~
~
~
u
THE STRINGS OF
THE GUITAR
played by the
musician Ziggy
Marley are formed
from a core steel
string covered with
an alloy wire wrap.
980
METALS
AND METALLURGY
WHAT'S AHEAD
23.1
Occurrence and Distribution of Metals
We begin by examining the occurrence of metals
in the lithosphere along with an introduction to
metallurgy, which is the technology associated
with extracting metals from their natural mineral
sources and preparing them for use.
23.2 Pyrometallurgy
We see how high temperatures can be used to
extract metals from minerals. Such processes play
a particularly important role in the production of
iron and steel.
23.3
Hydrometallurgy
We see how chemical processes that occur in
aqueous solutions can be used to separate metals,
with an emphasis on gold and aluminum.
23.4 Electrometallurgy
We investigate the use of electricity to reduce and
refine metals via electrolysis. We also learn the
importance of this approach for the production of
sodium, aluminum, and copper.
23.5
Metallic Bonding
We review the physical properties of metals and
then account for them in terms of two models for
metallic bonding-the electron-sea model and the
molecular-orbital model.
23.6 Alloys
We examine three different types of alloys, or
metals that contain more than one element:
solution alloys, heterogeneous alloys, and
intermetallic compounds.
23.7 Transition Metals
We examine the physical properties, electron
configurations, oxidation states, and magnetic
properties of transition metals and their
compounds.
23.8 Chemistry of Selected Transition Metals
To finish, we take a closer look at the chemistry
of three key transition metals: chromium, iron,
and copper.
METALS HAVE PLAYED A MAJOR ROLE IN THE DEVELOPMENT OF
and they continue to do so. For example, without
the contributions of metallurgy, rock music as we know it
would not exist. The strings of the guitar being played in the
chapter-opening photograph are the product of complex
CIVILIZATION,
technology. The core of the strings is made of mild steel, varying in hardness
and ductility, depending on the type of guitar on which it will be used.
The core string is wound with fine wrap wire, composed typically of an alloy
of copper and zinc for acoustic guitar use, or stainless steel for electric guitars.
Artists can be very passionate about the particular makes and models of
strings they use.
When we think of metals in everyday applications, we tend to think of iron
and aluminum, perhaps also chromium or nickel. But even metals of very low
natural abundance play vital roles in modern technology. To illustrate this
point, Figure 23.1 T shows the approximate composition of a high-performance
jet engine. Notice that iron, long the dominant metal of technology, is not
present to any significant extent.
In this chapter we will consider the chemical forms in which metallic
elements occur in nature and the means by which we obtain metals from these
sources. We will also examine how metals bond in solids and see how metals
and mixtures of metals, called alloys, are employed in modern technology.
Finally, we will look specifically at the properties of transition metals. As we
will see, metals have a varied and interesting chemistry.
981
982
CHAPTER 23
Metals and Metallurgy
I
I
3 8% Titanium
I
J
I
I J II
-
,
•
rf"'lh>~lt
'I
I'rl
I
- f, · ,
.'. '
'
_.,_ -'
3 7% Nickel
12% Chromium
h 0/n
~.Jc~t. ~ - - ' ' ·
~ ,1__/!,tH~~
r: - ·,. ~- . ~-::.:- I,I\ .
I...
I
'"
~ ·~ ~ ,,_>
~ ~ ~ Jj ·2·' frn~~--~-.~~~
~
~
-
1.·
'•
, ._
•/
,
,.. '\.
( II"' 1'"'),<'-ro ._..
.: ... .• ·
...,. ~
~~· '
'w,
~
Z
•
~'
'flo...ir-
~- "'!-.!-"'::--.. ~· .. ..::::·
. ·Mv~-.
...
... _
'-"
;
"""/ ~
~}j ,
~JJ.
II
... ,"'
1\\' ~
,•
!
I ,,..._
, ,,- ~
.--·
'• I
~
5% Alummum
1% Niobium - - - - - - - - - - - - - - - ' 1
0.02% Tantalum- - - - - - - - - - - - .A. Figure 23.1 Composition of a jet engine. The chart on the left shows the metallic elements employed in
manufacturing a modern jet engine.
23.1 OCCURRENCE AND DISTRIBUTION
OF METALS
The portion of our environment that constitutes the solid earth beneath our feet
is called the lithosphere. The lithosphere provides most of the materials we use
to feed, clothe, shelter, and entertain ourselves. Although the bulk of Earth is
solid, we have access to only a small region near the surface. Whereas Earth's
radius is 6370 km, the deepest mine extends only about 4 km into Earth.
Many of the metals that are most useful to us are not especially abundant
in that portion of the lithosphere to which we have ready access. Deposits
that contain metals in economically exploitable quantities are known as ores.
Usually, the desired compounds or elements must be separated from a large
quantity of unwanted material and then chemically processed to make them
useful. The costs, in terms of money, energy, and environmental impact, associated with extraction of metals from their ores is an important issue for society.
The occurrence and distribution of concentrated deposits of metals often play a
role in international politics as nations compete for access to these materials.
Advances in technology also drive demand for metals that are either rare or difficult to extract. As an illustration, consider the fact that over one-half of the
world's supply of cobalt comes from a politically unstable region of Africa
(Congo and Zambia). The supply and price of cobalt are increasingly important
because cobalt is an important component of superalloys used in aircraft
engines (Figure 23.1), as well as the cathodes of Li-ion batteries that power most
portable electronic devices. Hence the price of your cell phone may depend in
part on the policies of the government in Kinshasa.
Minerals
With the exception of gold and the platinum-group metals (Ru, Rh, Pd, Os, Ir,
and Pt), most metallic elements are found in nature in solid inorganic compounds called minerals. Table 23.1 ~ lists the principal mineral sources of several common metals, three of which are shown in Figure 23.2 ~ . Notice that
minerals are identified by common names rather than by chemical names.
Names of minerals are usually based on the locations where they were discovered, the person who discovered them, or some characteristic such as color.
Commercially, the most important sources of metals are oxide, sulfide, and
carbonate minerals. Silicate minerals are very abundant, but they are generally
difficult to concentrate and reduce. c:x:o (Section 22.10) Therefore, most silicates
are not economical sources of metals.
23.1
(a) Malachite
Occurrence and Distribution of Metals
(b) Magnetite
..6. Figure 23.2 Three common minerals.
TABLE 23.1 • Principal Mineral Sources of Some Common Metals
Metal
Mineral
Composition
Aluminum
Corundum
Gibbsite
Chromite
Chalcocite
Chalcopyrite
Malachite
Hematite
Magnetite
Galena
Pyrolusite
Cinnabar
Molybdenite
Cassiterite
Rutile
Ilmenite
Sphalerite
Al20 3
Al(OHh
FeCr204
Cu2S
CuFeS2
Cu2C03(0Hh
Fe20 3
Fe30 4
PbS
Mn02
HgS
MoS2
Sn02
Ti02
FeTi03
ZnS
Chromium
Copper
Iron
Lead
Manganese
Mercury
Molybdenum
Tin
Titanium
Zinc
Metallurgy
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, (2) concentrating the ore or otherwise preparing it for further
treatment, (3) reducing the ore to obtain the free metal, (4) refining or purifying
the metal, and (5) mixing the metal with other elements to modify its properties. This last process produces an alloy, a metallic material that is composed of
two or more elements (Section 23.6).
After being mined, an ore is usually crushed and ground and then treated
to concentrate the desired metal. The concentration stage relies on differences
in the properties of the mineral and the undesired material that accompanies it,
which is called gangue (pronounced "gang"). After an ore is concentrated, a variety of chemical processes are used to obtain the metal ' in suitable purity. In
Sections 23.2-23.4 we will examine some of the most common metallurgical
processes. You will see that these techniques depend in large part on basic
chemical concepts discussed earlier in the text.
(c) Cinnabar
983
984
CHAPTER 23
Metals and Metallurgy
23.2 PYROMETALLURGY
A large number of metallurgical processes utilize high temperatures to alter the
mineral chemically and to ultimately reduce it to the free metal. The use of heat
to alter or reduce the mineral is called pyrometallurgy. (Pyro means "at high
temperature.")
Calcination is the heating of an ore to bring about its decomposition and
the elimination of a volatile product. The volatile product could be, for example, C02 or H 20. Carbonates are often calcined to drive off C02, forming the
metal oxide. For example,
PbC03(s)
D.
~
PbO(s) + C0 2(g)
[23.1]
Most carbonates decompose reasonably rapidly at temperatures in the range of
400 octo 500 °C, although CaC03 requires a temperature of about 1000 °C. Most
hydrated minerals lose H 20 at temperatures on the order of 100 octo 300 °C.
Roasting is a thermal treatment that causes chemical reactions between the
ore and the furnace atmosphere. Roasting may lead to oxidation or reduction
and may be accompanied by calcination. An important roasting process is the
oxidation of sulfide ores, in which the metal sulfide is converted to the metal
oxide, as in the following examples:
2 ZnS(s) + 3 0 2(g)
~
2 ZnO(s) + 2 S0 2(g)
2 MoS2(s) + 7 02(g) ~ 2 Mo03(s) + 4 S02(g)
[23.2]
[23.3]
The sulfide ore of a less active metal, such as mercury, can be roasted to the free
metal:
[23.4]
HgS(s) + 0 2(g) ~ Hg(g) + S02(g)
In many instances the free metal can be obtained by using a reducing atmosphere during roasting. Carbon monoxide provides such an atmosphere
and is frequently used to reduce metal oxides:
PbO(s) + CO(g)
~
Pb(l) + C0 2(g)
[23.5]
This method of reduction is not always feasible, however, especially with active
metals, which are difficult to reduce.
GIVE
IT SOME THOUGHT
Would you expect CaO to reduce to the free metal if heated in the presence of CO?
What about Ag 20?
Smelting is a melting process in which the materials formed in the course
of chemical reactions separate into two or more layers. Smelting often involves
a roasting stage in the same furnace. Two of the important types of layers
formed in smelters are molten metal and slag. The molten metal may consist almost entirely of a single metal, or it may be a solution of two or more metals.
Slag consists mainly of molten silicate minerals, with aluminates, phosphates, and other ionic compounds as constituents. A slag is formed when a basic
metal oxide such as CaO reacts at high temperatures with molten silica (Si02):
CaO(Z) + Si02(l)
~
CaSi03(Z)
[23.6]
Pyrometallurgical operations may involve the concentration and reduction
of a mineral and the refining of the metal. Refining is the treatment of a crude,
relatively impure metal product from a metallurgical process to improve its
23.2
purity and to define its composition better. Sometimes the goal of the refining
process is to obtain the metal itself in pure form. The goal may also be to produce a mixture with a well-defined composition, as in the production of steels
from crude iron.
985
Pyrometallurgy
South Africa Canada
(40)
(33)
All others
United States
(145)
(54)
Ukraine
The Pyrometallurgy of Iron
The most important pyrometallurgical operation is the reduction of iron. Iron
occurs in many different minerals, but the most important sources are two iron
oxide minerals-hematite (Fe20 3) and magnetite (Fe30 4). Figure 23.3 ..,. shows
the world production of iron ore as a function of country. China, Brazil, and
Australia account for nearly two-thirds of the world's iron ore. As the highergrade deposits of iron ore have become depleted, lower-grade ores have been
tapped. Taconite, which consists of fine-grained silica with variable ratios of
iron, present mainly as magnetite, has increased in importance as a source of
iron from the great Mesabi Range west of Lake Superior. These deposits account for almost all of the iron ore production in the United States.
The reduction of iron oxides can be accomplished in a blast furnace, such as
the one illustrated in Figure 23.4 ..,.. A blast furnace is essentially a huge chemical reactor capable of continuous operation. The furnace is charged at the top
with a mixture of iron ore, coke, and limestone. Coke is coal
that has been heated in the absence of air to drive off volatile
components. It is about 85% to 90% carbon. Coke serves as
the fuel, producing heat as it is burned in the lower part of
the furnace. Through reactions with oxygen and water, coke
also serves as the source of the reducing gases CO and H 2.
Limestone (CaC03) serves as the source of the basic oxide
CaO, which reacts with silicates and other components of the
ore to form slag. Air, which enters the blast furnace at the botHot-air supply pipe
tom after preheating, is also an important raw material; it is
required for combustion of the coke. Production of 1 kg of
crude iron, called pig iron, requires about 2 kg of ore, 1 kg of Hot-air
blast nozzle
coke, 0.3 kg of limestone, and 1.5 kg of air.
(one of many)
In the furnace, oxygen reacts with the carbon in the coke
to form carbon monoxide:*
2 C(s)
+ 0 2(g)
~
2 CO(g)
~H
0
= -221 kJ
[23.7]
~
CO(g) + H2(g)
~H
0
Russia ---t:::::.-.
(105)
- c=
~
India
(150)
Usable ore (in metric tons X 106)
~ Figure 23.3 Worldwide mine
production of iron ores. The total
worldwide production of iron ore was
estimated to be 1 .69 x 10 12 kg in 2006.
Ore, limestone,
and coke
250 oc
600 oc
1000 oc
1600 oc
Slag
Molten-iron
outlet
Water vapor present in the air also reacts with carbon, forming
both carbon monoxide and hydrogen:
C(s) + H20(g)
(73)
= +131 kJ
[23.8]
The reaction of coke with oxygen is exothermic and provides heat for furnace
operation, whereas its reaction with water vapor is endothermic. Addition of
water vapor to the air thus provides a means of controlling furnace temperature.
In the upper part of the furnace, limestone decomposes to form CaO and
C02. Here, also, the iron oxides are reduced by CO and H 2. For example, the
important reactions for Fe30 4 are
Fe304(s) + 4 CO(g)
~ 3
Fe(s) + 4 C0 2(g)
~H
0
= - 15 kJ
[23.9]
Fe304(s) + 4 H2(g)
~ 3
Fe(s) + 4 H 20(g)
~H
0
= +150 kJ
[23.10]
*In the reactions shown the enthalpy changes are given as 6.H 0 values. Note, however, that the conditions in
the fu rnace are far from standard.
~ Figure 23.4 A blast furnace. The
furnace is used for reducing iron ore.
Notice the increasing temperatures
as the materials pass downward through
the furnace.
986
CHAPTER 23
Metals and Metallurgy
Reduction of other elements present in the ore also occurs in the hottest parts of
the furnace, where carbon is the major reducing agent.
Molten iron collects at the base of the furnace, as shown in Figure 23.4. It is
overlaid with a layer of molten slag formed by the reaction of CaO with the silica
present in the ore (Equation 23.6). The layer of slag over the molten iron helps to
protect it from reacting with the incoming air. Periodically the furnace is tapped to
drain off slag and molten iron. The iron produced in the furnace may be cast into
solid ingots. Most of the iron, however, is used directly in the manufacture of steel.
For this purpose, it is transported, while stillliquic;i, to the steelmaking shop
(Figure 23.5 ~ ). The production of,pig iron using blast furnaces has decreased in
recent years because of alternative reduction processes and the increased use of
iron scrap in steelmaking. Blast furnaces nevertheless remain a significant
means of reducing iron oxides.
.A Figure 23.5 Molten iron from a
blast furnace. The iron is poured for
transport to a basic oxygen converter.
Steelmakers convert iron to steel by
adding scrap steel and other metals as
alloying agents.
Steel shell
Refractory
brick liner
Formation of Steel
As we will see in Section 23.6, steel is an alloy of iron. To produce steel with favorable properties, it is necessary to remove undesired impurities from the crude iron.
Iron from a blast furnace typically contains 0.6-1.2% silicon, 0.4-2.0% manganese, and lesser amounts of phosphorus and sulfur. In addition, there is considerable dissolved carbon. In the production of steel these impurities are removed by
oxidation in a vessel called a converter. In modem steelmaking the oxidizing agent
is either pure 0 2 or 0 2 diluted with argon. Air cannot be used directly as the source
of 0 2 because N 2 reacts with the molten iron to form iron nitride, which causes the
steel to become brittle.
A cross-sectional view of one converter design appears in Figure 23.6 ~ .
In this converter 0 2, diluted with argon, is blown directly into the molten metal.
The oxygen reacts exothermically with carbon, silicon, and many metal impurities, reducing the concentrations of these elements in the iron. Carbon and sulfur
are expelled as CO and S02 gases, respectively. Silicon is oxidized to Si02 and
adds to whatever slag may have been present initially in the melt. Metal oxides
react with the Si02 to form silicates. The presence of a basic slag is also important
for removal of phosphorus:
Molten iron
and slag
.A Figure 23.6 Refining iron. In this
converter for refining iron, a mixture of
oxygen and argon is blown through the
molten iron and slag. The heat
generated by the oxidation of impurities
maintains the mixture in a molten state.
When the desired composition is
attained, the converter is tilted to pour
out its contents.
3 CaO(l) + P20s(l)
~
Ca3(P04h(Z)
[23.11]
Nearly all of the 0 2 blown into the converter is consumed in the oxidation reactions. By monitoring the 0 2 concentration in the gas coming from the converter,
it is possible to tell when the oxidation is essentially complete. Oxidation of
the impurities present in the iron normally requires about 20 min. When the desired composition is attained, the contents of the converter are dumped into a
large ladle. To produce steels with various kinds of properties, alloying elements are added as the ladle is being filled. The still-molten mixture is then
poured into molds, where it solidifies.
SAMPLE EXERCISE 23.1 I Enthalpy Changes in a Converter
Which produces more heat in a converter, oxidation of 1 mol of C or oxidation of
1 mol of Si?
-
SOLUTION
Analyze: We must compare the enthalpy change for 1 mol of C oxidized to form C02
with the enthalpy change for 1 mol of Si oxidized to form Si02.
Plan: The reactions occur under conditions far different from the standard-state conditions for the substances. Nevertheless, we can estimate the enthalpy changes by
using the thermodynamic values of Appendix C.
Solve: The oxidation reactions and enthalpy changes under standard conditions are
C(s) + Oz(g)
~
C02(g)
6-H~xn
Si(s) + 0 2(g)
~
Si0 2(s)
6-H~xn
= -393.5 kJ - (0 + 0) = -393.5 kJ
= -910.9 kJ - (0 + 0) = -910.9 kJ
23.3
The actual numbers will differ substantially from these because the temperature is far
from 298 K, because both C and Si are dissolved in the molten iron, and because Si02
will be incorporated in the slag. Nevertheless, the differences in the enthalpies of reaction are so great that it seems certain that the enthalpy change will be larger for Si.
-
PRACTICE EXERCISE
(a) What would you expect as the product when any dissolved manganese in the molten
mixture is oxidized? (b) Would the manganese oxidation be exothermic or endothermic?
Answers: (a) Mn02, (b) exothermic
23.3 HYDROMETALLURGY
Pyrometallurgical operations require large quantities of energy and are often a
source of atmospheric pollution, especially by sulfur dioxide. For some metals,
other techniques have been developed in which the metal is extracted from its
ore by use of aqueous reactions. These processes are called hydrometallurgy
(hydro means "water").
·
The most important hydrometallurgical process is leaching, in which the
desired metal-containing compound is selectively dissolved. If the compound is
water soluble, w ater by itself is a suitable leaching agent. More commonly, the
agent is an aqueous solution of an acid, a base, or a salt. Often the dissolving
process involves formation of a complex ion. CO" (Section 17.5) As an example,
we can consider the leaching of gold.
Hydrometallurgy of Gold
As noted in the" A Closer Look" box in Section 4.5, gold metal is often found relatively pure in nature. As concentrated deposits of elemental gold have been depleted, lower-grade sources have become more important. Gold from low-grade
ores can be concentrated by placing the crushed ore on large concrete slabs and
spraying a solution of NaCN over it. In the presence of CN- and air, the gold is
oxidized, forming the stable Au(CNh- ion, which is soluble in water:
4 Au(s) + 8 CN - (aq) + 0 2 (g) + 2 H 20(l)
~
4 Au(CNh - (aq) + 4 OH- (aq)
[23.12]
After a metal ion is selectively leached from an ore, it is precipitated from
solution as the free metal or as an insoluble ionic compound. Gold, for example,
is obtained from its cyanide complex by reduction with zinc powder:
2 Au(CNh - (aq)
+ Zn(s)
~ Zn(CN) 4 2- (aq)
+ 2 Au(s)
[23.13]
The cyanide process for gold recovery has been widely criticized because of
the potential for pollution of groundwater, rivers, and streams. For example,
leakage from containment ponds into nearby streams in Romania several years
ago resulted in poisoning of all aquatic life in small rivers that flow into the
Danube. Alternatives to cyanide are being investigated. Thiosulfate ion, S20l-,
cx:o (Section 22.6) has been studied as an alternative, but the costs are higher
than for the cyanide process.
GIVE IT SOME THOUGHT
Gold dissolves in the presence of thiosulfate as follows:
4 Au + 8 S20 32- (aq) + 2 H 20(Z) + 0 2(aq) ~ 4 Au(S20 3 h3- (aq) + 4 OH- (aq)
What is the oxidizing agent in this reaction, what is the reducing agent, and what
oxidation states of the reducing and oxidizing agents are involved?
Hydrometallurgy
987
988
CHAPTER 23
Metals and Metallurgy
Hydrometallurgy of Aluminum
Among metals, aluminum is second only to iron in commercial use. Worldwide
production of aluminum is about 2.4 X 1010 kg per year. The most useful ore of
aluminum is bauxite, in which Al is present as hydrated oxides, Al20 3 · xH 20.
The value of x varies, depending on the particular mineral present. Because
bauxite deposits in the Unit~d States are limited, most of the ore used in the
production of aluminum must be imported.
The major impurities found in bauxite are Si02 and Fe 20 3. It is essential to
separate Al20 3 from these impurities before the metal is recovered by electrochemical reduction, as described in Section 23.4. The process used to purify
bauxite, called the Bayer process, is a hydrometallurgical procedure. The ore is
first crushed and ground, then digested in a concentrated aqueous NaOH solution, about 30% NaOH by mass, at a temperature in the range of 150 oc to
230 °C. Sufficient pressure, up to 30 atm, is maintained to prevent boiling. The
Al20 3 dissolves in this solution, forming the complex aluminate ion, Al(OH) 4- :
Al203 · H20(s) + 2 H 20(l) + 2 OH- (aq)
~
2 Al(OH) 4- (aq)
[23.14]
The iron(III) oxides do not dissolve in the strongly basic solution. This difference in the behavior of the aluminum and iron compounds arises because Al 3+
is amphoteric, whereas Fe 3+ is not. cx:x:> (Section 17.5) Thus, the aluminate solution can be separated from the iron-containing solids by filtration. The pH of
the solution is then lowered, causing the aluminum hydroxide to precipitate.
After the aluminum hydroxide precipitate has been filtered, it is calcined in
preparation for electroreduction to the metal. The solution recovered from the
filtration is reconcentrated so that it can be used again. This task is accomplished by heating to evaporate water from the solution, a procedure that requires much energy and is the most costly part of the Bayer process.
GIVE IT SOME THOUGHT
What do you expect to be the product of the calcination of Al(OHh?
23.4 ELECTROMETALLURGY
Many processes that are used to reduce metal ores or to·refine metals are based
on electrolysis. cx:x:> (Section 20.9) Collectively these processes are referred to as
electrometallurgy. Electrometallurgical procedures can be broadly differentiated according to whether they involve electrolysis of a molten salt or of an aqueous solution.
Electrolytic methods are important for obtaining the more active metals, such
as sodium, magnesium, and aluminum. These metals cannot be obtained from
aqueous solution because water is more easily reduced than the metal ions. The
standard reduction potentials of water under both acidic and basic conditions are
more positive than those of Na + (E~ed = -2.71 V), Mg 2+ (E~ed = -2.37 V), and
Al3+ (E~ed = - 1.66 V):
2 H+(aq) + 2 e- ~ H 2 (g)
2 H 20( I)
+ 2 e-
~
H 2(g)
E ~ed
+ 2 OH- (aq)
= 0.00 V
E ~ed =
- 0.83 V
[23.15]
[23.16]
To form such metals by electrochemical reduction, therefore, we must employ a
nonaqueous, molten-salt medium in which the metal ion of interest is the most
readily reduced species.
Electrometallurgy of Sodium
In the commercial preparation of sodium, molten NaCl is electrolyzed in a specially designed cell called the Downs cell, illustrated in Figure 23.7 .... Calcium
chloride (CaC12) is added to lower the melting point of the molten NaCl from the
23.4
<IIIII
Electrometallurgy
989
Figure 23.7 Production of sodium.
The Downs cell is used in the commercia l
production of sodium .
MoltenNaCl
Iron screen to
prevent Na and
Cl2 from coming
together
normal melting point of 804 octo around 600 °C. The Na(l) and Cl2(g) produced
in the electrolysis are kept from coming in contact and re-forming NaCl. In addition, the Na must be prevented from contact with oxygen because the metal
would quickly oxidize under the high-temperature conditions of the cell reaction.
GIVE IT SOME THOUGHT
Which ions carry the current in a Downs cell?
Electrometallurgy of Aluminum
In Section 23.3 we discussed the Bayer process, in
which bauxite is concentrated to produce aluminum
Graphite
hydroxide. When this concentrate is calcined at ternanodes
peratures in excess of 1000 °C, anhydrous aluminum
oxide (Al20 3) is formed. Anhydrous aluminum oxide
melts at over 2000 °C. This is too high to permit its use
as a molten medium for electrolytic formation of free
_;---Al20 3 dissolved
in molten
aluminum. The electrolytic process commercially used
cryolite
to produce aluminum is known as the Hall-Heroult
M lt n
process, named after its inventors, Charles M. Hall and alu;i:um ,
Paul Herault (see the "A Closer Look" box in this sec~
/ ~arbon-lined
1ron
tion). The purified Al20 3 is dissolved in molten cryolite
~
~----~--~----------(Na3AlF6), which has a melting point of 1012 oc and is
.A Figure 23.8 The Haii-Heroult
an effective conductor of electric current. A schematic diagram of the electrolysis
process. An electrolysis cell used to
cell is shown in Figure 23.8 ...,. . Graphite rods are employed as anodes and are
form aluminum metal through
consumed in the electrolysis process. The electrode reactions are as follows:
reduction. Because molten aluminum is
Anode:
Cathode:
C(s) + 2 0 2- (l) ~ C0 2 (g) + 4 e3 e- + Al3+(l) ~ Al(l)
[23.17]
[23.18]
The amounts of raw materials and energy required to produce 1000 kg of aluminum metal from bauxite by this procedure are summarized in Figure 23.9 T .
Due to the large amount of electrical energy needed in the Hall-Herault process,
aluminum smelters are invariably located in areas with access to low cost sources
of electricity. In fact, the aluminum industry consumes about 10% of the world's
hydroelectric power. However, recycled aluminum only requires 5% of the
energy needed to produce "new" aluminum. Therefore, considerable energy
denser than the mixture of cryolite
(Na 3AIF 6) and Al 2 0 3, the metal collects
at the bottom of the cell.
990
CHAPTER 23
Metals and Metallurgy
savings can be realized by increasing the amount of aluminum that is
recycled. In the United States approximately 42% of post-consumer
aluminum is recycled, while in Europe the rate is estimated to be 52%.
Much higher rates of recycling are possible. In Norway, for example,
the aluminum recycling rate is 93%.
The electrodes represent another challenge surrounding aluminum
manufacture. The carbon anode is consumed in the process, generating
a variety of pollutants, including C02. Approximately 3.8 X 1010 kg of
C02 are emitted each year from aluminum manufacture. It is a significant challenge to find electrodes that
are conducting and chemically inert at elevated temperatures in molten cryolite.
GIVE IT SOME THOUGHT
Does molten cryolite conduct electricity through the movement of electrons
(electronic conductivity) or the movement of ions (ionic conductivity)?
.A. Figure 23.9 What it takes to
produce 1000 kg of aluminum.
Electrorefining of Copper
Copper is widely used to make electrical wiring and in other applications that
utilize its high electrical conductivity. Crude copper, which is usually obtained
by pyrometallurgical methods, is not suitable to serve in electrical applications
because impurities greatly reduce the metal's conductivity.
Purification of copper is achieved by electrolysis, as illustrated in Figure
23.11 .,.. . Large slabs of crude copper serve as the anodes in the cell, and thin
sheets of pure copper serve as the cathodes. The electrolyte consists of an acidic
solution of CuS04 . Application of a suitable voltage to the electrodes causes
oxidation of copper metal at the anode and reduction of Cu2+ to form copper
metal at the cathode. This strategy can be used because copper is both oxidized
and reduced more readily than water. The relative ease of reduction of Cu2+ and
H 20 is seen by comparing their standard reduction potentials:
Cu2+(aq) + 2 e- ~ Cu(s)
E~ed = +0.34 V [23.19]
2 H20(l) + 2 e-
~
H2(g) + 2 OH- (aq)
E~ed
= -0.83 V
[23.20]
CHARLES M. HALL
harles M. Hall (Figure 23.10 .,.. ) began to work on the problem of reducing aluminum in about 1885 after he had
learned from a professor of the difficulty of reducing ores of
very active metals. Before the development of Hall's electrolytic process, aluminum was obtained by a chemical reduction
using sodium or potassium as the reducing agent. Because the
procedure was very costly, aluminum metal was very expensive. As late as 1852, the cost of aluminum was $545 per pound,
far greater than the cost of gold. During the Paris Exposition in
1855, aluminum was exhibited as a rare metal, even though it is
the third most abundant element in Earth's crust.
Hall, who was 21 years old when he began his research,
utilized handmade and borrowed equipment in his studies
and used a woodshed near his Ohio home as his laboratory. In
about a year's time he was able to solve the problem of reducing aluminum. His procedure consisted of finding an ionic
compound that could be melted to form a conducting medium
that would dissolve Al20 3 but would not interfere in the electrolysis reactions. The relatively rare mineral cryolite,
(Na3AlF6), found in Greenland, met these criteria. Ironically,
<IIIII Figure 23.10 Charles
M. Hall (1863-1914) as
a young man.
Paul Herault, who was the same age as Hall, independently
made the same discovery in France at about the same time.
Because of the research of these two unknown young scientists,
large-scale production of aluminum became commercially feasible, and aluminum became a common and familiar metal.
23.5
Metallic Bonding
The impurities in the copper anode include lead, zinc,
nickel, arsenic, selenium, tellurium, and several precious
metals, including gold and silver. Metallic impurities that
are more active than copper are readily oxidized at the
anode but do not plate out at the cathode because their reduction potentials are more negative than that for Cu2+.
Less active metals, however, are not oxidized at the anode.
Instead they collect below the anode as a sludge that is
collected and processed to recover the valuable metals.
The anode sludges from copper-refining cells provide onefourth of U.S. silver production and about one-eighth of
U.S. gold production.
Anode sludge
GIVE IT SOME THOUGHT
Which ions carry the current in the electrorefining of copper?
.A Figure 23.11 Electrolysis cell for refining copper. As the
anodes dissolve away, the cathodes on which the pure metal is
deposited grow larger.
SAMPLE EXERCISE 23.2 I Impurities in Electrorefining
Nickel is one of the chief impurities in the crude copper that is subjected to electrorefining. What happens to this nickel in the course of the electrolytic process?
-
SOLUTION
Analyze: We are asked to predict whether nickel can be oxidized at the anode and
reduced at the cathode during the electrorefining of copper.
Plan: We need to compare the standard reduction potentials of Ni2+ and Cu2+. The
more negative the reduction potential, the less readily the ion is reduced but the more
readily the metal itself is oxidized. c::x::c (Section 20.4)
Solve: The standard reduction potential for Ni2+ is more negative than that for Cu2+:
~ Ni(s)
E~ed
= -0.28 V
Cu +(aq) + 2 e- ~ Cu(s)
E~ed
= +0.34 V
Ni 2+(aq)
+ 2 e-
2
As a result, nickel is more readily oxidized than copper, assuming standard conditions.
Although we do not have standard conditions in the electrolytic cell, we nevertheless
expect that nickel is preferentially oxidized at the anode. Because the reduction of Ni2+
occurs less readily than the reduction of Cu2 +, the Ni2+ accumulates in the electrolyte
solution, while the Cu2+ is reduced at the cathode. After a time it is necessary to recycle
the electrolyte solution to remove the accumulated metal ion impurities, such as Ni 2+.
-
PRACTICE EXERCISE
Zinc is another common impurity in copper. Using standard reduction potentials, determine whether zinc will accumulate in the anode sludge or in the electrolytic solution during the electrorefining of copper.
Answer: It is found in the electrolytic solution because the standard reduction potential of Zn2+ is more negative than that of Cu2+.
23.5 METALLIC BONDING
In our discussion of metallurgy we have confined ourselves to discussing the
methods employed for obtaining metals in pure form. Metallurgy is also concerned with understanding the properties of metals and with developing useful
new materials. As with any branch of science and engineering, our ability to
make advances is coupled with our understanding of the fundamental properties
of the systems with which we work. At several places in the text we have referred
to the differences in physical and chemical behavior between metals and nonmetals. Let's now consider the distinctive properties of metals and relate these
properties to a model for metallic bonding.
991
992
CHAPTER 23
Metals and Metallurgy
Physical Properties of Metals
• Figure 23.12 Malleability and
ductility. Gold leaf (left) demonstrates
the characteristic malleability of metals,
and copper wire (right) demonstrates
their ductility.
Metal ion (+)
• Figure 23.13 Electron-sea model.
Schematic illustration of the electron-sea
model of the electronic structure of
metals. Each gray sphere is a positively
charged metal ion.
You have probably held a length of copper wire or an iron bolt. Perhaps you have
even seen the surface of a freshly cut piece of sodium metal. These substances, although distinct from one another, share certain similarities that enable us to classify them as metallic. A clean metal surface has a characteristic luster. In addition,
metals that we can handle with bare hands have a characteristic cold feeling related to their high heat conductivity. Metals also have high electrical conductivities;
electrical current flows easily through them. Current flow occurs without any
displacement of atoms within the metal structure and is due to the flow of electrons within the metal. The heat conductivity of a metal usually parallels its electrical conductivity. Silver and copper, for example, which possess the highest
electrical conductivities among the elements, also possess the highest heat conductivities. This observation suggests that the two types of conductivity have the
same origin in metals, which we will soon discuss.
Most metals are malleable, which means that they can be hammered into thin
sheets, and ductile, which means that they can be drawn into wires (Figure 23.12 <01111 ).
These properties indicate that the atoms are capable of slipping with respect to
one another. Ionic solids or crystals of most covalent compounds do not exhibit
such behavior. These types of solids are typically brittle and fracture easily.
Consider, for example, the difference between dropping an ice cube and a block of
aluminum metal onto a concrete floor.
Most metals form solid structures in which the atoms are arranged as closepacked spheres. In these structures each atom is in contact with 12 neighboring
atoms. cx:o (Section 11.7) This arrangement of atoms is very different from the
structures of the nonmetallic elements. Consider for example the structures of
elements of the third period (Na-Ar). Argon with eight valence electrons has a
complete octet; as a result it does not form any bonds. Chlorine, sulfur, and phosphorous form molecules (Cl2, S8, and P4) where the atoms make one, two,
and three bonds, respectively. ex- (Chapter 22) Silicon forms an extended network
solid where each atom is bonded to four equidistant neighbors (Figure 12.3). Each
of these elements forms 8-N bonds, where N is the number of valence electrons.
This behavior can easily be understood through application of the octet rule.
If the trend were to continue as we move left across the periodic table, we
would expect aluminum to form five bonds. But, like many other metals including magnesium, titanium and gold, aluminum adopts a dose-packed structure
with 12 near neighbors. Clearly there is a change in the preferred bonding mechanism as the number of valence electrons decreases. As a general rule, metals do
not have enough valence-shell electrons to satisfy their bonding requirements
through the formation of localized electron pair bonds. In response to this deficiency, the valence electrons are collectively shared. A dose-packed lattice of
atoms facilitates a delocalized sharing of electrons among all atoms in the lattice.
Electron-Sea Model for Metallic Bonding
One very simple model that accounts for some of the most important characteristics of metals is the electron-sea model. c:x:o (Section 11.8) In this model the metal
is pictured as an array of metal cations in a "sea" of valence electrons, as illustrated in Figure 23.13 <01111 . The electrons are confined to the metal by electrostatic attractions to the cations, and they are uniformly distributed throughout the structure.
The electrons are mobile, however, and no individual electron is confined to any
particular metal ion. When a metal wire is connected to the terminals of a battery,
electrons flow through the metal toward the positive terminal and into the metal
from the battery at the negative terminal. The high heat conductivity of metals is
also accounted for by the mobility of the electrons, which permits ready transfer
of kinetic energy throughout the solid. The ability of metals to deform (their
malleability and ductility) can be explained by the fact that metal atoms form
bonds to many neighbors. Changes in the positions of the atoms brought about in
reshaping the metal are partly accommodated by a redistribution of electrons.
23.5
3500
6000 I
3000
~
, I ~
I
-
I
"
.5
~
<J.)
2500
~ 4000 I
......
2000
.50
..§ 1500
;§
·§
I
;--... 5000 I
"2
Q)
993
7000 ~~~--~~~--~~~--~~~~
4000
·~
Metallic Bonding
0...
~ 3000 I
OJ)
~
500
1/ /
T
I
I
I '...
IF
i
I
I
'\.I "\ I
I
·s
Q)
1000
I :'Ia I "\ I
It
~ 2000 r--t-f / 1
First series (K-Zn)
- e - Second series (Rb-Cd)
- e - Third series (Cs-Hg)
1000 I
I
O 1A 2A 3B 4B 5B 6B 7B 8B 8B 8B 1B 2B
Group
IT I
I
\\\--1
First series (K-Zn)
Second series (Rb-Cd)
- e - Third series (Cs-Hg)
1--i---+-•
I
O 1A 2A 3B 4B 5B 6B 7B 8B 8B 8B 1B 2B
Group
A Figure 23.14 The (a) melting
and (b) boiling points of metallic
elements.
The electron-sea model, however, does not adequately explain all properties.
According to the model, for example, the strength of bonding between metal
atoms should increase as the number of valence electrons increases, resulting in a
corresponding increase in the melting and boiling points. However, the elements
found near the middle of the transition metal series, rather than those at the end,
have the highest melting and boiling points in their respective periods (Figure
23.14 £ ), which implies that the strength of metallic bonding first increases with increasing number of electrons and then decreases. Similar trends are seen in other
physical properties of the metals, such as the heat of fusion and hardness.
Molecular-Orbital Model for Metals
To obtain a more accurate picture of the
bonding in metals we turn to molecularorbital theory. In Sections 9.7 and 9.8 we
learned how molecular orbitals are created
from the overlap of atomic orbitals. In
Section 12.2 we saw that in extended solids
(such as covalent network solids and metals)
the presence of a large number of atoms results in a large number of closely spaced
molecular orbitals. These closely spaced orbitals form continuous bands of allowed energy states. cx:o (Section 12.2)
The band structure for a typical metal
is shown schematically in Figure 23.15 ..... The
electron filling depicted corresponds to nickel metal, but the basic features of other metals
are similar. The electron configuration of a
nickel atom is [Ar]3d84s2, as shown on the left
side of the figure. The energy bands that
form from these orbitals are shown on the
right side of the figure. The 4s, 4p, and 3d orbitals are treated independently, each giving
rise to a band of molecular orbitals. In practice, these overlapping bands are not independent of each other, but for our purposes
this simplification is reasonable.
4pband
(holds 6e- /atom)
'OJ)
.5
"0
~
0
4p orbitals-- ~
El_1 _1 _H _H Jl
3d orbitals
H
-_(
Il l
I
··•
l
I
I~ro
I
<J.)
!-<
-
0
I
I~
......
OJ)
.5
"0
~
3d band
(holds 10e- /atom)
I~
0
11
A single Ni atom
(holds 2e - ;atom)
~
~
Many Ni atoms
.A Figure 23.15 The electronic band structure of nickel. The left side of
the figure shows the electron configuration of a single Ni atom, while the right-hand
side of the figure shows how these orbital energy levels broaden into energy bands
in bulk nickel. The horizontal dashed gray line denotes the position of the Fermi
Level, which separates the occupied molecular orbitals (shaded in blue) from the
unoccupied molecular orbitals.
994
CHAPTER 23
Metals and Metallurgy
When two atomic orbitals overlap they form two molecular orbitals: (1) a
bonding molecular orbital whose energy is lowered with respect to the parent
atomic orbitals and (2) an antibonding molecular orbital whose energy is raised
with respect to the parent atomic orbitals. c:x::o (Chapters 9 and 12) As the overlap
increases, the energy splitting between bonding and antibonding molecular orbitals increases. The 4s and 4p orbitals are considerably larger than the 3d orbitals,
which means that they overlap more effectively. This characteristic leads to a
large difference in energy between the molecular orbitals at the bottom of the
band, where the interactions are strongly bonding, and the molecular orbitals at
the top of the band, where the interactions are strongly antibonding. The result is
4s and 4p bands that span a relatively wide range of energy, but due to the Pauli
Exclusion Principle c-- (Section 6.7) these bands can hold only two and six electrons per atom, respectively.
The band that arises from overlap of the 3d orbitals differs in two important
ways. Because the 3d orbitals are smaller, yet the interatomic distance is the same,
they overlap less effectively, which reduces the strength of both the stabilizing
bonding interactions and the destabilizing antibonding interactions. The result is
a band that spans a much narrower range of energy. But because there are five 3d
orbitals on each atom, this band can hold up to ten electrons per atom. The filling
of bands with electrons is akin to filling a reservoir with water. In a solid the
energy of the highest filled molecular orbital is called the Fermi level. Only a tiny
energy separates it from the energy of the lowest empty orbital. The electrons
occupy the lowest energy molecular orbitals available to them, irrespective of the
orbitals populated in an individual atom. The occupied molecular orbitals are
shaded in blue in Figure 23.15.
Many properties of metals can be understood from the simple band structure
diagram shown in Figure 23.15. Because the electrons available for bonding do
not completely fill the available molecular orbitals, we can think of the energy
band as a partially filled container for electrons. The incomplete filling of the
energy band gives rise to characteristic metallic properties. The electrons in orbitals near the top of the occupied levels require very little energy input to be
"promoted" to still higher energy orbitals that are unoccupied. Under the influence of any source of excitation, such as an applied electrical potential or an input
of thermal energy, electrons move into previously vacant levels and are thus freed
to move through the lattice, giving rise to electrical and thermal conductivity.
Without the overlap of energy bands, the periodic properties of metals could
not be explained. In the absence of the d- and p-bands we would expect the s-band
to be half filled for the alkali metals (group 1A) and completely filled for the alkaline-earth metals (group 2A). If that were true, metals like magnesium, calcium,
and strontium would be expected to behave as semiconductors! While the conductivity of metals can be qualitatively understood using either the electron-sea model
or the molecular-orbital model, many of the physical properties of transition metals, such as the melting and boiling points plotted in Figure 23.14, can be explained
only by using the latter model. The molecular-orbital model predicts that bonding
will first become stronger as the number of valence electrons increases and
the bonding orbitals are populated. Upon moving past the middle elements of the
transition metal series, the bonds grow weaker as we fill the antibonding orbitals.
Strong bonds between atoms lead to metals with higher melting and boiling
points, higher heats of fusion, higher hardness, and so forth. Metals with a small
number of electrons per atom, such as Rb or Cs, have relatively few metal-metal
bonding orbitals occupied. In metals with a large number of electrons per atom,
such as Zn or Cd, the metal-metal bonding orbitals are filled, but at the same
time a high percentage of the metal-metal antibonding orbitals are also occupied.
In either case, the bonding will be much weaker compared to elements in the
middle of the transition series, dramatically impacting a number of physical
properties. However, we need to remember that factors other than the number of
electrons (such as atomic radius, nuclear charge, and the particular packing structure of the metal) also play a role in determining the properties of metals.
23.6
Alloys
995
GIVE IT SOME THOUGHT
Which element, W or Hg, has the greater number of electrons in antibonding orbitals? Which one would you expect to have a higher melting point?
23.6 ALLOYS
An alloy is a material that contains more than one element and has the character-
istic properties of metals. The alloying of metals is of great importance because it
is one of the primary ways of modifying the properties of pure metallic elements.
Nearly all the common uses of iron, for example, involve alloy compositions.
Bronze is formed by alloying copper and tin, while brass is an alloy of copper and
zinc. Copper is the majority element in both alloys. Pure gold is too soft to be
used in jewelry, but alloys of gold and copper are quite hard. Pure gold is termed
24 karat; the common alloy used in jewelry is 14 karat, meaning that it is 58%
gold (!~ X 100% ). A gold alloy of this composition has suitable hardness to be
used in jewelry. The alloy can be either yellow or white, depending on the elements added. Some further examples of alloys are given in Table 23.2 'Y .
Alloys can be classified as solution alloys, heterogeneous alloys, and intermetallic compounds. Solution alloys are homogeneous mixtures in which components are dispersed randomly and uniformly. Atoms of the solute can take positions
normally occupied by a solvent atom, thereby forming a substitutional alloy. Or they
can occupy interstitial positions in the "holes" between the solvent atoms, thereby
forming an interstitial alloy. These types are diagrammed in Figure 23.16'Y.
Substitutional alloys are formed when the two metallic components have similar atomic radii and chemical-bonding characteristics. For example, silver and
gold form such an alloy over the entire range of possible compositions. When two
metals differ in radii by more than about 15%, solubility is more limited.
For an interstitial alloy to form, the component present in the interstitial positions between the solvent atoms must have a much smaller bonding atomic radius
than the solvent atoms. Typically, an interstitial element is a nonmetal that bonds
to neighboring atoms. The presence of the extra bonds provided by the interstitial
component causes the metal lattice to become harder, stronger, and less ductile.
For example, steel, which is much harder and stronger than pure iron, is an alloy of
iron that contains up to 3% carbon. Mild steels contain less than 0.2% carbon; they are
malleable and ductile and are used to make cables, nails, and chains. Medium steels
contain 0.2-0.6% carbon; they are tougher than mild steels and are used to make
girders and rails. High-carbon steel, used in cutlery, tools, and springs, contains
0.6-1.5% carbon. In all three cases other elements may be added to form alloy steels.
Vanadium and chromium may be added to impart strength and to increase resistance to fatigue and corrosion. For example, a rail steel used in Sweden on lines bearing heavy ore carriers contains 0.7% carbon, 1% chromium, and 0.1 % vanadium.
GIVE IT SOME THOUGHT
Would you expect the alloy PdB 0.15 to be a substitutional alloy or an interstitial alloy?
TABLE 23.2 • Some Common Alloys
Primary Element N arne of Alloy
Composition by Mass
Properties
Bismuth
Wood's metal
50% Bi, 25% Pb, 12.5% Sn, 12.5% Cd Low melting point (70 °C)
Copper
Iron
Lead
Silver
Yellow brass
Stainless steel
Plumber's solder
Sterling silver
Dental amalgam
67% Cu, 33% Zn
80.6% Fe, 0.4% C, 18% Cr, 1% Ni
67% Pb,33% Sn
92.5% Ag, 7.5% Cu
70% Ag, 18% Sn, 10% Cu, 2% Hg
Uses
Fuse plugs, automatic
sprinklers
Ductile, takes polish
Hardware items
Resists corrosion
Tableware
Low melting point (275 oc) Soldering joints
Tableware
Bright surface
Easily worked
Dental fillings
996
CHAPTER 23
Metals and Metallurgy
THE ARRANGEMENT OF ATOMS IN TWO TYPES OF SOlUTION AllOYS
Solution alloys are homogenous mixtures in which the components are dispersed
randomly and uniformly.
In a Substitutional Alloy,
atoms of the solute take
positions normally occupied
by a solvent atom.
In an Interstitial Alloy, solute
atoms occupy interstitial
positions in the "holes"
between the solvent atoms.
A Figure 23.16 Substitutional and interstitial alloys. The blue spheres represent host metal; the yellow spheres represent
the other components of the alloy.
One of the most important iron alloys is stainless steel, which contains about
0.4% carbon, 18% chromium, and 1% nickel. The chromium is obtained by carbon reduction of chromite (FeCr20 4 ) in an electric furnace. The product of thereduction is ferrochrome (FeCr2 ), which is then added in the appropriate amount to
molten iron that comes from the converter, to achieve the desired steel composition. The ratio of elements present in the steel may vary over a wide range, imparting a variety of specific physical and chemical properties to the materials.
In a heterogeneous alloy the components are not dispersed uniformly. In
the form of steel known as pearlite, for example, two distinct phases-essentially pure iron and the compound Fe3C, known as cementite-are present in
alternating layers. In general, the properties of heterogeneous alloys depend on
both the composition and the manner in which the solid is formed from the
molten mixture. Rapid cooling leads to distinctly different properties than are
obtained by slow cooling.
lntermetallic Compounds
Intermetallic compounds are homogeneous alloys that have definite properties and compositions. Unlike substitutional and interstitial alloys, the different
types of atoms in an intermetallic compound are ordered rather than randomly
distributed. Examples of the atomic ordering seen in some important cubic
intermetallic phases are shown in Figure 23.17 ~ . The ordering of atoms in an
intermetallic compound generally leads to better structural stability and higher
23.6
Alloys
997"
Figure 23.17 Ordered intermetallic structures. (a) The
face-centered cubic unit cell of a cubic close-packed metal.
(b) The ordered structure of Ni 3AI with nickel atoms shown in gray
and aluminum atoms in blue. (c) The ordered structure of TiAI
with titanium atoms shown in red and aluminum atoms in blue.
(d) The ordered structure of ZnCu with copper atoms shown in
copper and zinc atoms in gray.
<1111111
Ni
Ni3Al
(a)
(b)
TiAl
CuZn
(c)
(d)
melting points than the constituent metals. These features can be attractive for
high temperature applications. On the other hand, intermetallic compounds are
often more brittle than substitutional alloys.
Intermetallic compounds play many important roles in modern society.
The intermetallic compound Ni3Al is a major component of jet aircraft engines
because of its strength at high temperature and its low density. Razor blades are
often coated with Cr3Pt, which adds hardness, allowing the blade to stay sharp
longer. Both compounds have the ordered structure shown in Figure 23.17(b).
The compound Co 5Sm is used in the permanent magnets in lightweight headsets and high-fidelity speakers because of its high magnetic strength per unit
weight. A related compound with the same structure, LaNi5, is used as the
anode in nickel-metal hydride batteries.
WDII!ratlktfl
SHAPE-MEMORY ALLOYS
n 1961 a naval engineer, William J. Buechler, made an unexpected and fortunate discovery. In searching for the best
metal to use in missile nose cones, he tested many metal alloys. One of them, an intermetallic compound of nickel and
titanium, NiTi, behaved very oddly. When he struck the cold
metal, the sound was a dull thud. When he struck the metal
at a higher temperature, however, it resonated like a bell.
Mr. Buechler knew that the way in which sound propagates
in a metal is related to its metallic structure. Clearly, the structure of the NiTi alloy had changed as it went from cold to
warm. As it turned out, he had discovered an alloy that has
shape memory.
Metals and metal alloys consist of many tiny crystalline
areas (crystallites). When a metal is formed into a certain shape
at high temperature, the crystallites are forced into a particular arrangement with respect to one another. Upon cooling
a normal metal, the crystallites are "locked" in place by the
I
bonds between them. When the metal is subsequently bent,
the resulting stresses are sometimes elastic, as in a spring.
Often, however, the metal merely deforms (for example,
when we bend a nail or crumple a sheet of aluminum foil). In
these cases bending weakens the bonds that tie the crystallites
together, and upon repeated flexing, the metal breaks apart.
In a shape-memory alloy, the atoms can exist in two
different bonding arrangements, representing two different
solid-state phases. The higher-temperature phase has strong,
fixed bonds between the atoms in the crystallites. By contrast,
the lower-temperature phase is quite flexible with respect to
the arrangements between the atoms. Thus, when the metal is
distorted at low temperature, the stresses of the distortion are
absorbed within the crystallites by changes in the atomic lattice. In the higher-temperature phase, however, the atomic
lattice is stiff, and the bonds between crystallites, as in a normal
metal, absorb stresses that are due to bending.
continues on next page
998
CHAPTER 23
Metals and Metallurgy
continued
To see how a shape-memory metal behaves, suppose we
bend a bar of NiTi alloy into a semicircle [Figure 23.18(a) ..,_]
and then heat it to about 500 oc. We then cool the metal to below
the transition temperature for the phase change to the lowtemperature, flexible form. Although the cold metal remains in
the semicircular form, as in Figure 23.18(b), it is now quite flexible and can readily be straightened or bent into another shape.
When the metal is subsequently warmed and passes through
the phase change to the "stiff" phase, it "remembers" its original
shape and immediately returns to it as shown in Figure 23.18(c).
There are many uses for such shape-memory alloys. The
curved shape in a dental brace, for example, can be formed at
high temperature into the curve that the teeth are desired to
follow. Then at low temperature, where the metal is flexible, it
can be shaped to fit the mouth of the wearer of the braces.
When the brace is inserted in the mouth and warms to body
temperature, the metal passes into the stiff phase and exerts a
force against the teeth as it attempts to return to its original
shape. Other uses for shape-memory metals include heat-actuated shutoff valves in industrial process lines, which need
no outside power source. Shape-memory alloys are also being
evaluated for use in metallic stents, inserted into narrow veins
to hold them open. The chilled NiTi alloy stent is in a collapsed form. When it warms in the body, the stent expands to
the original size that it "remembers."
Bend tube and heat-treat
to set shape
?
(a)
Cool metal to below
phase-transition
temperature
?
Straighten metal
tube, which bends
easily
(b)
Warm metal to above phasetransition temperature; it
immediately returns to bent
shape
?
(c)
.A Figure 23.18 Behavior of a shape-memory alloy.
23.7 TRANSITION METALS
Many of the most important metals of modern society are transition metals.
Transition metals, which occupy the d block of the periodic table (Figure
23.19 T ), include such familiar elements as chromium, iron, nickel, and copper.
They also include less familiar elements that have come to play important roles
in modern technology, such as those in the high-performance jet engine pictured
in Figure 23.1. In this section we consider some of the physical and chemical
properties of transition metals.
Physical Properties
Several physical properties of the elements of the first transition series are listed in Table 23.3 ..,_. Some of these properties, such as ionization energy and
atomic radius, are characteristic of isolated atoms. Others, including density
and melting point, are characteristic of the bulk solid metal.
I
1--I
I
I
I
----·
4B
5B
6B
7B
21
Sc
22
Ti
23
v
24
25
26
27
Cr
Mn
Fe
Co
28
Ni
1B
2B
29
30
Cu
Zn
40
42
43
Tc
44
45
Rh
46
47
48
Ru
Pd
Ag
Cd
77
78
79
Ir
Pt
Au
80
Hg
I
~~--·
y
Zr
41
Nb
Mo
71
72
Hf
73
Ta
w
39
Lu
~
8B
3B
74
75
Re
76
Os
--··-··-··-·····----
I
.A Figure 23.19 Transition elements in the periodic table. Transition metals are those elements that occupy the d
block of the periodic table.
23.7
Transition Metals
999
TABLE 23.3 • Properties of the First Transition-Series Elements
SB
Group
3B
4B
SB
6B
7B
Element:
Sc
Ti
v
Cr
Mn
Fe
Co
3d24s 2
658
1.47
4.5
1660
3d34s 2
650
1.35
6.1
1917
3d54s 1
653
1.29
7.9
1857
3d54s 2
717
1.37
7.2
1244
3d64s 2
759
1.26
7.9
1537
3d74s 2
758
1.25
8.7
1494
3d14s 2
First ionization energy (kJ/mol)
631
Radius in metallic substances (A) 1.64
3.0
Density (g/ cm3)
Melting point (°C)
1541
Electron configuration
The properties of metals vary in similar ways across each
series (row). We have already seen evidence for this in the
melting and boiling points of metals (Figure 23.14). Another
example is shown in Figure 23.20 ..,. where the atomic radius
observed in close packed metallic structures is plotted versus
group number*. The variation seen in Figure 23.20 is a result of
two competing forces. On the one hand, increasing effective
nuclear charge favors a steady decrease in the radius as we
proceed from left to right across each transition series. On the
other hand, the metallic bonding strength increases until we
reach the middle of each transition series and then decreases as
we fill antibonding orbitals (Section 23.5). As a general principle a bond shortens as it becomes stronger. c::co (Section 8.8)
For groups 3B through 6B these two effects work cooperatively and the result is a rapid decrease in radius. Upon moving
past group 6B the two effects counteract each other, slowing
the decrease and eventually leading to an increase in radius.
lB
2B
Ni
Cu
Zn
3d84s 2
737
1.25
8.9
1455
3d104s 1
745
1.28
8.9
1084
3d104s 2
906
1.37
7.1
420
1. 9 .----.,....-...,.----,--:--.,.---r-...,.-----,~--,.-----.
First series (Sc-Zn)
1.8 l--\-1--~1.7 •--- ~ \
~ Second series (Y-Cd)
~Third series (Lu-Hg)
----+----+·------+-···············+-
----+---+------+----+
"$
rJ)
~ 1.61--\-+- 4 -+ro
1-<
u
'§ 1.5
~
1.4 1------l- --~~••d
1.3 1----l-- +
1.2
3
4
GIVE IT SOME THOUGHT
Which of the following elements will have the largest bonding
atomic radius: Cr, Mn, Re? Which will have the smallest?
The incomplete screening of the nuclear charge by added electrons produces an interesting and important effect in the third transition-metal series.
In general, the radius increases as we move down in a family because of the
increasing principal quantum number of the outer-shell electrons. c:x:o (Section
7.3) Once we move beyond the group 3B elements, however, the second and
third transition-series elements have virtually the same radii. In group SB, for
example, tantalum has virtually the same radius as niobium. This effect has its
origin in the lanthanide series, the elements with atomic numbers 57 through
70, which occur between Ba and Lu (inside front cover). The filling of 4f orbitals
through the lanthanide elements causes a steady increase in the effective nuclear charge, producing a contraction in size called the lanthanide contraction.
This contraction just offsets the increase we would expect as we go from the
second to the third series. Thus, the second- and third-series transition metals
in each group have about the same radii all the way across the series. Consequently, the second- and third-series metals in a given group have great
similarity in their chemical properties. For example, the chemical properties of
zirconium and hafnium are remarkably similar. They always occur together in
nature, and they are very difficult to separate.
*Note that the radii defined in this way, often referred to as metallic radii, differ somewhat from the covalent
radii defined in Section 7.3.
I
I
~
.A"
I
~I
I ' - ' -!
5
6
7
~
1
B Group
.A Figure 23.20 Radii of transition
metals. The variation in the radius of
transition metal atoms in close packed
metallic substances as a function of
position in the periodic table.
2
1000
CHAPTER 23
Metals and Metallurgy
Electron Configurations and Oxidation States
Transition metals owe their location in the periodic table to the filling of the d
subshells. Many of the chemical and physical properties of transition metals,
result from the unique characteristics of the d orbitals. For a given transition
metal atom, the valence (n-l)d orbitals are smaller than the corresponding valence shell ns and np orbitals. In quantum mechanical terms the (n-l)d orbital
wave functions drop off more rapidly upon moving away from the nucleus
than do the ns and np orbital 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 they behave
like valence electrons in many instances but more like core electrons in other instances. The details depend upon the location of the element in the periodic
table and the environment of the atom.
When transition metals are oxidized, they lose their outer s electrons before
they lose electrons from the d subshell. cxo (Section 7.4) The electron configuration of Fe is [Ar]3d64s2, for example, whereas that of Fe 2+ is [Ar]3d 6 . Formation
of Fe 3+ requires loss of one 3d electron, giving [Ar]3d5 . Most transition-metal
ions contain partially occupied d subshells, which are responsible in large part
for several characteristics of transition metals:
1. They often exhibit more than one
stable oxidation state.
2. Many of their compounds are colored, as shown in Figure 23.21 ~ .
(We will discuss the origin of these
colors in Chapter 24.)
3. Transition metals and their compounds exhibit interesting and important magnetic properties.
Figure 23.22 ~ summarizes the common nonzero oxidation states for the
first transition series. The oxidation states shown as large blue circles are those
most frequently encountered either in
solution or in solid compounds. The
ones shown as small green circles are
less common. Notice that Sc occurs only
in the +3 oxidation state and Zn occurs
• Figure 23.21 Salts of transitiononly in the +2 oxidation state. The other metals, however, exhibit a variety of oxmetal ions and their solutions. From
idation states. For example, Mn is commonly found in solution in the + 2 (Mn2+)
left to right: Mn 2 +, Fe 2+, Co 2+, Ni 2+,
and +7 (Mn0 4 - )oxidation states. In the solid state the +4 oxidation state (as in
Cu 2 +, and Zn 2+.
Mn02) is common. The +3, +5, and +6 oxidation states are less common.
The +2 oxidation state, which commonly occurs for most of these
metals, is due to the loss of their two outer 4s electrons. This oxidalB 2B
tion state is found for all these elements except Sc, where the 3+ ion
with an [Ar] configuration is particularly stable.
+ 8 1--t
--+ -+--+Oxidation states above +2 are due to successive losses of 3d elec61
trons. From Sc through Mn the maximum oxidation state increases
+
tfrom +3 to +7, equaling in each case the total number of 4s plus 3d
electrons
in the atom. Thus, manganese has a maximum oxidation
+4 1-+ - · --· - · - ·
state of 2 + 5 = +7. As we move to the right beyond Mn in the first
+2
transition series, the maximum oxidation state decreases. This de0
I
crease
is due in part to the attraction of d orbital electrons to the nucleSc Ti V Cr Mn Fe Co Ni Cu Zn
us, which increases faster than the attraction of the s orbital electrons to
• Figure 23.22 Nonzero oxidation
the nucleus as we move left-to-right across the periodic table. Thus, in each peristates of the first transition series.
od the d electrons become more corelike as the atomic number increases. By the
The larger circles indicate the most
time we get to zinc, it is not possible to remove electrons from the 3d orbitals
common oxidation states.
·-- t l
.·-' -.--..··----.
I
II I
I I I I
II
23.7
through chemical oxidation. In the second and third transition series the increased size of the 4d and Sd orbitals
makes it possible to attain maximum oxidation states as
high as +8, which is achieved in Ru0 4 and Os04 . In general, the maximum oxidation states are found only when the
metals are combined with the most electronegative elements, especially 0, F, and in some cases Cl.
Magnetism
®0@0
080®
(j)080
0000
Transition Metals
1001
00000000
00000000
00000000
00000000
The magnetic properties of transition metals and their
(b)
(a)
compounds are both interesting and important. Measurements of magnetic properties provide information about
chemical bonding. In addition, many important uses are
made of magnetic properties in modern technology.
An electron possesses a "spin" that gives it a magnetic
moment, causing it to behave like a tiny magnet.
cx:o (Section 9.8) In a diamagnetic solid, one in which all of
the electrons in the solid are paired, the up-spin and downspin electrons effectively cancel each other. Diamagnetic
substances are generally described as nonmagnetic, but
when a diamagnetic substance is placed in a magnetic field,
(d)
(c)
the motions of the electrons cause the substance to be very
£ Figure 23.23 Types of magnetic behavior. (a) Simple
weakly repelled by the magnet.
paramagnetic: centers with magnetic moments not aligned unless
When an atom or ion possesses one or more un- the substance is in a magnetic field. (b) Ferromagnetic: where the
paired electrons, the substance is said to be paramagnetic.
spins of coupled centers align in a common direction. (c)
cx:o (Section 9.8) In a paramagnetic solid the electrons Antiferromagnetic: where the spins of coupled centers align in
opposite directions. (d) Ferrimagnetic: where the spins of coupled
on adjacent atoms or ions do not influence the unpaired
centers align in opposite directions, but because the two magnetic
electrons on neighboring atoms or ions of the solid. As a recenters have different numbers of unpaired electrons, the magnetic
sult the magnetic moments on the individual atoms or ions fields do not fully cancel.
are randomly oriented, as shown in Figure 23.23(a) ~ . When
placed in a magnetic field, however, the magnetic moments become aligned
roughly parallel to one another, producing a net attractive interaction with the
magnet. Thus, a paramagnetic substance is drawn into a magnetic field.
When you think of a magnet you probably conjure up an image of a simple
iron magnet (Figure 23.24 ~ ). Iron exhibits ferromagnetism, a much stronger form
of magnetism than paramagnetism. Ferromagnetism arises when the unpaired
electrons of the atoms or ions in a solid are influenced by the orientations of the
electrons of their neighbors. The most stable (lowest-energy) arrangement results
when the spins of electrons on adjacent atoms or ions are aligned in the same direction, as shown in Figure 23.23(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 simple paramagnetic substance. When the external magnetic field is removed, the interactions between the electrons cause the
solid as a whole to maintain a magnetic moment. We then refer to it as a permanent
magnet. The only transition metal elements that display ferromagnetism are Fe, Co,
and Ni. Many alloys also exhibit ferromagnetism, which is in some cases stronger
than the ferromagnetism of the pure metals. Particularly powerful magnetism is
found in certain intermetallic 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 23.23. In materials that exhibit
antiferromagnetism the unpaired electrons on a given atom align so that their
spins are oriented in the opposite direction as the spins on neighboring atoms, as
shown in Figure 23.23(c). In an antiferromagnetic substance the up-spin and
£ Figure 23.24 Permanent magnets.
down-spin electrons cancel each other. Examples of antiferromagnetism are
Permanent magnets are made from
found among metals (such as Cr), alloys (such as FeMn), and transition metal
ferromagnetic and ferrimagnetic
materials.
oxides (such as Fe203, LaFe03 and MnO).
00 (!) 00 (!)
(!) 00 (!) 00
00 (!) 00 (!)
(!) 00 (!) 00
00 0 00 0
0 00 0 00
00 0 00 0
0 00 0 00
1002
CHAPTER 23
Metals and Metallurgy
A substance that exhibits ferrimagnetism has characteristics of both a ferromagnet and an antiferromagnet. Like an antiferromagnet, the unpaired electrons
align themselves so that they are pointing in the opposite direction of their
neighbors. However, unlike an antiferromagnet the net magnetic moments of the
up-spin electrons are not fully cancelled by the down-spin electrons. This can
happen either because the magnetic centers have different numbers of unpaired
electrons (i.e. NiMn03), the number of magnetic sites aligned in one direction is
larger than in the other direction (i.e. Y3Fe50 12), or both (i.e. Fe30 4). Because the
magnetic moments do not cancel, the bulk properties of ferrimagnetic materials
are much like ferromagnetic materials.
All magnetically ordered materials-ferromagnets, ferrimagnets, and
antiferromagnets-become paramagnetic when heated above a critical temperature. This happens when the thermal energy is sufficient to overcome the forces
that are responsible for orienting the spins with respect to their neighbors.
This temperature is called the Curie temperature, Tc, for ferromagnets and ferrimagnets, and the Neel temperature, TN' for antiferromagnets. For Fe, Co, and Ni
the Curie temperatures are 770 oc, 1115 oc, and 354 °C, respectively.
23.8 CHEMISTRY OF SELECTED
TRANSITION METALS
Let's now briefly consider some of the chemistry of three common elements
from the first transition series: chromium, iron, and copper. As you read this
material, look for the trends that illustrate the generalizations outlined earlier.
Chromium
Chromium dissolves slowly in dilute hydrochloric or sulfuric acid, liberating
hydrogen. In the absence of air, the reaction results in the formation of a sky-blue
solution of the chromium(II) or chromous ion:
Cr(s) + 2 H+(aq) ~ Cr 2+(aq) + 2 H 2(g)
.A. Figure 23.25 Two forms of the
Cr(lll) ion. The tube on the left contains
the violet hydrated chromium(lll) ion,
Cr(H 20) 6 3+. The tube on the right
contains the green
[(H20)4Cr(OH)2Cr(H20)4] 4+ ion.
[23.21]
In the presence of air, the chromium(II) ion is rapidly oxidized by 0 2 to form the
chromium(III) ion. The reaction produces the green [(H 20) 4Cr(OHhCr(H 20) 4 ] 4+
ion (Figure 23.25 ~). In a strongly acidic solution, this ion reacts slowly with H +
ions to form the violet [Cr(H 20) 6 ] 3+ ion (Figure 23.25), which is often represented
simply as Cr 3+(aq). The overall reaction in acid solution is often given simply as
shown in Equation 23.22.
4 Cr 2+(aq)
+ 0 2 (g) + 4 H+(aq)
~ 4 Cr 3+(aq)
+ 2 H 20(l)
[23.22]
GIVE IT SOME THOUGHT
Would you expect the Cr 3+ ion to contain any unpaired electrons?
Chromium is frequently encountered in aqueous solution in the +6 oxidation state. In basic solution the yellow chromate ion (Cr0 4 2- ) is the most stable.
In acidic solution the dichromate ion (Cr2ol- ) is formed:
Cr0 4 2- (aq) + H+(aq) ~ HCr0 4 -(aq)
2 HCr0 4 - (aq) ~ Cr2ol-(aq)
.A. Figure 23.26 Chromate and
dichromate solutions. Solutions of
sodium chromate, Na 2Cr0 4 (left), and
potassium dichromate, K2Cr20 7 (right),
illustrate the difference in color of the
chromate and dichromate ions.
[23.23]
+ H 20(1)
[23.24]
Equation 23.24 is a condensation reaction in which water is split out from two
HCr0 4 - ions. Similar reactions occur among the oxyanions of other elements,
such as phosphorus. r:rc (Section 22.8) The equilibrium between the dichromate and chromate ions is readily observable because Cr0 4 2- is bright yellow
and Cr20l- is deep orange, as seen in Figure 23.26 ~. The dichromate ion in
acidic solution is a strong oxidizing agent, as evidenced by its large, positive
reduction potential. By contrast, the chromate ion in basic solution is not a
particularly strong oxidizing agent.
23.8
Chemistry of Selected Transition Metals
Iron
We have already discussed the metallurgy of iron in considerable detail in
Section 23.2. Here we consider some of its important aqueous solution chemistry. Iron exists in aqueous solution in either the +2 (ferrous) or +3 (ferric)
oxidation states. It often appears in natural waters because these waters come
into contact with deposits of FeC0 3 (Ksp = 3.2 X 10-11 ). Dissolved C02 in the
water can then help dissolve the mineral:
FeC0 3(s) + C0 2 (aq) + H 20(Z) ~ Fe 2+(aq) + 2 HC0 3- (aq) [23.25]
The dissolved Fe 2+, together with Ca 2+ and Mg 2+, contributes to water hardness.
c::co (Section
18.6)
The standard reduction potentials in Appendix E reveal much about the
kind of chemical behavior we should expect to observe for iron. The potential
for reduction from the +2 state to the metal is negative; however, the reduction
from the +3 to the +2 state is positive. Iron, therefore, should react with acids
that are not highly oxidizing such as dilute sulfuric acid or acetic acid to form
Fe 2+(aq), as indeed it does. In the presence of air, however, Fe 2+(aq) tends to be
oxidized to Fe 3+(aq), as shown by the positive standard emf for Equation 23.26:
4 Fe 2+(aq) + 02(g) + 4 H+(aq) ~ 4 Fe 3+(aq) + 2 H20(Z)
-
SAMPLE EXERCISE
Eo = +0.46 V
[23.26]
23.3 1Writing Half-Reactions for an OxidationReduction Reaction
Write the two balanced half-reactions for the reaction in Equation 23.26.
SOLUTION
Analyze: We are asked to write two oxidation-reduction half-reactions that together
make up a given oxidation-reduction reaction.
Plan: We need to separate the two half-reactions that make up the given balanced reaction. We can best start by noting that Fe appears in the +2 and +3 states and that
oxygen appears in the 0 and -2 oxidation states.
Solve: We note that iron is oxidized from the +2 to +3 oxidation state. The halfreaction for this process is
Fe 2+(aq) ~ Fe 3+(aq) + eOxygen in acidic medium is reduced:
02(g)
+ 4 H +(aq) + 4 e-
~ 2 H 20( I)
To achieve a balanced equation, we need four Fe 2+ for each 0 2 . Therefore, all we need
to do to obtain Equation 23.26 is to multiply the oxidation half-reaction through by 4.
Comment: As noted in the text, the potential for the oxidation of Fe 2+(aq) is negative, but the potential for the reduction of 0 2(g) in acidic aqueous solution is sufficiently positive to overcome that, leading to an overall positive potential for the
reaction of Equation 23.26.
-
PRACTICE EXERCISE
Calculate the standard potential for the reaction of Equation 23.22.
Answer: 1.64 V
You may have seen instances in which water dripping from a faucet or other
outlet has left a brown stain. The brown color is due to insoluble iron(III) oxide,
formed by oxidation of iron(II) present in the water:
4 Fe 2+(aq) + 8 HC03 - (aq) + 02(g) ~ 2 Fe203(s) + 8 C02(g) + 4 H20(Z)
[23.27]
When iron metal reacts with an oxidizing acid such as warm, dilute nitric
acid, Fe 3+(aq) is formed directly:
Fe(s) + N0 3- (aq) + 4 H+(aq) ~ Fe 3+(aq) + NO(g) + 2 H 20(Z) [23.28]
1003
1004
CHAPTER 23
Metals and Metallurgy
In the +3 oxidation state iron is soluble in acidic solution as the hydrated ion,
Fe(H 20) 63+. However, this ion hydrolyzes readily c:x:o (Section 16.11):
Fe(H 20) 63+(aq) ~ Fe(H 20) 5(0H) 2+(aq) + H+(aq)
.6. Figure 23.27 Precipitation of
Fe(OHh. Addition of NaOH solution to
an aqueous solution of Fe 3+ causes
Fe(OH) 3 to precipitate.
[23.29]
When an acidic solution of iron(III) is made more basic, a gelatinous red-brown
precipitate, most accurately described as a hydrous oxide, Fe 20 3 • nH 20, is
formed (Figure 23.27 ~ ). In this formulation n represents an indefinite number
of water molecules, depending on the precise conditions of the precipitation.
Usually, the precipitate that forms is represented merely as Fe(OH)3. The solubility of Fe(OHh is very low (Ksp = 4 X 10- 38). It dissolves in strongly acidic
solution but not in basic solution. The fact that it does not dissolve in basic solution is the basis of the Bayer process, in which aluminum is separated from
impurities, primarily iron(III). c:x:o (Section 23.3)
Copper
In its aqueous solution chemistry, copper exhibits two oxidation states: + 1
(cuprous) and +2 (cupric). In the + 1 oxidation state copper possesses a 3d 10
electron configuration. Salts of Cu+ are often water insoluble and are mostly
white in color. In aqueous solution the Cu+ ion readily disproportionates:
2 Cu+(aq) ~ Cu2+(aq) + Cu(s)
.6. Figure 23.28 Crystals of copper(ll)
sulfate pentahydrate, CuS04 • 5 H2 0.
K = 1.2 X 106
[23.30]
Because of this reaction and because copper(!) is readily oxidized to copper(II)
under most solution conditions, the + 2 oxidation state is by far the more common.
Many salts of Cu2+, including Cu(N03h, CuS04, and CuC12, are water soluble. Copper sulfate pentahydrate (CuS0 4 • 5 H 20), a widely used salt, has four
water molecules bound to the copper ion and a fifth held to the S0 42- ion by
hydrogen bonding. The salt is blue and is often called blue vitriol (Figure
23.28 ~ ). Aqueous solutions of Cu2+, in which the copper ion is coordinated by
water molecules, are also blue. Among the insoluble compounds of copper(II)
is Cu(OHh, which is formed when NaOH is added to an aqueous Cu2+
solution (Figure 23.29 ~ ). This blue compound readily loses water on heating to
form black copper(II) oxide:
Cu(OHh(s)
~
CuO(s) + H 20(l)
[23.31]
36
CuS is one of the least soluble copper(II) compounds (Ksp = 6.3 X 10- ). This
black substance does not dissolve in N aOH, NH3, or nonoxidizing acids such
as HCL It does dissolve in HN03, however, which oxidizes the sulfide to sulfur:
3 CuS(s)
+
8 H+(aq)
+ 2 N0 3-(aq)
~
3 Cu2+(aq) + 3 S(s) + 2 NO(g) + 4 H 20(l)
.6. Figure 23.29 Precipitation of
Cu(OH)l. Addition of NaOH solution to
an aqueous solution of Cu 2 + causes
Cu(OH) 2 to precipitate.
[23.32]
CuS04 is often added to water to stop algae or fungal growth, and other
copper preparations are used to spray or dust plants to protect them from
lower organisms and insects. Copper compounds are not generally toxic to
human beings, except in massive quantities. Our daily diet normally includes
from 2 to 5 mg of copper.
SAMPLE INTEGRATIVE EXERCISE I Putting Concepts Together
The most important commercial ore of chromium is chromite (FeCr20 4 ). (a) What is
the most reasonable assignment of oxidation states to Fe and Cr in this ore? (b) Below
74 K the FeCr2 0 4 magnetically orders so that the unpaired electrons on chromium
point in the opposite direction as those on iron. What type of magnetic state describes
FeCr20 4 below 74 K? Would you expect the magnetism to increase or decrease as
FeCr2 0 4 is cooled through the 74 K transition? (c) Chromite can be reduced in an
electric arc furnace (which provides the required heat) using coke (carbon). Write a
balanced chemical equation for this reduction, which forms ferrochrome (FeCr2).
(d) Two of the major forms of chromium in the +6 oxidation state are Cr0 4 2- and
Cr20 7 2- . Draw Lewis structures for these species. (Hint: You may find it helpful to
consider the Lewis structures of nonmetallic anions of the same formula.) (e) Chromium metal is used in alloys (for example, stainless steel) and in electroplating, but
Summary and Key Terms
1005
chromium is not widely used by itself, in part because it is not ductile at ordinary
temperatures. From what we have learned in this chapter about metallic bonding and
properties, suggest why chromium is less ductile than most metals.
SOLUTION
(a) Because each oxygen has an oxidation number of -2, the four oxygens represent a
total of -8. If the metals have whole-number oxidation numbers, our choices are
Fe = +4 and Cr = +2, or Fe = +2 and Cr = +3. The latter choice seems the more reasonable because a +4 oxidation number for iron is unusual. (Although one alternative
would be that Fe is +3 and the two Cr have different oxidation states of +2 and +3, the
properties of chromite indicate that the two Cr have the same oxidation number.)
(b) Fe 2+ and Cr 3+ have electron configurations of [Ar]3d6 and [Ar]3d3, respectively.
Therefore, Fe 2+ will have four unpaired electrons, and Cr 3+ will have three unpaired
electrons. Because they line up in opposite directions, FeCr20 4 will either be antiferromagnetic or ferrimagnetic. Because the ions possess different numbers of unpaired
electrons as well as the fact that there are twice as many Cr 3+ ions as there are Fe 2+
ions, their spins will not cancel each other out, and FeCr20 4 will be ferrimagnetic
below 74 K. The magnetic behavior of a ferrimagnet is similar to that of a ferromagnet,
so we would expect to see a large increase in the magnetism below 74 K.
(c) The balanced equation is
2 C(s) + FeCr20 4 (s) ~ FeCr2 (s) + 2 C0 2 (g)
(d) We expect that in Cr0 4 2- , the Cr will be surrounded tetrahedrally by four oxygens. The electron configuration of the Cr atom is [Ar]3d54s1, giving it six electrons
that can be used in bonding, much like the S atom in 50 4 2-. These six electrons must
be shared with four 0 atoms, each of which has six valence-shell electrons. In addition, the ion has a 2- charge. Thus, we have a total of 6 + 4( 6) + 2 = 32 valence electrons to place in the Lewis structure. Putting one electron pair in each Cr - 0 bond
and adding unshared electron pairs to the oxygens, we require precisely 32 electrons
to achieve an octet around each atom:
..". ·o:.. ]2-
:o·
/.
[ :~Cr':Q:
In Cr2 0 7 2- the structure is analogous to that of the diphosphate ion (P2 0 7 4- ),
which we discussed in Section 22.8. We can think of the Cr20 7 2- ion as formed by a
condensation reaction as shown in Equation 23.24.
..
:0:
.. ]2-
:0:
.. I .. I ..
0-Cr-0-Cr-0:
.. I .. I ..
:0:
:0:
..
..
[
(e) Recall that chromium, with six electrons available for bonding, has relatively
strong metallic bonding among the metals of the transition series, as evidenced by its
high melting point (Figure 23.14). This means that distortions of the metallic lattice of
the sort that occur when metals are drawn into wires will require more energy than
for other metals with weaker metallic bonding.
CHAPTER
REVIEW
SUMMARY AND KEY TERMS
Section 23.1 The metallic elements are extracted from
the lithosphere, the uppermost solid portion of our planet. Metallic elements occur in nature in minerals, which
are solid inorganic compounds found in various deposits, or ores. The desired components of an ore must be
separated from the undesired components, called gangue.
Metallurgy is concerned with obtaining metals from
these sources and with understanding and modifying the
properties of metals.
Section 23.2 Pyrometallurgy is the use of heat to bring
about chemical reactions that convert an ore from one
chemical form to another. In calcination an ore is heated
to drive off a volatile substance, as in heating a carbonate
ore to drive off C0 2• In roasting, the ore is heated under
conditions that bring about reaction with the furnace atmosphere. For example, sulfide ores might be heated to
oxidize sulfur to S02. In a smelting operation two or
more layers of mutually insoluble materials form in the
1006
CHAPTER 23
Metals and Metallurgy
furnace. One layer consists of molten metal, and the other
layer (slag) is composed of molten silicate minerals and
other ionic materials such as phosphates.
Iron, the most extensively used metal in modern society, is obtained from its oxide ores by reduction in a blast
furnace. The reducing agent is carbon, in the form of coke.
Limestone (CaC03) is added to react with the silicates present in the crude ore to form slag. The raw iron from the
blast furnace, called pig iron, is usually taken directly to a
converter, where refining takes place to produce various
kinds of steel. In the converter the molten iron reacts with
pure oxygen to oxidize impurity elements.
Section 23.3 Hydrometallurgy is the use of chemical
processes occurring in aqueous solution to separate a
mineral from its ore or one particular element from others.
In leaching, an ore is treated with an aqueous reagent to
dissolve one component selectively. In the Bayer process
aluminum is selectively dissolved from bauxite by treatment with concentrated NaOH solution.
Section 23.4 Electrometallurgy is the use of electrolytic
methods to prepare or purify a metallic element. Sodium is
prepared by electrolysis of molten NaCl in a Downs cell.
Aluminum is obtained in the Hall-Heroult process by
electrolysis of Al20 3 in molten cryolite (Na3AlF6). Copper
is purified by electrolysis of aqueous copper sulfate solution using anodes composed of impure copper.
Section 23.5 The properties of metals can be accounted
for in a qualitative way by the electron-sea model, in
which the electrons are visualized as being free to move
throughout the metal structure. In the molecular-orbital
model the valence atomic orbitals of the metal atoms interact to form energy bands that are incompletely filled by valence electrons. The bands that form from the valence shell
s, p, and d orbitals overlap so that aU metals have partially
filled bands. The orbitals that constitute the energy band
are delocalized over the atoms of the metal, and their energies are closely spaced. Because the energy differences between orbitals in the band are so small, promoting
electrons to higher-energy orbitals requires very little energy. This gives rise to high electrical and thermal conductivity, as well as other characteristic metallic properties.
Section 23.6 Alloys are materials that possess characteristic metallic properties and are composed of more than
one element. Usually, one or more metallic elements are
major components. Solution alloys are homogeneous alloys in which the components are distributed uniformly,
while in heterogeneous alloys the components are not distributed uniformly; instead, two or more distinct phases
with characteristic compositions are present. Solution alloys can be either substitutional alloys, where atoms of the
different metals randomly occupy metal atoms sites in the
lattice, or interstitial alloys, where smaller often nonmetallic
atoms occupy interstitial sites between metal atoms.
Intermetallic compounds are homogeneous alloys that
have definite properties and compositions.
Sections 23.7 and 23.8 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 a given series of
transition metals, the attraction between the nucleus and
the valence electrons increases faster for electrons occupying the d orbitals than for electrons occupying the s orbitals. As a result, the later transition elements in a given
row tend to adopt lower oxidation states. Although the
atomic and ionic radii increase in the second series as
compared to the first, the elements of tpe second and third
series are similar with respect to these and other properties. This similarity is due to the lanthanide contraction.
The presence of unpaired electrons in valence orbitals
leads to interesting magnetic behavior in transition metals
and their compounds. In ferromagnetic, ferrimagnetic,
and antiferromagnetic substances the unpaired electron
spins on atoms in a solid are affected by those 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
each other. 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.
The chapter concluded with a· concise overview of
the chemistry of three of the common transition metals:
chromium, iron, and copper.
KEY SKILLS
• Understand the differences between metals, minerals, and ores.
• Be able to write balanced chemical equations to illustrate the kinds of reactions that result when a mineral undergoes calcination or roasting.
• Understand and be able to write balanced chemical equations for the processes that are used to convert iron ore
into steel.
• Understand and be able to write balanced chemical equations for the Bayer process whereby aluminum is separated from other metals, most notably iron.
• Describe the electrometallurgical processes used to produce sodium, aluminum, and copper, and write balanced
chemical equations for them.
• Explain the relationship between the electronic structure of metals and their physical properties.
I
I
Visualizing Concepts
1007
• Use the molecular-orbital model to qualitatively predict the trends in melting point, boiling point, and hardness
of metals.
• Explain how solution alloys and heterogeneous alloys differ.
• Describe the differences between substitutional alloys, interstitial alloys, and intermetallic compounds.
• Be familiar with the periodic trends in radii and oxidation states of the transition metal ions, including the origin
and effect of the lanthanide contraction.
• Understand the differences between paramagnetism, ferromagnetism, antiferromagnetism, and ferrimagnetism.
VISUALIZING CONCEPTS
23.1 This diagram shows the approximate enthalpy change in
the roasting of ZnS. Does the roasting reaction cause an
increase or a decrease in the temperature of the roasting
oven? Explain. In light of your answer, is it necessary to
heat the ore to produce roasting? [Section 23.2]
2 ZnS(s)
+ 3 0 2(g)
590 kJ
2 ZnO(s)
+ S02(g)
23.4 Nb 3Sn and SmCo5 are two important intermetallic
compounds. Nb3Sn is a superconductor from which the
electromagnets of a magnetic resonance imaging (MRI)
instrument are built, while SmCo5 is used to make powerful permanent magnets. The unit cells of Nb3Sn and
SmCo5 are shown below. By counting the atoms in the
unit cell, determine the empirical formulas corresponding to each structure. Then use this information to determine which structure corresponds to which intermetallic
compound. (Hint: In both structures all of the atoms
shown are located either at the corners of the unit cell,
the body center of the unit cell, or the faces of the unit
cell.) [Section 23.6]
23.2 Magnesium is produced commercially by electmlysis
from a molten salt using a cell similar to the one
shown below. (a) What salt is used as the electrolyte?
(b) Which electrode is the anode, and which one is the cathode? (c) Write the overall cell reaction and individual halfreactions. (d) What precautions would need to be taken
with respect to the magnesium formed? [Section 23.4]
Compound A
Compound B
23.5 This chart shows the variation in an important property
of the metals from K through Ge. Is the property atomic
radius, electronegativity, or first ionization energy?
Explain your choice. [Section 23.7]
Electrolyte
23.3 The standard enthalpy of formation of the gaseous
element varies among some fourth-row metals as shown
in the figure below. Would you expect gaseous molybdenum to have a larger or smaller standard enthalpy of formation than zirconium? What about gaseous cadmium?
[Section 23.5]
LlHJ
Rbp
sr t==J
Yl
Zrl
23.6 (a) Except for scandium, chromium is the only element
in Figure 23.22 for which the +3 oxidation state is more
stable in general than the +2. Explain why the +3 oxidation state is most characteristic of scandium. (b) What
type of magnetism would you expect from SrCr04?
(c) By analogy with inorganic ions of the same formula
type, predict the geometrical structure of the permanganate ion, Mn0 4- .[Sections 23.7 and 23.8]
1008
CHAPTER 23
Metals and Metallurgy
EXERCISES
Metallurgy
23.7 Two of the most heavily utilized metals are aluminum
and iron. What are the most important natural sources
of these elements? In what oxidation state is each metal
found in nature?
23.8 (a) Pyrolusite (Mn02) is a commercially important mineral of manganese. What is the oxidation state of Mn in
this metal? (b) Name some reagents that might be used
to reduce this ore to the metal.
23.9 Explain in your own words what is meant by the
statement, "This ore consists of a small concentration of
chalcopyrite, together with considerable gangue."
23.10 What is meant by the following terms: (a) calcination,
(b) leaching, (c) smelting, (d) slag?
23.11 Complete and balance each of the following equations:
(a)
Cr203(s) + Na(l)
(b)
PbC03(s)
tJ.
+ 02(g) ~
tJ.
ZnO(s) + CO(g) ~
(c) CdS(s)
(d)
~
tJ.
~
PbS(s) + 02(g)
(b)
CoC03(s)
tJ.
~
tJ.
+ H2(g) ~
(d) VC13 (g) + K(Z) ~
(e) BaO(s) + P20s(l) ~
23.13 A sample containing PbS04 is to be refined to Pb metal
via calcination, followed by roasting. (a) What volatile
product would you expect to be produced by calcination? (b) Propose an appropriate atmosphere to accompany the roasting. (c) Write balanced chemical equations
for the two steps.
23.14 Consider the thermodynamics of calcination. (a) The
following equation is a generic reaction representing
calcination of a metal carbonate:
MC0 3(s)
~
0
23.16 Assess the feasibility of reducing Ti02 to titanium metal
by roasting in carbon monoxide. (a) Write a reaction for
this process. (b) Use the thermodynamic quantities
given in Appendix C to calculate ~c o , ~H o , and ~s o for
this reaction. Is this reaction spontaneous at 25 oc under
standard conditions? (c) If we assume that ~H o and s o
values do not change with temperature, at what temperature would this process become spontaneous? Do you
think this process would be practical?
23.17 What is the major reducing agent in the reduction of
iron ore in a blast furnace? Write a balanced chemical
equation for the reduction process.
23.18 Write balanced chemical equations for the reduction of
FeO and Fe203 by H2 and by CO.
---23.19 What role does each of the following materials play in
the chemical processes that occur in a blast furnace:
(a) air, (b) limestone, (c) coke, (d) water? Write balanced
chemical equations to illustrate your answers.
tJ.
~
(c) W03(s)
0
----
23.12 Complete and balance each equation:
(a)
spontaneous? If not, at what temperature does this reaction become spontaneous (assuming ~H o and s o values
do not change with temperature)?
---23.15 Use the thermodynamic quantities given in Appendix C
to calculate ~C , ~H , and ~s o for the reaction corresponding to roasting of PbO in a CO atmosphere
(Equation 23.5). Approximate the thermodynamic quantities of Pb(l) using the thermodynamic properties of
Pb(s). Is this reaction spontaneous at 25 °C under standard conditions? Is it exothermic or endothermic?
MO(s) + C02(g)
Would you expect this reaction to become more or less
spontaneous as the temperature increases? (b) What is
the standard free energy, ~c o, for the reaction corresponding to the calcination of PbC03 (Equation 23.1)
at 25 °C under standard conditions? Is this reaction
23.20 (a) In the basic oxygen process for steel formation, what
reactions cause the temperature in the converter to increase? (b) Write balanced chemical equations for the
oxidation of carbon, sulfur, and silicon in the converter.
23.21 Describe how electrometallurgy could be employed to
purify crude cobalt metal. Describe the compositions of
the electrodes and electrolyte, and write out all electrode reactions.
23.22 The element tin is generally recovered from deposits of
the ore cassiterite (Sn02). The oxide is reduced with carbon, and the crude metal is purified by electrolysis.
Write balanced chemical equations for the reduction
process and for the electrode reactions in the electrolysis.
(Assume that an acidic solution of SnS04 is employed as
an electrolyte in the electrolysis.)
Metals and Alloys
23.23 Sodium is a highly malleable substance, whereas sodium chloride is not. Explain this difference in properties.
23.25 Explain how the electron-sea model accounts for the
high electrical and thermal conductivity of metals.
23.24 Germanium has the same crystal structure as diamond
(Figure 11.41). Based on this fact, do you think germanium is likely to exhibit metallic properties? Explain
your answer.
23.26 (a) Compare the electronic structures of atomic chromium and atomic selenium. In what respects are they similar, and in what respects do they differ? (b) Chromium
is a metal, and selenium is a nonmetal. What factors are
important in determining this difference in properties?
t
t
Exercises
23.27 The densities of the elements K, Ca, Sc, and Ti are 0.86,
1.5, 3.2, and 4.5 g/ cm3, respectively. What factors are
likely to be of major importance in determining this
variation? Which factor do you think will be the most
important?
23.28 Explain this trend in melting points: Y 1522 oc, Zr
1852 °C, Nb 2468
oc, Mo 2617 oc.
1009
[23.32] The electrical conductivity of titanium is approximately
2500 times greater than that of silicon. Titanium has a
hexagonal close-packed structure, and silicon has the
diamond structure. Explain how the structures relate to
the relative electrical conductivities of the elements.
23.33 Define the term alloy. Distinguish among solution alloys,
heterogeneous alloys, and intermetallic compounds.
23.29 Which would you expect to be the more ductile element,
23.34 Distinguish between substitutional and interstitial alloys.
(a) Ag or Mo, (b) Zn or Si? In each case explain your
reasoning.
What conditions favor formation of substitutional alloys?
23.30 How do you account for the observation that the alkali
metals, like sodium and potassium, are soft enough to
be cut with a knife?
23.31 Tin exists in two allotropic forms: Gray tin has a dia-
mond structure, and white tin has a close-packed structure. One of these allotropic forms is a semiconductor
with a small band gap while the other is a metal. Which
one is which? Which form would you expect to have the
longer Sn -Sn bond distance?
23.35 For each of the following alloy compositions indicate
whether you would expect it to be a substitutional alloy,
an interstitial alloy, or an intermetallic compound:
(a) Feo.97Sio.o3, (b) Feo.6oNio.40, (c) Cu3Au.
23.36 For each of the following alloy compositions indicate
whether you would expect it to be a substitutional alloy,
an interstitial alloy, or an intermetallic compound:
(a) Cuo. 66 Zno. 34, (b) Ag3Sn, (c) Tio.990o.Ol·
Transition Metals
23.37 Which of the following properties are better considered
23.47 Which would you expect to be more easily oxidized,
Ti 2+ or Ni 2+?
characteristic of the free isolated atoms, and which are
characteristic of the bulk metal: (a) electrical conductivity, (b) first ionization energy, (c) atomic radius, (d) melting point, (e) heat of vaporization, (f) electron affinity?
23.48 Which would you expect to be the stronger reducing
23.38 Which of the following species would you expect
23.49 How does the presence of air affect the relative stabili-
to possess metallic properties: (a) TiC14, (b) NiCo alloy,
(c) W, (d) Ge, (e) Hg 2 2+? Explain in each case.
23.39 Zirconium and hafnium are the group 4B elements in
the second and third transition series. The radii of these
elements are virtually the same (Figure 23.20). Explain
this similarity.
23.40 What is meant by the term lanthanide contraction?
What properties of the transition elements are affected
by the lanthanide contraction?
23.41 Write the formula for the fluoride corresponding to the
highest expected oxidation state for (a) Sc, (b) Co, (c) Zn,
(d) Mo.
23.42 Write the formula for the oxide corresponding to the highest expected oxidation state for (a) Cd, (b) V, (c) W, (d) Ru.
23.43 Why does chromium exhibit several oxidation states in
its compounds, whereas aluminum exhibits only the +3
oxidation state?
agent, Cr 2+ or Fe 2 +?
ties of ferrous and ferric ions in solution?
23.50 (a) Give the chemical formulas and colors of the chro-
mate and dichromate ions. (b) Which of these ions is
more stable in acidic solution? (c) What type of reaction
is involved in their interconversion in solution?
23.51 Write balanced chemical equations for the reaction be-
tween iron and (a) hydrochloric acid, (b) nitric acid.
23.52 Mn0 2 reacts with aqueous HCl to yield MnC12(aq) and
chlorine gas. (a) Write a balanced chemical equation for
the reaction. (b) Is this an oxidation-reduction reaction?
If yes, identify the oxidizing and reducing agents.
23.53 On the atomic level, what distinguishes a paramagnetic
material from a diamagnetic one? How does each behave in a magnetic field?
23.54 On the atomic level, what distinguishes an antiferro-
magnetic material from a diamagnetic one?
23.44 The element vanadium exhibits multiple oxidation states
23.55 (a) On the atomic level, what distinguishes ferromagnet-
in its compounds, including +2. The compound VC1 2 is
known, whereas ScC12 is unknown. Use electron configurations and effective nuclear charges to account for this
difference in behavior.
23.56 The two most important iron oxide minerals are mag-
23.45 Write the expected electron configuration for (a) Cr 3+,
(b) Au3+, (c) Ru 2+, (d) Cu+, (e) Mn4+, (f) Ir+.
23.46 What is the expected electron configuration for (a) Ti 2+,
(b)
Co3+, (c) Pd 2+, (d) Mo3+, (e) Ru3+, (f) Ni4+?
ic, ferrimagnetic, and antiferromagnetic materials from
each other? (b) Which one of these types of magnetic
materials cannot be used to make a permanent magnet?
netite, Fe304, and hematite, Fe20 3. One is a ferrimagnetic
material, while the other is an antiferromagnetic material.
(a) Based on the oxidation states of iron, which one is
more likely to be ferrimagnetic? (b) Would it be possible
to use magnetic fields to separate these minerals?
1010
CHAPTER 23
Metals and Metallurgy
ADDITIONAL EXERCISES
23.57 Write a chemical equation for the reaction that occurs
23.65 Antimony and niobium both possess five valence-shell
when PbS is roasted in air. Why might a sulfuric acid
plant be located near a plant that roasts sulfide ores?
electrons and are metallic conductors. Which element
would you expect to be a better conductor of electricity?
23.58 Explain why aluminum, magnesium, and sodium met-
[23.66] Introduction of carbon into a metallic lattice generally
als are obtained by electrolysis instead of by reduction
with chemical reducing agents.
results in a harder, less ductile substance with lower
electrical and thermal conductivities. Explain why this
might be so.
23.59 Make a list of the chemical reducing agents employed in
the production of metals, as described in this chapter.
For each of them, identify a metal that can be formed
using that reducing agent.
23.60 Write balanced chemical equations for each of the fol-
lowing verbal descriptions: (a) Vanadium oxytrichloride
(VOC13 ) is formed by the reaction of vanadium(III) chloride with oxygen. (b) Niobium(V) oxide is reduced to
the metal with hydrogen gas. (c) Iron(III) ion in aqueous
solution is reduced to iron(II) ion in the presence of zinc
dust. (d) Niobium(V) chloride reacts with water to yield
crystals of niobic acid (HNb03).
23.61 Write a balanced chemical equation to correspond to
each of the following verbal descriptions: (a) NiO(s) can
be solubilized by leaching with aqueous sulfuric acid.
(b) After concentration, an ore containing the mineral
carrollite (CuCo2S4) is leached with aqueous sulfuric
acid to produce a solution containing copper ions and
cobalt ions. (c) Titanium dioxide is treated with chlorine
in the presence of carbon as a reducing agent to form
TiC14. (d) Under oxygen pressure ZnS(s) reacts at 150 oc
with aqueous sulfuric acid to form soluble zinc sulfate,
with deposition of elemental sulfur.
23.62 The crude copper that is subjected to electrorefining
contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common
oxidation state, Te 4+, is
Te 4+(aq) + 4 e- ~ Te(s)
E~ed = 0.57 V
Given this information, describe the probable fate of tellurium impurities during electrorefining.
23.63 Why is the +2 oxidation state common among the tran-
sition metals? Why do so many transition metals exhibit
a variety of oxidation states?
[23.64] Write balanced chemical equations that correspond to
the steps in the following brief account of the metallurgy
of molybdenum: Molybdenum occurs primarily as the
sulfide, MoS2 . On boiling with concentrated nitric acid,
a white residue of Mo03 is obtained. This is an acidic
oxide; when it is dissolved in excess hot concentrated
ammonia, ammonium molybdate crystallizes on cooling. On heating ammonium molybdate, white Mo03 is
obtained. On further heating to 1200 °C in hydrogen, a
gray powder of metallic molybdenum is obtained.
23.67 The thermodynamic stabilities of the three complexes
Zn(Hz0)4 2+, Zn(NH3)4 2 +, and Zn(CN)4 2- increase from
the H 20 to the NH3 to the CN- complex. How do you
expect the reduction potentials of these three complexes
to compare?
23.68 Indicate whether each of the following compounds is
expected to be diamagnetic or paramagnetic, and give a
reason for your answer in each case: (a) NbCls, (b) CrC12,
(c) CuCl, (d) Ru04, (e) NiC1 2 .
[23.69] Associated with every ferromagnetic solid is a temperature
known as its Curie temperature. When heated above its
Curie temperature, the substance no longer exhibits ferromagnetism but rather becomes paramagnetic. Use the kinetic-molecular theory of solids to explain this observation.
23.70 Associated with every antiferromagnetic solid is a tem-
perature known as its Neel temperature. When heated
above its Neel temperature the magnetic behavior
changes from antiferromagnetic to paramagnetic. In contrast diamagnetic substances do not generally become
paramagnetic upon heating. How do you explain this
difference in behavior?
23.71 Write balanced chemical equations for each of the following reactions characteristic of elemental manganese: (a) It
reacts with aqueous HN03 to form a solution of manganese(ll) nitrate. (b) When solid manganese(ll) nitrate is
heated to 450 K, it decomposes to Mn02 . (c) When Mn02
is heated to 700 K, it decomposes to Mn30 4. (d) When
solid MnC12 is reacted with F2(g), it forms MnF3 (one of
the products is ClF3).
[23.72] Based on the chemistry described in this chapter and others, propose balanced chemical equations for the following sequence of reactions involving nickel: (a) The ore
millerite, which contains NiS, is roasted in an atmosphere
of oxygen to produce an oxide. (b) The oxide is reduced to
the metal, using coke. (c) Dissolving the metal in hydrochloric acid produces a green solution. (d) Adding excess sodium hydroxide to the solution causes a gelatinous
green material to precipitate. (e) Upon heating, the green
material loses water and yields a green powder.
[23.73] Indicate whether each of the following solids is likely to
be an insulator, a metallic conductor, or a semiconductor:
(a) Ti02, (b) Ge, (c) Cu3Al, (d) Pd, (e) SiC, (f) Bi.
INTEGRATIVE EXERCISES
23.74 (a) A charge of 3.3 X 106 kg of material containing 27%
Cu2S and 13% FeS is added to a converter and oxidized.
What mass of S02(g) is formed? (b) What is the molar
ratio of Cu to Fe in the resulting mixture of oxides?
(c) What are the likely formulas of the oxides formed in
the oxidation reactions, assuming an excess of oxygen?
(d) Write balanced equations representing each of the
oxidation reactions.
Integrative Exercises
23.75 Using the concepts discussed in Chapter 13, indicate why
the molten metal and slag phases formed in the blast furnace shown in Figure 23.4 are immiscible.
very large copper buses that convey current to the cells.
Assuming that the cells are 96% efficient in producing
the desired products in electrolysis, what mass of Mg is
formed by passing a current of 97,000 A for a period of
24hr?
23.76 In an electrolytic process nickel sulfide is oxidized in a
two-step reaction:
Ni 3S2(s) ~ Ni 2+(aq) + 2 NiS(s) + 2 eNiS(s)
~ Ni 2+(aq)
+ S(s) + 2 e-
2
What mass of Ni + is produced in solution by passing a
current of 67 A for a period of 11.0 hr, assuming the cell is
90% efficient?
[23.77] (a) Using the data in Appendix C, estimate the freeenergy change for the following reaction at 1200 oe:
Si(s) + 2 MnO(s) ~ Si0 2(s) + 2 Mn(s)
(b) What does this value tell you about the feasibility of
carrying out this reaction at 1200 °C?
[23.78] (a) In the converter employed in steel formation (Figure
23.6), oxygen gas is blown at high temperature directly
into a container of molten iron. Iron is converted to rust
on exposure to air at room temperature, but the iron is
not extensively oxidized in the converter. Explain why
this is so. (b) The oxygen introduced into the converter
reacts with various impurities, particularly with carbon,
phosphorus, sulfur, silicon, and impurity metals. What
are the products of these reactions, and where do they
end up in the process?
23.79 Copper(!) is an uncommon oxidation state in aqueous
acidic solution because cu+(aq) disproportionates into
Cu2+ and Cu. Use data from Appendix E to calculate the
equilibrium constant for the reaction
2 Cu+(aq) ~ Cu 2+(aq) + Cu(s)
23.80 The reduction of metal oxides is often accomplished using
carbon monoxide as a reducing agent. Carbon (coke) and
carbon dioxide are usually present, leading to the following reaction:
C(s) + C0 2(g)
~
2 CO(g)
Using data from Appendix C, calculate the equilibrium
constant for this reaction at 298 K and at 2000 K, assuming that the enthalpies and entropies of formation do
not depend upon temperature.
23.81 An important process in the metallurgy of titanium is
the reaction between titanium dioxide and chlorine in
the presence of carbon, which acts as a reducing agent,
leading to the formation of gaseous TiC14. (a) Write a
balanced chemical equation for this reaction, and use it
with the values listed in Appendix C to calculate the
standard enthalpy change of this reaction. Is this reaction exothermic or endothermic? (b) Write a reaction for
the direct reaction between titanium dioxide and chlorine to form TiC14 and oxygen. Is this reaction exothermic or endothermic?
23.82 Magnesium is obtained by electrolysis of molten MgC12 .
(a) Why isn't an aqueous solution of MgC12 used in the
electrolysis? (b) Several cells are connected in parallel by
1011
23.83 Vanadium(V) fluoride is a colorless substance that melts
at 19.5 oc and boils at 48.3 °C. Vanadium(III) fluoride, on
the other hand, is yellow-green in color and melts at
800 oc. (a) Suggest a structure and bonding for VF5 that
accounts for its melting and boiling points. Can you identify a compound of a nonmetallic element that probably
has the same structure? (b) VF3 is prepared by the action
of HF on heated VC13 . Write a balanced equation for this
reaction. (c) While VF5 is a known compound, the other
vanadium(V) halides are unknown. Suggest why these
compounds might be unstable. (Hint: The reasons might
have to do with both size and electronic factors.)
23.84 The galvanizing of iron sheet can be carried out electrolytically using a bath containing a zinc sulfate solution.
The sheet is made the cathode, and a graphite anode is
used. Calculate the cost of the electricity required to deposit a 0.49-mm layer of zinc on both sides of an iron
sheet 2.0 m wide and 80 m long if the current is 30 A, the
voltage is 3.5 V, and the energy efficiency of the process is
90%. Assume the cost of electricity is $0.082 per kilowatt
hour. The density of zinc is 7.1 g/ cm3 .
23.85 As mentioned in the text, Ni3Al is used in the turbines of
aircraft engines because of its strength and low density.
Nickel metal has a cubic close-packed structure with a
face-centered cubic unit cell, while Ni3Al has the ordered
cubic structure shown in Figure 23.17(b). The length of
the cubic unit cell edge is 3.53 A for nickel and 3.56 A for
Ni3Al. Use this data to calculate and compare the densities of these two materials.
[23.86] Silver is found as Ag2S in the ore argentite. (a) By using
data in Table 17.1 and Appendix D.3, determine the equilibrium constant for the cyanidation of Ag2S to Ag(CNh - .
(b) Based on your answer to part (a), would you consider
cyanidation to be a practical means of leaching silver
from argentite ore? (c) Silver is also found as AgCl in the
mineral cerargyrite. Would it be feasible to use cyanidation as a leaching process for this ore?
[23.87] The heats of atomization, flHatom, in kJ/ mol, of the first
transition series of elements are as follows:
Element
Ca
Sc
Ti
V
Cr
Mn Fe
Co
Ni
Cu
flHatom
178 378 471 515 397 281 415 426 431 338
(a) Write an equation for the process involved in atomization, and describe the electronic and structural
changes that occur. (b) flHatom varies irregularly in the
series following V. How can this be accounted for, at
least in part, using the electronic configurations of the
gaseous atoms? (Hint: Recall the discussions of Sections
6.8 and 6.9.)