Chemistry part 35, Julia Burdge,2e (2009) docx

864 CHAPTER 22
Coordination
Chemistry
Think
About
It
Although ligands
are alphabetized in a compound's
name, they do not necessarily
appear in alphabetical order
in
the
compound's formula.
Figure
22
.5
Common geometries
of
complex ions. In each case M is a
metal and L is a monodentate ligand.
Strategy
If
you
can't
remember them yet, refer to
Tab
l
es
22.4 and 22.5 for the names
of
ligands

and anions containing metal atoms.
Setup (a) There are six ligands: five
NH
3 molecules and one
Cl-
ion.
Tb.e
oxidation state
of
cobalt
is
+3,
making the overall charge on the complex ion
+2.
Therefore, there are two chloride ions as
counter ions.
(b)
There
are four ligand
s:
two bidentate ethylenediamines and two Cl- ions.
The
oxidation state
of
platinum is
+4,
making the overall charge
on
the complex ion
+2

. Therefore, there are two nitrate
ions as counter ions.
Solution (a) [Co(NH
3
)sCl]CI
2
(b) [Pt(en)zCI
2
](N0
3
)2
Practice Problem Write the formulas for (a) tris(ethylenediamine)cobalt(lII) sulfate and (b) sodium
hexanitrocobaitate(III).
~
Checkpoint
22.1
Coordination Compounds
22
.1.1
22
.1.2
Select the correct name for the
compound
[Cu(
NH
3
)4]CI
2
.
a)

Copperteu'aammine dichloride
b)
Tetraamminecopper(II) chloride
c)
Tetraaminedichlorocuprate(II)
d)
Dichlorotetraaminecopper(II)
e)
Tetraaminedichlorocopper(II)
Select the correct name for the
compound K
3
[FeF
6
].
a)
Tripotassiumironhexaft uoride
b) Hexaftuorotripotassiumferrate(III)
c)
He
xaftuoroiron(III) potassium
d)
Potassium hexaftuoroferrate(lII)
e)
Potassium ironhexaftuorate
22
.1.3
22
.1.4
Select the correct formula for

pentaaminenitrocobalt(III).
a)
[Co(NH
3
)sN0
3
]
3+
b) [Co(NH
3
)sN0
3
f +
c)
Co(NH
3
)sN03
d)
[Co(NH
3
)s](N0
3
)
e)
[Co(NH
3
)S]
(N0
3
)2

Select the correct formula for
tetraaq uodichlorochromi u
rn
(III)
chloride.
a)
[Cr(H
2
O)4
CI
2]CI
3
b)
[Cr(H
2
O)4
CI2]CI
2
c)
[Cr(H
2
O)4
CI
2]C
l
d)
[Cr(H
2
O)4]CI
3

e)
[Cr(H
2
O)4]Cl
z
Structure
of
Coordination Compounds
The
geometry
of
a
coordination
compound
often
plays
a si
gnificant
role
in
determining
its
proper
-
ties.
Figure
22.5 s
how
s
four

different
geometric
arrangements
for
metal
atoms
with
monodentate
ligands.
In
these
diagrams
we
See
that
structure
and
the
coordination
number
of
the
meta
l
re
l
ate
to
each
other

as
follow
s:
L
M
Linear
Coordination Number
L
2
4
6
L
M
L ;

L
L
Tetrahedral
Structure
Linear
Tetrahedral
or
square
pla
n
ar
Octahedra
l
L
L L

~
/
L
M
/
/~L
-M-
/
L
L
Square planar Octahedral
L
•
SECTION 22.2
Structure
of
Coordination
Compounds
865
In
studying the geometry
of
coordination compounds, we sometimes find that there is more than
one way
to
an'ange the ligands around the central atom. Such compounds in which ligands are
•
••
••• •••
••••

<'
••••••••••••
•••
•
••
••
•••
•
••••
•
••
•••••••••
arranged differently, known
as
stereoisomers, have distinctly different physical and chemical prop-
erties. Coordination compounds may exhibit two types of stereoisomerism:
geometric and optical.
Geometric isomers are stereoisomers that cannot be interconverted without breaking chemi-
cal bonds. Geometric isomers come in pairs.
We
use the terms cis and trans
to
distinguish one '
geometric isomer
of
a compound from the other. Cis means that two palticular atoms (or groups
of
atoms) are adjacent to each other, and trans means that the atoms (or groups of atoms) are
on
opposite sides in the structural formula. The cis and trans isomers

of
coordination compounds
generally have quite different colors, melting points, dipole moments, and chemical reactivities.
Figure 22.6 shows the
cis and trans isomers
of
diamrninedichloroplatinum(II). Note that although
the types
of
bonds are the same in both isomers (two
Pt-N
and two
Pt-Cl
bonds), the spatial
arrangements are different. Another example is the tetraamminedichlorocobalt(III) ion, shown
in
Figure 22.7.
Optical isomers are nonsuperimposable mirror images. (Superimposable means that
if
one
structure is laid over the other, the positions
of
all the atoms will match.) Like geometric isomers,
optical isomers come in pairs. However, the optical isomers
of
a compound have identical physical
and chemical properties, such
as
melting point, boiling point, dipole moment, and chemical reac-
tivity toward molecules that are not

themselves optical isomers. Optical isomers differ from each
other, though, in their interactions with plane-polarized light,
as
we will see.
The structural relationship between two optical isomers is analogous to the relationship
between your left and right hands.
If you place your left hand in front
of
a mirror, the image you
see will look like your right hand (Figure 22.8). Your left hand and right hand are mirror images
of
each other. They are nonsuperimposable, however, because when you place your left hand over
your right hand (with both palms facing down), they do not match. This is why a right-handed
glove will not fit comfortably on your left hand.
Figure 22.9 shows the
cis and trans isomers
of
dichlorobis(ethylenediamine)cobalt(III) ion
and the mirror image
of
each. Careful examination reveals that the trans isomer and its mirror
Minor image
of
left hand Left hand
In
genera
l,
stereoisomers
are
compounds

that
are
made
up
of
the
same
types
and
numbers
of
atoms,
bonded
together
in
the
same
sequence,
but with different
spatial
arrangements.
Figure 22.6
The
(a) cis and (b)
trans
isomer
s
of
diamminedichloro-
platinum(II).

Note
that the two Cl
atoms are adjacent to each other in the
cis
isomer
and diagonally across from
each other in the
trans isomer.
Figure 22.7
The
(a) cis and (b)
trans iso
mer
s
of
tetraamminedichloro-
cobalt(III)
ion
, [Co(NH
3)4
CI
2t
.
The
ion
has only two geometric isomers.
Figure 22.8 A left
hand
and its
. .

II1lrror Image.
,
866 CHAPTER
22
Coordination
Chemistry
Figure 22.9 The (a) cis and (b)
trans isomers
of
dichlorobis(ethylene-
diamine )cobalt(III) ion and their mirror
images.
If
you could rotate the mirror
image in (b)
90° clockwise about the
vertical position and place the ion over
the
trans isomer, you would find that
the two are superimposable. No matter
how you rotate the
cis isomer and its ,
mirror image in (a), however, you
cannot superimpose one
on
the other.
Light
source
Fixed
polarizer

Mirror
(a)
CI
~co
~CI
Mirror
(b)
image are superimposable, but the cis isomer and its mirror image are not. Thus, the cis isomer and
its mirror image are
optical isomers.
Optical isomers are described
as
chiral (from the Greek word for "hand") because, like your
left and right hands, chiral molecules are nonsuperimposable. Isomers that are superimposable
with their mirror images are said to be
achiral. Chiral molecules
playa
vital role in enzyme reac-
tions in biological systems. Many drug molecules are chiral, although only one
of
a pair
of
chiral
isomers is biologically effective
[
~~
Section
10.4
].
Chiral molecules are said to be optically active because

of
their ability to rotate the plane
of
polarization
of
polarized light
as
it passes through them. Unlike ordinary light, which vibrates in
all directions, plane-polarized light vibrates only in a single plane.
We
use a polarimeter to mea-
sure the rotation
of
polarized light by optical isomers (Figure 22.10). A beam
of
unpolarized light
first passes through a
Polaroid sheet, called the polarizer, and then through a sample tube contain-
ing a solution of an optically active, chiral compound. As the polarized light passes through the
sample tube, its plane
of
polarization is rotated either
to
the right (clockwise) or
to
the left (coun-
terclockwise). This rotation can be measured directly by turning the analyzer in the appropriate
direction until minimal light transmission
is
achieved (Figure 22.11).

If
the plane
of
polarization is
rotated
to
the right, the isomer is said
to
be dextrorotatory and the isomer is labeled
d;
if
the rota-
tion is to the left, the isomer is
levorotatory and the isomer is labeled
l.
The d and l isomers of a
chiral substance, called
enantiomers, always rotate the plane
of
polarization by the same amount,
but in opposite directions. Thus, in an equimolar mixture
of
two enantiomers, called a racemic
mixture,
the net rotation is zero.
t
!
Polarimeter tube
Optically active
substance

in
solution
Analyzer
+ O·
Degree scale
+ 90'
Plane
of
polarization
-90'
•
Figure 22.10 Operation
of
a polarimeter. Initially, the tube is filled with
an
achiral compound.
The
analyzer is rotated so that its plane
of
polarization is perpendicular to that
of
the polarize
r.
Under this condition, no light reaches the observer. Next, a chiral compound is placed in the tube
as shown.
The
plane
of
polarization
of

the polarized light is rotated as it travels through the tube so that some light reaches the observer. Rotating the
analyzer (either to the left
or
to the right) until no light reaches the observer again allows the angle
of
optical rotation to be measured.
SECTION 22.3
Bonding
in
Coordination
Compounds: Crystal Field Theory
86
7
. 1
".
Bonding in Coordination Compounds:
Crystal Field Theory
A satisfactory theory
of
bonding in coordination compounds must account for properties such as
color and magnetism, as well as stereochemistry and bond strength. No single theory as yet does
all this for us. Rather, several different approaches have been applied to transition metal com-
plexes. We will consider only one
of
them here crystal field theory because it accounts for both
the color and magnetic properties
of
many coordination compounds.
We will begin our discussion
of

crystal field theory with the most straightforward
case-
namely, complex ions with octahedral geometry. Then we will see how it is applied to tetrahedral
and square-planar complexes.
Crystal Field Splitting
in
Octahedral Complexes
Crystal field theory explains the bonding in complex ions purely in terms
of
electrostatic forces. In
a complex ion, two types
of
electrostatic interaction come into play. One is the attraction between
the positive metal ion and the negatively charged ligand
or
the negatively charged end
of
a polar
ligand. This is the force that binds the ligands to the metal. The second type
of
interaction is the
electrostatic repulsion between the lone pairs on the ligands and the electrons in the
d orbitals
of
the metals.
The
d orbitals have different orientations
[
~~
Section

6.7], but in the absence
of
an exter-
nal disturbance, they all have the same energy. In an octahedral complex, a central metal atom
is surrounded by six lone pairs
of
electrons (on the six ligands), so all
fi
ve d orbitals experience
electrostatic repulsion. The magnitude
of
this repulsion depends on the orientation
of
the d orbital
that is involved. Take the
di _/ orbital as an example. In Figure 22.12, we see that the lobes
of
this
orbital point toward the comers
of
the octahedron along the x and y axes, where the lone-pair elec-
trons are positioned. Thus, an electron residing in this orbital would experience a greater repulsion
from the ligands than an electron would in the
d
X)
, d
yz
' or d
xz
orbitals. For this reason, the energy

of
the d
x
2
_
/ orbital is increased relative to the d
X)'
d
yZ
' and d
xz
orbitals. The d
z
2
orbital's energy
is also greater, because its lobes are pointed at the ligands along the z axis. As a result
of
these
metal-ligand interactions, the five
d orbitals in an octahedral complex are split between two sets
of
energy levels: a higher level with two orbitals
(d
} _ i and dl ) having the same energy, and a lower
level with three equal-energy orbitals
(d
X)'
d
yz
, and d

xz
) '
as shown in Figure 22.13. The crystal field
splitting
(Ll)
is the energy difference between two sets
of
d orbitals in a metal atom when ligands
are present. The magnitude
of
~
depends on the metal and the nature
of
the ligands; it has a direct
effect on the color and magnetic properties
of
complex ions.
Figure 22.11 Polarized lense
s.
0
light passes through the lenses when
they are rotated so that their planes
of
polarization are perpendicular .
•
868
CHAPTER
22
Coordination
Chemistry

Figure 22.12 The
fi
ve d orbitals in
an
octahedral environment. The metal
atom (or ion) is
at
the center of the
octahedron, and the six lone pairs on
the donor atoms of the ligands are at
the corners.
•
Figure 22.13 Cr
ys
tal
field
splitting
between
d orbitals
in
an
oc
tahed
ra
l
complex.
650
nm
580 nm
700

nm

400nm
1 5160
nm
430 nm 490 nm
Figure 22.14 A color wheel
with appropriate
wa
veleng
th
s.
Complementary color
s,
such as
red
and green, are
on
opposite sides of the
wheel.
z
•
d,
y
H
Color
. "
•
d
'

, d 2
x y-
Z
•
\ .
Cr
ys
tal
field splitting,
~
In
Chapter
6
we
learned that white light, such as sunlight, is a combination
of
all colors. A sub-
stance
appear
s
black
if
it absorbs all the visible light that strikes it.
If
it absorbs
no
visible light,
it is white
or
colorless.

An
object appears green
if
it
absorbs all light
but
reflects the green com-
ponent.
An
object also
look
s green
if
it reflects all colors except red, the complementary color
of
gre
en
(Figure 22.14).
What
ha
s
been
said
of
reflected light also applies to transmitted light (i.e., the light that
passe
s through the
medium
,
such

as a solution). Consider
the
hydrated cupric ion ([Cu(H
2
0)6]2+);
it absorbs light in the orange region
of
the s
pectrum
, so a solution
of
CUS04 appears
blue
to us.
Recall from
Chapter
6 that when
the
energy
of
a
photon
is equal to the difference between the
ground
state and an excited state, absorption occurs as
the
photon
strikes the
atom
(or

ion
or
com-
pound
), and an electron is
promoted
to a higher level.
Using
these concepts,
we
can
. calculate the
energy
change
in
volv
ed
in the electron transition.
The
energy
of
a photon is given
by
E=
hv
where h
repre
sents
Planck
's constant (6.63 X

10-
34
J . s) and v is the frequency
of
the radiation,
which is
5.00 X 10
14
S-
I for a
wa
velength
of
600
nm.
Here
E =
~,so
we
have
~
= hv
= (6.63 X
10-
34
J . s)(5.00 X
10
14
s)
= 3.32 X

10
-
19
J
This value is very small,
but
it is the energy absorbed
by
only one ion.
If
the wavelength
of
the
photon
absorbed by an ion lies outside the visible region, then the transmitted light looks
the
same
(to us) as
the
incident
light-white
-
and
the ion appears colorless.
The
be
st way to
mea
s
ure

crystal field splitting is to use spectroscopy to determine
the
wa
velength at which light is absorbed.
The
[Ti(H
2
0)6]
3+
ion
provides a straightforward exam-
ple, because Ti
3+
ha
s
only
one
3d
electron (
Figure
22.15).
The
[Ti(H
2
0)6]
3+
ion absorbs light
in
the visible region
of

the spectrum (Figure 22.16).
The
wavelength corresponding to
maximum
SECTION 22.3
Bonding
in
Coordination
Compounds: Crystal Field Theory 869
•
Photon
of
energy hv
~
DD
[JDD
400
500
[JD
•
DDD
(a)
600
700
Wavelength (nm)
(b)
absorption is 498
nm
[Figure 22.lS(b)].
To

calculate the crystal field splitting energy, we start by
writing
Next, recall that
~
= hv
e
v
=-
A
where e is the speed
of
light and A is the wavelength. Therefore,
A
he
(6.63 X
10-
34
J . s)(3.00 X
10
8
mls) -
19
I.l.
= = = 3.99 X 10 J
A (498
nm)(l
X 10-
9
mlnm)
This is the energy required to excite one [Ti(H

2
0)6]
3+
ion.
To
express this energy difference in the
more convenient units
of
kJ/mol, we write
~
= (3.99 X 10-
19
Jlion)(6.02 X 10
23
ionslmol)
= 240,000 J/mol
= 240 kJ/mol
Aided by spectroscopic data for a number
of
complexes, all having the same metal ion but
different ligands, chemists calculated the crystal field splitting for each ligand and established the
following
spectrochemical series, which is a list
of
ligands arranged in increasing order
of
their
abilities to split the
d orbital energy levels:
1- <

Br-
<
CI-
<
OH
- < F- < H
2
0 <
NH
3 < en <
CN
- < CO
These ligands are arranged in the order
of
increasing value
of
~.
CO and
CN-
are called strong-
field ligands, because they cause a large splitting
of
the d orbital energy levels.
The
halide ions and
hydroxide ion are
weak-field ligands, because they split the d orbitals to a lesser extent.
Magnetic Properties
The magnitude
of

the crystal field splitting also determines the magnetic properties
of
a com-
plex ion. The [Ti(H
2
0)6f
+ ion, having only one d electron, is always paramagnetic. However,
for an ion with several
d electrons, the situation is less immediately clear. Consider, for example,
the octahedral complexes [FeF
6
]3- and [Fe(CN)6]3- (Figure 22.17).
The
electron configuration
of
Fe
3+
is [Ar]3d
s
,
and there are two possible ways to distribute the five d electrons among
th
e
d orbitals. According to Hund's rule
[I
~~
Section
6.
8],
maximum stability is reached when the

Figure 22.15
(a)
The
process
of
photon absorption, and (b) a
graph
of
the absorption spectrum
of
[Ti(H
2
0)6l
H
.
The
energy
of
the
incoming photon is equal to the
crystal field splitting.
The
maximum
absorption peak in the visible region
occurs at 498
nm
.
Figure 22.16 Colors
of
some

of
the first-row transition metal ions
in
solutio
n.
From
left to right:
TiH,
Cr
3
+,
Mn
2+,
Fe
H
, C0
2
+ , Ni
2+
, Cu2+. The
3+ -+
Sc and
yO
ions are colorless.
•
870 CHAPTER
22
Coordination
Chemistry
Figure 22.17 Energy-level

diagrams for the
Fe
3+
ion
and
for the
[FeF
6
]3- and [Fe(CN)6]3- complex
•
IOns.
•
Figure 22.18 Orbital diagrams for
the high-spin and low-spin octahedral
complexes corresponding to the
electron configurations
of
(a) d
4
,
(b) d
5
,
(c) d
6
,
and (d) d
7
1 1
11 11

11
11
11
1
11
11 11
1
d
xy
fly
, d
Xl
d
2 2 d 2
X - y Z
IHIHI1
1
d
xy
d
yZ
d
xz
electrons are placed in five separate orbitals with parallel spins. This arrangement can be achieved
only at a cost, however, because two
of
the five electrons must be promoted to the higher-energy
d
i-
i and di orbitals. No such energy investment is needed

if
all five electrons enter the
dX}'
d
yz
'
and d
xz
orbitals. According
to
Pauli's exclusion principle
[
~~
Section
6.8], there will be only one
unpaired electron present
in
this case.
Figure 22
.1
8 shows the distribution
of
electrons among d orbitals that results in low- and
high-spin complexes. The actual arrangement
of
the electrons is determined by the amount
of
1
d 2 2
x

-y
d
z'
smaller
D.
d 2 '
d
z'
/ 1
1 1 1
x
-y.
1 1 1
d
4
larger
D.
d
ry
d
Yl
d
Xl
H 1 1
(a)
1 1
d 2 2 d '
x - y z·
d ' ,
d 2

/ 1
1 1
1 1
x

y
l
1 1 1
d
5
d\
y
d
Yl
d
xz
H H 1
(b)
1 1
d
X
2
-l
d
z'
d 2 '
d
z'
/ H 1
1 1 1

x
-y
.
H
1
1
d
6
d
X)
, d
yZ
d
xz
H H H
(c)
1
1 1
d 2 2
X
-y
d 2
l
d ' , d '
/ H
1~
~
7
1
1

x
y.
z·
H H
1
d
xy
d
yZ
d
xz
H H H
(d)
SECTION 22.3 Bonding in
Coordination
Compounds: Crystal Field Theory
871
stability gained by having maximum parallel spins versus the investment in energy required to
promote electrons to higher
d orbitals. Because
F-
is a weak-field ligand, the five d electrons enter
five separate
d orbitals with parallel spins to create a high-spin complex. The cyanide ion
is
a
strong-field ligand, though, so it is energetically preferable for all five electrons
to
be in the lower
orbitals, thus forming a low-spin complex. High-spin complexes are more paramagnetic than low-

spin complexes.
The
actual number
of
unpaired electrons (or spins) in a complex ion can be found by mag-
netic measurements, and in general, experimental findings support predictions based on crystal
field splitting. However, a distinction between low- and high-spin complexes can be made only
if
the metal ion contains more than three and fewer than eight d electrons,
as
shown in Figure
22.18. Sample
Problem 22.4 shows how to determine the number
of
spins in an octahedral
complex.
Predict the number
of
unpaired spins in the [Cr(en)3f + ion.
Strategy
The magnetic properties
of
a complex ion depend on the strength
of
the ligands. Strong-
field ligands, which cause a high degree
of
splitting among the d orbital energy levels, result in low-
spin complexes. Weak-field ligands, which cause only a small degree
of

splitting among the d orbital
energy levels, result in high-spin complexes.
Setup The electron configuration
of
Cr
2+
is
[Ar]3d
4;
and en
is
a strong-field ligand.
Solution Because en
is
a strong-field ligand, we expect [Cr(en)3f + to be a low-spin complex.
According to Figure 22.18, all four electrons will be placed in the lower-energy
d orbitals (d
X)
, d
yz
,
and d
xz
)
and there will be a total
of
two unpaired spins.
Practice Problem How many unpaired spins are in [Mn(H
2
0)6f

+? (Hint: H
2
0 is a weak-field
ligand.)
Tetrahedral and Square-Planar Complexes
So far we have concentrated on octahedral complexes.
The
splitting
of
the d orbital energy
levels in tetrahedral and square-planar complexes, though, can also
be
accounted for satisfac-
torily by the crystal field theory. In fact, the splitting pattern for a tetrahedral ion is
just
the
reverse
of
that for octahedral complexes.
In
this case, the d
xy>
d
yz
' and d
xz
orbitals are more
closely directed at the ligands and therefore have more energy than the
d
x

2
-l
and d
z
2
orbitals
(Figure 22.19).
Most
tetrahedral complexes are high-spin complexes. Presumably, the tet-
rahedral arrangement reduces the magnitude
of
the metal-ligand interactions, resulting in a
smaller
Ll
value. This is a reasonable assumption because the number
of
ligands is smaller in
a tetrahedral complex.
As Figure
22.20 shows, the splitting pattern for square-planar complexes is the most compli-
cated. The
d
x
2
-i orbital possesses the highest energy (as in the octahedral case), and the d
X)
, orbital
is the next highest. However, the relative placement
of
the di and the d

xz
and d
yZ
orbitals cannot be
determined simply by inspection and must be calculated.
/0
D D
/ d
ry
d
yZ
d
xz
/
/
[I
1(,
Crystal field splitting
L '-=
==,'==
-'=~~ J
,
,
,
,
"0
Think
About
It
It

is
easy to draw
the wrong conclusion regarding
high- and low-spin complexes.
Remember that the term
high
spin refers to the number
of
spins
(unpaired electrons), not
to
the
energy levels
of
the d orbitals. The
greater the energy gap between the
lower-energy and higher-energy
d
orbitals, the greater the chance that
the complex will be
low spin.
•
. Figure 22.19 Crystal field splitting
between
d orbitals in a tetrahedral
complex.
872
CHAPTER
22
Coordination

Chemistry
Figure 22.20 Energy-level
diagram for a square-planar complex.
Be
ca
use there are more than two energy
levels,
we
cannot
defi ne crystal field
splitting as we can for octahedral and
tetrahedral complexes.
•
22.3.1
Bonding in Coordination Compounds:
Crystal Field Theory
How
many
unpaired spins would you
expect
the
[Mn(
CO
)6
l
2+
ion to have?
a) 0
b)1
c) 2

d) 3
e) 5
22.3.2
Which
of
the following metal ions can
p
ote
ntia
ll
y form both l
ow
-spin and
high-sp
in
co
mple
xes? (Select all that
apply.)
a) Ti
2
+
b) Cu+
c)
Fe
2+
d) Ni
2+
e)
Cr

H
Reactions
of
Coordination Compounds
Complex ions undergo liga
nd
exchange (or substitution) reactions
in
solution. The rates
of
these
reactions vary widel
y,
depending on the nature
of
the metal ion and the ligands.
In studying ligand exchange reactions, it is often u
sef
ul
to distinguish between the stability
of
a complex ion and its tendency to react, which we call kinetic lability. Stability in this context
is a thermodynamic property, which is measured in terms
of
the species' formation constant K
f
[
~~
Section
17.5] . For exampl

e,
we s
ay
that the complex ion tetracyanonickelate(II) is stable
because it has a large
fo
rmation constant (K
f
= 1 X 10
30
):
By using
cy
anide ions labeled with the radioactive isotope carbon-14, chemists have shown that
[Ni
(CN)4]2- undergoes ligand exchange very rapidly in solution. The following equilibrium is
established almost
as
soon as the species are mixed:
where the asterisk denotes a
14
C atom. Complexes like the tetracyanonickelate(II) ion are termed
labile complexes because they undergo rapid ligand exchange reactions. Thus, a thermodynami-
cally
stable species (i.e., one that h
as
a large formation constant) is not necessarily unreactive.
A complex that is thermodynamically
unstable
in

acidic solution is [Co(NH
3
)6
]
3+
. The equi-
librium constant for the following reac
ti
on is about 1 X 10
20
:
When equilibrium is reached, the concentration
of
the [Co(NH
3
)6]
3+
ion is very
low.
This reaction
requires several days
to
complet
e,
however, because the [Co(NH
3
)6
]
3+
ion is so inert. This

is
an
example
of
an inert complex
-a
complex ion that undergoes very slow exchange reactions (on the
order
of
hours or even day
s)
.
It
shows that a thermodynamically unstable species is not necessarily
chemica
ll
y reactive. The rate
of
reaction is determined by the energy of activation, which is high
in this case.
Most compl
ex
ions containing Co
3+,
Cr
3+,
a
nd
Pt
2+

are kinetically inert. Because they
exchange ligands very slowl
y,
they are easy to study in solution. As a result, our knowledge
of
the
SECTION 22.S
Applications
of
Coordination
Compounds 873
bonding, structure, and isomerism
of
coordination compounds has come largely from studies
of
these compounds.
Applications
of
Coordination Compounds
Coordination compounds are found in living systems and have many uses in the home, in industry,
and in medicine.
We
briefly describe a few examples in this section.
Metallurgy
The extraction of silver and gold by the formation
of
cyanide complexes and the purification
of
nickel by converting the metal to the gaseous compound Ni(CO)4 are typical examples
of

the use
of
coordination compounds in metallurgical processes.
Chelation
Therapy
Earlier we mentioned that chelation therapy is used in the treatment of lead poisoning. Other met-
als, such
as
arsenic and mercury, can also be removed using chelating agents.
Chemotherapy
Several platinum-containing coordination compounds, including cisplatin [Pt(NH3)2
CI2J
and
carboplatin
[Pt(NH3
MOCO)
2
C4H
6
J,
can effectively inhibit the growth
of
cancerous cells. The
mechanism for the action
of
cisplatin is the chelation
of
DNA, the molecule that contains the
genetic code. During cell division, the double-stranded DNA unwinds into two single stra
nd

s,
which must be accurately copied in order for the new cells to be identical
to
their parent cell.
X-ray studies show that cisplatin binds to DNA by forming cross-links in which the two chlorides
on cisplatin are replaced by nitrogen atoms in the adjacent guanine bases on the same strand
of
the DNA. (Guanine is one
of
the four bases in DNA
[
~~
Section
10.6,
Figure
10.1SJ
.)
This
causes a bend
in
the double-stranded structure at the binding site. It is believed that this struc-
tural distortion is a key factor in inhibiting replication. The damaged cell is then destroyed by
the body's immune system. Because the binding
of
cisplatin to DNA requires both Cl atoms to
be on the same side
of
the complex, the trans isomer
of
the compound is totally ineffective as an

anticancer drug.
Cisplatin
Chemical Analysis
Although EDTA has a great affinity for a large number of metal ions (especially
2+
and
3+
ion
s),
other chelates are more selective in binding. Dimethylglyoxime, for example, forms an insoluble
brick-red solid with
Ni2+
and an insoluble bright-yellow solid with Pd
2
+. These characteristic
colors are used
in
qualitative analysis to identify nickel and palladium. Furthermore, the quanti-
ties
of
ions present can be determined by gravimetric analysis
[
~~
Section
4.6J
as
follows:
To
a
solution containing Ni

2
+ ions, say, we add
an
excess
of
dimethylglyoxime reagent, and a brick-red
precipitate forms. The precipitate is then filtered, dried, and weighed. Knowing the formula
of
the complex (Figure 22.21), we can readily calculate the amount
of
nickel present in the original
solution.
,
Figure 22.21 Structure
of
nickel
dimethylglyoxime. Note that the overall
structure is stabilized by hydrogen
bonds.
874 CHAPTER
22
Coordination Chemistry
,
.
~
Multimedia
Organic and Biochemistry-oxygen binding
in
hemoglobin.
Detergents

The cleansing action
of
soap in hard water is hampered by the reaction
of
the
Ca2+
ions in the
water with the soap molecules to form insoluble salts or curds. In the late
1940s the detergent
industry introduced a
"builder
,"
usually sodium tripolyphosphate, to circumvent this problem. The
tripolyphosphate ion is an effective chelating agent that forms stable, soluble complexes with
Ca
2
+ ions. Sodium tripolyphosphate revolutionized the detergent industry. Because phosphates
are plant nutrients, howeve
r,
wastewater containing phosphates discharged into rivers and lakes
causes algae to grow, resulting in oxygen depletion.
Under these conditions, most
or
all aquatic
life eventually succumbs. This process is called
eutrophication. Consequently, many states have
banned phosphate detergents since the 1970s, and manufacturers have reformulated their products
to eliminate phosphates.
Sequestrants
In addition to its use in medicine and chemical analysis, EDTA is used as a food additive to

sequester metal ions.
EDT
A sequesters copper, iron, and nickel ions that would otherwise catalyze
the oxidation reactions that cause food to spoil. EDTA is a common preservative in a wide variety
of
consumer products.
Bringing Chemistry
to
life
The Coordination Chemistry
of
Oxygen Transport
Because
of
its central function as an oxygen carrier for metabolic processes, hemoglobin is prob-
ably the most studied
of
all the protein
s.
The molecule contains four folded long chains called sub-
units. Hemoglobin carries oxygen in the blood from the lungs to the tissues, where it delivers the
oxygen molecules to myoglobin. Myoglobin, which
is
made up
of
only one subunit, stores oxygen
for metabolic processes in the muscle.
The porphine molecule forms an important part
of
the hemoglobin structure. Upon coordi-

nation to a metal, the H+ ions that are bonded to two
of
the four nitrogen atoms in porphine are
displaced. Complexes derived from porphine are called porphyrin
s,
and the iron-porphyrin com-
bination is called the heme group. The iron in the heme group has an oxidation state
of
+2;
it is
coordinated to the four nitrogen atoms in the porphine group and also to a nitrogen donor atom in a
ligand that is attached to the protein. The sixth ligand
is
a water molecule, which binds to the
Fe
2
+
ion on the other side
of
the ring to complete the octahedral complex. This hemoglobin molecule
is called deoxyhemoglobin and imparts a bluish tinge to venous blood. The water ligand can be
replaced readily by molecular oxygen (in the lungs) to form red oxyhemoglobin, which is found
in arterial blood. Each subunit contains a heme group, so each hemoglobin molecule can bind up
to four
O
2
molecules.
>r N, N ,f
Fe
,

)==N
N - ,<
HN {
Porphine
Protein
I
;
!
I
APPLYING
WHAT
YOU'VE LEARNED
Applying
What
You've Learned
Elevated
BLL
and other heavy metal poisoning can be treated with one of several chelat-
ing agents, including DMSA and EDTA. EDTA is administered intravenously
as
either
the sodium salt (Endrate) or
as
the calcium disodium salt (Versenate). Endrate is not
approved for the treatment
of
lead poisoning because
of
its high affinity for calcium.
It

is approved, however, for treating hypercalcemia, a condition in which there is excess
calcium in the blood usually
as
a result
of
bone cancer. The accidental use
of
Endrate
during treatment for lead poisoning resulted in the death of a 2-year-old girl in February
of
2005. The girl's death was attributed to sudden cardiac arrest caused by the removal
of
too much calcium from her blood.
Chelation therapy works by the administration
of
a ligand, which binds to metal
ions already in the body. Many drugs, including cisplatin
[
~.
Chapter
3,
Applying
What
You've
Learned],
are themselves coordination compounds in which the cen-
trill metal ion binds to electron-rich sites (such
as
oxygen or sulfur atoms) in biological
molecules.

Problems:
a) Determine the oxidation state
of
platinum in cisplatin, [Pt(NH
3
hCI
2
J.
[I ••
Sample
Problem
22.1]
b) Give the systematic name for cisplatin.
[
~.
Sample
Problem
22.2]
c) Write the formula for the compound potassium hexachloroplatinate(IV).
[
~.
Sample
Problem
22.3]
•
,
875
876
CHAPTER
22

Coordination
Chemistry
CHAPTER SUMMARY
Section 22.1
•
Coordination
compounds
contain coordinate covalent bonds between
a metal ion (often a transition metal ion) and two or more polar
molecules or ions.
• The molecules or anions that s
UlTound
a me
tal
in a coordination
compl
ex
are called ligands. I
• Many coordination compounds consist
of
a complex ion and a
counter
.
Ion.
• Trans
iti
on metals are those that have incompletely filled d subshell
s-
or that
give

ri
se
to ions with incompletely filled d subshell
s.
• Transition metals exhibit variable oxidation states ranging f
rom
+ 1 to
+7.
•
To
act as a ligand, a inolecule or ion must have at least one unshared
pair
of
electrons. The atom that bears the unshared pair
of
electrons is
the
donor
atom.
• Ligands are cla
ss
ified as
monodentate
, bidentate, or
pol
ydentate,
based on the number
of
donor atoms they contain. Bidentate and
polydentate ions are also known as

chelating
agents.
• The
coordination
number
is the number
of
donor atoms sUlTounding a
metal
in
a complex.
• Ionic coordination compounds are named by first naming the cation
and then the anion. Complex ions are named by listing the ligands
in
alphabeti
ca
l order, followed by the metal and its oxidation state (as
a Roman numeral
).
When the complex ion is the anion, the anion's
name ends in
-ate.
Section 22.2
• The coordination number largely determines the geometry
of
a
coordination complex.
• Coordination compounds containing different alTangements
of
the

same
li
gands are stereoisomers. The two types
of
stereoisomerism are
geometric and optical.
•
Geometric
isomers
contain the same atoms and bonds alTanged
differently in space.
!(EY
WORDS
Chelating agent,
861
Coordination compound, 858
Coord
in
ation number, 861
Crystal field splitting
(
/1),
867
Dextrorotatory, 866
Donor atom,
860
Enantiomers, 866
Geometric isomers, 865
QUESTIONS AND PROBLEMS
•

Optical
isomers
are nonsuperimposable min·or image
s.
We call a
pair
of
optical isomers
enantiomers.
The rotation
of
polarized
li
ght is
measured with
apolarimeter.
• Enantiomers rotate the plane
of
plane-polarized light in opposite
directions. The enantiomer that rotates it to the right is called
dextrorotatory and is labeled
d.
The enantiomer that rotates it to the
left is
ca
lled levorotatory and
is
labeled
l.
An equal mixture

of
a pair
of
enantiomers, called a
racemic
mixture,
does not cause any net
rotation
of
plane-polarized light.
Section 22.3
•
•
•
•
•
Ligands in a coordination compl
ex
cause the energy levels
of
the d
orbitals on a metal to split. The difference in energy between the lower
and higher
d orbital energy levels is called the crystal
field
splitting
(Ll).
The magnitude
of
/1

depends on the nature
of
the ligands in the
complex. The
spectrochemical
series orders some common
li
gands in
order
of
increa
si
ng field strength.
Strong-field
ligands give
ri
se to a larger
/1
value; weak-field
li
gands
yield a smaller
/1
value.
Crystal field splitting sometimes changes the number
of
unpaired
electron
s,
and therefore the magnetic properties,

of
a metal.
Complexes containing trans
iti
on metals with d
4
,
d
S
,
d
6
,
or d
7
configurations may be high spin or low spin. In high-spin complexes,
the number
of
unpaired electrons is maximized because
/1
is small;
in
low-spin complexes, the number
of
unpaired electrons is minimized
because
/1
is large.
Section 22.4
• Complex ions undergo ligand exchange

in
solution.
The
rate at
which ligand exchange occurs is a measure
of
a complex
's
kinetic
lability and does not necessarily correspond directly to the complex's
thermodynamic
stability.
Section 22.5
• Coordination chemistry is important
in
many biological, medical, a
nd
industrial processes.
Levorotatory, 866
Ligand,
860
Racemic mixture, 866
Spectrochemical series, 869
Stereo isomers, 865 Optical isomer
s,
865
Polarimeter, 866
============
==============~
Section 22.1: Coordination Compounds

Review Questions
22
.1
What
di
stinguishes a transition metal from a main group metal?
22.2
Why is zinc not considered a transition metal?
22.3
22.4
Explain why atomic radii decrease very gradually from scandium
to copper.
Without referring to the text, write the ground-state electron
configurations
of
the first-row transition metals. Explain any
irregularitie
s.
•
22.5
22.6
22.7
22.8
22.9
Write
the electron configurations
of
the following
ion
s: V

S
+,
C
3+
M
2+
F 3+ C 2+
SC
3
+
TI·
4
+
f ,
il,e,U,
,.
Why
do transition metals have
more
oxidation states than
other
elements?
Give the highest oxidation states for scandium to copper.
Define the following terms:
coordination compound, ligand,
donor atom, coordination number, chelating agent.
Describe the interaction
between
a
donor

atom and a metal atom
in terms
of
a Lewis acid-base reaction.
Problems
22.10
22.11
Complete
the following statements for the
complex
ion
[Co(enMH
2
0)CN]
H (a) en is
the
abbreviation for
__
(b)
The
oxidation
number
of
Co
is . (c)
The
coordination
number
of
Co is . (d) is a bidentate ligand.

Complete
the following statements for the
comple
x ion
[Cr(C
2
0
4
MH
2
0)
2
r.
(a)
The
oxidation
number
of
Cr
is
__
(b)
The
coordination
number
of
Cr
is . (c)
__
is a

bidentate ligand.
22.12
. Give the oxidation numbers
of
the metals in the following species:
(a) K
3
[Fe(CN)6]
(b) K
3
[Cr(C
2
0
4
)3]
(c) [Ni(CN)
4f

22.13 Give
the
oxidation
number
s
of
the metals in the following
species: (a)
Na
2
Mo0
4,

(b)
MgW0
4
, (c)
Fe(CO)
s.
22.14
What
are
the
systematic
name
s for the following
ion
s and
compound
s?
(a)
[Co(NH
3)4
Cl
2t
(b) [Cr(NH
3)3C
I3]
(c)
[Co(enhBr
2t
(d)
[Co(NH

3
)6]CI
3
22.15
What
are the systematic names for the following ions and
compound
s?
(a) [cis-Co(en)2Cl2]+
(b)
[Pt(NH
3
)s
CI]CI
3
(c)
[Co(NH
3
)sCl]CI
2
22.16
Write
the formulas for each
of
the following ions
and compounds: (a) tetrahydroxozincate(II),
(b) pentaaquochlorochromium(III) chloride, (c)
tetrabromocuprate(II), (d) ethy lenediaminetetraacetatoferrate(II).
22.17
Write

the
formulas for each
of
the following ions and
compounds: (a) bis(ethylenediamine)-dichlorochromium(III),
(b) pentacarbonyliron(O), (c)
pota
ss
ium
tetracyanocuprate(
II
),
(d) tetraammineaquochlorocobalt(III) chloride.
Section 22.2: Structure
of
Coordination Compounds
Review Questions
22.18 Define the following terms: stereoisomers, geometric isomers,
optical isomers, plane-pola
ri
zed light.
22.19 Specify which
of
the following structures can exhibit geometric
isomerism: (a) linear, (b) square planar, (c) tetrahedral, (d) octahedral .
22.20
What
determines
whether
a molecule is chiral?

How
does a
polarimeter measure the
ch
irality
of
a molecule?
22.21
Explain
the following terms: enantiomers, racemic mixtures.
QUESTIONS
AND
PROBLEMS 877
Problems
22.22
The
complex ion [Ni(
CN
)2
Br
2]2-
ha
s a square-planar geometry.
Dra
w
the
structures
of
the
geometric

isomer
s
of
this complex.
22.23
How
man
y geometric iso
mer
s are in the following species:
(a)
[Co(NH3)2CI
4r,
(b) [Co
(N
H
3)3
Cl3]?
22.24
Dra
w structures
of
all the geometric and optical isomers
of
each
of
the follow
ing
cobalt complexes:
(a)

[Co(NH
3
)6]
H
(b) [Co(NH
3
)sC
l]
2+
(c)
[CO
(C
2
0
4
)3]
3-
22.25
Dra
w structures of all the geometric and optical isomers
of
each
of
the following cobalt complexes: (a) [Co
(N
H
3)4
Cl2t ,
(b) [Co(
en

)3
]H
Section 22.3: Bonding in Coordination Compounds:
Crystal Field Theory
Review Questions
22.26 Briefly desc
ribe
crystal field theory.
22.27 Define the following terms:
crystal field splitting, high-spin
complex, low-spin complex, spectrochemical series.
22.28
What
is
the
origin
of
color
in
a coordination
compound?
22.29
Compounds
containing the Sc
H
ion are colorless, whereas those
containing the TiH ion are colored. Explain.
22.30 For the
same
ty

pe
of
ligand
s, explain
why
the
crystal field
splitting for an octahedral complex is always greater than that for
a tetrahedral complex.
Problems
22.31 Transition metal complexes containing
CN-
ligands are often
ye
llow in color, whereas those
con
taining H
2
0 ligands are often
green
or blue. Explain.
22
.32 Predict the
number
of
unpaired electrons
in
the following
complex
ion

s:
(a) [Cr(CN)6]4- , (b) [Cr(H
2
0
)6
]H
22.33
The
absorption
ma
x
imum
for the complex ion
[Co(NH
3
)6]H
occurs at
470 nm. (a)
Predict
the
color
of
the complex, and
(b) calculate
the
crystal field splitting
in
kl
lmo!.
22.34

From
each
of
the following pairs, choose the complex that absorbs
light at a longer wavelength: (a) [Co(NH
3
)6]2
+, [Co(H
2
0)6]
2+
;
(b)
[FeF
6
]3
-,
[Fe(C
N)6
]3
-;
(c) [Cu(NH
3
)4]2+,
[CuCI
4
]2

22.35 A solution
made

by dissolving 0.875 g
of
Co(NH3)4CI3 in 25.0 g
of
water freezes at - 0.S6°
C.
Calculate the number
of
moles
of
ions
produced when 1 mole
of
Co(NH
3
)4
Cl3 is dissolved in water, and
suggest a structure for the complex ion present in this compound.
22.36
In
biological system
s,
the
CU2 ~
ions
coo
rdinated with S atoms
tend to fo
rm
tetrahedral complexes, whereas those coordinated

with N atoms te
nd
to fo
rm
octahedral complexes. Explain.
22.37
Pla
stoc
ya
nin, a copper-containing protein found in
photo
synthetic
sys
tem
s, is involved in electron trans
port
, with the
copper
ion
switching
bet
wee
n the + 1 and
+2
oxidation states.
The
copper
ion is coordinated with t
wo
hi

stidine residues, a cysteine residue,
and a methionine res
idue
in
a tetrahedral configuration.
How
does
the crystal field splitting
(Ll)
c
hange
between
these two oxidation
states?
878 CHAPTER
22
Coordination
Chemistry
Section 22.4: Reactions
of
Coordination Compounds
Review Questions
22
.38 Define the terms labile complex and inert complex.
22
.39 Explain why a thermodynamically stable species may be
chemically reactive and a thermodynamically unstable species
may be unreactive.
Problems
•

22.40 Oxalic acid (H
Z
C
2
0
4
) is sometimes used to clean rust stains
from sinks and bathtubs. Explain the chemistry underlying this
cleaning action.
22.41
The
[Fe(CN)6]
3-
complex is more labile than the [Fe(CN)6]4-
complex. Suggest an experiment that would prove that
[Fe(CN)
6f
- is a labile complex.
22.42 Aqueous copper(II) sulfate solution is blue in color.
When
aqueous potassium fluoride is added, a green precipitate is
formed. When aqueous potassium chloride is added instead, a
bright-green solution is formed. Explain what is happening in
these two cases.
22.43 When aqueous potassium cyanide is added to a solution
of
copper(II)
sulfate, a white precipitate, soluble in
an
excess

of
potassium
cyanide, is formed. No precipitate is formed when hydrogen sulfide
is
bubbled through the solution at this point. Explain.
22.44 A concentrated aqueous copper(II) chloride solution is bright
green in color.
When
diluted with water, the solution becomes
light blue. Explain.
22.45
In
a dilute nitric acid solution,
Fe
3+
reacts with thiocyanate ion
(SCN- ) to form a dark-red complex:
[Fe(H
z
O)6]
3+
+
SCN
- . • H
2
0 + [Fe(H
z
O)sNCS]2+
The
equilibrium concentration

of
[Fe(HzO)5NCS]
2+
may be
determined by how darkly colored the solution is (measured
by a spectrometer).
In one such experiment, 1.0
mL
of
0.20 M
Fe(N0
3
)3
was mixed with 1.0
mL
of
1.0 X
10-
3
M KSCN and
8.0
mL
of
dilute
HN0
3
.
The
color
of

the solution quantitatively
indicated that the
[Fe(HzO)sNCSj"+ concentration was 7.3 X
10
-
5
M.
Calculate the formation constant for [Fe(H20)5NCS]
z+
.
Section 22.5: Applications
of
Coordination Compounds
Review Question
22.46 Describe and give examples
of
the applications
of
coordination
compounds.
Additional Problems
22.47 Suffocation victims usually look purple, but a person poisoned by
carbon monoxide often has rosy cheeks. Explain.
22.48 As
we
read across the first-row transition metals from left to
right, the
+2
oxidation state becomes more stable in comparison
with the + 3 state.

Why
is this so?
22.49 Which is a stronger oxidizing agent in aqueous solution,
Mn
3+
or
Cr
3+?
Explain your choice.
22.50 Carbon monoxide binds to
Fe
in hemoglobin some 200 times
more strongly than oxygen. This is the reason why
CO is a toxic
substance.
The
metal-to-ligand sigma bond is formed by donating
a lone pair from the donor atom to an empty
Sp3
d
2
orbital on Fe.
(a)
On the basis
of
electronegativities, would you expect the C or
o atom to form the bond to
Fe?
(b) Draw a diagram illustrating
the overlap

of
the orbitals involved in the bonding.
22.51
What
are the oxidation states
of
Fe
and Ti in the ore ilmenite
(Fe
Ti0
3
)?
(Hint:
Look
up the ionization energies
of
Fe
and
Ti
in
Table 22.1; the fourth ionization energy
of
Ti is 4180
kJ
/mol.)
22.52 A student has prepared a cobalt complex that has one
of
the
following three
st

ructures: [Co(NH
3
)6]CI
3
,
[Co(NH
3
)sCl]Cl
z
,
or
[CO(NH3)4ClZ]CI. Explain how the student would distinguish
between these possibilities by
an
electrical conductance
experiment.
At
the student's disposal are three strong
electrolytes- NaCl,
MgCl
z
,
and
FeCl
3
-which
may be used for
22.53
22.54
22.55

22.56
22.57
22.58
•
companson
purposes.
Chemical analysis shows that hemoglobin contains
0.34 percent
of
Fe
by mass.
What
is the minimum possible molar mass
of
hemoglobin? The actual molar mass
of
hemoglobin is about
65,000 g. How do you account for the discrepancy between your
minimum value and the actual value?
Explain the following facts: (a) Copper and iron have several
oxidation state
s,
whereas zinc has only one. (b) Copper and
ir
on
form colored ion
s,
whereas zinc does not.
A student in 1895 prepared three coordination compounds
containing chromium, with the following properties:

Formula
(a) CrCl
3
.
6H
z
O
(b) CrCl
3
.
6H
z
O
(c) CrCl
3
. 6H
2
0
Color
Violet
Light green
Dark
green
Cl
- Ions in Solution
per Formula
Unit
3
2
I

Write modem
fOIIDulas
for these compound
s,
and suggest a method
for confirming the number
of
Cl- ions present in solution in each
case.
(Hint: Some
of
the compounds may exist
as
hydrates.)
The formation constant for the reaction Ag + + 2NH3
~.=='
[Ag(NH
3
)z
]+ is
1.5
X 10
7
,
and that for the reaction Ag + +
2CN
- . • [Ag(CN)
zr
is
1.0 X

IO
Z1
at 25°C (see Table 17.5).
Calculate the equilibrium constant and
IlGo at 25°C for the reaction
[Ag(NH
3
)2]
+ +
2CW.
• [Ag(CN)
zr
+ 2NH3
From
the standard reduction potentials listed
in
Table 19.1
for ZnlZn
2
+ and Cu/Cu
2+
, calculate IlGo and the equilibri
um
constant for the reaction
Using the standard reduction potentials listed in Table 19.1 and
the
Handbook
of
ChemistlY
and

Physics, show that the following
reaction is favorable under standard-state conditions:
2Ag(s) +
Ptz+(aq)

2Ag
+(aq) + Pt(s)
What
is the equilibrium constant
of
this reaction at 25°
C?
22.59 The Co
2+
-porphyrin complex is more stable than the
Fe
z
+_
porphyrin complex. Why, then, is iron the metal ion in
hemoglobin (and other heme-containing proteins)?
22.60
What
are the differences between geometric isomers and optical
isomers?
22.61 Oxyhemoglobin is bright red, whereas deoxyhemoglobin is
purple. Show that the difference in color can be accounted for
qualitatively on the basis
of
high-spin and low-spin
comp

lexes.
(Hint: Oz is a strong-field ligand. See the Bringing Chemistry to
Life box on page 874.)
22.62 Hydrated
Mn
2+
ions are practically colorless (see Figure 22.16)
even though they possess five
3d
electrons. Explain. (Hint:
Electronic transitions in which there is a change in the number
of
unpaired electrons do
not
occur readily.)
22.63
22.64
22.65
22.66
22.67
22.68
22.69
Which
ofthe
following hydrated cations are colorless: Fe
2+
(aq),
Zn
2+
(aq),

Cu
+(aq),
Cu
2+
(aq), y
5+
(aq),
Ci
+(aq),
C0
2
+(
aq)
,
Sc
3+
(aq), Pb
2+
(aq)? Explain your choice.
Aqueous solutions
of
CoCl
2
are generally either light pink:
or
blue. Low concentrations and low temperatures fav
or
the
pink
form, whereas high concentrations and high temperatures fav

or
the blue form. Adding hydrochloric acid to a pink: solution
of
CoCl
2
causes the solution to turn blue; the pink: color is restored
by the addition
of
HgCI
2
.
Account for these obs
er
vations.
Sugge
st a method that would allow you to distinguish between
cis-Pt(NH3)2
CI
2 and trans-Pt(NH
3
)z
CI
2
.
You are given two solutions containing FeCl
2
and FeCI
3
at the
same

concentration.
One
solution is light yello
w,
and the other
one is brown. Identify these solutions
ba
s
ed
only on color.
The
la~el
of
a certain brand
of
mayonnaise lists EDTA as a
food preservative.
How
does EDTA prevent the spoilage
of
mayonnaise?
The
compound
1,1
,l-trifluoroacetylacetone (tfa) is a bidentate
ligand:
o
It
forms a tetrahedral complex with
Be

2
+ and a square-planar
complex with Cu
2+
.
Draw
structures
of
these complex ions,
and
identify the type
of
isomerism exhibited by these ions.
How
many geometric isomers
can
the following square-planar
complex have?
22.70
22.71
22.72
22.73
ANSWERS TO IN-CHAPTER MATERIALS 879
[Pt(NH
3)2
CI2] is found to exist
in
two geometric isomers
designated I and II, which react with oxalic acid as follows:
I +

H
2
C
2
0
4
•
[Pt(
NH
3
)z
C
2
0
4
]
Comment
on
the structures
of
I and II.
Commercial silver-plating operations frequently use a
solution containing the complex Ag(
CN
)2
ion. Because the
formation constant
(K
f
)

is quite
la
r
ge
, this procedure ensures
that the free
Ag
+ concentration
in
solution is low for uniform
electrodeposition.
In
one process, a chemist added 9.0 L
of
5.0 M
NaC
N to 90.0 L
of
0.20 M
AgN0
3
.
Calculate the concentration
of
free Ag +
ion
s at equilibrium.
See
Table 17.5 for K
f

value.
Dra
w qualitative diagrams for the
cr
ystal field splittings in (a) a
linear complex ion
MLz , (b) a trigonal-planar complex ion
ML
3
,
and (c) a trigonal-bipyramidal complex ion
ML
s.
(a) The free Cu(I) ion is unstable in solution and has a tendency
to dispropOt1ionate:
2Cu
+(
aq
) ( •
Cu
2+
(aq) + Cu(s)
Use the information in Table 19.1 (
page
767) to calculate the
equili
brium
con
stant for the reaction. (b)
Based

on y
our
results in
part (a), explain why
mo
st Cu(I) compounds are insoluble.
22.74 Consider the following two ligand exchange reactions:
[Co(H
2
0 )6]
3+
+
6NH
3 ( • [Co(
NH
3
)6]
3+
+
6H
2
0
(a) Which
of
the reactions should have a larger
t:.S
O? (b) Given
that the
Co-N
bond strength is approximately the

same
in
both complexes, which reaction will
ha
ve a larger equilibrium
constant? Explain your choices.
PRE-PROFESSIONAL PRACTICE
EXAM
PROBLEMS:
PHYSICAL
AND
BIOLOGICAL SCIENCES
The
following questions are not based on a passage.
l.
What
are the correct names for the complex ions [Ni(H
2
0
)6f
+ and
[Ni(NH
3
)6]
2+
?
a) Hexaaquonickel(YI) ion and hexaamminenickel(YI)
ion
b) Nickel(II) hexaaquo ion and nickel(II) hexaammine
ion

c)
He
xaaquonickel(II) ion and hexaamminenickel(II)
ion
d)
Hexaaquonickelate
and
hexaamminenickelate
2.
One
of
the complex ions
in
Question 1 is green, and the other is
purple.
Based
on the spectrochemical series, which is which?
1- <
Br
- < CI- <
OH
- < F- < H
2
0 <
NH
3 < en <
CN
- <
CO
a) [Ni(H

2
0
)6
f + is green and [Ni(
NH
3
)6
]
2+
is purple.
b) [Ni(H
2
0
)6
]
2+
is purple and [Ni(NH
3
)6
]
2+
is green.
c) Both ions should
be
the s
ame
color because they contain the same
metal.
d) There is not enough information to determine which is which.
3.

Ho
w many geometric is
omer
s are
po
ssible for the complex ion
[Ni(H
2
0 M
NH
3
)z
f +?
a) l b) 2 c) 4
d)6
4.
How
man
y ions
in
solution would
re
sult from dissolving
[Ni(H
2
0 )4(NH
3
)2]CI
2
?

a) 2 b
)3
c) 8 d)
11
ANSWERS TO
IN-CHAPTER
MATERIALS
•
Answers
to
Practice Problems
22.1 (a)
K:
+l,Au:
+ 3, (b)
K:
+
l,Fe:
+ 2.
22.2 (a) tetraaquodichlorochromium(III) chloride,
(b)
tris( ethy lenediamine
)chromium
(III) chloride.
22.3 [Co(en)3
lz(S0
4H 22.4
5.
Answers
to

Checkpoints
22.1.1
b.
22.1.2
d.
22.1.3
b.
22.1.4
c.
22.3.1
b.
22.3.2
c.
Answers
to
Applying
What
You've
learned
a) + 2. b) cis-Diamminedichloroplatinum(II). c) K
2
[PtCI
6
].
23.1
23.2
•
•
•
•

•
23.3
•
•
23.4
23.5
23.6
•
•
23.7
Occurrence
of
Metals
Metallurgical Processes
Preparation
of
the Ore
Production
of
Metals
The Metallurgy
of
Iron
Steelmaking
Purification
of
Metals
Band Theory
of
Conductivity

Conductors
Semiconductors
Periodic Trends in
Metallic Properties
The
Alkali Metals
The
Alkaline Earth
Metals
Magnesium
Calcium
Aluminum
eta
ur
~
•
emlstr
t e
eta
s
The
Importance
of
Metals
in
Human
Biology
Calcium, potassium, and sodium are the three most abundant metals in the human body.
Present in a much smaller amount, copper is nonetheless vitally important to human
health. Copper-dependent enzymes known

as
cuproenzymes
playa
critical role in a host
of
biochemical processes, including cellular energy production, connective tissue for-
mation, and reactions essential to normal functioning
of
the brain and central nervous
system. Two hereditary diseases that involve errors
of
copper metabolism are Menkes
disease and Wilson's disease.
In Menkes disease, copper is not absorbed in the intestine and is not available for dis-
tribution throughout the body. Consequently, cuproenzymes such as
cytochrome c oxi-
dase, lysyl oxidase,
and dopamine beta hydroxylase cannot function normally. Sufferers
exhibit severe developmental delay; subnormal body temperature; kinky, steel-colored
hair; seizures; and ultimately the degeneration
of
muscle, bone, and organs. Baby boys
born with the most severe form
of
Menkes disease typically do not live beyond their
third year
of
life.
Wilson's disease causes the body
to

retain copper. Although people with Wilson's dis-
ease are able to absorb copper in the intestine, the liver
of
a person with Wilson's disease
does not release copper into the bile as it should. Eventually, damaged by the buildup of
copper, the liver releases copper directly into the blood, where it is carried throughout
the body. Excess copper causes damage
to
the kidneys, brain, and eyes. Unlike Menkes
disease, Wilson's disease can be managed and treated. Sufferers must follow a strict low-
copper diet, avoiding such foods
as
mushrooms, nuts, and chocolate. In addition, they
must undergo lifelong periodic chelation therapy.
Menkes and Wilson's diseases illustrate the importance
of
metals
to
living systems.
Menkes
disease
is
an
X-
linked
recessive
condition
and
nearly
all

victims
are
ma
l
e.
In This Chapter, You Will Learn how some metals are produced and about their chemistry.
Before you begin, you should review
• Periodic trends and metallic character
[
~~
Section
7.4
]
• Electrolysis
[
~~
Sect
ion 19.7]
In 1943, because copper was n
ee
ded for the war effort, pe
nni
es were
made
from steel. A very small number
of
copper penni
es
were made in
1943, forty

of
which are
known
to still exist. The copper penny in this
photograph
is
a counterfeit. However, according to the United States
Mint, genuine copper
1943 pennies have sold for
as
much
as
$82,500
.
::::::.,
'

•
Media
Player/
MPEG
Content
Chapter
in
Review
I
881
882 CHAPTER 23
Metallurgy
and

the
Chemistry
of
Metals
•
Figure 23.2 Magnesium nodules
on the ocean floor.
Type
Uncombined metals
Carbonates
Halides
Oxides
Phosphates
Silicates
Sulfides
Sulfates
IA
I
Li
Na
K
Rb
Cs
2A
2
Be
Mg
Ca
Sr
Ba

Minerals
Ag, Au, Bi, Cu, Pd,
Pt
BaC0
3
(witherite),
CaC0
3
(calcite, limestone),
MgC0
3
(magnesite),
CaC0
3
•
MgC0
3
(dolomite),
PbC0
3
(cerussite),
ZnC0
3
(s
mithsonite)
CaF
2
(fluorite), N aCl (halite), KCI (sylvite),
Na
3AlF6 (cryolite)

Al
2
0
3
•
2H
?O (bauxite),
A1
2
0
3
(corundum), Fe203 (hematite),
Fe
30 4 (magnetite),
CU
20 (cuprite),
Mn02
(pyrolusite),
Sn02
(cassiterite), TiO? (rutile), ZnO (zincite)
Ca
3(
P0
4)?
(phosphate rock), CaS(P0
4)3
0H
(hydroxyapatite)
Be
3Al2Si60

1S
(beryl),
ZrSi0
4
(zircon),
NaAISi
3
0
s
(albite),
Mg
3
(Si
4
0
IO
)(OH)2 (talc)
Ag?S (argentite), CdS (greenockite),
CU2S
(chalcocite), FeS2
(pyrite), HgS (cinnabar),
PbS (galena), ZnS (sphalerite)
BaS0
4 (barite),
CaS04
(anhydrite),
PbS0
4
(anglesite),
SrS04

(celestite
),
MgS0
4 .
7H
2
0
(epsomite)
Sulfides
} j
Chlorides
Uncombined
} j
Other
compounds;
' '
see
caption
Oxide
s
3B
4B
5B
6B 7B
J;-
8B
~
IB
3 4
5

6 7 8
91011
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
La
Hf
Ta
W
Re Os
iT
Pt
Au
.
2B

12
Zn
Cd
Hg
3A
4A
SA
6A 7A
13
14
15
16
17
Al
Ga
In
Sn
Tl
Pb
Bi
8A
18
Figure 23.1 Metals and their best-known minerals. Lithium
is
found in spodumene (LeAISi
2
0
6
), and
beryllium in beryl

(s
ee Table 23.1). The rest
of
the alkaline earth metals are found in minerals that are
carbonates and sulfate
s.
The minerals for Sc,
Y,
and La are the phosphates. Some metals have more than
one type
of
important mineral. For example,
in
addition to the sulfide, iron is found
as
the oxides hematite
(F~0
3
)
and magnetite (Fe304); and aluminum, in addition
to
the oxide, is found in beryl (Be3AI2Si601S)'
Technetium (Tc) is a synthetic element.
Occurrence
of
Metals
Most metals come from minerals. A mineral is a naturally occurring substance with a range
of
chemical composition. A mineral deposit concentrated enough to allow economical recovery
of

a desired metal is known as ore. Table 23.1 lists the principal types
of
minerals, and Figure 23.1
shows a classification
of
metals according to their minerals.
The
most abundant metals, which exist as minerals in Earth's crust, are aluminum, iron, cal-
cium, magnesium, sodium, potassium, titanium, and manganese (see page 45). Seawater is a rich
source
of
some metal ions, including
Na
+,
Mg
2+,
and Ca
H
.
Furthermore, vast areas
of
the ocean
floor are covered with
manganese nodules, which are made up mostly
of
manganese, along with
iron, nickel, copper, and cobalt in a chemically combined state (Figure 23.2).