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Chemistry, Canadian Edition
Chapter 18: Student Study Guide
Chapter 18: The Transition Metals
Learning Objectives
Upon completion of this chapter you should be able to
• predict periodic properties of transition metals
• recognize and name transition metal coordination complexes
• use crystal field theory to explain the colour and magnetic properties of complexes
• explain the importance of transition metal complexes in biological processes
• explain the chemistry of essential steps in the production of pure metals from ores
• recognize the importance of transition metals in everyday life
Practical Aspects
This chapter uses concepts from Chapter 7 (relation of light energy to wavelength and color), Chapter 8,
(electron configurations and energy level diagrams), and Chapter 9 (shapes of molecules). In this chapter,
you will learn why transition metal compounds are usually vividly coloured. You will also learn about
the many important societal uses of transition metals as well as the essential roles they play within
biological systems.
18.1 OVERVIEW OF THE TRANSITION METALS
Skills to Master:
 Naming groups of transition metals.
 Knowing trends in melting point, density, and oxidation states.
 Determining oxidation states in compounds of transition metals.
Key Concepts:
 Transition metals are good conductors of heat and electricity.
 Most transition metals can exist in several different oxidation states.
 Most transition metals are found in nature as oxides or sulphides, rather than in their elemental form.
 In transition metal solids, electrons fill in bonding orbital bands up through Group 6. After Group 6,
electrons are added to delocalized antibonding orbital bands (a destabilizing factor).
 The physical properties of a transition element depend upon the element’s electron configuration.
For example, some periodic trends are:
 Melting point – increases across a row as electrons are added to bonding orbital bands, but
decreases sharply as soon as electrons are added to antibonding orbital bands.
 Density – increases within a given column (as elements become more massive), and increases
along a given row, while electrons are added to bonding orbital bands. When electrons start
adding to antibonding orbital bands, density decreases.
 Ionization energy – increases across a given row because Zeff increases. (Recall that the
definition of ionization energy is the energy required to remove a valence electron from an atom
in its gaseous state, so delocalized orbitals are not an option here.) Ionization energy decreases
down a column as electrons are held less tightly.
* Remember that there are exceptions to every trend.
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EXERCISE 1: Predict and explain. Which will have the greater: a) melting point, Mo or Tc? b) density,
Zr or Hf? c) ionization energy, Sc or V?
STRATEGY: Use periodic trends to compare the elements.
SOLUTION:
a) Tc is to the immediate right of Mo on the periodic table, so it contains one more proton and
electron than Mo. However, that extra electron is the first added to a destabilizing antibonding orbital
band. The melting point of Mo is therefore greater than Tc, because it will require more thermal
energy to disrupt Mo’s structure than Tc’s.
b) Hf is below Zr on the periodic table, making Hf more massive. Hf will therefore be more dense
than Zr.
c) Sc and V are in the same row, but V has two more protons than Sc. The ionization energy depends
upon the attraction of the electrons to the nucleus within a gaseous atom. V will therefore have a
higher ionization energy than Sc.
EXERCISE 2: Ag is an excellent electrical conductor. Use the electron configuration of Ag to explain why
it is such a good conductor.
STRATEGY: Electrical conductivity in a metal depends upon the ability of its electrons to delocalize.
SOLUTION: Ag is a d9 transition element. A transition element with over 6 d electrons has electrons
in delocalized antibonding orbitals. The electrons in Ag are highly delocalized, making Ag an
excellent electrical conductor.
Try It #1: Use periodic trends to explain why Hg is a liquid at room temperature.
18.2 COORDINATION COMPLEXES
Skills to Master:
 Recognizing and naming some common ligands.
 Drawing cis and trans as well as fac and mer isomers.
 Naming coordination compounds.
Key Terms:
 Geometric isomers – two complexes that have the same formula, but different 3-D orientations of
the ligands. Two main categories:

Cis vs. trans – applies to complexes that contain two ligands of one type, ML4X2:
“cis” indicates “same side”
“trans” indicates “across”
X
L
L
X
L
L
X
L
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M
M
L
L
X
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Chapter 18: Student Study Guide
Facial vs. meridianal – applies to complexes that contain three ligands of one type, ML3X3:
“fac” indicates facial and means
“mer” indicates meridianal and means that
that the three ligands lie on one
the three ligands lie in one plane that goes
face of the octahedron.
through the center of the complex.
L
L
X
X
L
L

X
M
M
X
L
X
L
X
Linkage isomers – two complexes that have the same formula, but show different connectivities of
the ligands to the metal center. This occurs only if a given ligand can attach to the metal using either
of two donor atoms.
Key Concepts:
 Ligands are classified by the number of coordination sites they can bind to: monodentate (1),
bidentate (2), tridentate (3), etc.
 In aqueous solution, water molecules will bind to any open coordination sites on a transition metal.
For example, Co2+ (aq) is really [Co(H2O)6]2+ (aq).
 The geometry of a coordination complex depends upon its coordination number.
 The most common coordination number is 6, which typically corresponds to an octahedral geometry.
 A coordination number of 4 will result in either a tetrahedral or square planar geometry.
Rules for Naming Transition Metal Complexes
Detailed rules for naming coordination compounds are provided in Section 18.2 of the text. Here is a
summary:
All coordination compounds follow this pattern for the root of the name:
1. Name the cation first and the anion second.
2. Write the complex as one long word: name the ligands in alphabetical order, then the metal, then
indicate the metal’s charge with Roman numerals. Note that these neutral ligands have special
names: H2O = aqua, NH3 = ammine (two “m”s), CO = carbonyl.
These suffixes and prefixes must be added accordingly:
 If the ligand is an anion, add the suffix “–o” to it.
 Use Greek prefixes to indicate how many ligands of a given type are present. If the
ligand name contains a Greek prefix, then use these alternate prefixes to describe the
number of ligands: “bis-“ for two, “tris-”for three, “tetrakis-” for four. When using this
alternate format, enclose the ligand name in parentheses.
 If the complex is an anion, add the suffix “–ate” to the metal name.
 Some complex anions take on the metal’s Greek name. For example, Ag  argentate.
EXERCISE 3: Determine the oxidation state, coordination number, and number of d valence electrons for
the metal in each: a) [Co(NH3)3(NO2)3]; b) Na2[MnCl4]; c) [Ni(en)(H2O)4]SO4.
STRATEGY: Use guidelines for assigning oxidation states (numbers). The coordination number equals
the steric number. The number of d valence electrons can be determined from the electron
configuration.
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SOLUTION:
a) [Co(NH3)3(NO2)3]: The overall charge is zero, each ammonia is neutral, each of the three nitrites
is –1, so the oxidation state of cobalt is +3. There are six monodentate ligands present, so the
coordination number is six. The Co3+ ion has the valence electron configuration d6.
b) Na2[MnCl4]: The overall charge is zero, each sodium is +1, each chloride is –1, so the oxidation
state of manganese is +2. There are four monodentate ligands present, so the coordination number is
four. The Mn2+ ion has the valence electron configuration d5.
c) [Ni(en)(H2O)4]SO4: The overall charge is zero, each ligand is neutral, and sulphate is –2, so the
oxidation state of nickel is +2. There is one bidentate ligand (en) and there are four monodentate
ligands, so the coordination number is six. The Ni2+ ion has an electron configuration of d8.
EXERCISE 4: Name each: a) K4[Fe(CN)6]; b) [Ni(en)(H2O)4]SO4; c) [Ni(en)3]Cl2
STRATEGY: Apply the rules for naming coordination complexes.
SOLUTION:
a) K4[Fe(CN)6]
Cation and anion: The cation is potassium and the anion is the complex ion. The suffix “-ate” will
need to be added.
Ligands: Six CN- ligands = “hexacyano.”
Metal charge/name: 4 K+ and 6 CN- = -2, so the charge on iron must be +2. (Fe2+) = ferrate(II).
Name: Potassium hexacyanoferrate(II)
b) [Ni(en)(H2O)4]SO4
Cation and anion: The cation is the complex ion and the anion is sulphate.
Ligands: One en and four waters. “aqua” is before “ethylenediamine” in the alphabet, so
“tetraaquaethylenediamine” is the collective name for the ligands.
Metal charge/name: All of the ligands are neutral, so SO42- is the only counter ion: nickel(II).
Name: Tetraaquaethylenediaminenickel(II) sulphate.
c) [Ni(en)3]Cl2
Cation and anion: The cation is the complex ion and the anion is chloride.
Ligands: Three ethylenediamine ligands. The Greek prefix “di-” is in the ligand name, so we must
denote “three” with the alternate prefix, tris. The ligand name is tris(ethylenediamine).
Metal charge/name: en is neutral, and there are 2 Cl- ions, so Ni2+ = nickel(II)
Name: Tris(ethylenediamine)nickel(II) chloride.
Try It #2: Name each compound: a) [Co(NH3)3(NO2)3];
b) Na2[MnCl4]
EXERCISE 5: Write the formula for each: a) Hexamminemolybdenum(III) chloride;
b) Triaquatrifluorocobalt(III); c) potassium dicarbonyltetracyanochromate(II).
STRATEGY: Follow the rules for naming complex ions.
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SOLUTION:
a) Hexamminemolybdenum(III) chloride: This contains 6 ammonia ligands attached to Mo3+, and
the counter ion is Cl-. Ammonia is neutral, so 3 Cl- ions are needed for the compound to be
electrically neutral: [Mo(NH3)6]Cl3.
b) Triaquatrifluorocobalt (III): Three waters and three fluorides are attached to a Co3+ ion. The three
F- ions counterbalance the Co3+, making the complex neutral overall: [Co(H2O)3F3].
c) Potassium dicarbonyltetracyanochromate(II): two CO ligands and four CN- ligands are attached to
a Cr2+ metal ion. The charge on the complex is –2, so two K+ ions are needed to counterbalance the
charge: K2[Cr(CO)2(CN)4].
Try It#3: Write the formulas for: a) Triaquatrichlororuthenium(III); b) Tetraamminetitanium(II) chloride
EXERCISE 6: Name the two isomers of [Pt(NH3)4(Cl)2]Cl2, then draw structures for their complex ions.
STRATEGY: This complex contains two chloride ligands and four ammonia ligands (ML4X2 pattern),
so the overall shape is octahedral. The two chloride ligands can be arranged next to each other (cis)
or across from each other (trans). The complex has an overall charge of +2.
SOLUTION:
cis-tetraamminedichloroplatinum (IV) chloride
H3N
trans-tetraamminedichloroplatinum (IV) chloride
2+
Cl
Cl
H3N
Pt
2+
Cl
NH3
Pt
H3N
NH3
H3N
NH3
NH3
Cl
EXERCISE 7: Name and draw structures for the two isomers that fit this formula: [Co(NH3)3(NO2)3].
STRATEGY: There are six ligands, so the geometry will be octahedral. When two sets of three ligands
are attached to a metal center (ML3X3 pattern), all ligands of the same type can be clustered together
on one face of the octahedron, or they can be arranged so that two of the same type are opposite each
other. These are termed “facial” and “meridianal,” respectively. The cobalt must have a +3 charge to
counter the three –1 charges of the nitrites. Notice that within the complex “NO2-” is named “nitro.”
SOLUTION:
fac-triamminetrinitrocobalt(III)
mer-triamminetrinitrocobalt(III)
NH3
NO2
H 3N
NO2
O2N
Co
H 3N
NH3
Co
NO2
NH3
O2N
NO2
NH3
Try It #4: Draw the structure for the dicarbonyltetracyanochromate(II) ion. If more than one isomer
exists, draw both isomers.
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18.3 BONDING IN COORDINATION COMPLEXES
Skills to Master:
 Constructing a crystal field energy-level diagram for an octahedral complex.
 Determining the electron configuration of a complex from magnetic properties.
 Determining the value of � from the absorption spectrum of a complex.
Key Terms:
 Crystal field splitting energy () – difference in energy between non-degenerate d orbitals.
 Pairing energy (P) – destabilizing energy resulting from repulsive forces between two electrons that
occupy the same orbital.
 High-spin – term used to describe electron arrangement in which the maximum unpaired spins are
present. This occurs when P > , and it is easier for the electrons to occupy the higher energy d
orbitals than to pair up in the lower energy orbitals.
 Low-spin – term used to describe electron arrangement in which the maximum paired spins are
present. This occurs when P < .
 Spectrochemical series – list of ligands in order of increasing energy level splitting ability:
I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < NO2- < CN- < CO
Key Concepts:
 All d orbitals are not degenerate. The set of d orbitals is split into two or more degenerate groups,
based on slight differences in energy arising from electron-electron repulsions.
 The magnitude of the crystal field splitting energy depends upon these factors:
 the charge on the metal ion – the greater the charge, the greater the splitting.
 the size of the metal ion – the greater the n, the greater the splitting.
 the coordination number – the higher the coordination number, the greater the splitting.
 the identity of the ligand – the more tightly the ligand binds, the greater the splitting.
 The pattern of orbital splitting depends upon the geometry of the metal cation:
Geometry:
Tetrahedral
Square Planar
t2g
t
Energy
eg
Energy
x2-y2
Energy
Crystal Field
Energy Level
Diagram:
Octahedral
xy
e
z2
xz
yz
Notes:

The t2g set includes the dxy,
dxz, and dyz orbitals. The eg
set includes the
d z 2 and d x 2  y 2 orbitals.
The two sets are the
reverse of octahedral.
Splitting is smaller than
in octahedral, so
tetrahedral complexes
are almost always
high-spin.
The d x 2  y 2 orbital is
greatly destabilized
relative to the other
orbitals because of its
orientation relative to the
ligands in this geometry.
The same rules apply for drawing energy level diagrams for transition metal cations as did for atoms Hund’s Rule, Aufbau Principle, Pauli Exclusion Principle.
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Helpful Hints
 One way to remember the spectrochemical series is to notice that it is broken down into groups of
what’s attaching to the metal: “halogens < oxygen < nitrogen < carbon.”
 The names “t2g” and “eg” are used to describe the two categories of d orbitals in an octahedral
geometry. The names “e” and “t” are used to describe d orbitals in a tetrahedral geometry.
 An energy-level diagram for a transition metal ion is sometimes called a crystal field energy diagram.
EXERCISE 8: Draw a crystal field energy diagram for each complex ion. Then, write the electron
configuration for each. a) [Co(NH3)3(NO2)3]; b) Na2[MnCl4], tetrahedral; c) K4[CrF6].
STRATEGY: Determine the splitting pattern from the orbital geometry. Then determine the relative
magnitude of the  by assessing the factors that affect it: the size and charge of the metal ion, the
coordination number, and the identity of the ligand.
SOLUTION:
a) [Co(NH3)3(NO2)3]: Co3+, 3d6 arrangement, octahedral geometry. Relatively high oxidation state,
fairly strong ligands. These factors indicate a high splitting energy. The complex will be low-spin.
b) Na2[MnCl4], tetrahedral: Mn2+, 3d5 arrangement. Tetrahedral complexes are almost always high
spin, because their splitting energies are small. The weak Cl- ligands shouldn’t increase the splitting
energy.
c) K4[CrF6]: Cr2+, 3d4 arrangement, octahedral geometry. Lower oxidation state, weak ligands.
These factors favour a low splitting energy. This complex will be high-spin.
[Co(NH3)3(NO2)3]
Complex:
Na2[MnCl4]
c) K4[CrF6]
Electron
configuration:
(t2g)6(eg)0
t
e
(e)2(t)3
eg
Energy
t2g
Energy
Energy
eg
Diagram:
t2g
(t2g)3(eg)1
Try It #5: Draw the crystal field splitting diagram and electron configuration for [Pt(NH3)6]Cl2.
Magnetism and Color
Key Term:
 Complementary colors – colors related to each other by their absorption and transmission
characteristics. Table 18-5 in the text lists complementary colors.
Key Concepts:
 Recall that paramagnetic substances can behave like magnets because they contain unpaired electron
spins, while diamagnetic substances don’t behave like magnets because all of their spins are paired.
 The crystal field splitting energy of a metal complex is related to its color.
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We can see reflected light (light that bounces off a substance) and transmitted light (light that passes
through a substance). We see the complement of the color that the substance absorbs.
Useful Relationship:
 Emolecule =  = h = hc/
This relationship can be used to relate  to the wavelength of light
that the complex absorbs.
EXERCISE 9: Which of the complexes listed in Exercise 8 will respond most strongly to a magnetic field?
Which will not respond at all?
STRATEGY: A paramagnetic substance will respond to a magnetic field, and a diamagnetic substance
will not. The substance with the highest overall spin will respond the most intensely.
SOLUTION: [Co(NH3)3(NO2)3] has all spins paired, so it is diamagnetic. It will not respond to a
magnetic field. Both Na2[MnCl4] and K4[CrF6] have unpaired spins, so they are both paramagnetic.
Na2[MnCl4], with five unpaired spins, will respond slightly more to a magnetic field than K4[CrF6],
with four unpaired spins.
EXERCISE 10: [Ni(en)3]SO4 is lavender, [Ni(en)(H2O)4]SO4 is blue, and [Ni(H2O)6]SO4 is teal. Explain
the color differences.
STRATEGY: The color we can see is the complement of the color that the substance absorbs. Use
Table 18-5 in the text to determine the color that each complex absorbed, then compare the relative
energies of the absorbed colors.
SOLUTION: Table 18-5 shows that:
Complex
[Ni(en)3]SO4
[Ni(en)(H2O)4]SO4
[Ni(H2O)6]SO4
Color observed
Lavender (light purple)
blue
Teal (blue-green)
Color absorbed
Yellow-green
Orange
Red-orange
Wavelength absorbed
560 nm
610 nm
680 nm
The only difference between each complex is the ligands. H2O is lower in the spectrochemical series
than en, so the splitting energy for the all-water complex should be the lowest. The all-en complex
should have the highest splitting energy. This is the trend observed. The complex with the shortest
wavelength of light absorbed corresponds to the complex with the greatest splitting energy.
EXERCISE 11: The splitting energy of [Rh(Cl)6]3- is 243 kJ/mol. a) Determine the wavelength of
maximum absorption for this complex. b) What color is it?
STRATEGY: Use the relationship E = hc/ to determine the wavelength. Use Table 18-5 in the text to
determine the color.
243x10 3 J
1 mol
x
 4.035x10 19J/photon
23
mol
6.022x10 photons
E  h 
hc
hc 6.626x10 34 J sec 2.9979x10 8 m 10 9 nm
; so  

x
x
 492 nm

E
sec
1m
4.035x10 19 J
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SOLUTION: a) The wavelength of maximum absorption will be 492 nm. b) The color of this complex
will be red-orange, which is the complement of the blue to blue-green light that is absorbed. The
answers seem reasonable because Rh’s d electrons are in n=4, and these large orbitals have larger
splitting energies than smaller orbitals. Metals with n=4 and 5 d electrons tend to be low-spin
because of their high splitting energies.
Try It #6: Approximate the crystal field splitting energy for a substance that appears green.
18.4 METALLURGY
Key Terms:
 Metallurgy – the process of purifying a metal from its ore. Metallurgy is a four-step process:
1. Separation of the desired ore from other metal ores.
2. Conversion of the pure ore to a form that is easily reduced.
3. Reduction of the metal compound to the pure metal with a reducing agent or electrolysis.
4. Refining the purity of the metal.
 Roasting – heating in the presence of air (oxygen), typically to convert sulphides to oxides.
 Coke – a form of carbon (charcoal) that has been heated to remove its impurities. Coke is used as the
reducing agent for many reduction processes.
Key Concepts:
 Electrolysis requires enormous amounts of electrical energy, so it is only used in the reduction step if
the metal is too reactive to use a reducing agent.
Helpful Hints
 Figure 18-20 in the text shows a general schematic diagram of the metallurgical process.
 Table 18-6 in the text summarizes the specific chemical processes used to separate a transition metal
from its ore.
 Section 18.4 of the text provides detailed descriptions of separation methods for several metals.
Fast Facts – Iron:
 Main natural source: hematite, Fe2O3, and magnetite, Fe3O4.
 Isolation process: See Figure 18-22 in the text for an illustration of the blast furnace.
 Main uses: steel (700 million tons per year, worldwide).
 Other:
 Iron is the most-used metal.
EXERCISE 12: Why are most transition metals found in an oxide or sulphide form, rather than in their
elemental form? To answer this question, compare the standard reduction potentials of several transition
metals to the positive standard reduction potentials for oxygen.
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STRATEGY: The greater a substance’s reduction potential, the more it wants to be in its reduced form.
SOLUTION:
Half-reaction
Cr3+ + 3e-  Cr
Cr2+ + 2e-  Cr
Fe2+ + 2e-  Fe
E (V)
-0.744
-0.913
-0.447
Half-reaction
Co2+ + 2e-  Co
Mn2+ + 2e-  Mn
Ni2+ + 2e-  Ni
E (V)
-0.28
-1.185
-0.257
All of these metals have negative standard reduction potentials. Reduction half-reactions for oxygen
all show positive standard reduction potentials, which means that oxygen in contact with these metals
will be the substance reduced, and the metal will be oxidized. In our oxygen-containing environment,
then, these metals will exist in their oxidized form.
18.5 APPLICATIONS OF TRANSITION METALS
Fast Facts – Titanium:
 Main natural source: rutile (TiO2) or ilmenite (FeTiO3).
 Properties: strong, low density, stable at high temperatures.
 Main uses: metal - aircraft frames, jet engines, supplies for chemical industry (pipes, pumps, etc).
 Other:
 9th most abundant element in earth’s crust.
 Alloys of titanium with tin or aluminum have the highest strength to weight ratio of all the
engineering metals.
 TiO2 – white pigment used for almost every white-coloured commercial product on the market.
Fast Facts – Chromium:
 Main natural source: chromite (FeCr2O4).
 Main uses: Metal alloys (Nichrome for heat radiating wires; stainless steel), chrome plating.
 Other:
 “Chromium” comes from the Greek “chroma” meaning color. Chromium compounds are often
vibrantly coloured.
 Chromium(VI) is highly toxic.
Fast Facts – Copper:
 Main natural source: chalcopyrite (FeCuS2) and other ores.
 Main uses: electrical wiring (50% of all Cu produced), pipes, alloys (bronze, brass).
 Other: The concentration of copper in the ore is often less than 1%, which makes copper refining
quite expensive.
Fast Facts – Silver:
 Main natural source: usually mixed in with copper or zinc ores.
 Main uses: photography (#1 use), silverware, jewelry, mirrors, batteries.
Fast Facts – Zinc:
 Main natural source: sphalerite (ZnS).
 Main uses: prevention of steel corrosion, batteries, and alloys (brass and bronze).
 Other: Zinc oxide is used as a catalyst in vulcanizing rubber and as a white pigment.
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Fast Facts – Mercury:
 Main natural source: cinnabar (HgS)
 Main uses: fluorescent lighting, thermometers, electrical switches, electrodes.
 Other: highly toxic; one of the two elements that is a liquid at room temperature.
Fast Facts – The Platinum Metals (Ru, Os, Rh, Ir, Pd, and Pt):
 Main natural source: contaminants in copper and nickel ores.
 Main uses: Catalysts – for example, automobile catalytic converters (Pt and some Rh and Pd) and
catalytic hydrogenation of vegetable oils (Pd).
EXERCISE 13: TiO2 is called both “titanium(IV) oxide” and “titanium dioxide.” Why does it have two
different names?
STRATEGY: Decipher the conventions for the names.
SOLUTION: “Titanium(IV) oxide” is the proper ionic compound name for TiO2. “Titanium dioxide”
is the proper covalent compound name for TiO2. The fact that there are two acceptable names must
be a result of TiO2’s ionic and covalent character.
EXERCISE 14: Why are catalytic converters so expensive?
STRATEGY AND SOLUTION: The platinum metals are used in making catalytic converters. They are
found in nature in small quantities as contaminants in copper (mostly) and nickel ores. Copper is
very expensive to refine, and the low abundance and large amount of work that goes into refining
platinum makes it an even more expensive material.
18.6 TRANSITION METALS IN BIOLOGY
Key Terms:
 Metalloprotein – protein molecule that contains a metal center. The three main roles of
metalloproteins are:
 Transport and storage – For example, hemoglobin’s function is oxygen transport; myoglobin’s
function is oxygen storage.
 Enzymes – For example, carboxypeptidase catalyzes the breakdown of proteins by removing
amino acids one at a time from a protein chain.
 Redox reagents – For example, the cytochromes, copper blue proteins, and iron-sulphur proteins
are all electron transport molecules that involve a change in oxidation state at their metal centers.
Key Concepts:
 A metalloprotein’s function depends upon its ability to bind and release ligands.
 The physical and chemical properties of the metalloproteins depend upon the environment of the
transition metal center.
Helpful Hint
 Section 18.6 of the text describes structures of several metalloproteins.
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EXERCISE 15: Hemoglobin in its deoxygenated form contains five nitrogen ligands (4 from the porphyrin
ring around it and one from a histidine). This form of hemoglobin is blue. When oxygen binds to the
sixth binding site in hemoglobin, the structure of the compound changes, and it turns red. In terms of
crystal field theory, what is causing this shift to red?
STRATEGY: Color depends upon the environment of the transition metal.
SOLUTION: The color we see is the complement of the color absorbed. Red light is lower in energy
than blue light. If we see red light, then a higher energy wavelength was absorbed than if we see
blue. When the color changes from blue to red, the structure of hemoglobin must change so that the
crystal field splitting energy increases. For an electronic transition to occur, then, the molecule must
absorb higher energy light, resulting in red light being reflected or transmitted to us.
© John Wiley and Sons Canada, Ltd.
355
Chemistry, Canadian Edition
Chapter 18: Student Study Guide
Chapter 18 Self-Test
You may use a periodic table and a calculator for this test.
1. The anticancer drug, cisplatin, has the formula: [Pt(NH3)2Cl2]. The molecule has a square planar
geometry. Draw the structure of cisplatin and name it according to IUPAC rules.
2. a) What is the highest possible spin a metal ion can have? Draw a crystal field splitting diagram to
illustrate this.
b) Propose a complex ion that would have this spin.
3. Write the formula, draw the structure, and construct a crystal field energy diagram for:
fac-triamminetriiodoplatinum(IV) bromide.
4. Why is it that zinc oxide (ZnO) and titanium(IV) oxide are white?
5. A typical blue copper protein contains a copper in a distorted tetrahedral environment. How does this
geometry affect its color?
6. What commonly-used items would potentially increase in price if a copper mine were shut down?
7. Predict the properties:
a) Rank these in order of increasing melting point: Ta, W, Pt
b) Rank these in order of increasing density: Cr, Mo, W
© John Wiley and Sons Canada, Ltd.
356
Chemistry, Canadian Edition
Chapter 18: Student Study Guide
Answers to Try Its:
1. Hg(l) consists of large atoms, each having ten electrons in its outermost d orbital. Many of these
electrons are in delocalized antibonding orbitals, which destabilizes the bonding structure of Hg
enough to make it a liquid at room temperature.
2. a) triamminetrinitrocobalt(III); b) sodium tetrachloromanganate(II)
3. a) [Ru(H2O)3Cl3]; b) [Ti(NH3)4]Cl2.
4.
5.
cis
trans
[Pt(NH3)6]Cl2, (t2g)6(eg)2
(don’t need to assign high/low
spin because it’s d8)
2-
CO
CO
NC
Cr
NC
eg
CN
Cr
CN
NC
CN
Energy
NC
2-
CO
CN
CO
t2g
6. From Table 18-5 in the text, green will be observed if the crystal field splitting energy is roughly 166
kJ/mol. (Could have calculated it too from E = hc/.)
Answers to Self-Test:
1. cis-diamminedichloroplatinum(II)
Cl
NH3
Pt
Cl
NH3
Energy
2. a) If all five d orbitals contain one electron, the overall spin will be 2.5. This is the highest possible
spin.
eg
t2g
Energy
b) A complex that would have this electron arrangement would contain a d5 metal cation in an
environment where P > . Factors that would favour this would include: a small metal (3d valence
electrons) with a low oxidation state, and relatively weak ligands, such as halides. A good choice
would be: [MnCl6]4-.
3.
eg
+
NH3
[Pt(NH3)3I3]Br
4+
I
NH3
Pt has six 5d electrons in an octahedral
Pt
environment. A large metal cation with a high
I
NH3
t2g
charge in an octahedral environment favours
I
low-spin.
4. Zn2+ is d10 and Ti4+ is d0. Neither of these have electrons that can undergo a transition between nondegenerate d orbitals, so neither can absorb light. All light is reflected, resulting in white.
5. Tetrahedral geometry is almost always high-spin because of the small crystal field splitting energy. If
we see blue, then the compound absorbs orange light (complementary color), which is relatively low
in energy and indicates a small crystal field splitting energy.
6. Cu: electrical wiring and pipes. The Pt metals: catalytic converters, processed vegetable oils. Ag:
photographs, mirrors, batteries, silverware, jewelry.
7. a) Pt < Ta < W: All are in same row, Pt has occupied anti-bonding orbital bands (lowest); W, from
Group 6, has the maximum occupied bonding orbitals (highest).
b) Cr < Mo < W. All are in the same column. Elements increase in density going down a column.
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357