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Chemistry, Student Solutions Manual
Chapter 18
Chapter 18 The Transition Metals
Solutions to Problems in Chapter 18
18.1 Oxidation states are determined by applying the rules given in Section 16.1 (Chapter
16). The procedure can be simplified when a polyatomic ion of known charge is
present:
(a) CO32– has a charge of –2, so Mn must be +2 to give overall neutrality.
(b) Cl (more electronegative) is –1, so Mo must be +5 to give overall neutrality.
(c) Na is +1, so VO4 has overall charge –3; each O is –2, so V is +5.
(d) O is –2, so Au must be +3 to give overall neutrality.
(e) H2O has zero charge, and SO42– has a charge of –2, so Fe must be +3 to give
overall neutrality.
18.3 Use the periodic table to locate and identify elements from their valence
configurations:
(a) There are six valence electrons, so the element is in Column 6; 3d orbital is
filling: Cr
(b) There are full s and d blocks, so the element is in Column 12; 4d has just filled:
Cd
(c) There are 11 valence electrons, so the element is in Column 11; 3d orbital is full:
Cu
18.5 Use the periodic table to locate a transition metal and determine the principal
quantum numbers of its valence electrons. Then remove electrons to give the
appropriate cation, remembering that when cations form, the valence s electrons are
the first ones removed:
(a) 3d5; (b) 5d6; (c) 3d8; and (d) 4d4.
18.7 The properties of transition metals vary regularly with their valence configurations,
so predict relative properties based on locations in the d block:
(a) Pd is in Column 10, Cd is in Column 12, both in the n = 4 row. Beyond the
middle of the d block, melting point decreases with Z because electrons are placed
in antibonding orbitals, so Pd melts at higher temperature.
(b) Cu and Au are both in Column 11, but Au has a higher molar mass, so Au has
higher density.
(c) Cr is in Column 6, Co is in Column 9, both in the n = 3 row. IE1 increases with
Z across a row because Zeff increases, so Co has the higher IE1.
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18.9 Use the charges of ligands and ions to determine the oxidation states of transition
metals in coordination complexes. Because s electrons are always removed first, the
count of d electrons is given by the number of valence electrons — the oxidation
state:
(a) Each Cl is –1, and NH3 is neutral, so Ru has oxidation state +2. Ru is in Column
8 (eight valence electrons), giving d6.
(b) Each I is –1, and en is neutral, so Cr has oxidation state +3. Cr is in Column 6
(six valence electrons), giving d3.
(c) Each Cl is –1, and trimethylphosphine is neutral, so Pd has oxidation state +2.
Pd is in Column 10 (10 valence electrons), giving d8.
(d) Each Cl is –1, and NH3 is neutral, so Ir has oxidation state +3. Ir is in Column 9
(nine valence electrons), giving d6.
(e) CO is a neutral molecule, so Ni has oxidation state 0. Ni is in Column 10 (10
valence electrons), giving s2 d8.
18.11 Compounds that contain coordination complexes are named following the six rules
stated in your textbook: name the cation first, name ligands in alphabetical order,
name the metal, add “o” for anions, use Greek prefixes, add “-ate” for anionic
complexes, give the oxidation number:
(a) Hexaammineruthenium(II) chloride
(b) trans-bis(Ethylenediamine)diiodochromium(III) iodide
(c) cis-Dichlorobis(trimethylphosphine)palladium(II)
(d) fac-Triamminetrichloroiridium(III)
(e) Tetracarbonylnickel(0)
18.13 The structure of a metal complex usually is octahedral (six ligands), tetrahedral
(four ligands), or square planar (four ligands):
(a) Six NH3 in an octahedron around the central Ru:
(b) Two I at opposite ends of one axis, two en in a square plane around the central
Cr:
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(c) Square planar arrangement about Pd, with two Cl adjacent to each other:
(d) Octahedral arrangement around Ir, with three Cl in a triangular face:
(e) Tetrahedral arrangement of C≡O about a central Ni:
18.15 The name of a complex contains the information needed to determine its chemical
formula. Determine the charge of the complex from charges on the ligands and the
oxidation number:
(a) cis-[Co(NH3)4ClNO2]+; (b) [PtNH3Cl3]–; (c) trans-[Cu(en)2(H2O)2]2+; (d)
[FeCl4]–.
18.17 (a) cis-Tetraamminechloronitrocobalt(III) has six ligands and octahedral geometry.
The cis indicates that the chlorine and nitro ligands will have a 90º angle between
them:
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(b) In amminetrichloroplatinate(II), platinum has eight d electrons in its valence
shell, resulting in square planar geometry:
(c) In trans-diaquabis(ethylenediamine)copper(II), ethylenediamine is a bidentate
ligand, giving a coordination number of 6 and octahedral geometry. The trans term
means that the water ligands are opposite each other on the complex:
(d) In tetrachloroferrate(III), iron(III) is a 3d metal with five valence electrons,
resulting in tetrahedral geometry:
18.19 The crystal field diagram for weak and strong octahedral fields is always the same,
with the populations changing depending on how many d electrons must be
accommodated. The valence configuration provides information about the number
of d electrons:
(a) Ti2+ (Column 4, two electrons) is d2; (b) Cr3+ (Column 6, three electrons) is d3.
For these two ions, the low-field and high-field configurations are the same:
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(c) Mn2+ (Column 7, two electrons) is d5.
(d) Fe3+ (Column 8, three electrons) is d5. These two ions have identical diagrams,
with high-spin when the splitting is small and low-spin when the splitting is large:
18.21 The magnetic properties of a complex are determined by its number of d electrons
and the extent of its crystal field splitting energy:
(a) Ir3+ (Column 9, three electrons) is d6, and NH3 generates relatively large
splitting. All the electrons will be paired, making the complex diamagnetic.
(b) Cr2+ (Column 6, two electrons) is d4, and water generates relatively small
splitting. The complex is paramagnetic with four unpaired electrons.
(c) Pt2+ (Column 10, two electrons) is d8, so regardless of the splitting energy, this
square planar complex is paramagnetic with two unpaired electrons.
(d) Pd has d10 configuration, so all orbitals are filled and this complex is
diamagnetic.
18.23 The colours of transition metal complexes are generally determined by d–d
transitions, but Zr4+ (Column 4, four electrons) has all its valence electrons
removed, so there is no valence electron that can undergo a transition involving the
absorption of visible light.
18.25 (a) The coordination number is the number of bonds formed between metal and
ligands. The en ligand is bidentate, so three of them form six bonds, CN = 6.
(b) The oxidation number is determined by examining the net charge and correcting
for charges on all species other than the transition metal. Here, en is neutral. Each
Cl– anion contributes –1 charge, so the oxidation number of Fe is +3. Fe is in
Column 8 of the periodic table, so its valence configuration is 3d5.
(c) Coordination number 6 means octahedral geometry.
(d) With an odd number of electrons, it is not possible for all of them to be paired,
so this complex is paramagnetic.
(e) The complex is low-spin, so all but one of the electrons is paired.
18.27 Complexes are high-spin and paramagnetic when the crystal field splitting is small,
and they become low-spin if the crystal field splitting is larger than the pairing
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energy. According to the spectrochemical series, CN– produces greater splitting
than H2O. Replacing the H2O ligands with CN– ligands increases the splitting, to
the point where it takes less energy to pair the electrons than to promote them to
the less stable d orbitals. Cr is in Column 6 of the periodic table, so Cr2+ has d4
valence configuration. There are four unpaired electrons in the high-spin
configuration but only two in the low-spin configuration:
18.29 The wavelength of light that is absorbed provides a measure of the crystal field
splitting energy:
 (6.626  034 J s)(2.998  108 m/s)(6.022  10 23 mol1 ) 103 kJ 




465  m

 1 J 
  257 kJ/mol

E
hcN A
The complex absorbs visible light around 465 nm, in the blue. The colour of the
complex will be the complementary colour to blue. Consult Table 18-5 to
determine that this colour is orange.
18.31 The structural difference between haemoglobin and myoglobin is that the former
has four subunits, whereas the latter has just one. As a consequence, haemoglobin
has much more complex cooperative chemical behaviour than myoglobin does.
18.33 An iron ion bonded to four sulphur atoms from cysteines is in a tetrahedral
environment, so the splitting pattern is the 2–3 pattern characteristic of tetrahedral
complexes. The iron cation loses one electron and is converted from Fe2+ to Fe3+:
18.35 Consult your textbook for the chemical reactions of various metallurgical
processes:
(a) CuFeS2 + 3 O2 → CuO + FeO + 2 SO2
(b) Si + O2 + CaO → CaSiO3
(c) TiCl4 + 4 Na → 4 NaCl + Ti
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18.37 This is a standard stoichiometry problem. Begin by analyzing the chemistry. The
reactants are Cu2S and air and the given products are Cu metal and SO2 gas.
Balanced reaction: Cu2S + O2 → 2 Cu + SO2.
Start by computing the number of moles of Cu2S, and then use the appropriate
mass–mole–mass and p–V–T calculations to determine the amounts of the
products:
mCu2S
 103 kg   2.37% 
6
 5.60  10 kg 

  1.327   g
 1 kg   100% 
nCu2S 
4
1.327  106 g
 8.34  3 mol nSO2
159.2 g/mol
 2 mol Cu 
6
mCu  8.34  103 mol 
 (63.55 g/mol)  1.06   g
1
mol
Cu
S

2 
1 bar


ptotal  755 Torr 
  1.006 bar
 750.06 Torr 
VSO2
nRT (8.34  103 mol)(0.08314 L bar mol1 K 1 )(273.15  23.5 K)


 2.04  5 L
p
1.006 bar
18.39 Standard free energy changes are calculated using standard free energies of
formation, found in Appendix D. Greaction  Gproducts  Greactants
ZnO (s) + C (s) → Zn (s) + CO (g)
Greaction = [1 mol(–137.2 kJ/mol) + 0] – [1 mol(–320.5 kJ/mol) + 0] = 183.3 kJ
ZnO (s) + CO (g) → Zn (s) + CO2 (g)
Greaction = [1 mol(–394.4 kJ/mol) + 1 mol(0)] – [1 mol(–320.5 kJ/mol) + 1
mol(–137.2 kJ/mol)] = 63.3 kJ
18.41 The coinage metals are those that have been used since antiquity for coins: copper,
silver, and gold. All are in Column 11 of the periodic table. They are characterized
by high electrical conductivity, good ductility, and low chemical reactivity, in
particular resistance to oxidation. Hence, they are used for money (a vanishing use
in technologically advanced countries), for electrical wire, for jewellery, and for
other decorative objects. See your text for special examples of uses for compounds
of these elements.
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18.43 Titanium is used as an engineering metal because of its relatively low density, high
bond strength, resistance to corrosion, and ability to withstand high temperatures,
all of which make it a favoured structural material.
18.45 The number of possible isomers of a complex is determined by its geometry and
the number of ligands of each type:
18.47 Bidentate ligands form complexes with two links. Each Fe ion forms an octahedral
complex with three oxalate anions. When oxalate is applied to rust, it complexes
and dissolves the Fe3+ cations. Here is a sketch showing the ligand–metal
orientations:
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18.49 To determine electron configurations, start from the position of the element in the
periodic table and remove s electrons preferentially.
(a) Cr is in Column 6, configuration [Ar] 4s1 3d5; Cr2+ is [Ar] 3d4; Cr3+ is [Ar] 3d3
(b) V is in Column 5, configuration [Ar] 4s2 3d3; V2+ is [Ar] 3d3; V3+ is [Ar] 3d2;
V4+ is [Ar] 3d1; V5+ is [Ar]
(c) Ti is in Column 4, configuration [Ar] 4s2 3d2; Ti2+ is [Ar] 3d2; Ti4+ is [Ar]
18.51 Compounds that contain coordination complexes are named following the six rules
stated in your textbook: name the cation first, name ligands in alphabetical order,
name the metal, add “o” for anions, use Greek prefixes, add “-ate” for anionic
complexes, list the oxidation number: (a) cis-tetraaquadichlorochromium(III)
chloride; (b) bromopentacarbonylmanganese(I); (c) cisdiamminedichloroplatinum(II).
18.53 Superoxide dismutase catalyzes the conversion of superoxide into molecular
oxygen and hydrogen peroxide. This reaction occurs at a metal site that contains
one Zn2+ ion and one Cu2+ ion, linked by a histidine ligand that bonds to both
metal ions. An O atom binds to the Zn2+ ion:
18.55 Cu(II) in water forms an aqua complex. Addition of fluoride produces the insoluble
green salt, CuF2, whereas addition of chloride produces the bright green
tetrachlorocopper(II) complex, [CuCl4]2–:
Cu2+ (aq) + 2 F– (aq) → CuF2 (s)
Cu2+ (aq) + 4 Cl– (aq) → [CuCl4]2– (aq)
18.57 Silver (Column 11 of the periodic table) has a filled set of d orbitals, making the
neutral metal difficult to oxidize. This gives silver good resistance to corrosion and
makes it suitable for jewellery. Vanadium (Column 5), with d3 configuration, is
readily oxidized, so it corrodes rapidly and is unsuited to jewellery.
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18.59 The mer isomer of an octahedral complex has three like-ligands arranged in a
meridian plane:
18.61 Silver tarnish is silver sulphide, Ag2S, formed by reaction with trace amounts of
H2S in the atmosphere:
4 Ag (s) + 2 H2S (g) + O2 (g) → 2 Ag2S (s) + 2 H2O (g)
18.63 In these complexes, chromium is in the +3 oxidation state. From its location in
Column 6 of the periodic table, we deduce that is has d3 valence electron
configuration. The three electrons occupy the three different t2g orbitals, regardless
of the magnitude of the splitting energy.
18.65 Orbital sketches show that the two orbitals experience quite different electron–
electron repulsion in a square planar environment:
18.67 (a) The charge on a complex is the charge on the transition metal cation, modified
by any charges on the ligands. Ammonia and water are neutral, so when L is either
of these, n+ = 3+; but chloride is –1, so for L = Cl–, n+ = 2+.
(b) The colour of each complex is the colour that is complementary to the one
corresponding to the wavelength of light that it absorbs. Consult Table 18-5 of your
textbook to determine these: L = Cl–, red to purple; L = H2O, orange; and L = NH3,
yellow-orange.
(c) The wavelength of light that is absorbed provides a measure of the crystal field
splitting energy:
E
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
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L = Cl–:
 (6.626  1034 J s)(2.998  108 m/s)(6.022  10 23 mol1 )  103 kJ 



515  9 m

 1 J  
 232 kJ/mol
L = H2O:
 (6.626  1034 J s)(2.998  108 m/s)(6.022  10 23 mol1 )  103 kJ 



  9 m

 1 J  
 249 kJ/mol
L = NH3:
 (6.626  1034 J s)(2.998  108 m/s)(6.022  10 23 mol1 )  103 kJ 



  9 m

 1 J  
 257 kJ/mol
The trend matches exactly the positions of these three ligands in the
spectrochemical series: Cl– < H2O < NH3.
18.69 Tetracarbonylnickel(0), [Ni(CO)4], has tetrahedral geometry since Ni is a first-row
transition element. Because of the strong field ligand, CO, ∆ is large, which will
result in the compound absorbing UV light and appearing colourless.
Tetracyanozinc(II), [Zn(CN)4]2–, has tetrahedral geometry (Zn is a first-row
transition metal) where Zn has a +2 charge (d10). This complex is colourless
because there are no possible d–d transitions.
18.71 Brass is an alloy of zinc and copper, and superoxide dismutase contains zinc and
copper ions in its reaction centre.
18.73 Four-coordinate complexes may be either tetrahedral or square planar. The splitting
patterns for these two geometries show that the d8 configuration can have zero spin
in the square planar case but not in the tetrahedral case. Thus, the magnetic
behaviour indicates that [Ni(CN)4]2– is square planar and [NiCl4]2– is tetrahedral:
18.75 When ferritin is neither empty nor filled to capacity, the protein has the capacity to
provide iron as needed for haemoglobin synthesis, or to store iron if an excess is
absorbed by the body.
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18.77 Visible spectroscopy is useful when a compound has an energy gap between the
highest occupied and lowest unoccupied orbital that matches the energy of visible
light. Because Zn2+ has d10 configuration, its d orbitals are completely filled, and
the lowest unoccupied orbital is quite high in energy. In contrast, Co2+ has d7
configuration, giving this cation unfilled d orbitals. Consequently, metalloproteins
that contain Co2+ absorb visible light, making it possible to study them with visible
spectroscopy.
18.79 Use standard reduction potentials from Appendix F to determine which metals can
be displaced by Zn. Any metal with a less negative standard potential can be
displaced:
Zn 2  2 e € Zn
E   0.7618 V
The following are a few examples of metals that can be displaced by Zn:
Co 2  2 e € Co
E   0.28 V
Cu 2  2 e € Cu
E   0.3419 V
Fe2  2 e  € Fe
E   0.447 V
Ni 2  2 e € Ni
E   0.257 V
When combined with the Zn half-reaction, the overall cell voltage is positive and
the process is spontaneous. For example:
Zn  Co2 € Co  Zn 2
E  (0.28 V)  (0.7618 V)  0.48 V
18.81 The colours of substances depend on what wavelengths of light they absorb.
According to Table 18-5, blue colour results when a substance absorbs 610 nm
light (196 kJ/mol), and red colour results when a substance absorbs 500 nm light
oxyhaemoglobin, suggesting that O2 causes larger energy-level splitting than H2O.
The hypothesis could be tested by replacing the H2O ligand in deoxyhaemoglobin
by a ligand that causes smaller splitting, such as Cl–, which should push the colour
toward the green. In addition, CO, which generates the largest energy-level
splitting, should push the colour to red-orange or orange.
18.83 The densities of transition metals increase with Z across each row of the d block,
up to about the middle of the block, as illustrated by Figure 18-2b in your
textbook. Thus, the upper-left-most transition metal in the periodic table,
scandium, would be the best replacement for Al. The density of scandium is only
about 10% greater than that of aluminum. The figure shows that osmium, in the
middle of the 5d block, is the densest transition metal. Its density is actually twice
that of lead.
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18.85 The chloride ions that form silver chloride precipitate are not bound to the complex
and should not be included in the chemical formula of the complex ion:
PtCl4•2NH3 is [PtCl4(NH3)2], diamminetetrachloroplatinum(IV), cis and trans:
PtCl4•3NH3 is [PtCl3(NH3)3]Cl, the complex ion being
triamminetrichloroplatinum(IV), mer and fac:
PtCl4•4NH3 is [PtCl2(NH3)4]Cl2, the complex ion being
tetraamminedichloroplatinum(IV), cis and trans:
PtCl4•5NH3 is [PtCl(NH3)5]Cl3, the complex ion being
pentaamminechloroplatinum(IV), and PtCl4•6NH3 is [Pt(NH3)6]Cl4, the complex
ion being hexaammineplatinum(IV):
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