CN)64
is potassium hexacyanoferrate(II), for example, and the ion 3CoCl442 - is
tetrachlorocobaltate(II) ion.
6.The oxidation number of the metal is given in parentheses in Roman numerals
following the name of the metal.
Three examples for applying these rules are
3Ni(NH3)64Br2
Hexaamminenickel(II) bromide
3Co(en)2(H2O)(CN)4Cl2
Aquacyanobis(ethylenediamine)cobalt(III) chloride
Na23MoOCl44
Sodium tetrachlorooxomolybdate(IV)
Table 23.5 Some Common Ligands and Their Names
Ligand
Azide, N3
Name in Complexes
-
Azido
Oxalate,
Name in Complexes
C2O42-
Oxalato
2-
-
Bromo
Oxide, O
Oxo
-
Chloro
Ammonia, NH3
Ammine
Cyano
Carbon monoxide, CO
Carbonyl
Fluoro
Ethylenediamine, en
Ethylenediamine
Hydroxo
Pyridine, C5H5N
Pyridine
Carbonato
Water, H2O
Aqua
Bromide, Br
Chloride, Cl
Cyanide, CN
-
Fluoride, F Hydroxide, OH
Carbonate,
Ligand
-
CO32-
S a mp l e
Exercise 23.3 Naming Coordination Compounds
Name the compounds (a) 3Cr(H2O)4Cl24Cl, (b) K43Ni(CN)44.
Solution
Analyze We are given the chemical formulas for two coordination compounds and assigned the task
of naming them.
Plan To name the complexes, we need to determine the ligands in the complexes, the names of the
ligands, and the oxidation state of the metal ion. We then put the information together following the
rules listed in the text.
Solve
(a) The ligands are four water molecules—tetraaqua—and
two chloride ions—dichloro. By assigning all the oxidation
numbers we know for this molecule, we see that the oxidation number of Cr is +3:
+3 + 4(0) + 2(−1) + (−1) = 0
[Cr(H2O)4Cl2]Cl
1013
1014
chapter 23 Transition Metals and Coordination Chemistry
Thus, we have chromium(III). Finally, the anion
is chloride. The name of the compound is
tetraaquadichlorochromium(III) chloride.
(b) The complex has four cyanide ion ligands, CN-, which
means tetracyano, and the oxidation state of the nickel
is zero:
4(+1) + 0 + 4(−1) = 0
Because the complex is an anion, the metal is indicated
as nickelate(0). Putting these parts together and naming
the cation first, we have potassium tetracyanonickelate(0).
K4[Ni(CN)4]
Practice Exercise 1
What is the name of the compound 3Rh(NH3)4Cl24Cl? (a) Rhodium(III)
tetraamminedichloro chloride, (b) Tetraammoniadichlororhodium(III) chloride,
(c) Tetraamminedichlororhodium(III) chloride, (d) Tetraamminetrichlororhodium(III),
(e) Tetraamminedichlororhodium(II) chloride.
Practice Exercise 2
Name the compounds (a) 3Mo(NH3)3Br34NO3, (b) (NH4)23CuBr44. (c) Write the formula for
sodium diaquabis(oxalato)ruthenate(III).
Isomerism
When two or more compounds have the same composition but a different arrangement
of atoms, we call them isomers.
(Section 2.9) Here we consider two main kinds of
isomers in coordination compounds: structural isomers (which have different bonds)
and stereoisomers (which have the same bonds but different ways in which the ligands
occupy the space around the metal center). Each of these classes also has subclasses, as
shown in ▼ Figure 23.19.
Structural Isomerism
Many types of structural isomerism are known in coordination chemistry, including
the two named in Figure 23.19: linkage isomerism and coordination-sphere isomerism.
Linkage isomerism is a relatively rare but interesting type that arises when a particular
ligand is capable of coordinating to a metal in two ways. The nitrite ion, NO2- , for example, can coordinate to a metal ion through either its nitrogen or one of its oxygens
NO
Do the molecules have
the same atoms?
Not
isomers
YES
Do the molecules have
the same bonds (atoms
linked to the same
partners)?
NO
YES
Structural
isomers
NO
Coordination
sphere
isomers
Do the same
ligands bind to
the metal ion?
Stereoisomers
YES
NO
Linkage
isomers
▲ Figure 23.19 Forms of isomerism in coordination compounds.
Geometrical
isomers
Are the
molecules nonsuperimposable
mirror images?
YES
Optical
isomers
section 23.4 Nomenclature and Isomerism in Coordination Chemistry
(▶ Figure 23.20). When it coordinates through the nitrogen atom, the NO2- ligand is
called nitro; when it coordinates through the oxygen atom, it is called nitrito and is generally written ONO- . The isomers shown in Figure 23.20 have different properties. The
nitro isomer is yellow, for example, whereas the nitrito isomer is red.
Another ligand capable of coordinating through either of two donor atoms is thiocyanate, SCN - , whose potential donor atoms are N and S.
1015
Go Figure
What is the chemical formula and
name for each of the complex ions in
this figure?
Give It Some Thought
Can the ammonia ligand engage in linkage
isomerism? Explain.
Coordination-sphere isomers are isomers that
differ in which species in the complex are ligands
and which are outside the coordination sphere in the
solid. For example, three isomers have the formula
CrCl3(H2O)6. When the ligands are six H2O and the
chloride ions are in the crystal lattice (as counterions),
we have the violet compound 3Cr(H2O)64Cl3. When the
ligands are five H2O and one Cl- , with the sixth H2O
and the two Cl- out in the lattice, we have the green
compound 3Cr(H2O)5Cl4Cl2 # H2O. The third isomer,
3Cr(H2O)4Cl24Cl # 2 H2O, is also a green compound.
In the two green compounds, either one or two water
molecules have been displaced from the coordination
sphere by chloride ions. The displaced H2O molecules
occupy a site in the crystal lattice.
2+
Nitro isomer
Bonding via ligand N atom
Stereoisomerism
Stereoisomers have the same chemical bonds but different spatial arrangements. In the square-planar complex
3Pt(NH3)2Cl24, for example, the chloro ligands can be either adjacent to or opposite
each other (▼ Figure 23.21). (We saw an earlier example of this type of isomerism in
the cobalt complex of Figure 23.8, and we will return to that complex in a moment.)
This form of stereoisomerism, in which the arrangement of the atoms is different but
the same bonds are present, is called geometric isomerism. The isomer on the left in
Go Figure
Which of these isomers has a nonzero dipole moment?
=N
= Cl
cis
Cl ligands adjacent to each other
NH3 ligands adjacent to each other
▲ Figure 23.21 Geometric isomerism.
=H
= Pt
trans
Cl ligands on opposite sides of central atom
NH3 ligands on opposite sides of central atom
2+
Nitrito isomer
Bonding via ligand O atom
▲ Figure 23.20 Linkage isomerism.
1016
chapter 23 Transition Metals and Coordination Chemistry
Figure 23.21, with like ligands in adjacent positions, is the cis isomer, and the isomer on
the right, with like ligands across from one another, is the trans isomer.
Geometric isomers generally have different physical properties and may also have
markedly different chemical reactivities. For example, cis@3Pt(NH3)2Cl24, also called
cisplatin, is effective in the treatment of testicular, ovarian, and certain other cancers,
whereas the trans isomer is ineffective. This is because cisplatin forms a chelate with
two nitrogens of DNA, displacing the chloride ligands. The chloride ligands of the trans
isomer are too far apart to form the N-Pt-N chelate with the nitrogen donors in DNA.
Geometric isomerism is also possible in octahedral complexes when two or more
different ligands are present, as in the cis and trans tetraamminedichlorocobalt(III) ion
in Figure 23.8. Because all the corners of a tetrahedron are adjacent to one another, cis–
trans isomerism is not observed in tetrahedral complexes.
S a mp l e
Exercise 23.4 Determining the Number of Geometric Isomers
The Lewis structure :C ‚ O: indicates that the CO molecule has two lone pairs of electrons.
When CO binds to a transition-metal atom, it nearly always does so by using the C lone pair.
How many geometric isomers are there for tetracarbonyldichloroiron(II)?
Solution
Analyze We are given the name of a complex containing only monodentate ligands, and we must
determine the number of isomers the complex can form.
Plan We can count the number of ligands to determine the coordination number of the Fe and
then use the coordination number to predict the geometry. We can then either make a series of
drawings with ligands in different positions to determine the number of isomers or deduce the
number of isomers by analogy to cases we have discussed.
Solve The name indicates that the complex has four carbonyl (CO) ligands and two chloro (Cl- )
ligands, so its formula is Fe(CO)4Cl2. The complex therefore has a coordination number of 6,
and we can assume an octahedral geometry. Like 3Co(NH3)4Cl24+ (Figure 23.8), it has four ligands of one type and two of another. Consequently, there are two isomers possible: one with the
Cl- ligands across the metal from each other, trans@3Fe(CO)4Cl24, and one with the Cl- ligands
adjacent to each other, cis@3Fe(CO)4Cl24.
Comment It is easy to overestimate the number of geometric isomers. Sometimes different orientations of a single isomer are incorrectly thought to be different isomers. If two structures can
be rotated so that they are equivalent, they are not isomers of each other. The problem of identifying isomers is compounded by the difficulty we often have in visualizing three-dimensional
molecules from their two-dimensional representations. It is sometimes easier to determine the
number of isomers if we use three-dimensional models.
Practice Exercise 1
Which of the following molecules does not have a geometric isomer?
[MX3Y]
[MX2Y2]
[MX4Y2]
[MX3Y3]
(a)
(b)
(c)
(d)
Practice Exercise 2
How many isomers exist for the square-planar molecule 3Pt(NH3)2ClBr4?
The second type of stereoisomerism listed in Figure 23.19 is optical isomerism.
Optical isomers, called enantiomers, are mirror images that cannot be superimposed
on each other. They bear the same resemblance to each other that your left hand bears
to your right hand. If you look at your left hand in a mirror, the image is identical to
your right hand (▶ Figure 23.22). No matter how hard you try, however, you cannot
section 23.4 Nomenclature and Isomerism in Coordination Chemistry
Mirror
Mirror
N
N
Co
Left hand
Enantiomers of [Co(en)3]3+
Mirror image of left hand
is identical to right hand
▲ Figure 23.22 Optical isomerism.
superimpose your two hands on each other. An example of a complex that exhibits this
type of isomerism is the 3Co(en)343 + ion. Figure 23.22 shows the two enantiomers of
this complex and their mirror-image relationship. Just as there is no way that we can
twist or turn our right hand to make it look identical to our left, so also there is no way to
rotate one of these enantiomers to make it identical to the other. Molecules or ions that
are not superimposable on their mirror image are said to be chiral (pronounced KY-rul).
S a mp l e
Exercise 23.5 Predicting Whether a Complex Has
Optical Isomers
Does either cis-3Co(en)2Cl24+ or trans@3Co(en)2Cl24- have optical isomers?
Solution
Analyze We are given the chemical formula for two geometric isomers and asked to determine
whether either one has optical isomers. Because en is a bidentate ligand, we know that both complexes are octahedral and both have coordination number 6.
Plan We need to sketch the structures of the cis and trans isomers and their mirror images. We
can draw the en ligand as two N atoms connected by an arc. If the mirror image cannot be superimposed on the original structure, the complex and its mirror image are optical isomers.
Solve The trans isomer of 3Co(en)2Cl24+ and its mirror image are:
Mirror
Cl
Cl
N
N
N
Co
N
N
Co
N
N
N
Cl
Cl
Notice that the mirror image of the isomer is identical to the original. Consequently
trans-3Co(en)2Cl24+ does not exhibit optical isomerism.
The mirror image of the cis isomer cannot be superimposed on the original:
Mirror
N
N
N
Co
N
Cl
N
N
Cl
Cl
Co
Co
Cl
N
N
1017
1018
chapter 23 Transition Metals and Coordination Chemistry
Thus, the two cis structures are optical isomers (enantiomers). We say that cis-3Co(en)2Cl24+ is a
chiral complex.
Practice Exercise 1
Which of the following complexes has optical isomers? (a) Tetrahedral 3CdBr2Cl242-,
(b) Octahedral 3CoCl4(en)42-, (c) Octahedral 3Co(NH3)4Cl242+, (d) Tetrahedral
3Co(NH3)BrClI4- .
Practice Exercise 2
Does the square-planar complex ion 3Pt(NH3)(N3)ClBr4- have optical isomers? Explain your
answer.
The properties of two optical isomers differ only if the isomers are in a chiral
environment—that is, an environment in which there is a sense of right- and lefthandedness. A chiral enzyme, for example, might catalyze the reaction of one optical isomer but not the other. Consequently, one optical isomer may produce a specific
physiological effect in the body, with its mirror image producing either a different effect
or none at all. Chiral reactions are also extremely important in the synthesis of pharmaceuticals and other industrially important chemicals.
Optical isomers are usually distinguished from each other by their interaction with plane-polarized light. If light is polarized—for example, by being passed
through a sheet of polarizing film—the electric-field vector of the light is confined to
a single plane (▼ Figure 23.23). If the polarized light is then passed through a solution containing one optical isomer, the plane of polarization is rotated either to the
right or to the left. The isomer that rotates the plane of polarization to the right is
dextrorotatory; it is the dextro, or d, isomer (Latin dexter, “right”). Its mirror image
rotates the plane of polarization to the left; it is levorotatory and is the levo, or l,
isomer (Latin laevus, “left”). The 3Co(en)343 + isomer on the right in Figure 23.22 is
found experimentally to be the l isomer of this ion. Its mirror image is the d isomer.
Because of their effect on plane-polarized light, chiral molecules are said to be
optically active.
Give It Some Thought
What is the similarity and what is the difference between the d and l isomers of
a compound?
When a substance with optical isomers is prepared in the laboratory, the chemical
environment during the synthesis is not usually chiral. Consequently, equal amounts of
the two isomers are obtained, and the mixture is said to be racemic. A racemic mixture
does not rotate polarized light because the rotatory effects of the two isomers cancel
each other.
Unpolarized
light
Polarizing
film
Angle of
rotation of
plane of
polarization
Rotated
polarized light
Light
source
Polarizer
axis
Polarized
light
Optically active
solution
Analyzer
▲ Figure 23.23 Using polarized light to detect optical activity.
section 23.5 Color and Magnetism in Coordination Chemistry
23.5 | Color and Magnetism in
Coordination Chemistry
Studies of the colors and magnetic properties of transition-metal complexes have played
an important role in the development of modern models for metal–ligand bonding. We
discussed the various types of magnetic behavior of the transition metals in Section 23.1,
and we discussed the interaction of radiant energy with matter in Section 6.3. Let’s
briefly examine the significance of these two properties for transition-metal complexes
before we develop a model for metal–ligand bonding.
Color
In Figure 23.4, we saw the diverse range of colors seen in salts of transition-metal ions
and their aqueous solutions. In general, the color of a complex depends on the identity
of the metal ion, on its oxidation state, and on the ligands bound to it. ▼ Figure 23.24,
for instance, shows how the pale blue color characteristic of 3Cu(H2O)442+ changes to
deep blue-violet as NH3 ligands replace the H2O ligands to form 3Cu(NH3)442+ .
For a substance to have color we can see, it must absorb some portion of the spectrum of visible light.
(Section 6.1) Absorption happens, however, only if the energy
needed to move an electron in the substance from its ground state to an excited state
corresponds to the energy of some portion of the visible spectrum.
(Section 6.3)
Thus, the particular energies of radiation a substance absorbs dictate the color we see
for the substance.
When an object absorbs some portion of the visible spectrum, the color we perceive is the sum of the unabsorbed portions, which are either reflected or transmitted
by the object and strike our eyes. (Opaque objects reflect light, and transparent ones
transmit it.) If an object absorbs all wavelengths of visible light, none reaches our eyes
and the object appears black. If it absorbs no visible light, it is white if opaque or colorless if transparent. If it absorbs all but orange light, the orange light is what reaches our
eye and therefore is the color we see.
Go Figure
Is the equilibrium binding constant of ammonia for Cu(II) likely to be larger or smaller than that of water for Cu(II)?
[Cu(H2O)4]2+(aq)
NH3(aq)
[Cu(NH3)4]2+(aq)
▲ Figure 23.24 The color of a coordination complex changes when the ligand changes.
1019
1020
chapter 23 Transition Metals and Coordination Chemistry
Only blue light absorbed;
eye perceives orange, blue’s
complementary color
Eye perceives orange since
only orange light reflected
650 nm
O
R
Y
750 nm
400 nm
560 nm
V
430 nm
580 nm
G
B
490 nm
▲ Figure 23.25 Two ways of perceiving the color orange. An object appears orange either
when it reflects orange light to the eye (left), or when it transmits to the eye all colors except
blue, the complement of orange (middle). Complementary colors lie opposite to each other on
an artist’s color wheel (right).
An interesting phenomenon of vision is that we also perceive an orange color when
an object absorbs only the blue portion of the visible spectrum and all the other colors strike our eyes. This is because orange and blue are complementary colors, which
means that the removal of blue from white light makes the light look orange (and the
removal of orange makes the light look blue).
Complementary colors can be determined with an artist’s color wheel, which
shows complementary colors on opposite sides (▲ Figure 23.25).
The amount of light absorbed by a sample as a function of wavelength is known as
the sample’s absorption spectrum. The visible absorption spectrum of a transparent sample can be determined using a spectrometer, as described in the “A Closer Look” box on
page 582. The absorption spectrum of the ion 3Ti(H2O)643 + is shown in ▶ Figure 23.26.
The absorption maximum is at 500 nm, but the graph shows that much of the yellow,
green, and blue light is also absorbed. Because the sample absorbs all of these colors, what
we see is the unabsorbed red and violet light, which we perceive as purple (the color purple is classified as a tertiary color located between red and violet on an artists color wheel).
S a mp l e
Exercise 23.6 Relating Color Absorbed to Color Observed
The complex ion trans-3Co(NH3)4Cl24+ absorbs light primarily in the red region of the visible
spectrum (the most intense absorption is at 680 nm). What is the color of the complex ion?
Solution
Plan For an object that absorbs only one color from the visible spec-
trum, the color we see is complementary to the color absorbed. We
can use the color wheel of Figure 23.25 to determine the complementary color.
Solve From Figure 23.25, we see that green is complementary to red,
so the complex appears green.
Comment As noted in Section 23.2, this green complex was one
of those that helped Werner establish his theory of coordination (Table 23.3). The other geometric isomer of this complex,
cis@3Co(NH3)4Cl24+, absorbs yellow light and therefore appears
violet.
Practice Exercise 1
A solution containing a certain transition-metal complex ion has
the absorption spectrum shown here.
Absorbance
Analyze We need to relate the color absorbed by a complex (red) to
the color observed for the complex.
400
500
600
700
Wavelength (nm)
What color would you expect a solution containing this ion to be?
(a) violet, (b) blue, (c) green, (d) orange, (e) red.
Practice Exercise 2
A certain transition-metal complex ion absorbs at 695 nm. Which
color is this ion most likely to be—blue, yellow, green, or red?
section 23.6 Crystal-field Theory
Magnetism of Coordination Compounds
Many transition-metal complexes exhibit paramagnetism, as described in Sections 9.8
and 23.1. In such compounds the metal ions possess some number of unpaired electrons. It is possible to experimentally determine the number of unpaired electrons per
metal ion from the measured degree of paramagnetism, and experiments reveal some
interesting comparisons.
Compounds of the complex ion 3Co(CN)643- have no unpaired electrons, for
example, but compounds of the 3CoF643- ion have four unpaired electrons per metal
ion. Both complexes contain Co(III) with a 3d6 electron configuration.
(Section 7.4)
Clearly, there is a major difference in the ways in which the electrons are arranged in
these two cases. Any successful bonding theory must explain this difference, and we
present such a theory in the next section.
1021
Go Figure
How would this absorbance spectrum
change if you decreased the
concentration of the 3Ti(H2O)643+ in
solution?
Blue, green, yellow
absorbed; a mixture of
violet and red light travel
to the eye, solution
appears purple
Give It Some Thought
What is the electron configuration for (a) the Co atom and (b) the Co3+ ion? How
many unpaired electrons does each possess? (See Section 7.4 to review electron
configurations of ions.)
Scientists have long recognized that many of the magnetic properties and colors of
transition-metal complexes are related to the presence of d electrons in the metal cation. In this section, we consider a model for bonding in transition-metal complexes,
crystal-field theory, that accounts for many of the observed properties of these substances.* Because the predictions of crystal-field theory are essentially the same as
those obtained with more advanced molecular-orbital theories, crystal-field theory
is an excellent place to start in considering the electronic structure of coordination
compounds.
The attraction of a ligand to a metal ion is essentially a Lewis acid–base interaction
in which the base—that is, the ligand—donates a pair of electrons to an empty orbital
on the metal ion (▶ Figure 23.27). Much of the attractive interaction between the metal
ion and the ligands is due, however, to the electrostatic forces between the positive
charge on the metal ion and negative charges on the ligands. An ionic ligand, such as
Cl- or SCN- , experiences the usual cation–anion attraction. When the ligand is a neutral molecule, as in the case of H2O or NH3, the negative ends of these polar molecules,
which contain an unshared electron pair, are directed toward the metal ion. In this case,
the attractive interaction is of the ion–dipole type.
(Section 11.2) In either case, the
ligands are attracted strongly toward the metal ion. Because of the metal–ligand electrostatic attraction, the energy of the complex is lower than the combined energy of the
separated metal ion and ligands.
Although the metal ion is attracted to the ligand electrons, the metal ion’s d electrons are repulsed by the ligands. Let’s examine this effect more closely, specifically the
case in which the ligands form an octahedral array around a metal ion that has coordination number 6.
In crystal-field theory, we consider the ligands to be negative points of charge
that repel the negatively charged electrons in the d orbitals of the metal ion. The
energy diagram in Figure 23.28 shows how these ligand point charges affect
the energies of the d orbitals. First we imagine the complex as having all the ligand
point charges uniformly distributed on the surface of a sphere centered on the metal
ion. The average energy of the metal ion’s d orbitals is raised by the presence of this
uniformly charged sphere. Hence, the energies of all five d orbitals are raised by
the same amount.
*The name crystal field arose because the theory was first developed to explain the properties of solid crystalline materials. The theory applies equally well to complexes in solution, however.
Absorbance
23.6 | Crystal-field Theory
400
500
600
700
Wavelength (nm)
▲ Figure 23.26 The color of [Ti(H2O)6]3+
A solution containing the 3Ti(H2O)643+
ion appears purple because, as its visible
absorption spectrum shows, the solution does
not absorb light from the violet and red ends
of the spectrum. That unabsorbed light is
what reaches our eyes.
M
n+
L
▲ Figure 23.27 Metal–ligand bond
formation. The ligand acts as a Lewis base by
donating its nonbonding electron pair to an
empty orbital on the metal ion. The bond that
results is strongly polar with some covalent
character.
1022
chapter 23 Transition Metals and Coordination Chemistry
Go Figure
Which d orbitals have lobes that point directly toward the ligands in an octahedral crystal field?
z
−
+
e set (dz2, dx2−y2)
Energy
+
−
−
−
−
In octahedral
crystal field
y
x
−
−
−
−
z
−
−
−
−
Free metal ion
dx2−y2
−
−
y
x
y
x
−
dz2
z
t2 set (dxy, dxz, dyz)
−
−
∆
Metal ion plus
ligands (negative
point charges
with spherical
symmetry)
z
−
z
−
−
−
−
−
y
x
−
−
−
−
−
−
−
dxy
dxz
dyz
y
x
▲ Figure 23.28 Energies of d orbitals in a free metal ion, a spherically symmetric crystal
field, and an octahedral crystal field.
This energy picture is only a first approximation, however, because the ligands are
not distributed uniformly on a spherical surface and, therefore, do not approach the
metal ion equally from every direction. Instead, we envision the six ligands approaching along x-, y-, and z-axes, as shown on the right in Figure 23.28. This arrangement
of ligands is called an octahedral crystal field. Because the metal ion’s d orbitals have
different orientations and shapes, they do not all experience the same repulsion from
the ligands and, therefore, do not all have the same energy under the influence of the
octahedral crystal field. To see why, we must consider the shapes of the d orbitals and
how their lobes are oriented relative to the ligands.
Figure 23.28 shows that the lobes of the dz 2 and dx2 - y2 orbitals are directed along
the x-, y-, and z-axes and so point directly toward the ligand point charges. In the
dxy , dxz , and dyz orbitals, however, the lobes are directed between the axes and so do not
point directly toward the charges. The result of this difference in orientation—dx2 - y2
and dz 2 lobes point directly toward the ligand charges; dxy, dxz, and dyz lobes do not—is
that the energy of the dx2 - y2 and dz 2 orbitals is higher than the energy of the dxy , dxz ,
and dyz orbitals. This difference in energy is represented by the red boxes in the energy
diagram of Figure 23.28.
It might seem like the energy of the dx2 - y2 orbital should be different from that of
the dz2 orbital because the dx 2 - y 2 has four lobes pointing at ligands and the dz2 has only
two lobes pointing at ligands. However, the dz2 orbital does have electron density in the
xy plane, represented by the ring encircling the point where the two lobes meet. More
advanced calculations show that two orbitals do indeed have the same energy in the presence of the octahedral crystal field.
Because their lobes point directly at the negative ligand charges, electrons in the
metal ion’s dz 2 and dx2 - y2 orbitals experience stronger repulsions than those in the
dxy , dxz , and dyz orbitals. As a result, the energy splitting shown in Figure 23.28 occurs.
The three lower-energy d orbitals are called the t2 set of orbitals, and the two higherenergy ones are called the e set.* The energy gap ∆ between the two sets is often called
the crystal-field splitting energy.
*The labels t2 for the dxy, dxz, and dyz orbitals and e for the dz2 and dx22y2 orbitals come from the application of
a branch of mathematics called group theory to crystal-field theory. Group theory can be used to analyze the
effects of symmetry on molecular properties.
1023
section 23.6 Crystal-field Theory
Crystal-field theory helps us account for the colors observed in transition-metal
complexes. The energy gap ∆ between the e and t2 sets of d orbitals is of the same order
of magnitude as the energy of a photon of visible light. It is therefore possible for a
transition-metal complex to absorb visible light that excites an electron from a lowerenergy (t2) d orbital into a higher-energy (e) d orbital. In 3Ti(H2O)643+ , for example, the
Ti(III) ion has an 3Ar43d1 electron configuration. (Recall from Section 7.4 that when
determining the electron configurations of transition-metal ions, we remove the s electrons first.) Ti(III) is thus called a d1 ion. In the ground state of 3Ti(H2O)643+ , the single
3d electron resides in an orbital in the t2 set (▶ Figure 23.29). Absorption of 495-nm
light excites this electron up to an orbital in the e set, generating the absorption spectrum shown in Figure 23.26. Because this transition involves exciting an electron from
one set of d orbitals to the other, we call it a d-d transition. As noted earlier, the absorption of visible radiation that produces this d-d transition causes the 3Ti(H2O)643+ ion to
appear purple.
Go Figure
How would you calculate the energy
gap between the t2 and e orbitals from
this diagram?
Energy
Light
e
of
495 nm
t2
▲ Figure 23.29 The d-d transition in
[Ti(H2O)6]3+ is produced by the absorption
of 495-nm light.
Give It Some Thought
Why are compounds of Ti(IV) colorless?
The magnitude of the crystal-field splitting energy and, consequently, the color of
a complex depend on both the metal and the ligands. For example, we saw in Figure 23.4
that the color of 3M(H2O)642+ complexes changes from reddish-pink when the metal ion is
Co2+ , to green for Ni2+ , to pale blue for Cu2+ . If we change the ligands in the 3Ni(H2O)642+
ion the color also changes. 3Ni(NH3)642+ has a blue-violet color, while 3Ni(en)342+ is purple
(▼ Figure 23.30). In a ranking called the spectrochemical series, ligands are arranged in
order of their abilities to increase splitting energy, as in this abbreviated list:
::Increasing ∆ ¡
Cl - 6 F - 6 H2O 6 NH3 6 en 6 NO2 - (N@bonded) 6 CN -
Go Figure
If you were to use a ligand L that was a stronger field ligand than ethylenediamine, what color would you expect the 3NiL642+
complex ion to have?
e
e
Energy
e
∆
∆
∆
t2
2+
[Ni(H2O)6]
Max absorption in
visible = 720 nm
t2
2+
[Ni(NH3)6]
Max absorption in
visible = 570 nm
t2
2+
[Ni(en)3]
Max absorption in
visible = 545 nm
▲ Figure 23.30 Effect of ligand on crystal-field splitting. The greater the crystal-field strength
of the ligand, the greater the energy gap ∆ it causes between the t2 and e sets of the metal
ion’s orbitals. This shifts the wavelength of the absorption maximum to shorter values.
chapter 23 Transition Metals and Coordination Chemistry
The magnitude of ∆ increases by roughly a factor of 2 from the far left to the far right of
the spectrochemical series. Ligands at the low@∆ end of the spectrochemical series are
termed weak-field ligands; those at the high@∆ end are termed strong-field ligands.
Let’s take a closer look at the colors and crystal field splitting as we vary the
ligands for the series of Ni2+ complexes discussed above. Because the Ni atom has an
3Ar43d84s2 electron configuration, Ni2+ has the configuration 3Ar43d8 and therefore is
a d8 ion. The t2 set of orbitals holds six electrons, two in each orbital, while the last two
electrons go into the e set of orbitals. Consistent with Hund’s rule, each e orbital holds
one electron and both electrons have the same spin.
(Section 6.8)
As the ligand changes from H2O to NH3 to ethylenediamine the spectrochemical
series tells us that the crystal field, ∆, exerted by the six ligands should increase. When
there is more than one electron in the d orbitals, interactions between the electrons make
the absorption spectra more complicated than the spectrum shown for 3Ti(H2O)643+ in
Figure 23.26, which complicates the task of relating changes in ∆ with color. With d8 ions
like Ni2+ three peaks are observed in the absorption spectra. Fortunately, for Ni2+ complexes
we can simplify the analysis because only one of these three peaks falls in the visible region of
the spectrum.* Because the energy separation ∆ is increasing, the wavelength of the absorption peak should shift to a shorter wavelength.
(Section 6.3) In the case of 3Ni(H2O)642+
the absorption peak in the visible part of the spectrum reaches a maximum near 720 nm, in
the red region of the spectrum. So the complex ion takes the complementary color—green.
For 3Ni(NH3)642+ the absorption peak reaches its maximum at 570 nm near the boundary between orange and yellow. The resulting color of the complex ion is a mixture of the
complementary colors—blue and violet. Finally for 3Ni(en)342+ the peak shifts to an even
shorter wavelength, 540 nm, which lies near the boundary between green and yellow. The
resulting color purple is a mixture of the complimentary colors red and violet.
Electron Configurations in Octahedral Complexes
Crystal-field theory helps us understand the magnetic properties and some important
chemical properties of transition-metal ions. From Hund’s rule, we expect electrons to
always occupy the lowest-energy vacant orbitals first and to occupy a set of degenerate
(same-energy) orbitals one at a time with their spins parallel.
(Section 6.8) Thus,
if we have a d1, d2, or d3 octahedral complex, the electrons go into the lower-energy t2
orbitals, with their spins parallel. When a fourth electron must be added, we have the
two choices shown in ▼ Figure 23.31: The electron can either go into an e orbital, where
Spin pairing energy larger than ∆
e
t2
e
Energy
1024
t2
e
t2
∆ larger than spin pairing energy
d 3 ion
d 4 ion
▲ Figure 23.31 Two possibilities for adding a fourth electron to a d 3 octahedral complex.
Whether the fourth electron goes into a t2 orbital or into an e orbital depends on the relative
energies of the crystal-field splitting energy and the spin-pairing energy.
*The other two peaks fall in the infrared (IR) and ultraviolet (UV) regions of the spectrum. For 3Ni(H2O)642+
the IR peak is found at 1176 nm and the UV peak at 388 nm.
1025
section 23.6 Crystal-field Theory
it will be the sole electron in the orbital, or become the second electron in a t2 orbital.
Because the energy difference between the t2 and e sets is the splitting energy ∆, the
energy cost of going into an e orbital rather than a t2 orbital is also ∆. Thus, the goal of
filling lowest-energy available orbitals first is met by putting the electron in a t2 orbital.
There is a penalty for doing this, however, because the electron must now be paired
with the electron already occupying the orbital. The difference between the energy
required to pair an electron in an occupied orbital and the energy required to place that
electron in an empty orbital is called the spin-pairing energy. The spin-pairing energy
arises from the fact that the electrostatic repulsion between two electrons that share
an orbital (and so must have opposite spins) is greater than the repulsion between two
electrons that are in different orbitals and have parallel spins.
In coordination complexes, the nature of the ligands and the charge on the metal
ion often play major roles in determining which of the two electron arrangements
shown in Figure 23.31 is used. In 3CoF643 - and 3Co(CN)643 - , both ligands have a 1charge. The F- ion, however, is on the low end of the spectrochemical series, so it is a
weak-field ligand. The CN- ion is on the high end and so is a strong-field ligand, which
means it produces a larger energy gap ∆ than the F- ion. The splittings of the d-orbital
energies in these two complexes are compared in ▶ Figure 23.32.
Cobalt(III) has an 3Ar43d6 electron configuration, so both complexes in Figure
23.32 are d 6 complexes. Let’s imagine that we add these six electrons one at a time to
the d orbitals of the 3CoF643 - ion. The first three go into the t2 orbitals with their spins
parallel. The fourth electron could pair up in one of the t2 orbitals. The F- ion is a weakfield ligand, however, and so the energy gap ∆ between the t2 set and the e set is small.
In this case, the more stable arrangement is the fourth electron in one of the e orbitals.
By the same energy argument, the fifth electron goes into the other e orbital. With all
five d orbitals containing one electron, the sixth must pair up, and the energy needed to
place the sixth electron in a t2 orbital is less than that needed to place it in an e orbital.
We end up with four t2 electrons and two e electrons.
Figure 23.32 shows that the crystal-field splitting energy ∆ is much larger in the
3Co(CN)643 - complex. In this case, the spin-pairing energy is smaller than ∆, so the
lowest-energy arrangement is the six electrons paired in the t2 orbitals.
The 3CoF643 - complex is a high-spin complex; that is, the electrons are arranged
so that they remain unpaired as much as possible. The 3Co(CN)643 - ion is a low-spin
complex; that is, the electrons are arranged so that they remain paired as much as possible
while still following Hund’s rule. These two electronic arrangements can be readily distinguished by measuring the magnetic properties of the complex. Experiments show that
3CoF643 - has four unpaired electrons and 3Co(CN)643 - has none. The absorption spectrum also shows peaks corresponding to the different values of ∆ in these two complexes.
∆ less than spinpairing energy,
electrons singly
occupy each e
orbital before
pairing in the t2
orbitals
∆ greater than
spin-pairing
energy, t2 orbitals
are filled with
electrons before
occupying the e
orbitals
e
Energy
e
t2
[CoF6]3−
High-spin
complex
t2
3−
[Co(CN)6]
Low-spin
complex
▲ Figure 23.32 High-spin and low-spin
complexes. The high-spin 3CoF643- ion has a
weak-field ligand and so a small ∆ value. The
low-spin 3Co(CN)643- ion has a strong-field
ligand and so a large ∆ value.
Give It Some Thought
In octahedral complexes, for which d electron configurations is it possible to have
high-spin and low-spin arrangements with different numbers of unpaired electrons?
In the transition metal ions of periods 5 and 6 (which have 4d and 5d valence electrons), the d orbitals are larger than in the period 4 ions (which have only 3d electrons).
Thus, ions from periods 5 and 6 interact more strongly with ligands, resulting in a
larger crystal-field splitting. Consequently, metal ions in periods 5 and 6 are invariably
low spin in an octahedral crystal field.
S a mp l e
Exercise 23.7 The Spectrochemical Series, Crystal Field Splitting, Color, and Magnetism
The compound hexaamminecobalt(III) chloride is diamagnetic and orange in color with a single
absorption peak in its visible absorption spectrum. (a) What is the electron configuration of the
cobalt(III) ion? (b) Is 3Co(NH3)643+ a high-spin complex or a low-spin complex? (c) Estimate the
wavelength where you expect the absorption of light to reach a maximum? (d) What color and
magnetic behavior would you predict for the complex ion 3Co(en)343+?
∆
∆
1026
chapter 23 Transition Metals and Coordination Chemistry
Solution
Analyze We are given the color and magnetic behavior of an octahedral complex containing Co with a + 3 oxidation number. We need to
use this information to determine its electron configuration, its spin
state (low-spin or high-spin), and the color of light it absorbs. In part
(d), we must use the spectrochemical series to predict how its properties will change if NH3 is replaced by ethylenediamine (en).
Plan (a) From the oxidation number and the periodic table we can
determine the number of valence electrons for Co(III) and from that
we can determine the electron configuration. (b) The magnetic behavior can be used to determine whether this compound is a low-spin
or high-spin complex. (c) Since there is a single peak in the visible
absorption spectrum the color of the compound should be complementary to the color of light that is absorbed most strongly. (d) Ethylenediamine is a stronger field ligand than NH3 so we expect a larger
∆ for 3Co(en)343+ than for 3Co(NH3)643+.
Solve
(a) Co has an electron configuration of 3Ar44s23d7 and Co3+ has
three fewer electrons than Co. Because transition-metal ions
always lose their valence shell s electrons, the electron configuration of Co3+ is 3Ar43d6.
(b) There are six valence electrons in the d orbitals. The filling of the
t2 and e orbitals for both high-spin and low-spin complexes is
shown below. Because the compound is diamagnetic we know all
of the electrons must be paired up, which allows us to determine
that 3Co(NH3)643+ is a low-spin complex.
e
e
(d) Ethylenediammine is higher in the spectrochemical series than
ammonia. Therefore, we expect a larger ∆ for 3Co(en)343+.
Because ∆ was already greater than the spin-pairing energy for
3Co(NH3)643+, we expect 3Co(en)343+ to be a low-spin complex
as well, with a d 6 configuration, so it will also be diamagnetic.
The wavelength at which the complex absorbs light will shift
to higher energy. If we assume a shift in the absorption maximum from blue to violet, the color of the complex will become
yellow.
Practice Exercise 1
Which of the following octahedral complex ions will have the
fewest number of unpaired electrons? (a) 3Cr(H2O)643+,
(b) 3V(H2O)643+, (c) 3FeF643-, (d) 3RhCl643-, (e) 3Ni(NH3)642+.
Practice Exercise 2
Consider the colors of the ammonia complexes of Co3+
given in Table 23.3. Based on the change in color would you
expect 3Co(NH3)5Cl42+ to have a larger or smaller value of
∆ than 3Co(NH3)643+? Is this prediction consistent with the
spectrochemical series?
∆
∆
t2
t2
Low spin
(c) We are told that the compound is orange and has a single absorption peak in the visible region of the spectrum. The compound
must therefore absorb the complementary color of orange, which
is blue. The blue region of the spectrum ranges from approximately 430 nm to 490 nm. As an estimate we assume that the
complex ion absorbs somewhere in the middle of the blue region,
near 460 nm.
High spin
Tetrahedral and Square-Planar Complexes
Thus far we have considered crystal-field theory only for complexes having an octahedral geometry. When there are only four ligands in a complex, the geometry is generally tetrahedral, except for the special case of d8 metal ions, which we will discuss in a
moment.
The crystal-field splitting of d orbitals in tetrahedral complexes differs from that
in octahedral complexes. Four equivalent ligands can interact with a central metal ion
most effectively by approaching along the vertices of a tetrahedron. In this geometry
the lobes of the two e orbitals point toward the edges of the tetrahedron, exactly in
between the ligands (▶ Figure 23.33). This orientation keeps the dx2 - y2 and dz2 as far
from the ligand point charges as possible. Consequently, these two d orbitals experience less repulsion from the ligands and lie at lower energy than the other three d orbitals. The three t2 orbitals do not point directly at the ligand point charges, but they do
come closer to the ligands than the e set and as a result they experience more repulsion
and are higher in energy. As we see in Figure 23.33, the splitting of d orbitals in a tetrahedral geometry is the opposite of what we find for the octahedral geometry, namely
the e orbitals are now below the t2 orbitals. The crystal-field splitting energy ∆ is much
smaller for tetrahedral complexes than it is for comparable octahedral complexes, in
part because there are fewer ligand point charges in the tetrahedral geometry, and in
part because neither set of orbitals have lobes that point directly at the ligand point
charges. Calculations show that for the same metal ion and ligand set, ∆ for the tetrahedral complex is only four-ninths as large as for the octahedral complex. For this reason,
section 23.6 Crystal-field Theory
z
z
z
y
t2 set (dxy, dxz, dyz)
y
x
dxy
∆
In spherical
crystal field
y
x
x
dxz
z
1027
dyz
z
e set (dz2, dx2−y2)
In tetrahedral
crystal field
y
y
x
dz2
x
dx2−y2
▲ Figure 23.33 Energies of the d orbitals in a tetrahedral crystal field. The splitting of the
e and t2 sets of orbitals is inverted with respect to the splitting associated with an octahedral
crystal field. The crystal field splitting energy ∆ is much smaller than it is in an octahedral
crystal field.
Give It Some Thought
Go Figure
For which d orbital(s) do the lobes
point directly at the ligands in a
square-planar crystal field?
dx2−y2
dxy
Energy
all tetrahedral complexes are high spin; the crystal-field splitting energy is never large
enough to overcome the spin-pairing energies.
In a square-planar complex, four ligands are arranged about the metal ion such
that all five species are in the xy plane. The resulting energy levels of the d orbitals
are illustrated in ▶ Figure 23.34. Note in particular that the dz2 orbital is considerably
lower in energy than the dx2 - y2 orbital. To understand why this is so, recall from Figure
23.28 that in an octahedral field the dz2 orbital of the metal ion interacts with the ligands
positioned above and below the xy plane. There are no ligands in these two positions
in a square-planar complex, which means that the dz2 orbital experiences less repulsion
and so remains in a lower-energy, more stable state.
Square-planar complexes are characteristic of metal ions with a d8 electron configuration. They are nearly always low spin, with the eight d electrons spin-paired to
form a diamagnetic complex. This pairing leaves the dx2 - y2 orbital empty. Such an electronic arrangement is particularly common among the d8 ions of periods 5 and 6, such as
Pd2+ , Pt2+ , Ir+ , and Au3+ .
d
2
dxz, dyz
Square planar
Why is the energy of the dxz and d yz orbitals in a square-planar complex lower
than that of the dxy orbital?
▲ Figure 23.34 Energies of the d orbitals in
a square-planar crystal field.
S a mp l e
Exercise 23.8 Populating d Orbitals in Tetrahedral and Square-Planar Complexes
Nickel(II) complexes in which the metal coordination number is 4 can have either square-planar
or tetrahedral geometry. 3NiCl442- is paramagnetic, and 3Ni(CN)442- is diamagnetic. One of these
complexes is square planar, and the other is tetrahedral. Use the relevant crystal-field splitting diagrams in the text to determine which complex has which geometry.
Solution
Analyze We are given two complexes containing Ni2+ and their mag-
netic properties. We are given two molecular geometry choices and
asked to use crystal-field splitting diagrams from the text to determine which complex has which geometry.
Plan We need to determine the number of d electrons in Ni2+ and
then use Figure 23.33 for the tetrahedral complex and Figure 23.34 for
the square-planar complex.
1028
chapter 23 Transition Metals and Coordination Chemistry
The tetrahedral complex has two unpaired electrons, and the square-planar
complex has none. We know from Section 23.1 that the tetrahedral complex must be paramagnetic and the square planar must be diamagnetic.
Therefore, 3NiCl442- is tetrahedral, and 3Ni(CN)442- is square planar.
Solve Nickel(II) has the electron configuration 3Ar43d8. Tetrahedral
complexes are always high spin, and square-planar complexes are almost always low spin. Therefore, the population of the d electrons in
the two geometries is
Comment Nickel(II) forms octahedral complexes more frequently
than square-planar ones, whereas d8 metals from periods 5 and 6 tend
to favor square-planar coordination.
dx2−y2
dxy
t2 (dxy, dyz, dxz)
dz2
e (dx2−y2, dz2)
Tetrahedral
Practice Exercise 1
How many unpaired electrons do you predict for the tetrahedral
3MnCl442 - ion? (a) 1, (b) 2, (c) 3, (d) 4, (e) 5.
Practice Exercise 2
Are there any diamagnetic tetrahedral complexes containing transition metal ions with partially filled d orbitals? If so what electron
count(s) leads to diamagnetism?
dxz, dyz
Square planar
Crystal-field theory can be used to explain many observations in addition to those
we have discussed. The theory is based on electrostatic interactions between ions and
atoms, which essentially means ionic bonds. Many lines of evidence show, however, that
the bonding in complexes must have some covalent character. Therefore, molecularorbital theory
(Sections 9.7 and 9.8) can also be used to describe the bonding in
complexes, although the application of molecular-orbital theory to coordination
compounds is beyond the scope of our discussion. Crystal-field theory, although not
entirely accurate in all details, provides an adequate and useful first description of the
electronic structure of complexes.
A Closer Look
Charge-Transfer Color
In the laboratory portion of your course, you have probably seen
many colorful transition-metal compounds, including those shown in
▼ Figure 23.35. Many of these compounds are colored because of
d-d transitions. Some colored complexes, however, including the violet
permanganate ion, MnO4-, and the yellow chromate ion, CrO42-, derive
their color from a different type of excitation involving the d orbitals.
The permanganate ion strongly absorbs visible light, with a maximum absorption at 565 nm. Because violet is complementary to yellow,
this strong absorption in the yellow portion of the visible spectrum is
responsible for the violet color of salts and solutions of the ion. What is
happening during this absorption of light? The MnO4- ion is a complex
KMnO4
of Mn(VII). Because Mn(VII) has a d 0 electron configuration, the absorption cannot be due to a d-d transition because there are no d electrons to excite! That does not mean, however, that the d orbitals are not
involved in the transition. The excitation in the MnO4- ion is due to a
charge-transfer transition, in which an electron on one oxygen ligand
is excited into a vacant d orbital on the Mn ion (▶ Figure 23.36). In
essence, an electron is transferred from a ligand to the metal, so this
transition is called a ligand-to-metal charge-transfer (LMCT) transition.
An LMCT transition is also responsible for the color of the
CrO42-, which is a d 0 Cr(VI) complex.
Also shown in Figure 23.35 is a salt of the perchlorate ion (ClO4-).
Like MnO4-, ClO4- is tetrahedral and has its central atom in the + 7 oxidation state. However, because the Cl atom does not have low-lying
K2CrO4
▲ Figure 23.35 The colors of compounds can arise from charge-transfer transitions. KMnO4
and K2CrO4 are colored due to ligand-to-metal charge-transfer transitions in their anions. The
perchlorate anion in KClO4 has no occupied d orbitals and its charge-transfer transition is at
higher energy, corresponding to ultraviolet absorption; therefore it appears white.
KClO4
section 23.6 Crystal-field Theory
Empty Mn 3d orbitals
Energy
t2 set
e set
Filled ligand orbitals
–
▲ Figure 23.36 Ligand-to-metal charge-transfer transition in MnO4 .
As shown by the blue arrow, an electron is excited from a nonbonding
pair on O into one of the empty d orbitals on Mn.
S a mp l e
Integrative Exercise d orbitals, exciting a Cl electron requires a more energetic photon
than does MnO4-. The first absorption for ClO4- is in the ultraviolet
portion of the spectrum, so all the visible light is transmitted and the
salt ­appears white.
Other complexes exhibit charge-transfer excitations in which
an electron from the metal atom is excited to an empty orbital on a
ligand. Such an excitation is called a metal-to-ligand charge-transfer
(MLCT) transition.
Charge-transfer transitions are generally more intense than d-d
transitions. Many metal-containing pigments used for oil painting, such
as cadmium yellow (CdS), chrome yellow (PbCrO4), and red ochre
(Fe2O3), have intense colors because of charge-transfer transitions.
Related Exercises: 23.82, 23.83
Putting Concepts Together
The oxalate ion has the Lewis structure shown in Table 23.4. (a) Show the geometry of the complex formed when this ion complexes with cobalt(II) to form 3Co(C2O4)(H2O)44. (b) Write
the formula for the salt formed when three oxalate ions complex with Co(II), assuming that the
­charge-balancing cation is Na+. (c) Sketch all the possible geometric isomers for the cobalt complex
formed in part (b). Are any of these isomers chiral? Explain. (d) The equilibrium constant for the
formation of the cobalt(II) complex produced by coordination of three oxalate anions, as in part (b),
is 5.0 * 109, and the equilibrium constant for formation of the cobalt(II) complex with three molecules of ortho-phenanthroline (Table 23.4) is 9 * 1019. From these results, what conclusions can
you draw regarding the relative Lewis base properties of the two ligands toward cobalt(II)?
(e) Using the approach described in Sample Exercise 17.14, calculate the concentration of free
­aqueous Co(II) ion in a solution initially containing 0.040 M oxalate (aq) and 0.0010 M Co2+(aq).
Solution
(a) The complex formed by coordination of one oxalate ion is octahedral:
(b) Because the oxalate ion has a charge of 2 -, the net charge of a complex with three oxalate
anions and one Co2+ ion is 4 - . Therefore, the coordination compound has the formula
Na43Co(C2O4)34.
(c) There is only one geometric isomer. The complex is chiral, however, in the same way the
3Co(en)343+ complex is chiral (Figure 23.22). The two mirror images are not superimposable,
so there are two enantiomers:
4−
1029
4−
1030
chapter 23 Transition Metals and Coordination Chemistry
(d) The ortho-phenanthroline ligand is bidentate, like the oxalate ligand, so they both exhibit
the chelate effect. Thus, we conclude that ortho-phenanthroline is a stronger Lewis base
toward Co2+ than oxalate. This conclusion is consistent with what we learned about bases
in Section 16.7, that nitrogen bases are generally stronger than oxygen bases. (Recall, for
­example, that NH3 is a stronger base than H2O.)
(e) The equilibrium we must consider involves 3 mol of oxalate ion (represented as Ox2 - ).
Co2+(aq) + 3 Ox2-(aq) ∆ 3Co(Ox)344-(aq)
The formation-constant expression is
Kf =
33Co(Ox)344-4
3Co2+43Ox2-43
Because Kf is so large, we can assume that essentially all the Co2+ is converted to the oxalato
complex. Under that assumption, the final concentration of 3Co(Ox)344- is 0.0010 M and that of
oxalate ion is 3Ox2-4 = (0.040) - 3(0.0010) = 0.037 M (three Ox2- ions react with each Co2+
ion). We then have
3Co2+4 = xM, 3Ox2-4 ≅ 0.037 M, 33Co(Ox)344-4 ≅ 0.0010 M
Inserting these values into the equilibrium-constant expression, we have
Kf =
(0.0010)
= 5 * 109
x(0.037)3
Solving for x, we obtain 4 * 10-9 M. From this, we see that the oxalate has complexed all but a
tiny fraction of the Co2+ in solution.
Chapter Summary and Key Terms
The Transition Metals (Section 23.1) Metallic elements are
obtained from minerals, which are solid inorganic compounds found
in nature. Metallurgy is the science and technology of extracting metals
from the earth and processing them for further use. Transition metals are characterized by incomplete filling of the d orbitals. The presence of d electrons in transition elements leads to multiple oxidation
states. As we proceed through the transition metals in a given row of
the periodic table, the attraction between the nucleus and the valence
electrons increases more markedly for d electrons than for s electrons.
As a result, the later transition elements in a period tend to have lower
oxidation states.
The atomic and ionic radii of period 5 transition metals are larger
than those of period 4 metals. The transition metals of periods 5 and
6 have comparable atomic and ionic radii and are also similar in other
properties. This similarity is due to the lanthanide contraction.
The presence of unpaired electrons in valence orbitals leads to magnetic behavior in transition metals and their compounds. In f­ erromagnetic,
ferrimagnetic, and antiferromagnetic substances, the unpaired electron
spins on atoms in a solid are affected by spins on neighboring atoms. In
a ferromagnetic substance the spins all point in the same direction. In an
antiferromagnetic substance the spins point in opposite directions and
cancel one another. In a ferrimagnetic substance the spins point in opposite directions but do not fully cancel. Ferromagnetic and ferrimagnetic
substances are used to make permanent magnets.
Transition-Metal Complexes (Section 23.2) Coordination
compounds are substances that contain metal complexes. Metal complexes contain metal ions bonded to several surrounding anions or
molecules known as ligands. The metal ion and its ligands make up
the coordination sphere of the complex. The number of atoms attached
to the metal ion is the coordination number of the metal ion. The most
common coordination numbers are 4 and 6; the most common coordination geometries are tetrahedral, square planar, and octahedral.
Common Ligands in Coordination Chemistry (Section 23.3) Ligands that occupy only one site in a coordination sphere are called
monodentate ligands. The atom of the ligand that bonds to the metal
ion is the donor atom. Ligands that have two donor atoms are bidentate
ligands. Polydentate ligands have three or more donor atoms. Bidentate
and polydendate ligands are also called chelating agents. In general,
chelating agents form more stable complexes than do related monodentate ligands, an observation known as the chelate effect. Many
biologically important molecules, such as the porphyrins, are complexes of chelating agents. A related group of plant pigments known
as c­ hlorophylls are important in photosynthesis, the process by which
plants use solar energy to convert CO2 and H2O into carbohydrates.
Nomenclature and Isomerism in Coordination Chemistry
(Section 23.4) In naming coordination compounds, the number
and type of ligands attached to the metal ion are specified, as is the
oxidation state of the metal ion. Isomers are compounds with the same
composition but different arrangements of atoms and therefore different properties. Structural isomers differ in the bonding arrangements of
the ligands. Linkage isomerism occurs when a ligand can coordinate to
a metal ion through different donor atoms. Coordination-sphere isomers
contain different ligands in the coordination sphere. Stereoisomers are
isomers with the same chemical bonding arrangements but different
spatial arrangements of ligands. The most common forms of stereoisomerism are geometric isomerism and optical isomerism. Geometric
isomers differ from one another in the relative locations of donor
atoms in the coordination sphere; the most common are cis–trans
isomers. Geometric isomers differ from one another in their chemical
and physical properties. Optical isomers are nonsuperimposable mirror
images of each other. Optical isomers, or enantiomers, are chiral, meaning that they have a specific “handedness” and differ only in the presence of a chiral environment. Optical isomers can be distinguished
from one another by their interactions with plane-polarized light;
solutions of one isomer rotate the plane of polarization to the right
(dextrorotatory), and solutions of its mirror image rotate the plane to
the left (levorotatory). Chiral molecules, therefore, are optically active. A
50950 mixture of two optical isomers does not rotate plane-polarized
light and is said to be racemic.
Exercises
Color and Magnetism in Coordination Chemistry
(Section 23.5) A substance has a particular color because it either
reflects or transmits light of that color or absorbs light of the complementary color. The amount of light absorbed by a sample as a function
of wavelength is known as its absorption spectrum. The light absorbed
provides the energy to excite electrons to higher-energy states.
It is possible to determine the number of unpaired electrons in
a complex from its degree of paramagnetism. Compounds with no
unpaired electrons are diamagnetic.
Crystal-Field Theory (Section 23.6) Crystal-field theory suc-
cessfully accounts for many properties of coordination compounds,
including their color and magnetism. In crystal-field theory, the
interaction between metal ion and ligand is viewed as electrostatic.
Because some d orbitals point directly at the ligands whereas others
point between them, the ligands split the energies of the metal d orbitals. For an octahedral complex, the d orbitals are split into a lowerenergy set of three degenerate orbitals (the t2 set) and a higher-energy
set of two degenerate orbitals (the e set). Visible light can cause a
1031
d-d transition, in which an electron is excited from a lower-energy
d orbital to a higher-energy d orbital. The spectrochemical series lists
ligands in order of their ability to increase the split in d-orbital energies in octahedral complexes.
Strong-field ligands create a splitting of d-orbital energies that
is large enough to overcome the spin-pairing energy. The d electrons
then preferentially pair up in the lower-energy orbitals, producing a
­low-spin complex. When the ligands exert a weak crystal field, the splitting of the d orbitals is small. The electrons then occupy the higherenergy d orbitals in preference to pairing up in the lower-energy set,
producing a high-spin complex.
Crystal-field theory also applies to tetrahedral and squareplanar complexes, which leads to different d-orbital splitting patterns.
In a tetrahedral crystal field, the splitting of the d orbitals results in a
higher-energy t2 set and a lower-energy e set, the opposite of the octahedral case. The splitting in a tetrahedral crystal field is much smaller
than that in an octahedral crystal field, so tetrahedral complexes are
always high-spin complexes.
Learning Outcomes After studying this chapter, you should be able to:
• Describe the periodic trends in radii and oxidation states of the
• Name coordination compounds given their formula and write their
• Determine the oxidation number and number of d electrons for
• Recognize and draw the geometric isomers of a complex. (Section 23.4)
• Recognize and draw the optical isomers of a complex. (Section 23.4)
• Use crystal-field theory to explain the colors and to determine
transition-metal ions, including the origin and effect of the lanthanide contraction. (Section 23.1)
metal ions in complexes. (Section 23.2)
• Identify common ligands and distinguish between chelating and
nonchelating ligands. (Section 23.3)
formula given their name. (Section 23.4)
the number of unpaired electrons in a complex. (Sections 23.5
and 23.6)
Exercises
Visualizing Concepts
23.1The three graphs below show the variation in radius, effective nuclear charge, and maximum oxidation state for the transition metals of period 4. In each part below identify
which property is being plotted. [Section 23.1]
Sc Ti
V
Cr Mn Fe Co Ni Cu Zn
(a)
Sc Ti
V
Cr Mn Fe Co Ni Cu Zn
(b)
23.2 Draw the structure for Pt(en)Cl2 and use it to answer the following questions:
(a) What is the coordination number for platinum in this complex? (b) What
is the coordination g­ eometry? (c) What is the oxidation state of the platinum?
(d) How many unpaired electrons are there? [Sections 23.2 and 23.6]
Sc Ti
V
Cr Mn Fe Co Ni Cu Zn
(c)
1032
chapter 23 Transition Metals and Coordination Chemistry
23.3Draw the Lewis structure for the ligand shown here.
(a) Which atoms can serve as donor atoms? Classify this ligand
as monodentate, bidentate, or polydentate. (b) How many of
these ligands are needed to fill the coordination sphere in an
octahedral complex? [Section 23.2]
NH2CH2CH2NHCH2CO2−
23.4The complex ion shown here has a 1 - charge. Name the
complex ion. [Section 23.4]
=N
= Cl
23.8 Which of these crystal-field splitting diagrams represents:
(a) a weak-field octahedral complex of Fe3+, (b) a strong-field
octahedral complex of Fe3+, (c) a tetrahedral complex of Fe3+,
(d) a tetrahedral complex of Ni2+? (The diagrams do not indicate the relative magnitudes of ∆.) [Section 23.6]
=H
= Pt
23.5There are two geometric isomers of octahedral complexes of
the type MA3X3, where M is a metal and A and X are monodentate ligands. Of the complexes shown here, which are
identical to (1) and which are the geometric isomers of (1)?
[Section 23.4]
(1)
(2)
(3)
(4)
(1)
(2)
(3)
(4)
23.9In the linear crystal field shown here, the negative charges are
on the z-axis. Using Figure 23.28 as a guide, predict which of
the following choices most accurately describes the splitting
of the d orbitals in a linear crystal field? [Section 23.6]
(5)
23.6 Which of the complexes shown here are chiral? [Section 23.4]
= Cr
(1)
= NH2CH2CH2NH2
(2)
(3)
= Cl
= NH3
(4)
23.7The solutions shown here each have an absorption spectrum
with a single absorption peak like that shown in Figure 23.26.
What color does each solution absorb most strongly?
[Section 23.5]
(a)
(b)
(c)
(d)
(e)
23.10Two Fe(II) complexes are both low spin but have different ligands. A solution of one is green and a solution of the other is
red. Which solution is likely to contain the complex that has
the stronger-field ligand? [Section 23.6]
Exercises
The Transition Metals (Section 23.1)
23.11 The lanthanide contraction explains which of the following
periodic trends? (a) The atomic radii of the transition metals first decrease and then increase when moving horizontally
across each period. (b) When forming ions the transition metals lose their valence s orbitals before their valence
d ­o rbitals. (c) The radii of the period 5 transition metals
(Y–Cd) are very similar to the radii of the period 6 transition
metals (Lu–Hg).
23.12Which periodic trend is responsible for the observation that
the maximum oxidation state of the transition-metal elements
peaks near groups 7B and 8B? (a) The number of valence
electrons reaches a maximum at group 8B. (b) The effective
nuclear charge increases on moving left across each period.
(c) The radii of the transition-metal elements reaches a minimum for group 8B and as the size of the atoms decreases it
becomes easier to remove electrons.
23.13 For each of the following compounds determine the electron
configuration of the transition metal ion. (a) TiO, (b) TiO2,
(c) NiO, (d) ZnO.
23.14Among the period 4 transition metals (Sc–Zn), which elements
do not form ions where there are partially filled 3d orbitals?
23.15 Write out the ground-state electron configurations of
(a) Ti3+, (b) Ru2+, (c) Au3+, (d) Mn4+.
23.16How many electrons are in the valence d orbitals in these
transition-metal ions? (a) Co3+, (b) Cu+, (c) Cd2+, (d) Os3+.
23.17 Which type of substance is attracted by a magnetic field, a
diamagnetic substance or a paramagnetic substance?
23.18Which type of magnetic material cannot be used to make
permanent magnets, a ferromagnetic substance, an antiferromagnetic substance, or a ferrimagnetic substance?
23.19 What kind of magnetism is exhibited by this diagram:
Magnetic field
Put in vertical
magnetic field
23.20The most important oxides of iron are magnetite, Fe3O4, and
hematite, Fe2O3. (a) What are the oxidation states of iron in
these compounds? (b) One of these iron oxides is ferrimagnetic, and the other is antiferromagnetic. Which iron oxide is
likely to show which type of magnetism? Explain.
Transition-Metal Complexes (Section 23.2)
23.21 (a) Using Werner’s definition of valence, which property is
the same as oxidation number, primary valence or secondary
valence? (b) What term do we normally use for the other type
of valence? (c) Why can the NH3 molecule serve as a ligand
but the BH3 molecule cannot?
23.22(a) What is the meaning of the term coordination number as
it applies to metal complexes? (b) Give an example of a ligand
that is neutral and one that is negatively charged. (c) Would
you expect ligands that are positively charged to be common?
Explain. (d) What type of chemical bonding is characteristic
of coordination compounds? Illustrate with the compound
1033
Co(NH3)6Cl3. (e) What are the most common coordination
numbers for metal complexes?
23.23 A complex is written as NiBr2 # 6 NH3. (a) What is the oxidation state of the Ni atom in this complex? (b) What is the
likely coordination number for the complex? (c) If the complex is treated with excess AgNO3(aq), how many moles of
AgBr will precipitate per mole of complex?
23.24Crystals of hydrated chromium(III) chloride are green, have
an empirical formula of CrCl3 # 6 H2O, and are highly soluble, (a) Write the complex ion that exists in this compound.
(b) If the complex is treated with excess AgNO3(aq), how many
moles of AgCl will precipitate per mole of CrCl3 # 6 H2O dissolved in solution? (c) Crystals of anhydrous chromium(III)
chloride are violet and insoluble in aqueous solution. The coordination geometry of chromium in these crystals is octahedral as is almost always the case for Cr3+. How can this be the
case if the ratio of Cr to Cl is not 1:6?
23.25 Indicate the coordination number and the oxidation number
of the metal for each of the following complexes:
(a) Na23CdCl44
(b) K23MoOCl44
(c) 3Co(NH3)4Cl24Cl
(d) 3Ni(CN)543 -
(e) K33V(C2O4)34
(f) 3Zn(en)24Br2
23.26Indicate the coordination number and the oxidation number
of the metal for each of the following complexes:
(a) K33Co(CN)64
(b) Na23CdBr44
(c) 3Pt(en)34(ClO4)4
(d) 3Co(en)2(C2O4)4+
(e) NH43Cr(NH3)2(NCS)44
(f) 3Cu(bipy)2I]I
Common Ligands in Coordination Chemistry
(Section 23.3)
23.27 (a) What is the difference between a monodentate ligand and
a bidentate ligand? (b) How many bidentate ligands are necessary to fill the coordination sphere of a six-coordinate complex? (c) You are told that a certain molecule can serve as a
tridentate ligand. Based on this statement, what do you know
about the molecule?
23.28For each of the following polydentate ligands, determine
(i) the maximum number of coordination sites that the ligand can occupy on a single metal ion and (ii) the number
and type of donor atoms in the ligand: (a) ethylenediamine
(en), (b) bipyridine (bipy), (c) the oxalate anion (C2O42- ),
(d) the 2 - ion of the porphine molecule (Figure 23.13),
(e) 3EDTA44-.
23.29 Polydentate ligands can vary in the number of coordination
positions they occupy. In each of the following, identify the
polydentate ligand present and indicate the probable number
of coordination positions it occupies:
(a) 3Co(NH3)4(o@phen)]Cl3
(b) 3Cr(C2O4)(H2O)44Br
1034
chapter 23 Transition Metals and Coordination Chemistry
(c) 3Cr(EDTA)(H2O)4-
(d) potassium diaquatetrabromovanadate(III)
(d) 3Zn(en)24(ClO4)2
(e) bis(ethylenediamine)zinc(II) tetraiodomercurate(II)
23.30Indicate the likely coordination number of the metal in each
of the following complexes:
23.36Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere:
(a) 3Rh(bipy)34(NO3)3
(a) tetraaquadibromomanganese(III) perchlorate
(d) Na23Co(EDTA)Br4
(d) cesium diamminetetracyanochromate(III)
(b) Na33Co(C2O4)2Cl24
(b) bis(bipyridyl)cadmium(II) chloride
(c) 3Cr(o@phen)34(CH3COO)3
(c) potassium tetrabromo(ortho-phenanthroline)-cobaltate (III)
23.31 (a) What is meant by the term chelate effect? (b) What thermodynamic factor is generally responsible for the chelate
­effect? (c) Why are polydentate ligands often called sequestering agents?
23.32Pyridine (C5H5N), abbreviated py, is the molecule
N
(a) Why is pyridine referred to as a monodentate ligand? (b) For
the equilibrium reaction
3Ru(py)4(bipy)42+ + 2 py ∆ 3Ru(py)642+ + bipy
what would you predict for the magnitude of the equilibrium
constant? Explain your answer.
23.33 True or false? The following ligand can act as a bidentate
ligand?
N
23.37 Write the names of the following compounds, using the standard nomenclature rules for coordination complexes:
(a) 3Rh(NH3)4Cl24Cl
(b) K23TiCl64
(c) MoOCl4
(d) 3Pt(H2O)4(C2O4)]Br2
23.38Write names for the following coordination compounds:
(a) 3Cd(en)Cl24
(b) K43Mn(CN)64
(c) 3Cr(NH3)5(CO3)]Cl
(d) 3Ir(NH3)4(H2O)24(NO3)3
23.39 Consider the following three complexes:
(Complex 1) 3Co(NH3)4Br24Cl
(Complex 2) 3Pd(NH3)2(ONO)24
(Complex 3) 3V(en)2Cl24+,
N
23.34When silver nitrate is reacted with the molecular base
ortho-phenanthroline, colorless crystals form that contain the
transition-metal complex shown below. (a) What is the coordination geometry of silver in this complex? (b) Assuming
no oxidation or reduction occurs during the reaction, what
is charge of the complex shown here? (c) Do you expect any
nitrate ions will be present in the crystal? (d) Write a formula
for the compound that forms in this reaction. (e) Use the accepted nomenclature to write the name of this compound.
N
(e) tris(ethylenediamine)rhodium(III) tris(oxalato)cobaltate(III)
N
Ag
N N
Nomenclature and Isomerism in Coordination
Chemistry (Section 23.4)
23.35 Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere:
(a) hexaamminechromium(III) nitrate
(b) tetraamminecarbonatocobalt(III) sulfate
(c) dichlorobis(ethylenediamine)platinum(IV) bromide
Which of the three complexes can have (a) geometric isomers,
(b) linkage isomers, (c) optical isomers, (d) coordinationsphere isomers?
23.40Consider the following three complexes:
(Complex 1) 3Co(NH3)5SCN42+
(Complex 2) 3Co(NH3)3Cl342+
(Complex 3) CoClBr # 5NH3
Which of the three complexes can have (a) geometric isomers,
(b) linkage isomers, (c) optical isomers, (d) coordinationsphere isomers?
23.41 A four-coordinate complex MA2B2 is prepared and found to
have two different isomers. Is it possible to determine from
this information whether the complex is square planar or
tetrahedral? If so, which is it?
23.42Consider an octahedral complex MA3B3. How many geometric isomers are expected for this compound? Will any of the
isomers be optically active? If so, which ones?
23.43 Sketch all the possible stereoisomers of (a) tetrahedral
3Cd(H2O)2Cl24, (b) square-planar 3IrCl2(PH3)24-, (c) octahedral 3Fe(o@phen)2Cl24+.
23.44Sketch all the possible stereoisomers of (a) 3Rh(bipy)
(o-phen)243+, (b) 3Co(NH3)3(bipy)Br42+, (c) square-planar
3Pd(en)(CN)24.
Color and Magnetism in Coordination
Chemistry; Crystal-Field Theory
(Sections 23.5 and 23.6)
23.45 (a) If a complex absorbs light at 610 nm, what color would
you expect the complex to be? (b) What is the energy in Joules
Additional Exercises
of a photon with a wavelength of 610 nm? (c) What is the energy
of this absorption in kJ>mol?
23.46(a) A complex absorbs photons with an energy of
4.51 * 10-19 J. What is the wavelength of these photons? (b) If
this is the only place in the visible spectrum where the complex
absorbs light, what color would you expect the complex to be?
23.47 Identify each of the following coordination complexes as
either diamagnetic or paramagnetic:
(a) 3ZnCl442-
(b) 3Pd(NH3)2Cl24
(c) 3V(H2O)643+
(d) 3Ni(en)342+
23.48Identify each of the following coordination complexes as
­either diamagnetic or paramagnetic:
(a) 3Ag(NH3)24+
(b) square planar 3Cu(NH3)442+
(c) 3Ru(bipy)342+
(d) 3CoCl442-
23.49 In crystal-field theory, ligands are modeled as if they are point
negative charges. What is the basis of this assumption, and
how does it relate to the nature of metal–ligand bonds?
23.50The lobes of which d orbitals point directly between the ligands in (a) octahedral geometry, (b) tetrahedral geometry?
23.51 (a) Sketch a diagram that shows the definition of the crystal-field splitting energy (∆) for an octahedral crystal field.
(b) What is the relationship between the magnitude of ∆ and the
energy of the d-d transition for a d1 complex? (c) Calculate ∆ in
kJ>mol if a d1 complex has an absorption maximum at 545 nm.
23.52As shown in Figure 23.26, the d-d transition of 3Ti(H2O)643+
produces an absorption maximum at a wavelength of about
500 nm. (a) What is the magnitude of ∆ for 3Ti(H2O)643+ in
kJ>mol? (b) How would the magnitude of ∆ change if the
H2O ligands in 3Ti(H2O)643+ were replaced with NH3 ligands?
23.53 The colors in the copper-containing minerals malachite
(green) and azurite (blue) come from a single d-d transition
in each compound. (a) What is the electron configuration of
the copper ion in these minerals? (b) Based on their colors in
which compound would you predict the crystal field splitting
∆ is larger?
23.54The color and wavelength of the absorption maximum for
3Ni(H2O)642+, 3Ni(NH3)642+, and 3Ni(en)342+ are given in
Figure 23.30. The absorption maximum for the 3Ni(bipy)342+
ion occurs at about 520 nm. (a) What color would you expect
Additional Exercises
23.65 The Curie temperature is the temperature at which a ferromagnetic solid switches from ferromagnetic to paramagnetic,
and for nickel, the Curie temperature is 354 °C. Knowing this,
you tie a string to two paper clips made of nickel and hold the
paper clips near a permanent magnet. The magnet attracts the
paper clips, as shown in the photograph on the left. Now you
heat one of the paper clips with a cigarette lighter, and the clip
drops (right photograph). Explain what happened.
1035
for the 3Ni(bipy)342+ ion? (b) Based on these data, where
would you put bipy in the spectrochemical series?
23.55 Give the number of (valence) d electrons associated with the central metal ion in each of the following complexes: (a) K33TiCl64,
(b) Na33Co(NO2)64, (c) 3Ru(en)34Br3, (d) 3Mo(EDTA)]ClO4,
(e) K33ReCl64.
23.56Give the number of (valence) d electrons associated with
the central metal ion in each of the following complexes:
(a) K33Fe(CN)64, (b) 3Mn(H2O)64(NO3)2, (c) Na3Ag(CN)24,
(d) 3Cr(NH3)4Br24ClO4, (e) 3Sr(EDTA)42 - .
23.57 A classmate says, “A weak-field ligand usually means the
complex is high spin.” Is your classmate correct? Explain.
23.58A classmate says, “A strong-field ligand means that the ligand
binds strongly to the metal ion.” Is your classmate correct?
Explain.
23.59 For each of the following metals, write the electronic configuration of the atom and its 2 + ion: (a) Mn, (b) Ru, (c) Rh.
Draw the crystal-field energy-level diagram for the d orbitals
of an octahedral complex, and show the placement of the d
electrons for each 2 + ion, assuming a strong-field complex.
How many unpaired electrons are there in each case?
23.60For each of the following metals, write the electronic configuration of the atom and its 3 + ion: (a) Fe, (b) Mo, (c) Co.
Draw the crystal-field energy-level diagram for the d orbitals of an octahedral complex, and show the placement of the
d electrons for each 3 + ion, assuming a weak-field complex.
How many unpaired electrons are there in each case?
23.61 Draw the crystal-field energy-level diagrams and show
the placement of d electrons for each of the following:
(a) 3Cr(H2O)642+ (four unpaired electrons), (b) 3Mn(H2O)642+
(high spin), (c) 3Ru(NH3)5(H2O)42+ (low spin), (d) 3IrCl642(low spin), (e) 3Cr(en)343+, (f) 3NiF644-.
23.62Draw the crystal-field energy-level diagrams and show
the placement of electrons for the following complexes: (a) 3VCl643-, (b) 3FeF643- (a high-spin complex),
(c) 3Ru(bipy)343+ (a low-spin complex), (d) 3NiCl442- (tetrahedral), (e) 3PtBr642-, (f) 3Ti(en)342+.
23.63 The complex 3Mn(NH3)642+ contains five unpaired electrons.
Sketch the energy-level diagram for the d orbitals, and indicate the placement of electrons for this complex ion. Is the ion
a high-spin or a low-spin complex?
23.64The ion 3Fe(CN)643- has one unpaired electron, whereas
3Fe(NCS)643- has five unpaired electrons. From these results,
what can you conclude about whether each complex is high
spin or low spin? What can you say about the placement of
NCS- in the spectrochemical series?
1036
chapter 23 Transition Metals and Coordination Chemistry
23.66Explain why the transition metals in periods 5 and 6 have
nearly identical radii in each group.
23.67 Based on the molar conductance values listed here for the
series of platinum(IV) complexes, write the formula for each
complex so as to show which ligands are in the coordination sphere of the metal. By way of example, the molar conductances of 0.050 M NaCl and BaCl2 are 107 ohm-1 and
197 ohm-1, respectively.
Complex
Molar Conductance (ohm−1)*
of 0.050 M Solution
Pt(NH3)6Cl4
523
Pt(NH3)4Cl4
228
Pt(NH3)3Cl4
97
Pt(NH3)2Cl4
KPt(NH3)Cl5
0
108
*The ohm is a unit of resistance; conductance is the
inverse of resistance.
23.68(a) A compound with formula RuCl3 # 5 H2O is dissolved
in water, forming a solution that is approximately the same
color as the solid. Immediately after forming the solution,
the addition of excess AgNO3(aq) forms 2 mol of solid AgCl
per mole of complex. Write the formula for the compound,
showing which ligands are likely to be present in the coordination sphere. (b) After a solution of RuCl3 # 5 H2O has stood
for about a year, addition of AgNO3(aq) precipitates 3 mol of
AgCl per mole of complex. What has happened in the ensuing time?
23.69 Sketch the structure of the complex in each of the following
compounds and give the full compound name:
(a) cis@3Co(NH3)4(H2O)24(NO3)2
(b) Na23Ru(H2O)Cl54
(c) trans@NH43Co(C2O4)2(H2O)24
(d) cis@3Ru(en)2Cl24
−
H
Trifluoromethyl
acetylacetonate CF
3
(tfac)
C
C
C
O
O
Sketch all possible isomers for the complex with three tfac ligands on cobalt(III). (You can use the symbol
to represent the ligand.)
23.74 Which transition metal atom is present in each of the following biologically important molecules: (a) hemoglobin,
(b) chlorophylls, (c) siderophores.
23.75 Carbon monoxide, CO, is an important ligand in coordination chemistry. When CO is reacted with nickel metal the
product is 3Ni(CO)44, which is a toxic, pale yellow liquid.
(a) What is the oxidation number for nickel in this compound? (b) Given that 3Ni(CO)44 is diamagnetic molecule
with a tetrahedral geometry, what is the electron configuration
of nickel in this compound? (c) Write the name for 3Ni(CO)44
using the nomenclature rules for coordination compounds.
23.76 Some metal complexes have a coordination number of 5.
One such complex is Fe(CO)5, which adopts a trigonal bipyramidal geometry (see Figure 9.8). (a) Write the name
for Fe(CO)5, using the nomenclature rules for coordination
compounds. (b) What is the oxidation state of Fe in this compound? (c) Suppose one of the CO ligands is replaced with a
CN- ligand, forming 3Fe(CO)4(CN)4-. How many geometric
isomers would you predict this complex could have?
23.77Which of the following objects is chiral: (a) a left shoe,
(b) a slice of bread, (c) a wood screw, (d) a molecular model
of Zn(en)Cl2, (e) a typical golf club?
23.78 The complexes 3V(H2O)643+ and 3VF643- are both known.
(a) Draw the d-orbital energy-level diagram for V(III) octahedral complexes. (b) What gives rise to the colors of these
complexes? (c) Which of the two complexes would you expect
to absorb light of higher energy?
[23.79] One of the more famous species in coordination chemistry is
the Creutz–Taube complex:
23.70(a) Which complex ions in Exercise 23.69 have a mirror
plane? (b) Will any of the complexes have optical isomers?
23.71 The molecule dimethylphosphinoethane 3(CH3)2PCH2CH2
P(CH3)2, which is abbreviated dmpe] is used as a ligand for
some complexes that serve as catalysts. A complex that contains this ligand is Mo(CO)4(dmpe). (a) Draw the Lewis
structure for dmpe, and compare it with ethylenediamine as
a coordinating ligand. (b) What is the oxidation state of Mo
in Na23Mo(CN)2(CO)2(dmpe)]? (c) Sketch the structure of
the 3Mo(CN)2(CO)2(dmpe)42- ion, including all the possible
isomers.
23.72Although the cis configuration is known for 3Pt(en)Cl24, no
trans form is known. (a) Explain why the trans compound is
not possible. (b) Would NH2CH2CH2CH2CH2NH2 be more
likely than en (NH2CH2CH2NH2) to form the trans compound? Explain.
23.73The acetylacetone ion forms very stable complexes with many
metallic ions. It acts as a bidentate ligand, coordinating to
the metal at two adjacent positions. Suppose that one of the
CH3 groups of the ligand is replaced by a CF3 group, as shown
here:
CH3
5+
(NH3)5RuN
NRu(NH3)5
It is named for the two scientists who discovered it and initially studied its properties. The central ligand is pyrazine, a
planar six-membered ring with nitrogens at opposite sides.
(a) How can you account for the fact that the complex, which
has only neutral ligands, has an odd overall charge? (b) The
metal is in a low-spin configuration in both cases. Assuming
octahedral coordination, draw the d-orbital energy-level diagram for each metal. (c) In many experiments the two metal
ions appear to be in exactly equivalent states. Can you think
of a reason that this might appear to be so, recognizing that
electrons move very rapidly compared to nuclei?
23.80 Solutions of 3Co(NH3)642+, 3Co(H2O)642+ (both octahedral),
and 3CoCl442- (tetrahedral) are colored. One is pink, one is
blue, and one is yellow. Based on the spectrochemical series
and remembering that the energy splitting in tetrahedral
complexes is normally much less than that in octahedral ones,
assign a color to each complex.
Integrative Exercises
23.81Oxyhemoglobin, with an O2 bound to iron, is a low-spin
Fe(II) complex; deoxyhemoglobin, without the O2 molecule,
is a high-spin complex. (a) Assuming that the coordination
environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case?
(b) What ligand is coordinated to the iron in place of O2 in
deoxyhemoglobin? (c) Explain in a general way why the two
forms of hemoglobin have different colors (hemoglobin is red,
whereas deoxyhemoglobin has a bluish cast). (d) A 15-minute exposure to air containing 400 ppm of CO causes about
10% of the hemoglobin in the blood to be converted into the
carbon monoxide complex, called carboxyhemoglobin. What
does this suggest about the relative equilibrium constants for
binding of carbon monoxide and O2 to hemoglobin? (e) CO
is a strong-field ligand. What color might you expect carboxyhemoglobin to be?
[23.82] Consider the tetrahedral anions VO43- (orthovanadate ion),
CrO42- (chromate ion), and MnO4- (permanganate ion).
(a) These anions are isoelectronic. What does this statement
mean? (b) Would you expect these anions to exhibit d-d transitions? Explain. (c) As mentioned in “A Closer Look” on
charge-transfer color, the violet color of MnO4- is due to a
ligand-to-metal charge transfer (LMCT) transition. What is
meant by this term? (d) The LMCT transition in MnO4- occurs at a wavelength of 565 nm. The CrO42- ion is yellow. Is
the wavelength of the LMCT transition for chromate larger
or smaller than that for MnO4-? Explain. (e) The VO43- ion
is colorless. Do you expect the light absorbed by the LMCT
to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.
23.83(a) Given the colors observed for VO43- (orthovanadate ion),
CrO42- (chromate ion), and MnO4- (permanganate ion) (see
Exercise 23.82), what can you say about how the energy separation between the ligand orbitals and the empty d orbitals
changes as a function of the oxidation state of the transition
metal at the center of the tetrahedral anion?
[23.84] The red color of ruby is due to the presence of Cr(III) ions at
octahedral sites in the close-packed oxide lattice of Al2O3. Draw
the crystal-field splitting diagram for Cr(III) in this environment. Suppose that the ruby crystal is subjected to high pressure. What do you predict for the variation in the wavelength of
absorption of the ruby as a function of pressure? Explain.
23.85 In 2001, chemists at SUNY-Stony Brook succeeded in synthesizing the complex trans@3Fe(CN)4(CO)242-, which could be a
1037
model of complexes that may have played a role in the origin
of life. (a) Sketch the structure of the complex. (b) The complex is isolated as a sodium salt. Write the complete name of
this salt. (c) What is the oxidation state of Fe in this complex?
How many d electrons are associated with the Fe in this complex? (d) Would you expect this complex to be high spin or
low spin? Explain.
[23.86] When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was
because of the presence of carbon atoms in the molecule. To
disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able
to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that
contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we will not worry about that
for now.)
23.87Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in
the +3 rather than in the + 2 oxidation state (for metals that
form stable + 3 ions in the first place). Suggest an explanation,
keeping in mind the Lewis acid–base nature of the metal–
ligand bond.
23.88Many trace metal ions exist in the blood complexed with
amino acids or small peptides. The anion of the amino acid
glycine (gly),
O
H2NCH2C
O−
can act as a bidentate ligand, coordinating to the metal
through nitrogen and oxygen atoms. How many isomers are
possible for (a) 3Zn(gly)24 (tetrahedral), (b) 3Pt(gly)24 (square
planar), (c) 3Co(gly)34 (octahedral)? Sketch all possible isomers. Use the symbol to represent the ligand.
23.89The coordination complex 3Cr(CO)64 forms colorless, diamagnetic crystals that melt at 90 °C. (a) What is the oxidation number of chromium in this compound? (b) Given that
3Cr(CO)64 is diamagnetic, what is the electron configuration
of chromium in this compound? (c) Given that 3Cr(CO)64 is
colorless, would you expect CO to be a weak-field or strongfield ligand? (d) Write the name for 3Cr(CO)64 using the nomenclature rules for coordination compounds.
Integrative Exercises
[23.90] Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains Zn2+ in its active site, is responsible for
rapidly interconverting dissolved CO2 and bicarbonate ion,
HCO3-. The zinc in carbonic anhydrase is tetrahedrally coordinated by three neutral nitrogen-containing groups and a
water molecule. The coordinated water molecule has a pKa of
7.5, which is crucial for the enzyme’s activity. (a) Draw the active site geometry for the Zn(II) center in carbonic anhydrase,
just writing “N” for the three neutral nitrogen ligands from
the protein. (b) Compare the pKa of carbonic anhydrase’s active site with that of pure water; which species is more acidic?
(c) When the coordinated water to the Zn(II) center in carbonic anhydrase is deprotonated, what ligands are bound
to the Zn(II) center? Assume the three nitrogen ligands are
unaffected. (d) The pKa of 3Zn(H2O)642+ is 10. Suggest an
explanation for the difference between this pKa and that of
carbonic anhydrase. (e) Would you expect carbonic anhydrase to have a deep color, like hemoglobin and other metalion containing proteins do? Explain.
23.91 T w o d i f f e r e n t c o m p o u n d s h a v e t h e f o r m u l a t i o n
CoBr(SO4) # 5 NH3. Compound A is dark violet, and compound B is red-violet. When compound A is treated with
AgNO3(aq), no reaction occurs, whereas compound B
1038
chapter 23 Transition Metals and Coordination Chemistry
reacts with AgNO3(aq) to form a white precipitate. When
compound A is treated with BaCl2(aq), a white precipitate is formed, whereas compound B has no reaction with
BaCl2(aq). (a) Is Co in the same oxidation state in these complexes? (b) Explain the reactivity of compounds A and B with
AgNO3(aq) and BaCl2(aq). (c) Are compounds A and B isomers of one another? If so, which category from Figure 23.19
best describes the isomerism observed for these complexes?
(d) Would compounds A and B be expected to be strong electrolytes, weak electrolytes, or nonelectrolytes?
23.92A manganese complex formed from a solution containing
potassium bromide and oxalate ion is purified and analyzed.
It contains 10.0% Mn, 28.6% potassium, 8.8% carbon, and
29.2% bromine by mass. The remainder of the compound
is oxygen. An aqueous solution of the complex has about
the same electrical conductivity as an equimolar solution
of K43Fe(CN)64. Write the formula of the compound, using
brackets to denote the manganese and its coordination sphere.
23.93The E° values for two low-spin iron complexes in acidic solution are as follows:
3Fe(o@phen)343+(aq) + e- ∆
Fe(o@phen)342+(aq)
3Fe(CN)643-(aq) + e- ∆
Fe(CN)644-(aq)
E° = 1.12 V
E° = 0.36 V
(a) Is it thermodynamically favorable to reduce both Fe(III)
complexes to their Fe(II) analogs? Explain. (b) Which complex, 3Fe(o@phen)343+ or 3Fe(CN)643-, is more difficult to reduce? (c) Suggest an explanation for your answer to (b).
23.94 A palladium complex formed from a solution containing bromide ion and pyridine, C5H5N (a good electron-pair donor),
is found on elemental analysis to contain 37.6% bromine,
28.3% carbon, 6.60% nitrogen, and 2.37% hydrogen by mass.
The compound is slightly soluble in several organic solvents;
its solutions in water or alcohol do not conduct electricity. It
is found experimentally to have a zero dipole moment. Write
the chemical formula, and indicate its probable structure.
23.95(a) In early studies it was observed that when the complex
3Co(NH3)4Br24Br was placed in water, the electrical conductivity
of a 0.05 M solution changed from an initial value of 191 ohm-1
to a final value of 374 ohm-1 over a period of an hour or so. Suggest an explanation for the observed results. (See Exercise 23.67
for relevant comparison data.) (b) Write a balanced chemical
equation to describe the reaction. (c) A 500-mL solution is made
up by dissolving 3.87 g of the complex. As soon as the solution
is formed, and before any change in conductivity has occurred,
a 25.00-mL portion of the solution is titrated with 0.0100 M
AgNO3 solution. What volume of AgNO3 solution do you expect to be required to precipitate the free Br-(aq)? (d) Based on
the response you gave to part (b), what volume of AgNO3 solution would be required to titrate a fresh 25.00-mL sample of
3Co(NH3)4Br24Br after all conductivity changes have occurred?
23.96 The total concentration of Ca2+ and Mg2+ in a sample of hard
water was determined by titrating a 0.100-L sample of the water with a solution of EDTA4-. The EDTA4- chelates the two
cations:
Mg2+ + 3EDTA44- ¡ 3Mg(EDTA)42Ca2+ + 3EDTA44- ¡ 3Ca(EDTA)42-
It requires 31.5 mL of 0.0104 M 3EDTA44- solution to reach
the end point in the titration. A second 0.100-L sample was
then treated with sulfate ion to precipitate Ca2+ as calcium
sulfate. The Mg2+ was then titrated with 18.7 mL of 0.0104 M
3EDTA44-. Calculate the concentrations of Mg2+ and Ca2+ in
the hard water in mg>L.
23.97Carbon monoxide is toxic because it binds more strongly to
the iron in hemoglobin (Hb) than does O2, as indicated by
these approximate standard free-energy changes in blood:
Hb + O2 ¡ HbO2
∆G° = - 70 kJ
Hb + CO ¡ HbCO
∆G° = - 80 kJ
Using these data, estimate the equilibrium constant at 298 K
for the equilibrium
HbO2 + CO ∆ HbCO + O2
[23.98] The molecule methylamine (CH3NH2) can act as a monodentate ligand. The following are equilibrium reactions and the
thermochemical data at 298 K for reactions of methylamine
and en with Cd2+(aq):
Cd2+(aq) + 4 CH3NH2(aq) ∆ 3Cd(CH3NH2)442+(aq)
∆H ° = -57.3 kJ; ∆S° = -67.3 J>K; ∆G° = - 37.2 kJ
Cd2+(aq) + 2 en(aq) ∆ 3Cd(en)242+(aq)
∆H ° = -56.5 kJ; ∆S° = +14.1 J>K; ∆G° = - 60.7 kJ
(a) Calculate ∆G° and the equilibrium constant K for the following ligand exchange reaction:
3Cd(CH3NH2)442+(aq) + 2 en(aq) ∆
3Cd(en)242+(aq) + 4 CH3NH2(aq)
Based on the value of K in part (a), what would you conclude
about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic (∆H °) and the entropic
( - T∆S°) contributions to ∆G° for the ligand exchange reaction.
Explain the relative magnitudes. (c) Based on information in this
exercise and in the “A Closer Look” box on the chelate effect,
predict the sign of ∆H° for the following hypothetical reaction:
3Cd(CH3NH2)442+(aq) + 4 NH3(aq) ∆
3Cd(NH3)442+(aq) + 4 CH3NH2(aq)
23.99 The value of ∆ for the 3CrF643- complex is 182 kJ>mol.
­Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lower-energy
to the higher-energy d-orbital set in this complex. Should the
complex absorb in the visible range?
[23.100]A Cu electrode is immersed in a solution that is 1.00 M in
3Cu(NH3)442+ and 1.00 M in NH3. When the cathode is a
standard hydrogen electrode, the emf of the cell is found to be
+0.08 V. What is the formation constant for 3Cu(NH3)442+?
[23.101]The complex 3Ru(EDTA)(H2O)4- undergoes substitution
reactions with several ligands, replacing the water molecule
with the ligand. In all cases, the ruthenium stays in the + 3
oxidation state and the ligands use a nitrogen donor atom to
bind to the metal.
3Ru(EDTA)(H2O)4- + L ¡ 3Ru(EDTA)L4 - + H2O
The rate constants for several ligands are as follows:
Ligand, L
k (M −1s −1)
Pyridine
6.3 * 103
SCN-
2.7 * 102
CH3CN
3.0 * 10
Design an Experiment
(a) One possible mechanism for this substitution reaction
is that the water molecule dissociates from the Ru(III) in
the rate-determining step, and then the ligand L binds to
Ru(III) in a rapid second step. A second possible mechanism is that L approaches the complex, begins to form a new
bond to the Ru(III), and displaces the water molecule, all in
1039
a single concerted step. Which of these two mechanisms is
more consistent with the data? Explain. (b) What do the results suggest about the relative donor ability of the nitrogens
of the three ligands toward Ru(III)? (c) Assuming that the
complexes are all low spin, how many unpaired electrons
are in each?
Design an Experiment
Following a procedure found in a scientific paper you go into the lab and attempt to prepare crystals of dichlorobis(ethylenediamine)cobalt(III) chloride. The paper states that this
compound can be made by reacting CoCl2 # 6 H2O, an excess of ethylenediamine, O2 from
the air (which acts as an oxidizing agent), water, and concentrated hydrochloric acid. At the
end of the reaction you filter off the solution and are left with a green, crystalline product.
(a) What experiment(s) could you perform to confirm that you have prepared 3CoCl2(en)24Cl
and not 3Co(en)34Cl3? (b) How could you verify that cobalt was present as Co3+ and determine the spin state of the cobalt complex in your product? (c) How many geometric isomers
exist for 3CoCl2(en)24Cl? How could you determine if your product contains a single geometric
isomer or a mixture of geometric isomers? (d) If the product does contain a single geometric
isomer how would you determine which one was present? (Hint: You may find the information in
Table 23.3 helpful.)
24
The Chemistry of Life:
Organic and Biological
Chemistry
We are all familiar with how chemical substances can influence
our health and behavior. Aspirin, also known as acetylsalicylic acid, relieves
aches and pains. Cocaine, whose full chemical name is methyl (1R,2R,3S,5S)3-(benzoyloxy)-8-methyl-8-azabicyclo[3.2.1]octane-2-carboxylate, is a plantderived substance that is used in clinical situations as an anesthetic,
but also is used illegally to experience extreme euphoria (a “high”).
Understanding how these molecules exert their effects, and developing new molecules
that can target disease and pain, is an enormous part of the modern chemical enterprise.
This chapter is about the molecules, composed mainly of carbon, hydrogen, oxygen,
and nitrogen, that bridge chemistry and biology.
More than 16 million carbon-containing compounds are known. Chemists make
thousands of new compounds every year, about 90% of which contain carbon. The
study of compounds whose molecules contain carbon constitutes the branch of chemistry known as organic chemistry. This term arose from the eighteenth-century belief
that organic compounds could be formed only by living (that is, organic) systems. This
idea was disproved in 1828 by the German chemist Friedrich Wöhler when he synthesized urea 1H2NCONH22, an organic substance found in the urine of mammals, by
heating ammonium cyanate 1NH4OCN2, an inorganic (nonliving) substance.
What’s
Ahead
24.1 General Characteristics of Organic
Molecules We begin with a review of the structures and
properties of organic compounds.
24.2 Introduction to Hydrocarbons We consider
hydrocarbons, compounds containing only C and H, including
the hydrocarbons called alkanes, which contain only single
C ¬ C bonds. We also look at isomers, compounds with identical
compositions but different molecular structures.
▶ BRAIN ON COCAINE. These positron-
emitting tomography (PET) scans of the
human brain show how rapidly glucose is
metabolized in different parts of the brain.
The top row of images show the brain of
a normal person; the bottom two rows of
images show the brain of a person who
has taken cocaine, after 10 days and after
100 days (red = high glucose metabolism,
yellow = medium, blue = low). Notice that
glucose metabolism is suppressed in the
brain of the person who has taken cocaine.
24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons We next explore hydrocarbons with one or more C “ C bonds,
called alkenes, and those with one or more C ‚ C bonds, called
alkynes. Aromatic hydrocarbons have at least one planar ring with
delocalized p electrons.
24.4 Organic Functional Groups We recognize that a
central organizing principle of organic chemistry is the functional
group, a group of atoms at which most of the compound’s
chemical reactions occur.
24.5 Chirality in Organic Chemistry We learn that
compounds with nonsuperimposable mirror images are chiral and
that chirality plays important roles in organic and biological chemistry.
24.6 Introduction to Biochemistry We introduce the
chemistry of living organisms, known as biochemistry, biological
chemistry, or chemical biology. Important classes of compounds
that occur in living systems are proteins, carbohydrates, lipids,
and nucleic acids.
24.7 Proteins We learn that proteins are polymers of amino
acids linked with amide (also called peptide) bonds. Proteins
are used by organisms for structural support, as molecular
transporters and as catalysts for biochemical reactions.
24.8 Carbohydrates We observe that carbohydrates
are sugars and polymers of sugars used primarily as fuel
by organisms (glucose) or as structural support in plants
(cellulose).
24.9 Lipids We find that lipids are a large class of molecules
used primarily for energy storage in organisms.
24.10 Nucleic Acids We learn that nucleic acids are polymers
of nucleotides that contain an organism’s genetic information.
Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are
polymers composed of nucleotides.
1042
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
The study of the chemistry of living species is called biological chemistry, chemical
biology, or biochemistry. In this chapter, we present some of the elementary aspects of
both organic chemistry and biochemistry.
24.1 | General Characteristics
of Organic Molecules
What is it about carbon that leads to the tremendous diversity in its compounds and
allows it to play such crucial roles in biology and society? Let’s consider some general
features of organic molecules and, as we do, review principles we learned in earlier
chapters.
The Structures of Organic Molecules
Because carbon has four valence electrons (3He42s22p2), it forms four bonds in virtually all its compounds. When all four bonds are single bonds, the electron pairs are
(Section 9.2) In the hybridization model,
disposed in a tetrahedral arrangement.
(Section 9.5) When there is
the carbon 2s and 2p orbitals are then sp3 hybridized.
one double bond, the arrangement is trigonal planar (sp2 hybridization). With a triple
bond, it is linear (sp hybridization). Examples are shown in ▼ Figure 24.1.
Almost every organic molecule contains C ¬ H bonds. Because the valence shell
of H can hold only two electrons, hydrogen forms only one covalent bond. As a result,
hydrogen atoms are always located on the surface of organic molecules whereas the
C ¬ C bonds form the backbone, or skeleton, of the molecule, as in the propane molecule:
H
H
H
H
C
C
C
H
H
H
Go Figure
What is the geometry around the bottom carbon atom in acetonitrile?
180°
109.5°
Tetrahedral
4 single bonds
sp3 hybridization
120°
Trigonal planar
2 single bonds
1 double bond
sp2 hybridization
Linear
1 single bond
1 triple bond
sp hybridization
▲ Figure 24.1 Carbon geometries. The three common geometries around carbon are
tetrahedral as in methane 1CH42, trigonal planar as in formaldehyde 1CH2O2, and linear as in
acetonitrile 1CH3CN2. Notice that in all cases each carbon atom forms four bonds.
H
section 24.1 General Characteristics of Organic Molecules
1043
The Stabilities of Organic Substances
Carbon forms strong bonds with a variety of elements, especially H, O, N, and the halo (Section 8.8) Carbon also has an exceptional ability to bond to itself, formgens.
ing a variety of molecules made up of chains or rings of carbon atoms. Most reactions
(Section 14.5) begin when a region of high
with low or moderate activation energy
electron density on one molecule encounters a region of low electron density on another molecule. The regions of high electron density may be due to the presence of a
multiple bond or to the more electronegative atom in a polar bond. Because of their
strength (the C ¬ C single bond enthalpy is 348 kJ>mol, the C ¬ H bond enthalpy is
Table 8.4) and lack of polarity, both C ¬ C single bonds and C ¬ H
413 kJ>mol
bonds are relatively unreactive. To better understand the implications of these facts,
consider ethanol:
H
H
H
C
C
H
H
O
Go Figure
How would replacing OH groups on
ascorbic acid with CH3 groups affect
the substance’s solubility in (a) polar
solvents and (b) nonpolar solvents?
H
The differences in the electronegativity values of C (2.5) and O (3.5) and of O and
H (2.1) indicate that the C ¬ O and O ¬ H bonds are quite polar. Thus, many reactions of ethanol involve these bonds while the hydrocarbon portion of the molecule
remains intact. A group of atoms such as the C ¬ O ¬ H group, which determines
how an organic molecule reacts (in other words, how the molecule functions), is called
a functional group. The functional group is the center of reactivity in an organic
molecule.
Glucose (C6H12O6)
Give It Some Thought
Which bond is most likely to be the location of a chemical reaction:
C “ N, C ¬ C, or C ¬ H?
Solubility and Acid–Base Properties
of Organic Substances
In most organic substances, the most prevalent bonds are carbon–carbon and carbon–
hydrogen, which are not polar. For this reason, the overall polarity of organic molecules
is often low, which makes them generally soluble in nonpolar solvents and not very
(Section 13.3) Organic molecules that are soluble in polar solvents
soluble in water.
are those that have polar groups on the molecule surface, such as glucose and ascorbic
acid (▶ Figure 24.2). Organic molecules that have a long, nonpolar part bonded to a
polar, ionic part, such as the stearate ion shown in Figure 24.2, function as surfactants
(Section 13.6) The nonpolar part of the moland are used in soaps and detergents.
ecule extends into a nonpolar medium such as grease or oil, and the polar part extends
into a polar medium such as water.
Many organic substances contain acidic or basic groups. The most important acidic organic substances are the carboxylic acids, which bear the functional
(Sections 4.3 and 16.10) The most important basic organic subgroup ¬ COOH.
stances are amines, which bear the ¬ NH2, ¬ NHR, or ¬ NR2 groups, where R is an
(Section 16.7)
organic group made up of carbon and hydrogen atoms.
As you read this chapter, you will find many concept links (
) to related materials in earlier chapters. We strongly encourage you to follow these links and review
the earlier material. Doing so will enhance your understanding and appreciation of
organic chemistry and biochemistry.
Ascorbic acid (HC6H7O6)
−
Stearate (C17H35COO−)
▲ Figure 24.2 Organic molecules that are
soluble in polar solvents.
1044
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
24.2 | Introduction to Hydrocarbons
Because carbon compounds are so numerous, it is convenient to organize them into
families that have structural similarities. The simplest class of organic compounds is
the hydrocarbons, compounds composed of only carbon and hydrogen. The key structural feature of hydrocarbons (and of most other organic substances) is the presence
of stable carbon–carbon bonds. Carbon is the only element capable of forming stable,
extended chains of atoms bonded through single, double, or triple bonds.
Hydrocarbons can be divided into four types, depending on the kinds of carbon–
carbon bonds in their molecules. ▼ Table 24.1 shows an example of each type.
Alkanes contain only single C ¬ C bonds. Alkenes, also known as olefins, contain
at least one C “ C double bond, and alkynes contain at least one C ‚ C triple bond.
In aromatic hydrocarbons the carbon atoms are connected in a planar ring structure,
(Section 8.6)
joined by both s and delocalized p bonds between carbon atoms.
Benzene 1C6H62 is the best-known example of an aromatic hydrocarbon.
Each type of hydrocarbon exhibits different chemical behaviors, as we will see
shortly. The physical properties of all four types, however, are similar in many ways.
Because hydrocarbon molecules are relatively nonpolar, they are almost completely
insoluble in water but dissolve readily in nonpolar solvents. Their melting points and
(Section 11.2) As a result, hydroboiling points are determined by dispersion forces.
carbons of very low molecular weight, such as C2H61bp = -89 °C2, are gases at room
temperature; those of moderate molecular weight, such as C6H141bp = 69 °C2, are
liquids; and those of high molecular weight, such as C22H461mp = 44 °C2, are solids.
▶ Table 24.2 lists the ten simplest alkanes. Many of these substances are familiar
because they are used so widely. Methane is a major component of natural gas. Propane is the major component of bottled gas used for home heating and cooking in areas
where natural gas is not available. Butane is used in disposable lighters and in fuel canisters for gas camping stoves and lanterns. Alkanes with 5 to 12 carbon atoms per molecule are used to make gasoline. Notice that each succeeding compound in Table 24.2
has an additional CH2 unit.
Table 24.1 The Four Hydrocarbon Types with Molecular Examples
Type
Example
H
Alkane
Ethane
109.5°
CH3CH3
H
C
1.54 Å
H
C
H
H
Alkene
Alkyne
Ethylene
Acetylene
CH2
CH
H
CH2
C
H
1.34 Å
1.21 Å
C
H
C
H
C
C6H6
H
180°
H
Aromatic Benzene
H
C
122°
H
CH
H
H
C
120°
C
H
C
C
C
1.39 Å
H
H
section 24.2 Introduction to Hydrocarbons
1045
Table 24.2 First Ten Members of the Straight-Chain Alkane Series
Molecular
Formula
Condensed Structural Formula
Name
Boiling
Point 1 °C 2
CH4
CH4
Methane
C2H6
CH3CH3
Ethane
- 89
C3H8
CH3CH2CH3
Propane
- 44
C4H10
CH3CH2CH2CH3
Butane
- 0.5
C5H12
CH3CH2CH2CH2CH3
Pentane
36
C6H14
CH3CH2CH2CH2CH2CH3
Hexane
68
C7H16
CH3CH2CH2CH2CH2CH2CH3
Heptane
98
C8H18
CH3CH2CH2CH2CH2CH2CH2CH3
Octane
125
C9H20
CH3CH2CH2CH2CH2CH2CH2CH2CH3
Nonane
151
C10H22
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
Decane
174
-161
The formulas for the alkanes given in Table 24.2 are written in a notation called
condensed structural formulas. This notation reveals the way in which atoms are bonded
to one another but does not require drawing in all the bonds. For example, the structural formula and the condensed structural formulas for butane 1C4H102 are
H
H
H
H
H
C
C
C
C
H
H
H
H
H3C
H
CH2
CH2
109.5°
s orbital of H
sp3 orbital of C
CH3
or
CH3CH2CH2CH3
Give It Some Thought
▲ Figure 24.3 Bonds about carbon in
methane. This tetrahedral molecular geometry
is found around all carbons in alkanes.
How many C ¬ H and C ¬ C bonds are formed by the middle carbon atom of
propane?
Structures of Alkanes
According to the VSEPR model, the molecular geometry about each carbon atom in
(Section 9.2) The bonding may be described as involvan alkane is tetrahedral.
ing sp3@hybridized orbitals on the carbon, as pictured in ▶ Figure 24.3 for methane.
(Section 9.5)
Rotation about a carbon–carbon single bond is relatively easy and occurs rapidly
at room temperature. To visualize such rotation, imagine grasping either methyl group
of the propane molecule in ▶ Figure 24.4 and rotating the group relative to the rest of
the molecule. Because motion of this sort occurs rapidly in alkanes, a long-chain alkane
molecule is constantly undergoing motions that cause it to change its shape, something
like a length of chain that is being shaken.
Structural Isomers
The alkanes in Table 24.2 are called straight-chain or linear hydrocarbons because
all the carbon atoms are joined in a continuous chain. Alkanes consisting of four or
more carbon atoms can also form branched chains, and when they do, they are called
branched-chain hydrocarbons. (The branches in organic molecules are often called side
chains.) Table 24.3, for example, shows all the straight-chain and branched-chain alkanes containing four and five carbon atoms.
Compounds that have the same molecular formula but different bonding arrangements (and hence different structures) are called structural isomers. Thus, C4H10 has
two structural isomers and C5H12 has three. The structural isomers of a given alkane
differ slightly from one another in physical properties, as the melting and boiling points
in Table 24.3 indicate.
▲ Figure 24.4 Rotation about a C ¬ C bond
occurs easily and rapidly in all alkanes.
1046
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Table 24.3 Isomers of C 4h 10 and C 5h 12
Systematic Name
(Common Name)
Butane
(n-butane)
2-Methylpropane
(isobutane)
Condensed Structural
Formula
Structural Formula
H
H C
H H
C C
H
C H
H
H H
H
H
H
H
H C
C
C H
Space-filling
Model
CH3CH2CH2CH3
CH3
H
H
H C H
CH
CH3
CH3
Melting
Point (°C)
Boiling
Point (°C)
−138
−0.5
−159
−12
−130
+36
−160
+28
−16
+9
H
Pentane
(n-pentane)
H
H
H H
H C
C
C
H
H
H H
H
C H
C
CH3CH2CH2CH2CH3
H
H
2-Methylbutane
(isopentane)
H C H
H H
H
H C
C
C
C H
CH3
CH3
CH
CH2
CH3
H H
H H
H
2,2-Dimethylpropane
(neopentane)
H
H C
H
H C H
C
H C H
CH3
H
C H
H
CH3
C
CH3
CH3
H
The number of possible structural isomers increases rapidly with the number of
carbon atoms in the alkane. There are 18 isomers with the molecular formula C8H18,
for example, and 75 with the molecular formula C10H22.
Give It Some Thought
What evidence can you cite to support the fact that although isomers have the
same molecular formula they are in fact different compounds?
Nomenclature of Alkanes
In the first column of Table 24.3, the names in parentheses are called the common
names. The common name of the isomer with no branches begins with the letter n (indicating the “normal” structure). When one CH3 group branches off the major chain,
the common name of the isomer begins with iso-, and when two CH3 groups branch
off, the common name begins with neo-. As the number of isomers grows, however,
it becomes impossible to find a suitable prefix to denote each isomer by a common
name. The need for a systematic means of naming organic compounds was recognized
section 24.2 Introduction to Hydrocarbons
1047
as early as 1892, when an organization called the International Union of Chemistry met
in Geneva to formulate rules for naming organic substances. Since that time the task of
updating the rules for naming compounds has fallen to the International Union of Pure
and Applied Chemistry (IUPAC). Chemists everywhere, regardless of their nationality,
subscribe to a common system for naming compounds.
The IUPAC names for the isomers of butane and pentane are the ones given first
in Table 24.3. These systematic names, as well as those of other organic compounds,
have three parts to them:
prefix
What
substituents?
base suffix
How many
carbons?
What family?
The following steps summarize the procedures used to name alkanes, which all
have names ending with -ane. We use a similar approach to write the names of other
organic compounds.
1.Find the longest continuous chain of carbon atoms, and use the name of this
chain (given in Table 24.2) as the base name. Be careful in this step because the
longest chain may not be written in a straight line, as in the following structure:
CH3
2
1
CH
CH3
CH2
CH2
3
CH2
4
5
CH3
6
2-Methylhexane
Because the longest continuous chain contains six C atoms, this isomer is named as a
substituted hexane. Groups attached to the main chain are called substituents because
they are substituted in place of an H atom on the main chain. In this molecule, the CH3
group not enclosed by the blue outline is the only substituent in the molecule.
2.Number the carbon atoms in the longest chain, beginning with the end nearest
a substituent. In our example, we number the C atoms beginning at the upper
right because that places the CH3 substituent on C2 of the chain. (If we had numbered from the lower right, the CH3 would be on C5.) The chain is numbered from
the end that gives the lower number to the substituent position.
3.Name each substituent. A substituent formed by removing an H atom from an
alkane is called an alkyl group. Alkyl groups are named by replacing the -ane ending of the alkane name with -yl. The methyl group 1CH32, for example, is derived
from methane 1CH42 and the ethyl group 1C2H52 is derived from ethane 1C2H62.
▶ Table 24.4 lists six common alkyl groups.
4.Begin the name with the number or numbers of the carbon or carbons to which
each substituent is bonded. For our compound, the name 2-methylhexane indicates the presence of a methyl group on C2 of a hexane (six-carbon) chain.
5.When two or more substituents are present, list them in alphabetical order.
The presence of two or more of the same substituent is indicated by the prefixes
di- (two), tri- (three), tetra- (four), penta- (five), and so forth. The prefixes are
ignored in determining the alphabetical order of the substituents:
7 CH
3
CH3
5
4
CH
6
CH2
CH
3
CH
CH3 2 CH
1
Table 24.4 Condensed Structural
Formulas and Common Names
for Several Alkyl Groups
Group
Name
CH3 ¬
Methyl
CH3CH2 ¬
Ethyl
CH3CH2CH2 ¬
Propyl
CH3CH2CH2CH2 ¬
Butyl
Isopropyl
CH3
HC
CH2CH3
CH3
CH3
3-Ethyl-2,4,5-trimethylheptane
CH3
CH3
CH3
C
CH3
tert-Butyl
1048
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
S a mpl e
Exercise 24.1 Naming Alkanes
Give the systematic name for the following alkane:
CH3
CH2
CH
CH3
CH3
CH
CH2
CH3
CH2
Solution
Analyze We are given the condensed structural formula of an alkane
and asked to give its name.
Plan Because the hydrocarbon is an alkane, its name ends in -ane.
Practice Exercise 1
What is the proper name of this compound?
CH3
The name of the parent hydrocarbon is based on the longest continuous chain of carbon atoms. Branches are alkyl groups, named after the
number of C atoms in the branch and located by counting C atoms
along the longest continuous chain.
CH3
CH2
left CH3 group to the lower left CH3 group and is seven C atoms long:
CH3
2
CH2
3
CH3
4
CH
CH3
CH
5
CH3
6
7
CH2
CH2
The parent compound is thus heptane. There are two methyl groups
branching off the main chain. Hence, this compound is a dimethylheptane. To specify the location of the two methyl groups, we must
number the C atoms from the end that gives the lower two numbers
to the carbons bearing side chains. This means that we should start
numbering at the upper left carbon. There is a methyl group on C3
and one on C4. The compound is thus 3,4-dimethylheptane.
CH3
CH2
Solve The longest continuous chain of C atoms extends from the upper
1
C
CH3
(a) 3-ethyl-3-methylbutane, (b) 2-ethyl-2-methylbutane,
(c)3,3-dimethylpentane, (d) isoheptane,
(e)1,2-dimethyl-neopentane.
Practice Exercise 2
Name the following alkane:
CH3
CH3
CH
CH
CH2
CH3
CH3
S a mpl e
Exercise 24.2 Writing Condensed Structural Formulas
Write the condensed structural formula for 3-ethyl-2-methylpentane.
Solution
Analyze We are given the systematic name for a hydrocarbon and
asked to write its condensed structural formula.
Plan Because the name ends in -ane, the compound is an alkane,
meaning that all the carbon–carbon bonds are single bonds. The parent
hydrocarbon is pentane, indicating five C atoms (Table 24.2). There are
two alkyl groups specified, an ethyl group (two carbon atoms, C2H5)
and a methyl group (one carbon atom, CH3). Counting from left to
right along the five-carbon chain, the name tells us that the ethyl group
is attached to C3 and the methyl group is attached to C2.
Solve We begin by writing five C atoms attached by single bonds.
These represent the backbone of the parent pentane chain:
C¬C¬C¬C¬C
We next place a methyl group on the second C and an ethyl group
on the third C of the chain. We then add hydrogens to all the other
C atoms to make four bonds to each carbon:
CH3
CH3
CH
CH
CH2
CH2CH3
CH3
The formula can be written more concisely as
CH3CH1CH32CH1C2H52CH2CH3
where the branching alkyl groups are indicated in parentheses.
Practice Exercise 1
How many hydrogen atoms are in 2,2-dimethylhexane?
(a) 6, (b) 8, (c) 16, (d) 18, (e) 20.
Practice Exercise 2
Write the condensed structural formula for 2,3-dimethylhexane.
section 24.2 Introduction to Hydrocarbons
Cycloalkanes
Alkanes that form rings, or cycles, are called cycloalkanes. As ▼ Figure 24.5 illustrates, cycloalkane structures are sometimes drawn as line structures, which are polygons in which each corner represents a CH2 group. This method of representation is
(Section 8.6) (Remember from our bensimilar to that used for benzene rings.
zene discussion that in aromatic structures each vertex represents a CH group, not a
CH2 group.)
Carbon rings containing fewer than five carbon atoms are strained because the
C ¬ C ¬ C bond angles must be less than the 109.5 ° tetrahedral angle. The amount
of strain increases as the rings get smaller. In cyclopropane, which has the shape of an
equilateral triangle, the angle is only 60 °; this molecule is therefore much more reactive
than propane, its straight-chain analog.
Give It Some Thought
Are the C ¬ C bonds cyclopropane weaker than those in cyclohexane?
Reactions of Alkanes
Because they contain only C ¬ C and C ¬ H bonds, most alkanes are relatively unreactive. At room temperature, for example, they do not react with acids, bases, or strong
oxidizing agents. Their low chemical reactivity, as noted in Section 24.1, is due primarily
to the strength and lack of polarity of C ¬ C and C ¬ H bonds.
Alkanes are not completely inert, however. One of their most commercially important
(Section 3.2) For
reactions is combustion in air, which is the basis of their use as fuels.
example, the complete combustion of ethane proceeds according to this highly exothermic
reaction:
2 C2H61g2 + 7 O21g2 ¡ 4 CO21g2 + 6 H2O1l2
∆H° = -2855 kJ
Go Figure
The general formula for straight-chain alkanes is CnH2n +2. What is the general formula for cycloalkanes?
H2C
H2C
CH2
CH2
CH2
CH2
H2C
H2C
CH2
CH2
CH2
CH2
H2C
CH2
Cyclohexane
Cyclopentane
Cyclopropane
Each vertex
represents one
CH2 group
Five vertices =
five CH2 groups
Three vertices =
three CH2 groups
▲ Figure 24.5 Condensed structural formulas and line structures for three cycloalkanes.
1049
1050
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Chemistry Put to Work
Gasoline
Petroleum, or crude oil, is a mixture of hydrocarbons plus smaller
quantities of other organic compounds containing nitrogen, oxygen,
or sulfur. The tremendous demand for petroleum to meet the world’s
energy needs has led to the tapping of oil wells in such forbidding
places as the North Sea and northern Alaska.
The usual first step in the refining, or processing, of petroleum is to
separate it into fractions on the basis of boiling point (▼ Table 24.5).
Because gasoline is the most commercially important of these fractions,
various processes are used to maximize its yield.
Gasoline is a mixture of volatile alkanes and aromatic hydrocarbons. In a traditional automobile engine, a mixture of air and gasoline
vapor is compressed by a piston and then ignited by a spark plug. The
burning of the gasoline should create a strong, smooth expansion of
gas, forcing the piston outward and imparting force along the driveshaft of the engine. If the gas burns too rapidly, the piston receives
a single hard slam rather than a strong, smooth push. The result is a
“knocking” or “pinging” sound and a reduction in the efficiency with
which energy produced by the combustion is converted to work.
The octane number of a gasoline is a measure of its resistance to
knocking. Gasolines with high octane numbers burn more smoothly
and are thus more effective fuels ( ▶ Figure 24.6 ). Branched alkanes and aromatic hydrocarbons have higher octane numbers than
straight-chain alkanes. The octane number of gasoline is obtained
by comparing its knocking characteristics with those of isooctane
(2,2,4-trimethylpentane) and heptane. Isooctane is assigned an octane number of 100, and heptane is assigned 0. Gasoline with the same
Table 24.5 Hydrocarbon Fractions from Petroleum
Fraction
Size Range of
Molecules
Uses
C1 to C5
Boiling-Point
Range 1 °C 2
Gas
-160 to 30
Gaseous fuel,
production
of H2
Straight-run
gasoline
C5 to C12
30 to 200
Motor fuel
Kerosene,
fuel oil
C12 to C18
180 to 400
Diesel fuel,
furnace fuel,
cracking
Lubricants
C16 and up
350 and up
Lubricants
Paraffins
C20 and up
Low-melting
solids
Candles,
matches
Asphalt
C36 and up
Gummy
residues
Surfacing
roads
▲ Figure 24.6 Octane rating. The octane rating of gasoline measures
its resistance to knocking when burned in an engine. The octane rating
of the gasoline in the foreground is 89.
knocking characteristics as a mixture of 91% isooctane and 9% heptane, for instance, is rated as 91 octane.
The gasoline obtained by fractionating petroleum (called
straight-run gasoline) contains mainly straight-chain hydrocarbons
and has an octane number around 50. To increase its octane rating, it
is subjected to a process called reforming, which converts the straightchain alkanes into branched-chain ones.
Cracking is used to produce aromatic hydrocarbons and to convert
some of the less-volatile fractions of petroleum into compounds suitable
for use as automobile fuel. In cracking, the hydrocarbons are mixed with a
catalyst and heated to 4009500 °C. The catalysts used are either clay minerals or synthetic Al2O3 9SiO2 mixtures. In addition to forming molecules
more suitable for gasoline, cracking results in the formation of such lowmolecular-weight hydrocarbons as ethylene and propene. These substances
are used in a variety of reactions to form plastics and other chemicals.
Adding compounds called either antiknock agents or octane enhancers increases the octane rating of gasoline. Until the mid-1970s
the principal antiknock agent was tetraethyl lead, 1C2H524Pb. It is no
longer used, however, because of the environmental hazards of lead
(Section 14.7, “Cataand because it poisons catalytic converters.
lytic Converters”) Aromatic compounds such as toluene 1C6H5CH32
and oxygenated hydrocarbons such as ethanol 1CH3CH2OH2 are now
generally used as antiknock agents.
Related Exercises: 24.19 and 24.20
24.3 | Alkenes, Alkynes, and Aromatic
Hydrocarbons
Because alkanes have only single bonds, they contain the largest possible number of hydrogen atoms per carbon atom. As a result, they are called saturated hydrocarbons. Alkenes,
alkynes, and aromatic hydrocarbons contain multiple carbon–carbon bonds (double,
section 24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons
Go Figure
How many isomers are there for propene, C3H6?
Methyl groups on same
side of double bond
Red numbers mark longest
chain containing C=C
H
CH3
3
CH3
C
2
4
C
1
H
CH3CH2
H
Methylpropene
bp −7 °C
Methyl group branching
off longest chain
3
C
H
2
C
1
H
1-Butene
bp −6 °C
CH3
4
CH3
C
H
3
C
2
1
H
cis-2-Butene
bp +4 °C
Methyl groups on opposite
sides of double bond
H
CH3
4
C
H
3
C
2
CH3
1
trans-2-Butene
bp +1 °C
No methyl group
branching off
▲ Figure 24.7 The alkene C4H8 has four structural isomers.
triple, or delocalized p bonds). As a result, they contain less hydrogen than an alkane with
the same number of carbon atoms. Collectively, they are called unsaturated hydrocarbons.
On the whole, unsaturated molecules are more reactive than saturated ones.
Alkenes
Alkenes are unsaturated hydrocarbons that contain at least one C “ C bond. The simplest alkene is CH2 “ CH2, called ethene (IUPAC) or ethylene (common name), which
plays important roles as a plant hormone in seed germination and fruit ripening. The
next member of the series is CH3 ¬ CH “ CH2, called propene or propylene. Alkenes
with four or more carbon atoms have several isomers. For example, the alkene C4H8
has the four structural isomers shown in ▲ Figure 24.7. Notice both their structures
and their names.
The names of alkenes are based on the longest continuous chain of carbon atoms
that contains the double bond. The chain is named by changing the ending of the
name of the corresponding alkane from -ane to -ene. The compound on the left in
Figure 24.7, for example, has a double bond as part of a three-carbon chain; thus, the
parent alkene is propene.
The location of the double bond along an alkene chain is indicated by a prefix
number that designates the double-bond carbon atom that is nearest an end of the
chain. The chain is always numbered from the end that brings us to the double bond
sooner and hence gives the smallest-numbered prefix. In propene, the only possible
location for the double bond is between the first and second carbons; thus, a prefix
indicating its location is unnecessary. For butene (Figure 24.7), there are two possible
positions for the double bond, either after the first carbon (1-butene) or after the second carbon (2-butene).
Give It Some Thought
How many distinct locations are there for a double bond in a five-carbon linear
chain?
If a substance contains two or more double bonds, the location of each is indicated by a numerical prefix, and the ending of the name is altered to identify the
number of double bonds: diene (two), triene (three), and so forth. For example,
CH2 “ CH ¬ CH2 ¬ CH “ CH2 is 1,4-pentadiene.
The two isomers on the right in Figure 24.7 differ in the relative locations of
their methyl groups. These two compounds are geometric isomers, compounds
that have the same molecular formula and the same groups bonded to one another
1051
1052
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Rotation around
double bond requires
considerable energy
to break π bond
Overlapping p
orbitals create
one π bond
H
H
cis isomer
R
H
R
R
Rotation has
destroyed
orbital overlap
R
H
R
R
H
H
trans isomer
This end of molecule
rotated 90°
▲ Figure 24.8 Geometric isomers exist because rotation about a carbon–carbon double bond
requires too much energy to occur at ordinary temperatures.
(Section 23.4) In the cis
but differ in the spatial arrangement of these groups.
isomer the two methyl groups are on the same side of the double bond, whereas
in the trans isomer they are on opposite sides. Geometric isomers possess distinct
physical properties and can differ significantly from each other in their chemical
behavior.
Geometric isomerism in alkenes arises because, unlike the C ¬ C bond, the C “ C
bond resists twisting. Recall from Section 9.6 that the double bond between two carbon
atoms consists of a s and a p bond. ▲ Figure 24.8 shows a cis alkene. The carbon–
carbon bond axis and the bonds to the hydrogen atoms and to the alkyl groups (designated R) are all in a plane, and the p orbitals that form the p bond are perpendicular
to that plane. As Figure 24.8 shows, rotation around the carbon–carbon double bond
requires the p bond to be broken, a process that requires considerable energy (about
250 kJ>mol). Because rotation does not occur easily around the carbon–carbon bond,
the cis and trans isomers of an alkene cannot readily interconvert and, therefore, exist
as distinct compounds.
S a mpl e
Exercise 24.3 Drawing Isomers
Draw all the structural and geometric isomers of pentene, C5H10, that have an unbranched
hydrocarbon chain.
Solution
Analyze We are asked to draw all the isomers (both structural and geometric) for an alkene with
a five-carbon chain.
Plan Because the compound is named pentene and not pentadiene or pentatriene, we know that
the five-carbon chain contains only one carbon–carbon double bond. Thus, we begin by placing the double bond in various locations along the chain, remembering that the chain can be
numbered from either end. After finding the different unique locations for the double bond, we
consider whether the molecule can have cis and trans isomers.
Solve There can be a double bond after either the first carbon (1-pentene) or second carbon
(2-pentene). These are the only two possibilities because the chain can be numbered from either
end. Thus, what we might erroneously call 3-pentene is actually 2-pentene, as seen by numbering
the carbon chain from the other end:
1
C
1
C
1
C
1
C
2
C
2
C
2
C
2
C
3
C
3
C
3
C
3
C
4
C
4
C
4
C
4
C
5
C
5
C
5
C
5
C
5
renumbered as
C
renumbered as
C
5
4
C
4
C
3
C
3
C
2
C
2
C
1
C
1
C
section 24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons
Because the first C atom in 1-pentene is bonded to two H atoms, there are no cis–trans isomers.
There are cis and trans isomers for 2-pentene, however. Thus, the three isomers for pentene are
CH2
CH
CH2
CH2
CH3
CH3
CH2
C
1-Pentene
H
H
CH3
C
H
cis-2-Pentene
C
CH2
H
CH3
C
CH3
trans-2-Pentene
(You should convince yourself that cis-3-pentene is identical to cis-2-pentene and trans-3-pentene
is identical to trans-2-pentene. However, cis-2-pentene and trans-2-pentene are the correct
names because they have smaller numbered prefixes.)
Practice Exercise 1
Which compound does not exist?
(a) 1,2,3,4,5,6,7-octaheptaene, (b) cis-2-butane, (c) trans-3-hexene,
(d) 1-propene, (e) cis-4-decene.
Practice Exercise 2
How many straight-chain isomers are there of hexene, C6H12?
Alkynes
Alkynes are unsaturated hydrocarbons containing one or more C ‚ C bonds. The simplest alkyne is acetylene 1C2H22, a highly reactive molecule. When acetylene is burned
in a stream of oxygen in an oxyacetylene torch, the flame reaches about 3200 K. Because alkynes in general are highly reactive, they are not as widely distributed in nature
as alkenes; alkynes, however, are important intermediates in many industrial processes.
Alkynes are named by identifying the longest continuous chain containing the triple bond and modifying the ending of the name of the corresponding alkane from -ane
to -yne, as shown in Sample Exercise 24.4.
S a mpl e
Exercise 24.4 Naming Unsaturated Hydrocarbons
Name the following compounds:
CH3
(a) CH3CH2CH2
CH
CH3
C
H
C
H
(b) CH3CH2CH2CH
C
CH
CH2CH2CH3
Solution
Analyze We are given the condensed structural formulas for an alkene and an alkyne and asked
to name the compounds.
Plan In each case, the name is based on the number of carbon atoms in the longest continu-
ous carbon chain that contains the multiple bond. In the alkene, care must be taken to indicate
whether cis–trans isomerism is possible and, if so, which isomer is given.
Solve
(a) The longest continuous chain of carbons that contains the double bond is seven carbons
long, so the parent hydrocarbon is heptene. Because the double bond begins at carbon 2
(numbering from the end closer to the double bond), we have 2-heptene. With a methyl
group at carbon atom 4, we have 4-methyl-2-heptene. The geometrical configuration at
the double bond is cis (that is, the alkyl groups are bonded to the double bond on the same
side). Thus, the full name is 4-methyl-cis-2-heptene.
1053
1054
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
(b) The longest continuous chain containing the triple bond has six carbons, so this compound
is a derivative of hexyne. The triple bond comes after the first carbon (numbering from the
right), making it 1-hexyne. The branch from the hexyne chain contains three carbon atoms,
making it a propyl group. Because this substituent is located on C3 of the hexyne chain, the
molecule is 3-propyl-1-hexyne.
Practice Exercise 1
If a compound has two carbon–carbon triple bonds and one carbon–carbon double bond,
what class of compound is it?
(a) an eneyne, (b) a dieneyne, (c) a trieneyne, (d) an enediyne, (e) an enetriyne.
Practice Exercise 2
Draw the condensed structural formula for 4-methyl-2-pentyne.
Addition Reactions of Alkenes and Alkynes
The presence of carbon–carbon double or triple bonds in hydrocarbons markedly increases their chemical reactivity. The most characteristic reactions of alkenes and alkynes are addition reactions, in which a reactant is added to the two atoms that form
the multiple bond. A simple example is the addition of elemental bromine to ethylene
to produce 1,2,-dibromoethane:
H2C
CH2 + Br2
H2C
CH2
Br Br
[24.1]
The p bond in ethylene is broken and the electrons that formed the bond are used to form
two s bonds to the two bromine atoms. The s bond between the carbon atoms is retained.
Addition of H2 to an alkene converts it to an alkane:
Ni, 500 °C
CH3CH “ CHCH3 + H2 —
¡ CH3CH2CH2CH3[24.2]
The reaction between an alkene and H2, referred to as hydrogenation, does not occur
readily at ordinary temperatures and pressures. One reason for the lack of reactivity of
H2 toward alkenes is the stability of the H2 bond. To promote the reaction, the reaction
temperature must be raised 1500 °C2, and a catalyst (such as Ni) is used to assist in rupturing the H ¬ H bond. We write such conditions over the reaction arrow to indicate
they must be present in order for the reaction to occur. The most widely used catalysts
(Section 14.7)
are finely divided metals on which H2 is adsorbed.
Hydrogen halides and water can also add to the double bond of alkenes, as in these
reactions of ethylene:
CH2 “ CH2 + HBr ¡ CH3CH2Br[24.3]
H2SO4
CH2 “ CH2 + H2O ¡ CH3CH2OH[24.4]
The addition of water is catalyzed by a strong acid, such as H2SO4.
The addition reactions of alkynes resemble those of alkenes, as shown in these
examples:
Cl
CH3C
CCH3 + Cl2
CH3
C
C
2-Butyne
[24.5]
Cl
CH3
trans-2,3-Dichloro-2-butene Cl Cl
CH3C
CCH3 + 2 Cl2
CH3
C
C
CH3
Cl Cl
2-Butyne
2,2,3,3-Tetrachlorobutane [24.6]
+ HBr
section 24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons
1055
S a mpl e
Exercise 24.5 Predicting the Product of an Addition Reaction
Write the condensed structural formula for the product of the hydrogenation of
3-methyl-1-pentene.
Solution
Analyze We are asked to predict the compound formed when a par-
ticular alkene undergoes hydrogenation (reaction with H2) and to
write the condensed structural formula of the product.
Plan To determine the condensed structural formula of the product,
we must first write the condensed structural formula or Lewis structure of the reactant. In the hydrogenation of the alkene, H2 adds to the
double bond, producing an alkane.
Solve The name of the starting compound tells us that we have a
chain of five C atoms with a double bond at one end (position 1) and
a methyl group on C3:
CH3
CH2
CH
CH
CH2
Hydrogenation—the addition of two H atoms to the carbons of the
double bond—leads to the following alkane:
CH3
CH3
CH2
CH
CH2
CH3
Comment The longest chain in this alkane has five carbon atoms; the
product is therefore 3-methylpentane.
Practice Exercise 1
What product is formed from the hydrogenation of 2-methylpropene? (a) propane, (b) butane, (c) 2-methylbutane, (d) 2-methylpropane, (e) 2-methylpropyne.
Practice Exercise 2
Addition of HCl to an alkene forms 2-chloropropane. What is the
alkene?
CH3
A Closer Look
Mechanism of Addition Reactions
The second, faster step is addition of Br- to the positively charged
carbon. The bromide ion donates a pair of electrons to the carbon,
forming the C ¬ Br bond:
As the understanding of chemistry has grown, chemists have advanced from simply cataloging reactions that occur to explaining
+
δ+
how they occur, by drawing the individual steps of a reaction based
CH3CH CH2CH3 + Br−
CH3CH CH2CH3
CH3CHCH2CH3
upon experimental and theoretical evidence. The sum of these steps is
δ−
(Section 14.6)
termed a reaction mechanism.
Br
Br
The addition reaction between +
HBr and an alkene, for inδ+
−
stance, is thought to proceed in two
the2CH
first3 step,
is CH3CH CH2CH3
+ Brwhich
CHsteps.
CH3CHCH2CH3
3CH InCH
[24.8]
(Section 14.6), HBr attacks the electron-rich
rate determining
δ−
Br
Br
double bond, transferring a proton to one of the double-bond car
bons. In the reaction of 2-butene with HBr, for example, the first
Because the rate-determining step involves both the alkene and the
step is
acid, the rate law for the reaction is second order, first order in the
alkene and first order in HBr:
δ+
+
+
CH3CH CHCH3 HBr
CH3CH CHCH3
CH3CH CH2CH3 + Br−
∆3CH3CH “ CHCH34
Rate = = k3CH3CH “ CHCH343HBr4
H
∆t
δ−
[24.9]
Br
δ+
CH3CH
CHCH3
H
+
CH3CH
CH2CH3 + Br−
[24.7]
−
The
pair that formed the p bond is used to form the new
Br δelectron
C ¬ H bond.
The energy profile for the reaction is shown in Figure 24.9. The
first energy maximum represents the transition state in the first step,
and the second maximum represents the transition state in the second
step. The energy minimum represents the energies of the intermediate
+
species, CH3CH ¬ CH2CH3 and Br-.
1056
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Go Figure
Energy
What features of an energy profile allow you to distinguish
between an intermediate and a transition state?
To show electron movement in reactions like these, chemists often use curved arrows pointing in the direction of electron flow. For
the addition of HBr to 2-butene, for example, the shifts in electron
positions are shown as
First transition
state
Second transition
state
CH3CH
+
slow
CH3CH
CH2CH3
CH3CH
CHCH3 + H Product
Br
+ Br−
CH3CH
CHCH3
CH3CHCH2CH3
Intermediates
+ HBr
Reactants
Br
CH3
H
H
C
C
+
CHCH3 + H
CH3 + Br−
fast
slow
Br
CH3
H
CH3
H
H
C
C
H
H
C
C
+
H
CH3
Br H
Reaction pathway
▲ Figure 24.9 Energy profile for addition of HBr to 2-butene. The two
maxima tell you that this is a two-step mechanism.
Aromatic Hydrocarbons
The simplest aromatic hydrocarbon, benzene 1C6H62, is shown in ▼ Figure 24.10 along
with some other aromatic hydrocarbons. Benzene is the most important aromatic hydrocarbon, and most of our discussion focuses on it.
CH3
Benzene
Anthracene
Naphthalene
Toluene
(Methylbenzene)
Pyrene
▲ Figure 24.10 Line formulas and common names of several aromatic compounds.
The aromatic rings are represented by hexagons with a circle inscribed inside to denote delocalized
p bonds. Each vertex represents a carbon atom. Each carbon is bound to three other atoms—either
three carbons, or two carbons and a hydrogen—so that each carbon has the requisite four bonds.
Stabilization of p Electrons by Delocalization
If you draw one Lewis structure for benzene, you draw a ring that contains three CC
(Section 8.6) You might therefore expect
double bonds and three single CC bonds.
benzene to resemble the alkenes and to be highly reactive. Benzene and the other aromatic hydrocarbons, however, are much more stable than alkenes because the p elec (Section 9.6)
trons are delocalized in the p orbitals.
We can estimate the stabilization of the p electrons in benzene by comparing the
energy required to form cyclohexane by adding hydrogen to benzene, to cyclohexene
(one double bond) and to 1,4-cyclohexadiene (two double bonds):
+ 3 H2
∆H° = −208 kJ/mol
+ H2
∆H° = −120 kJ/mol
+ 2 H2
∆H° = −232 kJ/mol
CH3 + Br−
fast
section 24.3 Alkenes, Alkynes, and Aromatic Hydrocarbons
From the second and third reactions, it appears that the energy required to hydrogenate each double bond is roughly 118 kJ/mol for each bond. Benzene contains the
equivalent of three double bonds. We might expect, therefore, the energy required to
hydrogenate benzene to be about 3 times -118, or -354 kJ>mol, if benzene behaved
as though it were “cyclohexatriene,” that is, if it behaved as though it had three isolated double bonds in a ring. Instead, the energy released is 146 kJ less than this,
indicating that benzene is more stable than would be expected for three double
bonds. The difference of 146 kJ>mol between the expected heat of hydrogenation
1-354 kJ>mol2 and the observed heat of hydrogenation, 1-208 kJ>mol2 is due to
stabilization of the p electrons through delocalization in the p orbitals that extend
around the ring. Chemists call this stabilization energy the resonance energy.
Substitution Reactions
Although aromatic hydrocarbons are unsaturated, they do not readily undergo addition reactions. The delocalized p bonding causes aromatic compounds to behave
quite differently from alkenes and alkynes. Benzene, for example, does not add Cl2
or Br2 to its double bonds under ordinary conditions. In contrast, aromatic hydrocarbons undergo substitution reactions relatively easily. In a substitution reaction,
one hydrogen atom of the molecule is removed and replaced (substituted) by another atom or group of atoms. When benzene is warmed in a mixture of nitric and
sulfuric acids, for example, one of the benzene hydrogens is replaced by the nitro
group, NO2:
+ HNO3
Benzene
NO2
H2SO4
+ H2O
Nitrobenzene
[24.10]
More vigorous treatment results in substitution of a second nitro group into the
molecule:
NO2
NO2
H2SO4
+ HNO3
+ H2O
NO2
[24.11]
There are three isomers of benzene that contain two nitro groups—the 1,2or ortho- isomer, the 1,3- or meta-isomer, and the 1,4- or para-isomer of dinitro­
benzene:
NO2
6
5
1
4
2
NO2
NO2
3
6
5
1
4
NO2
6
2
5
3
NO2
1
4
2
3
NO2
ortho-Dinitrobenzene
1,2-Dinitrobenzene
mp 118 °C
meta-Dinitrobenzene
1,3-Dinitrobenzene
mp 90 °C
para-Dinitrobenzene
1,4-Dinitrobenzene
mp 174 °C
1057
1058
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
In the reaction of Equation 24.11, the principal product is the meta isomer.
Bromination of benzene, carried out with FeBr3 as a catalyst, is another substitution reaction:
Br
+ Br2
Benzene
FeBr3
+ HBr
Bromobenzene
[24.12]
In a similar substitution reaction, called the Friedel–Crafts reaction, alkyl groups
can be substituted onto an aromatic ring by reacting an alkyl halide with an aromatic
compound in the presence of AlCl3 as a catalyst:
+ CH3CH2Cl
Benzene
AlCl3
CH2CH3
+ HCl
Ethylbenzene
[24.13]
Give It Some Thought
When the aromatic hydrocarbon naphthalene, shown in Figure 24.10, reacts
with nitric and sulfuric acids, two compounds containing one nitro group are
formed. Draw the structures of these two compounds.
24.4 | Organic Functional Groups
The C “ C double bonds of alkenes and C ‚ C triple bonds of alkynes are just two
of many functional groups in organic molecules. As noted earlier, these functional
groups each undergo characteristic reactions, and the same is true of all other functional groups. Each kind of functional group often undergoes the same kinds of reactions in every molecule, regardless of the size and complexity of the molecule. Thus,
the chemistry of an organic molecule is largely determined by the functional groups it
contains.
▶ Table 24.6 lists the most common functional groups. Notice that, except for
C “ C and C ‚ C, they all contain either O, N, or a halogen atom, X.
We can think of organic molecules as being composed of functional groups
bonded to one or more alkyl groups. The alkyl groups, which are made of C ¬ C and
C ¬ H single bonds, are the less reactive portions of the molecules. In describing general features of organic compounds, chemists often use the designation R to represent
any alkyl group: methyl, ethyl, propyl, and so on. Alkanes, for example, which contain
no functional group, are represented as R ¬ H. Alcohols, which contain the functional
group ¬ OH, are represented as R ¬ OH. If two or more different alkyl groups are
present in a molecule, we designate them R, R′, R″, and so forth.
Alcohols
Alcohols are compounds in which one or more hydrogens of a parent hydrocarbon have been replaced by the functional group ¬ OH, called either the
section 24.4 Organic Functional Groups
1059
Table 24.6 Common Functional Groups
Example
Functional
Group
Compound
Type
Suffix or
Prefix
Structural
Formula
H
C
C
Alkene
-ene
Alkyne
C
-yne
H
Systematic Name
(common name)
H
C
C
H
C
Ball-and-stick
Model
Ethene
(Ethylene)
H
C
C
H
O
H
Ethyne
(Acetylene)
H
C
O
H
Alcohol
-ol
H
C
Methanol
(Methyl alcohol)
H
H
C
O
C
Ether
ether
H
C
H
O
(X
X
Alkyl halide
or haloalkane
halogen)
C
N
Amine
-ide
H
Aldehyde
H
Chloromethane
(Methyl
chloride)
Cl
H
-amine
H
O
C
C
Dimethyl ether
H
H
H
H
C
C
-al
H
H
H
C
C
N
H
H
H
H
O
C
C
H
H
O
H
C
C
C
H
Ethylamine
Ethanal
(Acetaldehyde)
H
O
C
C
C
Ketone
-one
H
H
O
C
O
H
Carboxylic
acid
-oic acid
H
O
C
Ester
-oate
H
O
N
Amide
H
Ethanoic acid
(Acetic acid)
H
H
O
C
C
O
H
O
C
C
H
O
H
C
Propanone
(Acetone)
H
O
C
H
-amide
H
H
H
O
C
C
H
C
N
H
H
H
Methyl ethanoate
(Methyl acetate)
Ethanamide
(Acetamide)
1060
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
CH3
CH3
CH
CH3
CH3
OH
C
CH3
OH
2-Propanol
Isopropyl alcohol;
rubbing alcohol
CH2
CH2
OH
OH
1,2-Ethanediol
Ethylene glycol
2-Methyl-2-propanol
t-Butyl alcohol
H3C
H 3C
OH
Phenol
CH2
CH
CH2
OH
OH
OH
1,2,3-Propanetriol
Glycerol; glycerin
H3C
H
H
CH3
CH3
HO
2,15-dimethyl-14(1,5-dimethylhexyl)tetracyclo[8.7.0.02,7.011,15]heptacos7-en-5-ol
Cholesterol
▲ Figure 24.11 Condensed structural formulas of six important alcohols. Common names are
given in blue.
hydroxyl group or the alcohol group. Note in ▲ Figure 24.11 that the name for an alcohol ends in -ol. The simple alcohols are named by changing the last letter in the name of
the corresponding alkane to -ol—for example, ethane becomes ethanol. Where necessary, the location of the OH group is designated by a numeric prefix that indicates the
number of the carbon atom bearing the OH group.
The O ¬ H bond is polar, so alcohols are more soluble in polar solvents than are
hydrocarbons. The ¬ OH functional group can also participate in hydrogen bonding. As a result, the boiling points of alcohols are higher than those of their parent
alkanes.
◀ Figure 24.12 shows several commercial products that consist entirely or in large
part of an alcohol.
The simplest alcohol, methanol (methyl alcohol), has many industrial uses and is
produced on a large scale by heating carbon monoxide and hydrogen under pressure in
the presence of a metal oxide catalyst:
CO1g2 + 2 H21g2
200 - 300 atm
¡ CH3OH1g2
400 °C
[24.14]
Because methanol has a very high octane rating as an automobile fuel, it is used as a
gasoline additive and as a fuel in its own right.
Ethanol (ethyl alcohol, C2H5OH) is a product of the fermentation of carbohydrates
such as sugars and starches. In the absence of air, yeast cells convert these carbohydrates into ethanol and CO2:
▲ Figure 24.12 Everyday alcohols. Many
of the products we use every day—from
rubbing alcohol to hair spray and antifreeze—
are composed either entirely or mainly of
alcohols.
yeast
C6H12O61aq2 ¡ 2 C2H5OH1aq2 + 2 CO21g2[24.15]
In the process, the yeast cells derive energy necessary for growth. This reaction is carried
out under carefully controlled conditions to produce beer, wine, and other beverages in
which ethanol (called just “alcohol” in everyday language) is the active ingredient.
The simplest polyhydroxyl alcohol (an alcohol containing more than one OH
group) is 1,2-ethanediol (ethylene glycol, HOCH2CH2OH), the major ingredient in
automobile antifreeze. Another common polyhydroxyl alcohol is 1,2,3-propanetriol
1glycerol, HOCH2CH1OH2CH2OH2, a viscous liquid that dissolves readily in water
and is used in cosmetics as a skin softener and in foods and candies to keep them moist.
Phenol is the simplest compound with an OH group attached to an aromatic ring.
One of the most striking effects of the aromatic group is the greatly increased acidity of
the OH group. Phenol is about 1 million times more acidic in water than a nonaromatic
alcohol. Even so, it is not a very strong acid 1Ka = 1.3 * 10-102. Phenol is used industrially to make plastics and dyes, and as a topical anesthetic in throat sprays.
section 24.4 Organic Functional Groups
Cholesterol, shown in Figure 24.11, is a biochemically important alcohol. The OH
group forms only a small component of this molecule, so cholesterol is only slightly
soluble in water (2.6 g/L of H2O). Cholesterol is a normal and essential component of
our bodies; when present in excessive amounts, however, it may precipitate from solution. It precipitates in the gallbladder to form crystalline lumps called gallstones. It may
also precipitate against the walls of veins and arteries and thus contribute to high blood
pressure and other cardiovascular problems.
Ethers
Compounds in which two hydrocarbon groups are bonded to one oxygen are called
ethers. Ethers can be formed from two molecules of alcohol by eliminating a molecule
of water. The reaction is catalyzed by sulfuric acid, which takes up water to remove it
from the system:
H2SO4
CH3CH2 ¬ OH + H ¬ OCH2CH3 ¡ CH3CH2 ¬ O ¬ CH2CH3 + H2O
[24.16]
A reaction in which water is eliminated from two substances is called a condensation
(Sections 12.8 and 22.8)
reaction.
Both diethyl ether and the cyclic ether tetrahydrofuran, shown below, are common
solvents for organic reactions. Diethyl ether was formerly used as an anesthetic (known
simply as “ether” in that context), but it had significant side effects.
CH3CH2
O
CH2
CH2CH3
Diethyl ether
CH2
CH2 CH2
O
Tetrahydrofuran (THF)
Aldehydes and Ketones
Several of the functional groups listed in Table 24.6 contain the carbonyl group,
C “ O. This group, together with the atoms attached to its carbon, defines several important functional groups that we consider in this section.
In aldehydes, the carbonyl group has at least one hydrogen atom attached:
O
H
C
O
H
Methanal
Formaldehyde
CH3
C
H
Ethanal
Acetaldehyde
In ketones, the carbonyl group occurs at the interior of a carbon chain and is therefore
flanked by carbon atoms:
OH
H3C
H3C
O
CH3
C
O
CH3
Propanone
Acetone
CH3
C
CH2CH3
2-Butanone
Methyl ethyl ketone
O
C
Testosterone
The systematic names of aldehydes contain -al and that ketone names contain -one.
Notice that testosterone has both alcohol and ketone groups; the ketone functional
group dominates the molecular properties. Therefore, testosterone is considered a
ketone first and an alcohol second, and its name reflects its ketone properties.
1061
1062
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Many compounds found in nature contain an aldehyde or ketone functional
group. Vanilla and cinnamon flavorings are naturally occurring aldehydes. Two isomers of the ketone carvone impart the characteristic flavors of spearmint leaves and
caraway seeds.
Ketones are less reactive than aldehydes and are used extensively as solvents. Acetone, the most widely used ketone, is completely miscible with water, yet it dissolves a
wide range of organic substances.
Carboxylic Acids and Esters
Carboxylic acids contain the carboxyl functional group, often written COOH.
(Section 16.10) These weak acids are widely distributed in nature and are com (Section 4.3) They are also important in the manufacture of
mon in citrus fruits.
polymers used to make fibers, films, and paints. ▼ Figure 24.13 shows the formulas of
several carboxylic acids.
The common names of many carboxylic acids are based on their historical origins. Formic acid, for example, was first prepared by extraction from ants; its name is
derived from the Latin word formica, “ant.”
Carboxylic acids can be produced by oxidation of alcohols. Under appropriate
conditions, the aldehyde may be isolated as the first product of oxidation, as in the
sequence
O
CH3CH2OH + (O)
Ethanol
Acetaldehyde
CH3CH
[24.17]
O
O
+ (O)
Acetaldehyde
CH3CH + H2O
CH3COH
[24.18]
Acetic acid where (O) represents any oxidant that can provide oxygen atoms. The air oxidation of
ethanol to acetic acid is responsible for causing wines to turn sour, producing vinegar.
Go Figure
Which of these substances have both a carboxylic acid functional group and an alcohol functional group?
O
CH3
CH
C
O
OH
H
OH
Lactic acid
C
HO
O
O
HO
C
C
C
CH2
C
CH2
OH
Methanoic acid
Formic acid
O
OH
OH
Citric acid
O
C
OH
O
C
CH3
O
Acetylsalicylic acid
Aspirin
O
CH3
C
O
OH
Ethanoic acid
Acetic acid
C
OH
Phenyl methanoic acid
Benzoic acid
▲ Figure 24.13 Structural formulas of common carboxylic acids. The monocarboxylic acids
are generally referred to by their common names, given in blue type.
section 24.4 Organic Functional Groups
The oxidation processes in organic compounds are related to those oxidation reactions we studied in Chapter 20. Instead of counting electrons, the number of C ¬ O
bonds is usually considered to show the extent of oxidation of similar compounds. For
example, methane can be oxidized to methanol, then formaldehye (methanal), then
formic acid (methanoic acid):
H
H
C
H
H
H
H
C
OH
O
H
H
Methane
Methanol
C
O
H
H
Formaldehyde
C
O
H
Formic acid
From methanol to formic acid, the number of C ¬ O bonds increases from 0 to 3
(double bonds are counted as two). If you were to calculate the oxidation state of carbon in these compounds, it would range from -4 in methane (if H’s are counted as +1)
to +2 in formic acid, which is consistent with carbon being oxidized. The ultimate oxidation product of any organic compound, then, is CO2, which is indeed the product of
combustion reactions of carbon-containing compounds (CO2 has 4 C ¬ O bonds, and
C has the oxidation state +4).
Give It Some Thought
What chemical process is happening when formic acid is converted back to
methane?
Aldehydes and ketones can be prepared by controlled oxidation of alcohols. Complete oxidation results in formation of CO2 and H2O, as in the burning of methanol:
CH3OH1g2 +
3
2
O21g2 ¡ CO21g2 + 2 H2O1g2
Controlled partial oxidation to form other organic substances, such as aldehydes and
ketones, is carried out by using various oxidizing agents, such as air, hydrogen peroxide
1H2O22, ozone 1O32, and potassium dichromate 1K2Cr2O72.
Give It Some Thought
Write the condensed structural formula for the acid that would result from
oxidation of the alcohol
CH2 CHOH
CH2
CH2
CH2
Acetic acid can also be produced by the reaction of methanol with carbon monoxide in the presence of a rhodium catalyst:
O
CH3OH + CO
catalyst
CH3
C
OH
[24.19]
This reaction is not an oxidation; it involves, in effect, the insertion of a carbon monoxide molecule between the CH3 and OH groups. A reaction of this kind is called
carbonylation.
Carboxylic acids can undergo condensation reactions with alcohols to form esters:
O
O
CH3
C
OH + HO
Acetic acid
CH2CH3
Ethanol
CH3
C
O
CH2CH3 + H2O
Ethyl acetate
[24.20]
1063
1064
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Esters are compounds in which the H atom of a carboxylic acid is replaced by a carboncontaining group:
O
C
O
C
Give It Some Thought
What is the difference between an ether and an ester?
The name of any ester consists of the name of the group contributed by the alcohol followed by the name of the group contributed by the carboxylic acid, with the -ic
replaced by -ate. For example, the ester formed from ethyl alcohol, CH3CH2OH, and
butyric acid, CH31CH222 COOH, is
O
CH3CH2CH2C
OCH2CH3
Ethyl butyrate
Notice that the chemical formula generally has the group originating from the acid
written first, which is opposite of the way the ester is named. Another example is isoamyl acetate, the ester formed from acetic acid and isoamyl alcohol. Isoamyl acetate
smells like bananas or pears.
O
(CH3)2CHCH2CH2
O
Isoamyl
C
CH3
Acetate
Many esters such as isoamyl acetate have pleasant odors and are largely responsible for the pleasing aromas of fruit.
An ester treated with an acid or a base in aqueous solution is hydrolyzed; that is, the
molecule is split into an alcohol and a carboxylic acid or its anion:
O
CH3CH2
C
O
CH3 + OH−
Methyl propionate
O
CH3CH2
C
Propionate
O− + CH3OH
Methanol [24.21]
The hydrolysis of an ester in the presence of a base is called saponification, a term
that comes from the Latin word for soap, sapon. Naturally occurring esters include
fats and oils, and in making soap an animal fat or a vegetable oil is boiled with a
strong base. The resultant soap consists of a mixture of salts of long-chain carboxylic
acids (called fatty acids), which form during the saponification reaction.
section 24.4 Organic Functional Groups
S a mpl e
Exercise 24.6 Naming Esters and Predicting Hydrolysis
Products
In a basic aqueous solution, esters react with hydroxide ion to form the salt of the carboxylic
acid and the alcohol from which the ester is constituted. Name each of the following esters, and
indicate the products of their reaction with aqueous base.
O
C
(a)
O
OCH2CH3
(b) CH3CH2CH2
C
O
Solution
Analyze We are given two esters and asked to name them and to predict the products formed
when they undergo hydrolysis (split into an alcohol and carboxylate ion) in basic solution.
Plan Esters are formed by the condensation reaction between an alcohol and a carboxylic acid.
To name an ester, we must analyze its structure and determine the identities of the alcohol and
acid from which it is formed. We can identify the alcohol by adding an OH to the alkyl group
attached to the O atom of the carboxyl (COO) group. We can identify the acid by adding an H to
the O atom of the carboxyl group. We have learned that the first part of an ester name indicates
the alcohol portion and the second indicates the acid portion. The name conforms to how the
ester undergoes hydrolysis in base, reacting with base to form an alcohol and a carboxylate anion.
Solve
(a) This ester is derived from ethanol 1CH3CH2OH2 and benzoic acid 1C6H5COOH2. Its name
is therefore ethyl benzoate. The net ionic equation for reaction of ethyl benzoate with
hydroxide ion is
O
C
OCH2CH3(aq) + OH−(aq)
O
C
O−(aq)
+
HOCH2CH3(aq)
The products are benzoate ion and ethanol.
(b) This ester is derived from phenol 1C6H5OH2 and butanoic acid (commonly called butyric
acid) 1CH3CH2CH2COOH2. The residue from the phenol is called the phenyl group. The
ester is therefore named phenyl butyrate or phenyl butanoate. The net ionic equation for the
reaction of phenyl butyrate with hydroxide ion is
O
CH3CH2CH2C
O
(aq)
+
OH−(aq)
O
CH3CH2CH2C
O−(aq) +
HO
(aq)
The products are butyrate ion and phenol.
Practice Exercise 1
For the generic ester RC(O)OR’, which bond will hydrolyze under basic conditions?
(a) the R ¬ C bond (b) the C “ O bond (c) the C ¬ O bond (d) the O ¬ R> bond
(e) more than one of the above
Practice Exercise 2
Write the condensed structural formula for the ester formed from propyl alcohol and propionic acid.
1065
1066
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
Amines and Amides
Amines are compounds in which one or more of the hydrogens of ammonia 1NH32 are
replaced by an alkyl group:
NH2
CH3CH2NH2
(CH3)3N
Ethylamine
Trimethylamine
Phenylamine
Aniline
(Section 16.7) As we saw in the
Amines are the most common organic bases.
Chemistry Put to Work box in Section 16.8, many pharmaceutically active compounds
are complex amines:
HO
N
O
CH3
O
H3C
H3C
O
O
O
O
N
O
N
CH3
HO
Cocaine
CH3
HO
Morphine
Codeine
An amine with at least one H bonded to N can undergo a condensation reaction
with a carboxylic acid to form an amide, which contains the carbonyl group 1C “ O2
attached to N (Table 24.6):
O
O
OH + H
CH3C
N(CH3)2
N(CH3)2 + H2O
CH3C
[24.22]
We may consider the amide functional group to be derived from a carboxylic acid with
an NRR′, NH2 or NHR′ group replacing the OH of the acid, as in these examples:
O
CH3C
O
CH3C
NH
O
NH2
Ethanamide
Acetamide
NH2
C
Phenylmethanamide
Benzamide
OH
N-(4-hydroxyphenyl)ethanamide
Acetaminophen
The amide linkage
O
R
C
N
R′
H
where R and R′ are organic groups, is the key functional group in proteins, as we will
see in Section 24.7.
section 24.5 Chirality in Organic Chemistry
1067
24.5 | Chirality in Organic Chemistry
A molecule possessing a nonsuperimposable mirror image is termed chiral (Greek
(Section 23.4) Compounds containing carbon atoms with four difcheir, “hand”).
ferent attached groups are inherently chiral. A carbon atom with four different attached
groups is called a chiral center. For example, consider 2-bromopentane:
Br
CH3
C
CH2CH2CH3
H
All four groups attached to C2 are different, making that carbon a chiral center.
▼ Figure 24.14 illustrates the nonsuperimposable mirror images of this molecule. Imagine
moving the molecule shown to the left of the mirror over to the right of the mirror. If you
then turn it in every possible way, you will conclude that it cannot be superimposed on the
molecule shown to the right of the mirror. Nonsuperimposable mirror images are called
(Section 23.4) Organic chemists use the labels R
either optical isomers or enantiomers.
and S to distinguish the two forms. We need not go into the rules for deciding on the labels.
Go Figure
If you replace Br with CH3, will the compound be chiral?
Mirror
▲ Figure 24.14 The two enantiomeric forms of 2-bromopentane. The mirror-image isomers are
not superimposable on each other.
The two members of an enantiomer pair have identical physical properties and identical
chemical properties when they react with nonchiral reagents. Only in a chiral environment
do they behave differently from each other. One interesting property of chiral substances is
that their solutions may rotate the plane of polarized light, as explained in Section 23.4.
Chirality is common in organic compounds. It is often not observed, however,
because when a chiral substance is synthesized in a typical reaction, the two enantiomers are formed in precisely the same quantity. The resulting mixture is called a racemic mixture, and it does not rotate the plane of polarized light because the two forms
(Section 23.4)
rotate the light to equal extents in opposite directions.
Many drugs are chiral compounds. When a drug is administered as a racemic
mixture, often only one enantiomer has beneficial results. The other is often inert,
or nearly so, or may even have a harmful effect. For example, the drug (R)-albuterol
(▶ Figure 24.15) is a bronchodilator used to relieve the symptoms of asthma. The
enantiomer (S)-albuterol is not only ineffective as a bronchodilator but also actually
counters the effects of (R)-albuterol. As another example, the nonsteroidal analgesic
ibuprofen is a chiral molecule usually sold as the racemic mixture. However, a preparation consisting of just the more active enantiomer, (S)-ibuprofen (Figure 24.16),
relieves pain and reduces inflammation more rapidly than the racemic mixture. For
this reason, the chiral version of the drug may in time come to replace the racemic one.
Give It Some Thought
What are the requirements on the four groups attached to a carbon atom in order
that it be a chiral center?
OH
HOH2C
H
N
HO
▲ Figure 24.15 (R)-Albuterol. This
compound, which acts as a bronchodilator
in patients with asthma, is one member of
an enantiomer pair. The other member,
(S)-albuterol, has the OH group pointing
down, and does not have the same
physiological effect.
1068
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
COOH
H
C
CH3
CH3
CH
CH3
CH2
▲ Figure 24.16 (S)-Ibuprofen. For relieving
pain and reducing inflammation, the ability
of this enantiomer far outweighs that of the
(R) isomer. In the (R) isomer, the positions of
the H and CH3 group on the far-right carbon
are switched.
24.6 | Introduction to Biochemistry
The functional groups discussed in Section 24.4 generate a vast array of molecules with
very specific chemical reactivities. Nowhere is this specificity more apparent than in
biochemistry—the chemistry of living organisms.
Before we discuss specific biochemical molecules, we can make some general
observations. Many biologically important molecules are quite large, because organisms build biomolecules from smaller, simpler substances readily available in the biosphere. The synthesis of large molecules requires energy because most of the reactions
are endothermic. The ultimate source of this energy is the Sun. Animals have essentially no capacity for using solar energy directly, and so depend on plant photosynthesis
(Section 23.3)
to supply the bulk of their energy needs.
In addition to requiring large amounts of energy, living organisms are highly
organized. In thermodynamic terms, this high degree of organization means that the
entropy of living systems is much lower than that of the raw materials from which the
systems formed. Thus, living systems must continuously work against the spontaneous
tendency toward increased entropy.
(Section 19.3)
In the “Chemistry and Life” essays that appear throughout this text, we have introduced some important biochemical applications of fundamental chemical ideas. The
remainder of this chapter will serve as only a brief introduction to other aspects of biochemistry. Nevertheless, you will see some patterns emerging. Hydrogen bonding,
(Section 11.2), for example, is critical to the function of many biochemical systems, and
the geometry of molecules
(Section 9.1) can govern their biological importance and
activity. Many of the large molecules in living systems are polymers
(Section 12.8)
of much smaller molecules. These biopolymers can be classified into three broad categories:
proteins, polysaccharides (carbohydrates), and nucleic acids. Lipids are another common
class of molecules in living systems, but they are usually large molecules, not biopolymers.
24.7 | Proteins
Proteins are macromolecules present in all living cells. About 50% of your body’s dry
mass is protein. Some proteins are structural components in animal tissues; they are a
key part of skin, nails, cartilage, and muscles. Other proteins catalyze reactions, transport oxygen, serve as hormones to regulate specific body processes, and perform other
tasks. Whatever their function, all proteins are chemically similar, being composed of
smaller molecules called amino acids.
Amino Acids
An amino acid is a molecule containing an amine group, ¬ NH2, and a carboxylic acid
group, ¬ COOH. The building blocks of all proteins are a@amino acids, where the a
(alpha) indicates that the amino group is located on the carbon atom immediately adjacent to the carboxylic acid group. Thus, there is always one carbon atom between the
amino group and the carboxylic acid group.
The general formula for an a@amino acid is represented by
One of about 20 different groups
R
H2N
C
H
R
α carbon
COOH
or
+H N
3
C
COO−
H
The doubly ionized form, called a zwitterion, usually predominates at near-neutral pH values.
This form is a result of the transfer of a proton from the carboxylic acid group to the amine
(Section 16.10: “The Amphiprotic Behavior of Amino Acids”)
group.
Amino acids differ from one another in the nature of their R groups. Twenty-two
amino acids have been identified in nature, and ▶ Figure 24.17 shows the 20 of these
22 that are found in humans. Our bodies can synthesize 11 of these 20 amino acids in
section 24.7 Proteins
1069
Go Figure
Which group of amino acids has a net positive charge at pH 7?
Nonpolar amino acids
CH2
CH3
CH3
CH3 CH
CH3
H
CH3 CH
+
+
H
Glycine
(Gly; G)
H
Alanine
(Ala; A)
H
Valine
(Val; V)
S
CH2
CH2
CH3 CH
CH2
+
+
CH3
CH2
+
+
H3N C COO− H3N C COO− H3N C COO− H3N C COO− H3N C COO− H3N C COO−
H
Leucine
(Leu; L)
Polar amino acids
H
Isoleucine
(Ile; I)
Aromatic amino acids
CH2 H
2
C
CH2
H2N C COO−
+
H
Methionine
(Met; M)
H
Proline
(Pro; P)
OH
H
N
OH
CH2
+
H3N
CH2
COO−
C
CH3
SH
+
H3N
C
HC
+
COO−
H3N
C
OH
CH2
CH2
COO−
+
H3N
C
+
COO− H3N
CH2
COO−
C
+
H3N
C
COO−
H
H
H
H
H
H
Serine
(Ser; S)
Cysteine
(Cys; C)
Threonine
(Thr; T)
Phenylalanine
(Phe; F)
Tyrosine
(Tyr; Y)
Tryptophan
(Trp; W)
Basic amino acids
NH2
+
+
+
HN
NH
CH2
+
NH3
C NH2
CH2
NH
CH2
CH2
CH2
CH2
CH2
+
Acidic amino acids and their amide derivatives
H3N C COO− H3N C COO− H3N C COO−
C
O−
O
C
CH2
+
O−
O
O
CH2
CH2
CH2
+
+
NH2
C
NH2
C
CH2
CH2
+
O
CH2
+
H3N C COO− H3N C COO− H3N C COO− H3N C COO−
H
H
H
H
H
H
H
Histidine
(His; H)
Lysine
(Lys; K)
Arginine
(Arg; R)
Aspartic acid
(Asp; D)
Glutamic acid
(Glu; E)
Asparagine
(Asn; N)
Glutamine
(Gln; Q)
▲ Figure 24.17 The 20 amino acids found in the human body. The blue shading shows the
different R groups for each amino acid. The acids are shown in the zwitterionic form in which
they exist in water at near-neutral pH values. The amino acid names shown in bold are the nine
essential ones.
sufficient amounts for our needs. The other 9 must be ingested and are called essential
amino acids because they are necessary components of our diet.
The a@carbon atom of the amino acids, which is the carbon between the amino and
carboxylate groups, has four different groups attached to it. The amino acids are thus
chiral (except for glycine, which has two hydrogens attached to the central carbon).
For historical reasons, the two enantiomeric forms of amino acids are often distinguished by the labels d (from the Latin dexter, “right”) and l (from the Latin
laevus, “left”). Nearly all the chiral amino acids found in living organisms have the
1070
chapter 24 The Chemistry of Life: Organic and Biological Chemistry
l configuration at the chiral center. The principal exceptions are the proteins that make
up the cell walls of bacteria, which can contain considerable quantities of the d isomers.
Go Figure
How many chiral carbons are there in
one molecule of aspartame?
Polypeptides and Proteins
Amino acids are linked together into proteins by amide groups (Table 24.6):
O
R
C
N
R
H
[24.23]
Each amide group is called a peptide bond when it is formed by amino acids. A peptide
bond is formed by a condensation reaction between the carboxyl group of one amino
acid and the amino group of another amino acid. Alanine and glycine, for example,
form the dipeptide glycylalanine:
+
H3N
H
O
C
C
H O
H
O− + H
H
Glycine (Gly; G)
+
N
C
H
CH3
H3N
H2N C
C
H O
CH2
C
N
C
H
CH2
C O
H
O
C
C
H
CH3
OH
O
Aspartic acid
(Asp)
O−
Alanine (Ala; A)
+
H O
C
Phenylalanine
(Phe)
▲ Figure 24.18 Sweet stuff. The artificial
sweetener aspartame is the methyl ester of a
dipeptide.
H O
O− + H2O
C
N
C
H
CH3
Glycylalanine (Gly–Ala; GA)
The amino acid that furnishes the carboxyl group for peptide-bond formation is named
first, with a -yl ending; then the amino acid furnishing the amino group is named. Using
the abbreviations shown in Figure 24.18, glycylalanine can be abbreviated as either
Gly-Ala or GA. In this notation, it is understood that the unreacted amino group is on
the left and the unreacted carboxyl group on the right.
The artificial sweetener aspartame (◀ Figure 24.18) is the methyl ester of the
dipeptide formed from the amino acids aspartic acid and phenylalanine.
S a mpl e
Exercise 24.7 Drawing the Structural Formula of a Tripeptide
Draw the structural formula for alanylglycylserine.
Solution
Analyze We are given the name of a substance with peptide bonds and
asked to write its structural formula.
Plan The name of this substance suggests that three amino acids—
alanine, glycine, and serine—have been linked together, forming a
tripeptide. Note that the ending -yl has been added to each amino acid
except for the last one, serine. By convention, the sequence of amino
acids in peptides and proteins is written from the nitrogen end to the
carbon end: The first-named amino acid (alanine, in this case) has a
free amino group and the last-named one (serine) has a free carboxyl
group.
Solve We first combine the carboxyl group of alanine with the
amino group of glycine to form a peptide bond and then the
carboxyl group of glycine with the amino group of serine to form
another peptide bond:
Amino group
+
H3N
H
O
C
C
CH3
Ala
A
Carboxyl group
H O
H O
N
C
H
H
Gly
G
C
C
N
C
H
CH2OH
O−
Ser
S
We can abbreviate this tripeptide as either Ala-Gly-Ser or AGS.
section 24.7 Proteins
Practice Exercise 1
How many nitrogen atoms are in the tripeptide Arg-Asp-Gly?
(a) 3, (b) 4, (c) 5, (d) 6, (e) 7.
Practice Exercise 2
Name the dipeptide and give the two ways of writing its abbreviation.
+
H3N
H
O
C
C
HOCH2
Polypeptides are formed when a large number of amino acids 17 302 are linked
together by peptide bonds. Proteins are linear (that is, unbranched) polypeptide molecules with molecular weights ranging from about 6000 to over 50 million amu. Because
up to 22 different amino acids are linked together in proteins and because proteins consist of hundreds of amino acids, the number of possible arrangements of amino acids
within proteins is virtually limitless.
Protein Structure
The sequence of amino acids along a protein chain is called its primary structure and
gives the protein its unique identity. A change in even one amino acid can alter the
biochemical characteristics of the protein. For example, sickle-cell anemia is a genetic
disorder resulting from a single replacement in a protein chain in hemoglobin. The
chain that is affected contains 146 amino acids. The substitution of an amino acid with
a hydrocarbon side chain for one that has an acidic functional group in the side chain
alters the solubility properties of the hemoglobin, and normal blood flow is impeded.
(Section 13.6, “Sickle-Cell Anemia”)
Proteins in living organisms are not simply long, flexible chains with random
shapes. Rather, the chains self-assemble into structures based on the intermolecular forces we learned about in Chapter 11. This self-assembling leads to a protein’s
secondary structure, which refers to how segments of the protein chain are oriented in
a regular pattern, as seen in Figure 24.19.
One of the most important and common secondary structure arrangements is
the A1alpha2@helix. As the a@helix of Figure 24.19 shows, the helix is held in position by hydrogen bonds between amide H atoms and carbonyl O atoms. The pitch
of the helix and its diameter must be such that (1) no bond angles are strained and
(2) the N ¬ H and C “ O functional groups on adjacent turns are in proper position
for hydrogen bonding. An arrangement of this kind is possible for some amino acids
along the chain but not for others. Large protein molecules may contain segments of
the chain that have the a@helical arrangement interspersed with sections in which the
chain is in a random coil.
The other common secondary structure of proteins is the B (beta) sheet. Beta
sheets are made of two or more strands of peptides that hydrogen-bond from an amide
H in one strand to a carbonyl O in the other strand (Figure 24.19).
Give It Some Thought
If you heat a protein to break the intramolecular hydrogen bonds, will you
maintain the a@helical or b@sheet structure?
Proteins are not active biologically unless they are in a particular shape in solution.
The process by which the protein adopts its biologically active shape is called folding.
The shape of a protein in its folded form—determined by all the bends, kinks, and sections of rodlike a@helical, b@sheet, or flexible coil components—is called the tertiary
structure.
H
O
N
C
C
H
CH2
COOH
O−
1071