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Shriver & Atkins: Inorganic Chemistry 5e
Answers to self-tests and exercises
CHAPTER 1
1.7
Take the summation of the rest masses of all
the nuclei of the products minus the masses of
the nuclei of the reactants. If you get a
negative number, energy will be released.
But what you have calculated is the mass
difference, which in the case of a nuclear
reaction is converted to energy.
1.8
0.25
1.9
–13.2 eV
1.10
1524nm, 1.524 X 104 cm-1
Self-tests
1
0
+ γ
S1.1
80
35 Br
S1.2
d orbitals, 5 orbitals
S1.3
4
S1.4
3p
S1.5
The added p electron is in a different (p)
orbital, so it is less shielded.
S1.6
Ni :[Ar]3d 8 4s 2 , Ni 2+ :[Ar]3d 8
S1.7
Period 4, Group 2, s block
S1.8
Going down a group the atomic radius
increases and the first ionization energy
generally decreases.
λ
S1.9
Group 16. The first four electrons are
removed with gradually increasing values.
Removing the fifth electron requires a large
increase in energy, indicating breaking into a
complete subshell.
S1.10
S1.11
+
81
35 Br
n
1.11
⎛1
1 ⎞
7 −1
−
⎟ = 1.0974 X10 m
2
⎝ 1 ∞2 ⎠
1
= R⎜
1
= R⎜
1
= R⎜
1
= R⎜
λ
λ
λ
⎛1
1 ⎞
7 −1
−
⎟ = 1.0288X10 m
2
2
⎝1
⎠
4
⎛1 1⎞
− ⎟ = 9.7547X106 m −1
⎝ 12 32 ⎠
⎛1 1⎞
− ⎟ = 8.2305X106 m −1
⎝ 12 22 ⎠
Adding another electron to C would result in
the stable half filled p subshell.
1.12
0 up to n-X
Cs+
1.13
n2
1.14
Exercises
1.1
(a)
14
7
N+ 42 He→178 O+11p + γ
(b)
12
6
(c)
14
7
C+11p→137 N + γ
1
0
3
1
12
6
N+ n→ H+ C
1
Cm+126 C→257
112 Uub+ 0 n
1.2
246
96
1.3
The higher value of I2 for Cr relative to Mn is
a consequence of the special stability of halffilled subshell configurations and the higher
Zeff of a 3d electron verses a 4s electron.
1.15
1.16
Ne + He → Mg + n
1.4
22
10
4
2
1.5
9
4
1.6
Since helium-4 is the basic building block,
most additional fusion processes will produce
nuclei with even atomic numbers.
9
4
12
6
25
12
1
0
4
2
1
0
N
l
ml
Orbital
designation
2
1
2p
3
2
3d
5
4
4
0
3
+1, 0,
−1
+2, +1,
…, −2
0
+3, +2,
…, −3
Number
of
orbitals
3
4s
4f
1
7
n=5, l = 3, and ml = -3,-2,-1,0,1,2,3
Li: σ = Z – Zeff ; σ = 3-1.28 = 1.72
Be: σ = Z – Zeff ; σ = 4-1.19 = 2.09
B: σ = Z – Zeff ; σ = 5-2.42 = 2.58
C: σ = Z – Zeff ; σ = 6-3.14 = 2.86
Be+ Be→ C+ He +2 n
N: σ = Z – Zeff ; σ = 7-3.83 = 3.17
O: σ = Z – Zeff ; σ = 8-4.45 = 3.55
F: σ = Z – Zeff ; σ = 9-5.10 = 3.90
1.17
1
The 1s electrons shield the positive charge
form the 2s electrons.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
2
1.18
See Figs 1.11 through 1.16
1.19
See Table 1.6 and discussion.
1.20
Table 1.6 shows SR > Ba < Ra. Ra is
anomalous because of higher Zeff due to
lanthanide contraction.
1.21
Anomalously high value for Cr is associated
with the stability of a half filled d shell.
1.22
(a) [He]2s22p2
1.30
(b) [He]2s22p5
(c) [Ar]4s2
(d) [Ar]3d10
(e) [Xe]4f145d106s26p3
(f) [Xe]4f145d106s2
1.23
(a) [Ar]3d14s2
CHAPTER 2
Self-tests
S2.1
(b) [Ar]3d2
(c) [Ar]3d5
(d) [Ar]3d4
(e) [Ar]3d
S2.2
6
(b) square planar
(f) [Ar]
10
(g) [Ar]3d 4s
1
S2.3
Linear
S2.4
S22– : 1σg22σu23σg21πu42πg4 ;
(h) [Xe]4f 7
1.24
(a) Angular
Cl2– : 1σg22σu23σg21πu42πg44σu1.
(a) Xe]4f145d46s2
S2.5
1σg22σu23σg21πu42πg4
(b) [Kr]4d6
S2.6
½[2-2+4+2] = 3
S2.7
Bond order: C≡N, C=N, and C–N; Bond
strength: C≡N > C=N > C–N.
S2.8
If it contains 4 or fewer electrons.
S2.9
–21 kJ mol–1
S2.10
(a) +1/2
6
(c) [Xe]4f
(d) [Xe]4f7
(e) [Ar]
(f) [Kr]4d
1.25
2
(a) S
(b) Sr
(b) +5
(c) V
(d) Tc
Exercises
(e) In
(f) Sm
2.1
(a) angular
1.26
See Figure 1.4.
(b) tetrahedral
1.27
(a) I1 increases across the row except for a dip
at S; (b) Ae tends to increase except for Mg
(filled subshell), P (half filled subshell), and
at AR (filled shell).
(c) tetrahedral
1.28
Radii of Period 4 and 5 d-metals are similar
because of lanthanide contraction.
1.29
2s2 and 2p0
2.2
(a) trigonal planar
(b) trigonal pyramidal
(c) square pyramidal
2.3
(a) T-shaped
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
2.4
(b) square planar
(b) one
(c) linear
(c) none
(a)
(d) two
2.15
3
(a) 1σg22σu2
(b) 1σg22σu21πu2
Cl
Cl
(c) 1σg22σu21πu43σg1
I
Cl
Cl
(d) 1σg22σu23σg21πu42πg3
2.16
The configuration for the neutral C2 would be
1σg21σu2 1πu4. The bond order would be ½[22+4] = 2.
2.17
(a) 1σg22σu23σg21πu42πg4
(b)
(b) 1
F
(c) There is no bond between the two atoms.
F
S
1800
2.18
120 0
2.5
2.6
(b) 1
(c) 2
F
F
(a) 2
2.19
(a) +0.5
(a) tetrahedral
(b) –0.5
(b) octahedral
(c) +0.5
2.20
(a) 176 pm
(b) 217 pm
(c) 221 pm
2.7
2(Si–O) = 932kJ > Si=O = 640kJ; therefore
two Si–O are preferred and SiO2 should (and
does) have four single Si–O bonds.
2.8
Multiple bonds are much stronger for period 2
elements than heavier elements
2.9
–483 kJ difference is smaller than expected
because bond energies are not accurate.
2.10
2.21
(a) 0
(a-c)
-1
(b) 205 kJ mol
2.11
Difference in electronegativities are AB 0.5,
AD 2.5, BD 2.0, and AC 1.0. The increasing
covalent character AD < BD < AC < AB.
2.12
(a) covalent
(b) ionic
(d) Possibly stable in isolation (only bonding
and nonbonding orbitals are filled; not stable
in solution because solvents would have
higher proton affinity than He.
(c) ionic
2.13
(a) sp2
(b) sp3
(c) sp3d or spd3
2 2
2
(d) p d or sp d.
2.14
(a) one
2.22
1
2.23
HOMO exclusively F; LUMO mainly S.
2.24
(a) electron deficient
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
4
(b) electron precise
CHAPTER 3
3.4
3.5
XA2
K3C60
3.6
281 pm
3.7
429 pm
3.8
CuAu. Primitive 12 carat.
3.9
Zintl phase region
3.10
(a) 6:6 and 8:8
Self-tests
S3.1
S3.2
See Fig. 3.7 and Fig. 3.32.
See Figure 3.35.
S3.3
52%
S3.4
rh = ((3/2)1/2 – 1) r = 0.225 r
(b) CsCI
S3.5
409 pm
3.11
6
S3.6
401 pm
3.12
S3.7
FeCr3
2B type and 4A type – in a distorted
octahedral arrangement.
S3.8
X2A3.
3.13
S3.9
Ti CN = 6 (thought these are as two slightly
different distances it is often described as 4 +
2) and O CN = 3.
(a) ρ = 0.78, so fluorite
(b) FrI? ρ = 0.94, so CsCl
(c) BeO? ρ = 0.19, so ZnS
(d) InN? ρ = 0.46, so NaCl
3.14
CsCl.
S3.10
LaInO3
3.15
S3.11
2421 kJ mol–1
Lattice enthalpies for the di- and the trivalent
ions. Also, the bond energy and third electron
gain enthalpy for nitrogen will be large.
S3.12
Unlikely
3.16
4 four times the NaCl value or 3144 kJmol–1.
S3.13
MgSO4 < CaSO4 < SrSO4 < BaSO4.
3.17
(a) 10906 kJmol–1
S3.14
NaClO4
(b) 1888 kJ mol–1
S3.15
Schottky defects.
(c) 664 kJ mol–1
S3.16
Phosphorus and aluminium.
S3.17
The dx2-y2 and dz2 have lobes pointing along
the cell edges to the nearest neighbor metals.
S3.18
3.18
(b) NaBF4
(a) n-type
(b) p-type
Exercises
3.1
a ≠ b ≠ c and α = 90°,
3.2
Points on the cell corners at (0,0,0), (1,0,0),
(0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and
(1,1,1) and in the cell faces at (½,½,0),
(½,1,½), (0,½,½) (½,½,1), (½,1,½), and
(1,½,½).
3.3
(a) MgSO4
= 90°,
= 90°
(c) and (f) are not as they have neighbouring
layers of the same position.
3.19
CsI < RbCl < LiF < CaO < NiO < AlN
3.20
Ba2+; solubilities decrease with increasing
radius of the cation.
3.21
(a) Schottky defects
(b) Frenkel defects
3.22
Solids have a greater number of defects as
temperatures approaches their melting points.
3.23
The origin of the blue color involves electron
transfer from cationic centres.
Vanandium carbide and manganese oxide.
3.24
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
3.25
3.26
3.27
Yes.
A semiconductor is a substance with an
electrical conductivity that decreases with
increasing temperature. It has a small,
measurable band gap. A semimetal is a solid
whose band structure has a zero density of
states and no measurable band gap.
Ag2S and CuBr: p-type; VO2: n-type.
CHAPTER 4
S4.9
5
Identify the acids and bases?
(a) FeCl3 + Cl– → [FeCl4]–, acid is FeCl3,
base is Cl–.
(b) I– + I2 → I3–, acid is I2, base is I–.
S4.10
The difference in structure between
(H3Si)3N and (H3C)3N? The N atom of
(H3Si)3N is trigonal planar, whereas the N
atom of (H3C)3N is trigonal pyramidal.
S4.11
Draw the structure of BF3·OEt2?
Self-tests
S4.1
(a) HNO3 + H2O → H3O+ + NO3–
HNO3, acid. Nitrate ion, conjugate base. H2O,
base. H3O+, conjugate acid.
(b) CO32– + H2O → HCO3– + OH–
carbonate ion, base; hydrogen carbonate, or
bicarbonate, conjugate acid; H2O, acid;
hydroxide ion, conjugate base.
Exercises
4.1
(c) NH3 + H2S → NH4+ + HS–
Ammonia, base; NH4+, conjugate acid;
hydrogen sulphide, acid; HS–, conjugate base.
S4.2
What is the pH of a 0.10 M HF solution?
pH= 2.24
S4.3
Calculate the pH of a 0.20 M tartaric acid
solution?
Sketch an outline of the s and p blocks of
the periodic table, showing the elements
that form acidic, basic, and amphoteric
oxides?
The elements that form basic oxides are in
plain type, those forming acidic oxides are in
outline type, and those forming amphoteric
oxides are in boldface type.
pH=1.85
S4.4
Which solvent?
dimethylsulfoxide (DMSO) and ammonia.
S4.5
Is aKBrF4 an acid or a base in BrF3?
A base.
S4.6
Arrange in order of increasing acidity?
The order of increasing acidity is [Na(H2O)6]+
<
[Ni(H2O)6]2+
<
<
[Mn(H2O)6]2+
3+
[Sc(H2O)6] .
S4.7
4.2
Predict pKa values? (a) H3PO4 pKa ≈ 3. The
actual value, given in Table 4.1, is 2.1.
(b)
pKa(2) ≈ 8. The actual value,
given in Table 4.1, is 7.4.
of
the
(b) HSO4–? The conjugate base is SO4–.
(c) CH3OH? The conjugate base is CH3O–.
(c) HPO42– pKa(3) ≈ 13. The actual value,
given in Table 4.1, is 12.7.
(d) H2PO4–? The conjugate base is HPO42–.
(e) Si(OH)4?
SiO(OH)3–.
What happens to Ti(IV) in aqueous
solution as the pH is raised?
Treatment with ammonia causes the
precipitation of TiO2. Further treatment with
NaOH causes the TiO2 to redissolve.
bases
(a) [Co(NH3)5(OH2)]3+, conjugate base is
[Co(NH3)5(OH)]2+.
H2PO4–
S4.8
Identify the conjugate
following acids?
(f) HS−?
4.3
The
conjugate
base
is
The conjugate base is S2–.
Identify the conjugate
following bases?
acids
of
the
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
6
ClO4–, cannot be studied in sulfuric acid.
(a) C5H5N (pyridine)? The conjugate acid
is pyridinium ion, C5H6N+.
4.9
(b) HPO42–? The conjugate acid is H2PO42–.
(c) O2–? The conjugate acid is OH–.
electron withdrawing
(d) CH3COOH? The conjugate acid is
CH3C(OH)2+.
(e) [Co(CO)4]–? The conjugate acid is
HCo(CO)4.
(f) CN–?
4.4
Calculate the [H3O+] and pH of a 0.10 M
butanoic acid solution? pH=2.85
4.5
What is the Kb of ethanoic acid?
4.6
4.7
4.11
Is the pKa for HAsO42– consistent with
Pauling’s rules? No. Pauling’s rules are
only approximate.
4.12
What is the order of increasing acid
strength for HNO2, H2SO4, HBrO3, and
HClO4?
The conjugate acid is HCN.
Kb = 5.6 × 10
the order is HClO4 > HBrO3 > H2SO4 >
HNO2.
4.13
–10
What is the Ka for C5H5NH+?
Ka = 5.6 × 10–6
-
Predict if F will behave as an acid or a
base in water?
4.14
F will behave as a base in water.
What are the structures and the pKa values
of chloric (HClO3) and chlorous (HClO2)
acid?
O
O
the aluminum-containing species is more
acidic.
(c) Si(OH)4 or Ge(OH)4?
Cl
H
O
Which of the following is the stronger acid?
(a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+? The
Fe(III) complex, [Fe(OH2)6]3+, is the stronger
acid.
(b) [Al(OH2)6]3+ or [Ga(OH2)6]3+?
O
Cl
Account for the trends in the pKa values of
the conjugate acids of SiO44–, PO43–, SO42–,
and ClO4–?
The acidity of the four conjugate acids
increases in the order HSiO43– < HPO42– <
HSO4– < HClO4.
-
4.8
Is the –CN group electron donating or
withdrawing?
H
Si(OH)4, is more acidic.
O
(d) HClO3 or HClO4?
chloric acid
HClO4 is a stronger acid.
chlorous acid
(e) H2CrO4 or HMnO4?
Chloric acid, the predicted pKa = –2; actual
value = –1.
HMnO4 is the stronger acid.
(f) H3PO4 or H2SO4?
Chlorous acid, the predicted pKa = 3; actual
value = 2.
4.8
Which bases are too strong or too weak to
be studied experimentally? (a) CO32– O2–,
ClO4–, and NO3– in water?
H2SO4 is a stronger acid.
4.15
CO32– is of directly measurable base strength.
O2–, is too strong to be studied experimentally
in water.
ClO42– and NO3– are too weak to be studied
experimentally.
order of increasing basicity is Cl2O7 < SO3 <
CO2 < B2O3 < Al2O3 < BaO.
4.16
(b) HSO4–, NO3–, and ClO4–, in H2SO4?
HSO4–, not too
experimentally.
–
strong
to
be
Arrange the following in order of
increasing acidity? HSO4−, H3O+, H4SiO4,
CH3GeH3, NH3, and HSO3F?
increasing acidity is NH3 < CH3GeH3 <
H4SiO4 < HSO4– < H3O+ < HSO3F.
studied
NO3 is of directly measurable base strength
in liquid H2SO4.
Arrange the following oxides in order of
increasing basicity? Al2O3, B2O3, BaO,
CO2, Cl2O7, and SO3?
4.17
Which aqua ion is the stronger acid, Na+ or
Ag+? Ag+(aq).
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
4.18
(d) AsF3(g) + SbF5(g) → [AsF2][SbF6]?
The very strong Lewis acid SbF5 displaces the
Lewis acid [AsF2]+ from the Lewis base F–.
Which of the following elements form oxide
polyanions or polycations? Al, As, Cu, Mo,
Si, B, Ti?
polycations: Al, Cu, and Ti.
(e) EtOH readily dissolves in pyridine? A
Lewis acid–base complex formation reaction
between EtOH (the acid) and py (the base)
produces the adduct EtOH–py.
polyoxoanions (oxide polyanions): Mo
polyoxoanions: As, B, and Si.
4.19
The change in charge upon aqua ion
polymerization?
4.26
Polycation formation reduces the average
positive charge per central M atom by +1 per
M.
4.20
2[Fe(OH2)6]3+ →
2-Me-py or 4-Me-py?
4.27
4.24
4.25
–
2HCO 3-
Give the equations for HF in H2SO4 and
HF in liquid NH3?
4.28
H2SO4 + HF ⇔ H3SO4+ + F-
4.23
<1
(d) [AgCl2] (aq) + 2 CN (aq) → [Ag (CN)2]–
(aq) + 2Cl–(aq)? >1
–
(b) CO2 and CaCO3?
NH3 + HF
Which of the following reactions have Keq >
1? (a) R3P–BBr3 + R3N–BF3→ R3P–BF3 +
R3N–BBr3? <1
(c) CH3HgI + HCl → CH3HgCl + HI?
H3PO4 + HPO42- ÅÆ 2H2PO4-
4.22
4-Me-py.
(b) SO2 + Ph3P–HOCMe3 → Ph3P–SO2 +
HOCMe3? > 1.
(a) H3PO4 and Na2HPO4?
Ca2+ +
B(n-Bu)3.
(b) More basic toward BMe3: NMe3 or
NEt3? NMe3.
More balanced equations?
CO 2 + CaCO 3 + H 2O
Boron trichloride.
B(n-Bu)3 or B(t-Bu)3?
[(H2O)4Fe(OH)2Fe(OH2)4+ + 2H3O+
4.21
Select the compound with the named
characteristic? (a) Strongest Lewis acid:
BF3, BCl3, or BBr3? BBr3.
BeCl2 or BCl3?
Write balanced equations for the formation
of P4O124– from PO43– and for the formation
of [(H2O) 4Fe(OH)2Fe(OH2)4]4+ from
[Fe(OH2)6]3+?
4PO43– + 8H3O+ → P4O124– + 12H2O
7
The phosphorus atom in Me2NPF2 is the
softer of the two basic sites, so it will bond
more strongly with the softer Lewis acid BH3
NH2- + H2F+
Why is H2Se a stronger acid than H2S? As
you go down a family in the periodic chart,
the acidy of the homologous hydrogen
compounds increases.
Identifying elements that form Lewis
acids? All of the p-block elements except
nitrogen, oxygen, fluorine, and the lighter
noble gases form Lewis acids in one of their
oxidation states.
The hard nitrogen atom will bond more
strongly to the hard Lewis acid BF3.
4.29
Why is trimethylamine out of line?
Trimethyl amine is sterically large enough to
fall out of line with the given enthalpies of
reaction.
4.30
Discuss relative basicities? (a) Acetone and
DMSO?
DMSO is the stronger base regardless of how
hard or how soft the Lewis acid is. The
ambiguity for DMSO is that both the oxygen
atom and sulfur atom are potential basic sites.
Identifying acids and bases: (a) SO3 + H2O
→ HSO4– + H+? The acids in this reaction
are the Lewis acids SO3 and H+ and the base
is the Lewis base OH–.
(b) Me[B12]– + Hg2+ → [B12] +
MeHg+? The Lewis acid Hg2+ displaces the
Lewis acid [B12] from the Lewis base CH3–.
(c) KCl + SnCl2 → K+ + [SnCl3]–? The
Lewis acid SnCl2 displaces the Lewis acid K+
from the Lewis base Cl–.
Choose between the two basic sites in
Me2NPF2?
(b) Me2S and DMSO? Depending on the
EA and CA values for the Lewis acid, either
base could be stronger.
4.31
Write a balanced equation
dissolution of SiO2 by HF?
for
the
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
8
SiO2 + 6HF
2H2O + H2SiF6
or
SiO2 + 4HF
C6H5COOH + HF
2H2O + SiF4
Both a Brønsted acid–base reaction and a
Lewis acid–base reaction.
4.32
(c) C6H5COOH?
4.37
The dissolution of silicates by HF? both
4.38
Are the f-block elements hard? yes.
4.39
Calculate the enthalpy change for I2 with
phenol?
ΔfH = -20.0kJ/mol
Al2 O 3 + 3H 2 S
Al2 S 3 + 3H 2 O
Describe solvent properties? (a) Favor
displacement of Cl– by I– from an acid
center? If you choose a solvent that
decreases the activity of chloride relative to
iodide, you can shift the following
equilibrium to the right:
acid-Cl- + I-
φ
Write a balanced equation to explain the
foul odor of damp Al2S3? The foul odor
suggests H2S formation.
4.33
CHAPTER 5
Self-tests
S5.1
acid-I- + Cl-
(b) Favor basicity of R3As over
R3N? Alcohols such as methanol or ethanol
would be suitable.
+
Propose a mechanism for the acylation of
benzene? An alumina surface, such as the
partially dehydroxylated one shown below,
would provide Lewis acidic sites that could
abstract Cl–:
4.35
Why does Hg(II) occur only as
HgS? Mercury(II) is a soft Lewis acid, and
so is found in nature only combined with soft
Lewis bases, the most common of which is
S2–.
4.36
Write Brønsted acid–base reactions in
liquid HF?
(a) CH3CH2OH?
is:
CH 3CH 2OH + HF
(b) NH3? The equation is
NH
3
+
HF
S5.2
Does Cu metal dissolve in dilute HCl? No.
S5.3
Can Cr2O72– be used to oxidize Fe2+, and
would Cl– oxidation be a problem? Yes.
Cl- oxidation is not a problem.
S5.4
Fuel cell emf with oxygen and hydrogen
gases at 5.0 bar?
E = 1.25 V.
S5.5
The fate of SO2 emitted into clouds? The
aqueous solution of SO42– and H+ ions
precipitates as acid rain.
S5.6
Can Fe2+ disproportionate under standard
conditions? No.
S5.7
bpy binding to Fe(III) or Fe(II)? Fe(II)
preferentially.
S5.8
Potential of AgCl/Ag,Cl– couple?
Eox= – 1.38 V
S5.9
Latimer diagram for Pu? (a) Pu(IV)
disproportionates to Pu(III) and Pu(V) in
aqueous solution; (b) Pu(V) does not
disproportionate into Pu(VI) and Pu(IV).
S5.10
Frost diagram for thallium in aqueous
acid?
The balanced equation
CH 3CH 2OH 2+ + FNH
4
+
+
F-
oxidation
2MnO4− (aq) + 5Zn(s) + 16H+(aq) →
5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l)
3+
(d) Promote the reaction 2FeCl3 + ZnCl2
→Zn2+ + 2[FeCl4]−? A suitable solvent is
acetonitrile, MeCN.
Half-reactions and balanced reaction for
oxidation of zinc metal by permanganate
ions?
2[MnO4− (aq) + 8H+ (aq) + 5e– → Mn2+(aq)
reduction
+ 4H2O(l)]
5 [ Zn(s) → Zn2+(aq) + 2e– ]
(c) Favor acidity of Ag over Al ? An
example of a suitable solvent is diethyl ether.
Another suitable solvent is H2O.
4.34
C6H5COO- + H2F+
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
5.3
9
Write balanced equations, if a reaction
occurs, for the following species in aerated
aqueous acid? (a) Cr2+?
4Cr2+(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq)
Eº = 1.65 V
+ 2H2O(l)
(b) Fe2+?
4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq)
Eº = 0.46 V
+ 2H2O(l)
(c) Cl–? no reaction.
(d) HOCl? No reaction.
S5.11
The oxidation number of manganese?
(e) Zn(s)?
2Zn(s) + O2(g) + 4H+(aq) → 2Zn2+(aq)
Eº = 1.99 V
+2H2O(l)
Mn2+(aq)
S5.12
Compare the strength of
oxidizing agent in acidic
solution?
NO3–
and
as an
basic
A competing reaction is:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (Eº =
0.763 V).
Nitrate is a stronger oxidizing agent in acidic
solution than in basic solution.
S5.13
S5.14
The possibility of finding Fe(OH)3 in a
waterlogged soil? Fe(OH)3 is not stable.
5.4
Balanced equations for redox reactions?
(a) Fe2+?
(i) Fe2+ will not oxidize water.
The minimum temperature for reduction
of MgO by carbon? 1800ºC or above.
(ii) Fe2+ will not reduce water.
Exercises
(iii) Fe2+ will reduce O2 and in doing so will
be oxidized to Fe3+.
5.1
(iv) disproportionation will not occur.
Oxidation numbers?
2 NO(g) + O2(g) → 2 NO2(g)
+1 -2
0
+4 -2
(b) Ru2+? Ru2+ will not oxidize or reduce
water. Ru2+ will reduce O2 and in doing so
will be oxidized to Ru3+. Ru2+ will
disproportionate in aqueous acid to Ru3+ and
metallic ruthenium.
2Mn3+(aq) + 2H2O → MnO2 + Mn2+ + 4H+(aq)
+3
+1 -2
+4 -2
+2
+1
LiCoO2(s) + C(s) → LiC(s) + CoO2(s)
+1 +3 -2
0
+1-1
+4 -2
Ca(s) + H2(g)
0
0
5.2
(c) HClO2? HClO2 will oxidize water, will
not reduce water. HClO2 will reduce O2 and
in doing so will be oxidized to ClO3–. HClO2
will disproportionate in aqueous acid to ClO3–
and HClO.
→ CaH2(s)
+2 -1
Suggest chemical reagents for redox
transformations? (a) Oxidation of HCl to
Cl2? S2O82–, H2O2, or α–PbO2 to oxidize Cl–
to Cl2.
(d) Br2? Br2 will not oxidize or reduce water.
Br2 will not
Br2 will not reduce O2.
disproportionate in aqueous acid to Br– and
HBrO.
(b) Reducing Cr3+(aq) to Cr2+(aq)? metallic
manganese, metallic zinc, or NH3OH+.
5.5
Standard potentials vary with
temperature in opposite directions? The
amino and cyano complexes must have
different equilibrium shifts with respect to
changes in temperature that results in the
opposite directions of change for the cell
potential.
(c) Reducing Ag+(aq) to Ag(s)? The reduced
form of any couple with a reduction potential
less than 0.799 V.
(d) Reducing I2 to I–? The reduced form of
any couple with a reduction potential less than
0.535 V.
5.6
Balance redox reaction in acid solution:
MnO4– + H2SO3 → Mn2+ + HSO4–? pH
dependence?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
10
2MnO4− (aq) + 5H2SO3(aq) + H+(aq) →
2Mn2+(aq) +5HSO3− (aq) + 3H2O(l)
The potential decreases as the pH increases.
5.7
5.13
Calculate the equilibrium constant for
Au+(aq) + 2CN–(aq) → [Au(CN)2]–(aq)?
K = 5.7 × 1038
5.14
Find the approximate potential of an
aerated lake at pH = 6, and predict the
predominant species? (a) Fe? 0.5 – 0.6 V
Write the Nernst equation for (a) The
reduction of O2?
(b) Mn? E = 0.55 V
Q = 1/(p(O2)[H+]4)
(c) S? At pH 0, 0.387 V. At pH 14, SO42–
would again predominate. HSO4– is the
predominant sulfur species at pH 6.
and
E = Eº – [(0.059V)/4][log(1/(p(O2)[H+]4)]
(b) The reduction of Fe2O3(s)?
5.15
Frost diagram and standard potential for
the HSO4−/S8(s) couple? 0.387 V
5.16
Equilibrium constant for the reaction
Pd2+(aq) + 4 Cl–(aq) ≡ [PdCl4]2–(aq) in 1 M
HCl(aq)? K = 4.37 × 1010
5.17
Reduction potential for MnO4– to MnO2(s)
at pH = 9.00? E = 0.98 V
5.18
Tendency of mercury species to act as an
oxidizing agent, a reducing agent, or to
undergo disproportionation? Hg2+ and
Hg22+ are both oxidizing agents. None of
these species are likely to be good reducing
agents. Hg22+ is not likely to undergo
disproportionation.
Thermodynamic tendency of HO2 to
undergo disproportionation? E = +1.275 V.
(is
positive),
HO2
will
undergo
disproportionation.
Dissolved carbon dioxide corrosive towards
iron? Carbon dioxide and water generate
carbonic acid which encourages the corrosion
process by lowering solution pH.
What is the maximum E for an anaerobic
environment rich in Fe2+ and H2S? –0.1 V.
+ 6
Q = 1/[H ] and E = Eº – (RT/nF)(13.8 pH)
5.8
Using Frost diagrams? (a) What happens
when Cl2 is dissolved in aqueous basic
Cl2 is thermodynamically
solution?
susceptible to disproportionation to Cl– and
ClO4– when it is dissolved in aqueous base.
The oxidation of ClO– is slow, so a solution of
Cl– and ClO– is formed when Cl2 is dissolved
in aqueous base.
(b) What happens when Cl2 is dissolved in
aqueous acid solution? Cl2 will not
disproportionate. Cl2 is thermodynamically
capable of oxidizing water.
(c) Should HClO3 disproportionate in
aqueous acid solution? Kinetic.
5.9
Write equations for the following
reactions: (a) N2O is bubbled into aqueous
NaOH solution?
5N2O(aq) + 2OH–(aq) → 2NO3–(aq) +
4N2 (g) + H2O(l)
(b) Zinc metal is added to aqueous acidic
sodium triiodide?
Zn(s) + I3 (aq) → Zn (aq) + 3I (aq)
–
2+
3I2(s) + 5ClO3– (aq) + 3H2O(l) → 6IO3–
(aq) + 5Cl– (aq) + 6H+(aq)
5.10
Electrode potential for Ni2+/Ni couple at pH
= 14? E =– 0.21 V
5.11
Will acid or base most favour the following
half-reactions? (a) Mn2+ → MnO4–? Base
(b) ClO4– → ClO3–? Acid
Base
(d) I2 → 2I–? Acid or base, no difference.
5.12
5.20
5.21
5.22
–
(c) I2 is added to excess aqueous acidic
HClO3?
(c) H2O2 → O2?
5.19
Determine the standard potential for the
reduction of ClO4– to Cl2? 1.392 V
How will edta4– complexation affect M2+ →
M0 reductions? The reduction of a M(edta)2–
complex will be more difficult than the
reduction of the analogous M2+ aqua ion.
5.23
Which of the boundaries depend on the
choice of [Fe2+]? Any boundary between a
soluble species and an insoluble species will
change as the concentration of the soluble
species changes. The boundaries between the
two soluble species, and between the two
insoluble species, will not depend on the
choice of [Fe2+].
5.24
Under what conditions will Al reduce
MgO? Above about 1400ºC.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 11
CHAPTER 6
SF6 has Oh symmetry. Analysis of the
stretching vibrations leads to:
Γstr = A1g (Raman, polarized) + Eg
Self-tests
S6.1
S6.2
Sketch the S4 axis of an NH4+ ion. How
many of these axes are there in the ion?
Three S4 axes.
(Raman) + T1u (IR).
SF5Cl has C4v symmetry.
stretching vibrations leads to:
Γstr = 3A1 (IR and Raman, polarized)
(a) BF3 point group? D3h.
2–
(b) SO4 point group? Td.
S6.3
S6.4
S6.5
S6.14 Symmetries of all the vibration modes of
[PdCl4]2-? A1g + B1g + B2g + A2u + B2u + 2Eu
A conformation of the ferrocene molecule
that lies 4 kJ mol–1 above the lowest energy
configuration is a pentagonal antiprism. Is
it polar? No.
6.1
Is the skew form of H2O2 chiral? Yes.
S6.7
Can the bending mode of N2O be Raman
active? Yes.
S6.8
Confirm that the symmetric mode is Ag?
D2h character table, which is the Ag symmetry
type.
Show that the four CO displacements in
the square-planar (D4h) [Pt(CO)4]2+ cation
transform as A1g + B1g + Eu. How many
bands would you expect in the IR and
Raman spectra for the [Pt(CO)4]2+ cation?
The reducible representation:
D4h
E
2C4
C2
2C2’
2C2″
i
2S4
σh
2 σv
2σd
Γ3N
4
0
0
2
0
0
0
4
2
0
Reduces to A1g + B1g + Eu
A1g + B1g are Raman active. Eu is IR active.
S6.10
Orbital symmetry for a tetrahedral array
of H atoms in methane? A1
S6.11
Orbital symmetry for a square-planar
array of H atoms? B2g.
S6.12
Which Pt atomic orbitals can combine with
which of these SALCs? The atomic orbitals
much have matching symmetries to
generate SALCs. 5s and 4dz2 have A1g
symmetry; the dx2-y2 has B1g symmetry; and
5px and 5py have Eu symmetry.
S6.13
+ 2B1 (Raman) + E (IR, Raman).
Symmetry species of all five d orbitals of
the central Xe atom in XeF4 (D4h, Fig. 6.3)?
dx2-y2 is B1g;
dxy is B2g;
dxz and dyz are Eg;
dz2 is A1g.
What is the maximum possible degeneracy
for an Oh molecule? 3.
S6.6
S6.9
Analysis of the
Predict how the IR and Raman spectra of
SF5Cl differ from that of SF6?
S6.15
+T1u.
SALCs for sigma bonding in O? A1g + Eg
Exercises
Symmetry elements? (a) a C3 axis and a σv
plane in the NH3 molecule?
N
H
N
H
H
H
H
H
σv
C3
(b) a C4 axis and a σh plane in the squareplanar [PtCl4]2– ion?
Cl
Cl
Cl
Cl
Cl
Cl
Cl
C4
6.2
Cl
Pt
Pt
σh
S4 or i? (a) CO2? i
(b) C2H2? i.
(c) BF3? neither.
(d) SO42–? three different S4.
6.3
Assigning point groups: (a) NH2Cl? Cs
(b) CO32–? D3h
(c) SiF4? Td
(d) HCN? C∞v.
(e) SiFClBrI? C1.
(f) BrF4–? D4h.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
12
6.4
6.5
How many planes of symmetry does a
benzene molecule possess? What chlorosubstituted benzene has exactly four planes
of symmetry? 7,and C6H3Cl3.
The symmetry elements of orbitals? (a) An
s orbital? Infinite number of Cn axes, plus
an infinite number of mirror planes of
symmetry, plus center of inversion, i.
6.12
and irreducible representations?
A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu
6.13
(b) A p orbital? An infinite number of mirror
planes that pass through both lobes and
include the long axis of the orbital. In
addition, the long axis is a Cn axis, where n
can be any number from 1 to ∞.
(c) A dxy orbital? Center of symmetry, three
mutually perpendicular C2 axes, three
mutually perpendicular mirror planes of
symmetry, two planes that are rotated by 45º
about the z axis from the xz plane and the yz
plane.
(d) A dz2 orbital? In addition to the
symmetry elements possessed by a p orbital:
(i) a center of symmetry, (ii) a mirror plane
that is perpendicular to the C∞ axis, (iii) an
infinite number of C2 axes that pass through
the center of the orbital and are perpendicular
to the C∞ axis, and (iv) an S∞ axis.
6.6
For the 90o-twisted form of B2F4 (D2d)
The vibrations are: 3A1 (Raman, polarized) +
B1 (Raman) + 2B2 (IR and Raman ) + 3E (IR
and Raman).
6.14
(a) Take the 4 hydrogen 1s orbitals of CH4
and determine how they transform under
Td. (b) Confirm that it is possible to reduce
this representation to A1 + T2. (c) Which
atomic orbitals on C can form MOs with
H1s SALCs?
Using symmetry Td, Γ3N reduces to: A1 + T2.
The MOs would be constructed from SALCs
with H1s and 2s and 2p atomic orbitals on C.
6.15
Use the projection operator method to
construct the SALCs of A1 + T2 symmetry
that derive from the four H1s orbitals in
methane..
SO32– ion? (a) Point group? C3v
(c) Which s and p orbitals have the
maximum degeneracy? 3px and 3py orbitals
are doubly degenerate.
PF5? (a) Point group? D3h.
(b) Degenerate MOs? 2.
(c) Which p orbitals have the maximum
degeneracy? 3px and 3py atomic orbitals are
doubly degenerate
6.8
AsCl5 Raman spectrum consistent with a
trigonal bipyamidal geometry? No.
6.9
Vibrational modes of SO3? (a) In the plane
of the nuclei? 5
(b) Perpendicular to the molecular plane?
1
6.10
Vibrations that are IR and Raman active?
(a) SF6? None.
(b) BF3? The E′ modes are active in both IR
and Raman.
6.11
IR and Raman to distinguish between: (a)
planar and pyramidal forms of PF3, (b)
planar and 90o-twisted forms of B2F4 (D2h
and D2d respectively)?
(a) Planar PF3, D3h, vibrations are: A1’
(Raman, polarized) + 2E’ (IR and Raman) +
A2” (IR).
Pyramidal PF3, C3v, vibrations are: 2A1 (IR
and Raman, polarized) + 2E’ (IR and Raman)
(b) For the planar form of B2F4 (D2h):
The vibrations are:
3Ag (Raman, polarized) + 2B2g (Raman) +
B3g (Raman) + Au(inactive) + 2B1u (IR) +
B2u (IR) + 2B3u (IR).
(b) Degenerate MOs? 2
6.7
[AuCl4]− ion? Γ of all 3N displacements
Vibrations of a C6v molecule that are
neither IR nor Raman active? Any A2, B1,
or B2 vibrations of a C6v molecule will not be
observed in either the IR spectrum or the
Raman spectrum.
s = (1/2)(ϕ1 + ϕ2 + ϕ3 + ϕ3) (= A1)
px = (1/2)(ϕ1 – ϕ2 + ϕ3 – ϕ3) (= T2)
py = (1/2)(ϕ1 – ϕ2 – ϕ3 + ϕ3) (= T2)
pz = (1/2)(ϕ1 + ϕ2 – ϕ3 – ϕ3) (= T2)
SALCs for σ-bonds
(a) BF3?
(1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’)
(1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (=
E’)
(b) PF5?
(axial F atoms are ϕ4 + ϕ5)
(1/√2)(ϕ4 + ϕ5) (= A1’)
(1/√2)(ϕ4 − ϕ5) (= A2”)
(1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’)
(1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (=
E’)
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 13
Exercises
CHAPTER 7
7.1
Self-tests
S7.1
Name and draw the structures of the
Nickel
complexes? (a) [Ni(CO)4]?
tetracarbonyl or tetracarbonyl nickel(0).
Give formulas corresponding to the
following names? (a) Cisdiaquadichloroplatinum(II)? cis[PtCl2(OH2)2],
CO
Ni
trans-diaquadichloroplatinum(II), trans[PtCl2(OH2)2].
OC
(b)
Diamminetetra(isothiocyanato)
chromate(III)?
[Cr(NCS)4(NH3)2] –. ,can exist as, cis[Cr(NCS)4(NH3)2] – or trans[Cr(NCS)4(NH3)2] –.
(c) Tris(ethylenediamine)rhodium
[Rh(en)3]3+.
(b) [Ni(CN)4]2–? Tetracyanonickelate (II).
2-
NC
(III)?
Ni
CN
CN
NC
(d)
Bromopentacarbonylmanganese
(I)? [MnBr (CO)5].
(c) [CoCl4]2–? Tetrachlorocobaltate (II)
(e) Chlorotris(triphenylphosphine)rhodium
(I)? [RhCl(PPh3)3].
S7.2
CO
CO
2-
Cl
What type of isomers are possible for
[Cr(NO2)2•6H2O]? The hydrate isomers
and linkage isomers of the NO2 group. Also,
[Cr(ONO)(H2O)5]NO2 •H2O.
Co
Cl
Cl
Cl
S7.3
Identifying isomers? Note that the two
phosphine ligands in the trans isomer are
related, therefore, they exhibit the same
chemical shift.
(d) [Mn(NH3)6]2+?
Hexaamminemanganesium (II)
2+
S7.4
Sketches of the mer and fac isomers of
[Co(gly)3]?
H3N
NH3
Mn
H3N
NH3
NH3
NH3
7.2
Write the formulas for the following
complexes?
(a) [CoCl(NH3)5]Cl2
(b) [Fe(OH2)6](NO3)3
(c) cis-[FeCl2(en)2]
S7.5
–
(b) trans-[CrCl2(ox)2]3 ? Not chiral.
S7.6
(d) [Cr(NH3)5μ–OH–Cr(NH3)5]Cl5
Which of the following are chiral? (a) cis[CrCl2 (ox)2]3–? Chiral .
7.3
Name the following complexes?
(c) cis-[RhH(CO)(PR3)2]? Not chiral.
(a) cis-[CrCl2(NH3)4]+? cistetra(ammine)di(chloro)chromium(III)
Calculate all of the stepwise formation
constants? Kf1 = 1 X 105. Kf2 will be 30%
less or 30000, Kf3 = 9000, Kf4 = 2700, Kf5 =
810, and finally Kf6 = 243.
(b) trans-[Cr(NCS)4(NH3)2] ? transdi(ammine)tetrakis(isothiocyanato)chromate
(III)
-
(c) [Co(C2O4)(en)2]+?
bis(ethylenediamine)oxalatocobalt(III).
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
14
7.4
Four-coordinate complexes? (a) Sketch the
two observed structures?
L
M
M
L
L
L
L
L
L
L
square planar
tetrahedral
(b)
Isomers expected for MA2B2?
tetrahedral complex, no isomers, for a squareplanar complex, two isomers, cis and trans.
7.5
For five-coordinate complexes, sketch the
two observed structures?
A
A
B
B
E
M
M
E
B
E
B
A
Trigonal Bipyramidal
Square based pyramid
A = axial ligands
A = axial
B = basal
E = equatorial ligands
7.6
Six-coordinate complexes? (a) Sketch the
two observed structures?
[Mg(edta)(OH2)]2–
7.11
What types of isomers are [RuBr(NH3)5]Cl
and [RuCl(NH3)5]Br? Ionization isomers.
7.12
Which complexes have isomers?
[CoBrClI(OH2)]
7.13
Which complexes have isomers?
(a) [Pt(ox)(NH3)2] no isomers
(b) [PdBrCl(PEt3)2] has two isomers.
(b) Which one of these is rare? Trigonal
prism.
7.7
7.8
7.9
7.10
Explain
the
difference
between
monodentate,
bidentate,
and
quadridentate? A monodentate ligand can
bond to a metal atom only at a single atom, a
bidentate ligand can bond through two atoms,
a quadridentate ligand can bond through four
atoms.
What type of isomers do you get with
ambidentate ligands? linkage isomers.
Which ligand could act like a chelating
ligand? (a) Triphenylphosphite, no/
(b) Bis(dimethyl)phosphino ethane (dmpe)
yes.
(c) Bipyridine (bipy), yes.
(d) Pyrazine, no.
Draw structures of complexes that contain
the ligands (a) en, (b) ox, (c) phen, and (d)
edta? .
(c) [IrHCO(PR3)2] has two isomers.
(d) [Pd(gly)2] has two isomers.
7.14
How many isomers are possible for the
following complexes?
(a) [FeCl(OH2)5]2+? None.
(b) [IrCl3(PEt3)2]? 2
(c) [Ru(biby)3]2+? 2
(d) [CoCl2(en)(NH3)2]+? 4
(e) [W(CO)4(py)2] 2
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 15
7.15
Draw all possible isomers for [MA2BCDE]?
Including optical isomers, 15 isomers are
possible! .
A
B
A
A
B
D
E
M
C
A
M
E
C
A
E
B
M
M
C
C
C
B
D
D
A
A
A
A
A
D
D
E
M
A
C
C
D
M
C
S8.4
E
A
E
S8.3
M
A
B
7.16
B
D
A
E
D
A
C
D
A
B
S8.2
A
M
E
identical reflections to those of rutile TiO2 but
shifted to slightly higher diffraction angles.
S8.5
A
M
B
14% of naturally occurring tungsten is 183W,
which has I = ½. Thus, the signal is split into
2 lines.
B
E
Which of the following complexes are
chiral? (a) [Cr(ox)3]3–? Chiral
S8.6
(b) cis-[PtCl2(en)]? Chiral (The en is not
planar).
S8.7
(c) cis-[RhCl2(NH3)4]+? not chiral.
(d) [Ru(bipy)3]2+? chiral
7.17
7.18
Draw both isomers, Λ or Δ, of the complex
[Ru(en)3]+2 ?
How would you determine crystalline
components in mineral sample? Powder Xray diffraction.
8.2
Why are there no diffraction maxima in
borosilicate glass? Glass has no long-range
periodicity or order.
8.3
What is the minimum size of a cubic
crystal that can be studied? 0.5 μm by 0.5
μm by 0.5 μm.
8.4
Wavelength of neutron at 2.20 km/s? 1.80 ×
10–12 m or 180 pm.
8.5
Order of stretching frequencies? The
smaller effective mass of the oscillator for
CN− causes the molecule to have the higher
stretching frequency. The bond order for NO
is 2.5, and N is heavier than C, hence CO has
a higher stretching frequency than NO.
8.6
Wavenumber for O–O in O2+? In the region
of 1800 cm 1.
8.7
UV photoelectron spectrum of NH3? The
band at 11 eV is due to the lone pair and the
pyramidal angle. The ionised molecule has
greater planarity, thus the long progression.
8.8
Raman bands assignments? N(SiH3)3 is
planar. N(CH3) 3 is pyramidal.
N
N
N
N
N
Ru
Ru
N
N
8.1
Chiral (dien is
Which isomer, Λ or Δ, is the complex
Mn(acac)3, shown in the exercise? The Λ
isomer.
N
N
N
N
N
Λ isomer
Δ isomer
7.19
Suggest a reason why Kf5 is so different?
Because of a change in coordination.
7.20
Compare these values with those of
ammonia given in exercise 7.19 and suggest
why they are different? The chelate effect.
CHAPTER 8
Self-tests
S8.1
Isomer shift for iron in Sr2FeO4? The
Smaller and less positive.
Why does the mass spectrum of ClBr3 have
five peaks separated by 2 u? Halogen
isomers.
Exercises
(e) fac-[Co(NO2)3(dien)]? Not chiral.
(f) mer-[Co(NO2)3(dien)]?
not planar).
TiO2 in sunscreens?
Titania articles absorb this ultraviolet
radiation
Molecular shape and vibrational modes for
XeF2? Trigonal bipyramidal, 4 total
vibrational modes.
(a) 77Se-NMR spectrum consists of a triplet
of triplets? The triplet of triplets.
(b) Proton resonance of the hydrido ligand
consist of eight equal intensity lines? yes.
EPR signal of new material arises from W
sites?
Main features of the CrO2 powder XRD
pattern? XRD pattern for CrO2 will show
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
16
8.9
Single 13C peak in NMR? Chemically
distinct carbonyls are exchanging position
sufficiently quickly.
8.10
Form of 19F-NMR and 77Se-NMR spectra
of 77SeF4? 19F NMR spectrum reveals two
1:3:3:1 quartets. The 77Se-NMR spectrum is a
triplet of triplets.
NMR spectral features for XeF5−? All 5 of
the F atoms are chemically equivalent.
Approximately 25% is present as 129Xe, I =
1/2, and in this case the 19F resonance is a
doublet. The final result is a composite: two
lines of 12.5% intensity from the 19F coupled
to the 129Xe, and one remaining line of 75%.
8.11
8.12
g-values? 1.94, 1.74, and 1.56.
8.13
Slower process, NMR or EPR? NMR.
8.14
Differences in EPR spectrum for d-metal
with one electron in solution versus frozen?
In aqueous solution at room temperature,
molecular tumbling removes the effect of the
g-value anisotropy. In frozen solution, gvalue anisotropy can be observed.
8.15
8.16
8.17
8.18
8.19
8.20
8.21
8.22
Isomer shift for iron in BaFe(VI)O4? A
positive shift for Fe(VI) well below +0.2 mm
s-1.
Charge on Fe atoms in Fe4[Fe(CN)6]3? EPR
and Mössbauer.
No quadrupole splitting in Mössbauer
spectrum of SbF5? The geometry must be
close to cubic in the solid state.
No peak in the mass spectrum of Ag at 108
u? Two isotopes, 107Ag (51.82%) and 109Ag
(48.18%). Compounds that contain silver will
have two mass peaks.
Peaks
in
mass
spectrum
of
Mo(C6H6)(CO)3? 258, 230, 200, 186, and
174.
Cyclic voltammogram of Fe(III) complex?
The complex undergoes a reversible oneelectron reduction with a reduction potential
of 0.21 V. Above 0.720 V the complex is
oxidized.
S9.1
Found in aluminosilicate minerals or
sulfides?
Cd and Pb will be found as sulfides.
Rb and Sr can be found in aluminosilicate
minerals.
Cr and Pd can be found in both oxides and
sulfides.
S9.2
Sulfur forms catenated polysulfides
whereas polyoxygen anions are unknown?
Owing to a strong tendency to form strong
double bonds, it is more likely that
polyoxygen anions will form pi bonds that
limit extended bonding owing to restrictions
on pi orbital overlap through multiple
bridging centres.
S9.3
Shape of XeO4 and identify the Z + 8
compound with the same structure?
A tetrahedral geometry. The same structure is
SmO4.
S9.4
Comment on ΔfHө values? It is evident from
the values that as we move down the group,
steric crowding of the fluorines is minimized.
S9.5
Further data useful when drawing
comparisons with the value for V2O5? We
would have to know the products formed
upon decomposition.
Exercises
9.1
Maximum stable oxidation state? (a) Ba;
+2, (b) As; +5, (c) P; +5, (d) Cl; +7.
9.2
Form saline hydrides, oxides and
peroxides, and all the carbides react with
water to liberate a hydrocarbon?
the alkaline earth metals or Group 2 elements.
9.3
Elements vary from metals through
metalloids to non-metals; form halides in
oxidation states +5 and +3 and toxic
gaseous hydrides? Elements in Group 15.
9.4
Born–Haber cycle for the formation of the
hypothetical compound NaCl2? Which
thermochemical step is responsible for the
fact that NaCl2 does not exist?
The second ionization energy of
sodium is 4562 kJ mol-1 and is responsible for
the fact that the compound does not exist.
9.5
Inert pair effect beyond Group 15? The
relative stability of an oxidation state in which
the oxidation number is 2 less than the group
number is an example of the inert pair effect.
Zeolite of composition CaAl2Si6O16.nH2O,
determine n.? n=7.2 As an integer, n = 7.
Ratio of cobalt to acetylacetonate in the
product? The ratio is 3:1.
CHAPTER 9
Self-tests
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 17
9.6
Ionic radii, ionization energy, and metallic
character? Metallic character, ionic radii
decrease across a period and down a group.
Ionization energy increases across a period
and decreases down a group.
9.7
10.3
(a) H2S? H = +1, S = –2.
(b) KH? H = –1, K = +1.
(c) [ReH9]2–?
Names of ores? (a) Mg; MgCO3 magnesite,
(b) Al; Al2O3 bauxite, and (c) Pb; PbS galena.
9.8
Calculate ΔfHө for SeF6?
ΔfHө = −1397 kJ mol-1.
(e) H2PO(OH)? H bonded to an oxygen
atom = +1. H bonded to the phosphorus atom?
If they are assigned an oxidation number of
+1, and O = –2, then P = +1.
10.4
(ii) C(s) + H2O → CO(g) + H2(g) (1000°C)
(iii) CO(g) + H2O → CO2(g) + H2 (g)
10.5
Self-tests
S10.2
(c)
Different
molecular
hydrides? Molecular hydrides are found in
groups 13/III through 17/VII.
GeH4
Reactions of hydrogen compounds?
(a) Ca(s) + H2(g) → CaH2(s).
10.6
What are the physical properties of water
without hydrogen bonding? It most likely
would be a gas at room temperature; ice
would be denser than water.
10.7
Which molecule has the stronger hydrogen
bonds? S–H···O has a weaker hydrogen bond
than O–H···S.
10.8
Name and classify the following?
(b) NH3(g) + BF3(g) → H3N–BF3(g).
(c) LiOH(s) + H2(g) → NR.
S10.3
A procedure for making Et3MeSn?
2Et3SnH + 2Na → 2Na+Et3Sn– + H2
Na+Et3Sn– + CH3Br → Et3MeSn + NaBr
Properties of hydrides of the elements? (a)
Position in the periodic table? See Figure
10.2.
(b) Trends in ΔfGº? See Table 10.1.
Which of the following CH4, SiH4,, or GeH4
would best H+ or H donor?
CH4, the strongest Bronsted acid.
would be the best hydride donor.
Preparation of hydrogen gas?
(i) CH4(g) + H2O → CO(g) + 3H2(g) (1000°C)
CHAPTER 10
S10.1
H = –1, Re = +7.
(d) H2SO4? H = +1, O = –2, S = +6.
Identify the Z + 8 element for P.
Similarities? V (vanadium).
9.9
Assign oxidation numbers to elements?
(a) BaH2? barium hydride.
Exercises
(b) SiH4?
silane.
10.1
(c) NH3?
ammonia,
Where does Hydrogen fit in the periodic
chart? (a) Hydrogen in group 1? Hydrogen
has one valence electron like the group 1
metals and is stable as H+, especially in
aqueous media.
(b) Hydrogen in group 17? Hydrogen can
fill its 1s orbital and make a hydride H–. The
halogens are diatomic gases just like
hydrogen, but chemically it fits well in both
group 1 and group 17.
(d) AsH3? Arsine.
(e) PdH0.9? palladium hydride.
(f) HI? hydrogen iodide.
10.9
(a) Hydridic character? Barium hydride
(b) Brønsted acidity? Hydrogen iodide.
(c) Variable composition? PdH0.9 .
(c) Hydrogen in group 14? There is no
reason for hydrogen to be placed in this
group.
10.2
Low reactivity of hydrogen? Hydrogen
exists as a diatomic molecule (H2). It has a
high bond enthalpy. It also only has two
electrons shared between two protons.
Chemical characteristics of hydrides?
(d) Lewis basicity? Ammonia.
10.10
Phases of hydrides of the elements? BaH2
and PdH0.9 are solids, none is a liquid, and
SiH4, NH3, AsH3, and HI are gases.
10.11
Structures of H2Se, P2H4, and H3O+? The
Lewis structures of these three species are:
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
18
10.21.1 Dihydrogen as an oxidizing agent? It’s
reaction with an active s-block metal such as
sodium.
H3O+ should be trigonal pyramidal.
10.12
The reaction that will give the highest
proportion of HD? Reaction (b) will
produce 100% HD and no H2 or D2.
10.13
Most
likely
to
undergo
radical
reactions? (CH3)3SnH, the tin compound is
the most likely to undergo radical reactions
with alkyl halides.
10.14
Arrange H2O, H2S, and H2Se in order?
(a) Increasing acidity? H2O < H2S < H2Se.
CHAPTER 11
Self-tests
S11.1
Change in cell parameter for CsCl? At 445
°C the CsCl structure changes to rock-salt and
assumes the face centered cubic.
S11.2
Lattice enthalpies of formation? LiF is 625
kJ mol−1 and for NaF is 535 kJ mol−1.
S11.3
Trend is stability of Group 1 ozonides?
Group 1 ozonides are less stable compared to
the superoxides.
(b) Increasing basicity toward a hard
acid? H2Se < H2S < H2O.
10.15
The synthesis of binary hydrogen
compounds? (i) direct combination of the
elements, (ii) protonation of a Brønsted base,
and (iii) metathesis using a compound such as
LiH, NaBH4, or LiAlH4.
10.16
Compare BH4–, AlH4–, and GaH4–? Since
AlH4– is more “hydride-like,” it is the
strongest reducing agent.
10.17
Compare period 2 and period 3 hydrogen
compounds? Period 2 compounds:
- except for B2H6, are all exoergic
- tend to be weaker Brønsted acids and
stronger Brønsted bases
- bond angles in period 2 hydrogen
compounds reflect a greater degree of sp3
hybridization
S11.4 Sketch the thermodynamic cycle of Group
1 carbonate.
M2CO3(s)
CO2(g)
2M+(g) + CO32−(g)
O2−(g) + CO2(g)
S11.5
- Several period 2 compounds exhibit strong
hydrogen bonding.
10.18
M2O(s) +
2M+(g) +
Explain the differences in temperature of
decomposition of LiNO3 and KNO3? KNO3
decomposes in two steps at two different
temperatures.
KNO3(s) → KNO2(s) + 1/2O2(g)
- boiling points of HF, H2O, and NH3 are all
higher than their respective period 3
homologues.
2KNO2(s) → K2O(s) + 2NO2(g) + 1/2O2(g)
Suggest a method for the preparation of
BiH3?
LiNO3 decomposes in one step.
The redistribution
BiH2Me.
of
LiNO3(s) → 1/2 Li2O(s) + NO2(g) + 1/4O2(g)
methylbismuthine,
3BiH2Me → 2BiH3 + BiMe3
10.19
Describe the compound formed between
water and Kr? A clathrate hydrate.
10.20
Potential energy surfaces for hydrogen
bonds? (See Figure 10.9) The surface for
the H2O, Cl– system has a double minimum,
while the surface for the bifluoride ion has a
single minimum.
S11.6
Predicted 7Li NMR of Li3N? Two peaks in
the NMR spectrum at low temperature. Only
one resonance in the NMR at high
temperature.
Exercises
11.1
(a) Why are group 1 metals good reducing
agents? They have one valence electron in
the ns1 subshell, and relatively low first
ionization energies.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 19
11.2
(b)
Why are group 1 metals poor
They are large,
complexing agents?
electropositive metals and have little tendency
to act as Lewis acids.
CHAPTER 12
Trends of the fluorides and chlorides of the
group 1 metals?
Fluoride is a hard Lewis base and will form
strong complexes with hard Lewis acids. The
trends reverse for the chloride ion.
S12.1
11.3
Synthesis of group 1 alkyls? Most alkyl
lithiums are made using elemental lithium
with the corresponding alkyl chlorides.
11.4
Which is more likely to lead to the desired
result? (a) Cs+ or Mg2+, form an acetate
complex? Mg2+.
(b) Be or Sr, dissolve
ammonia? Strontium.
+
in
S12.2
S12.3
↓
S12.4
Trends in solubility? Higher for LiF and
CsI, lower for CsF and LiI.
11.7
Thermal stability of hydrides versus
decompose
to
carbonates? Hydrides
elements. Carbonates decompose to oxides.
11.8
The structures of CsCl and NaCl?
6-coordinate Na+, 8-coordinate Cs+.
11.9
The effect of the alkyl group on the
structure of lithium alkyls? Whether a
molecule is monomeric or polymeric is based on
the streric size of the alkyl group – less bulky
alkyl groups lead to polymerization.
Exercises
12.1
12.2
Why are the properties of beryllium more
similar to aluminium and zinc than to
magnesium?
Because of a diagonal relationship between
Be and Al.
12.3
Identify the compounds A, B, C, and D of
the group 2 element M?
M + H2O → M(OH)2; A = M(OH)2
M(OH)2 + CO2 → MCO3; B = MCO3
2MCO3 + 5C → 2MC2 + 3CO2; C = MC2
MC2 + 2H2O → M(OH)2 + C2H2
M(OH)2 + 2HCl → MCl2 + 2H2O; D = MCl2.
12.4
(b) MgCl2 + LiC2H5 → Mg(C2H5)Br + LiBr
(c) C2H5Li + C6H6 → LiC6H5 + C2H6
Why are compounds of beryllium covalent
whereas those of the other group 2
elements are predominantly ionic?
Be has large polarizing power and a high
charge density due.
11.10 Predict the products of the following
reactions?
(a) CH3Br + Li → Li(CH3) + LiBr
Calculate the lattice enthalpy of MgF2 and
comment on how it will affect the solubility
compared to MgCl2?
MgF2 is 2991 kJ mol–1, will reduce solubility
compared to MgCl2.
NaNH2
11.6
Use ionic radii to predict a structure type
of BeSe?
According to Table 3.6 should be close to
ZnS-like structure.
Identify the compounds?
+ NH3
Calculate the lattice enthalpies for CaO
and CaO2 and check that the above trend is
confirmed?
Calcium oxide and calcium peroxide are 3465
kJ mol–1 and 3040 kJ mol–1.
+
NaOH ← H2O + Sodium metal + O2 → Na2O2 +
heat → Na2O
Predict whether (a) BeCl2 and (b) BaCl2
are predominantly ionic or covalent?
BeCl2 is covalent; BaCl2 is ionic.
liquid
(c) Li or K , form a complex with
C2.2.2? Potassium ion.
11.5
Self-tests
Why does beryllium fluoride form a glass
when cooled from a melt?
BeF2 adopts SiO2 like arrangement.
12.5
Why is magnesium hydroxide a much more
effective antacid than calcium or barium
hydroxide? Mg(OH)2 is sparingly soluble
and mildly basic.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
20
12.6
Explain why Group 1 hydroxides are much
more corrosive to metals than Group 2
hydroxides?
B nuclei have I = 3/2. Predict the number
of lines and their relative intensities in the
1
H-NMR spectrum of BH4–? 4. Relative
intensity ratio is 1:3:3:1.
S13.2
Which of the salts MgSeO or BaSeO would
be expected to be more soluble in water?
Write an equation for the reaction of
LiBH4 with propene in ether solvent and a
1:1 stoichiometry and another equation for
its reaction with ammonium chloride in
THF with the same stoichiometry?
MgSeO4
Simple alkenes are inert towards LiBH4.
Group 1 hydroxides are more soluble than
group 2 hydroxides, and therefore have higher
OH− concentrations.
12.7
12.8
4
4
Which Group 2 salts are used as drying
agents and why?
Anhydrous Mg, and Ca sulphates are
preferred as drying agents, because of the
higher affinity of Mg and Ca sulphates for
water.
12.9
How do group 2 salts give rise to scaling
from hard water?
LiBH4 + NH4Cl
S13.3
Use the data in Table 1.7 and the Ketelaar
triangle in Fig. 2.38 to predict the nature of
the bonding in BeBr , MgBr , and BaBr .
2
BCl3(g) + py(l) → Cl3B − py(s)
2
(c) BBr3 and F3BN(CH3)3?
BBr3(l) + F3BN(CH3)3(s) → BF3(g) +
Br3BN(CH3)3(s)
S13.4
BeBr2 should be covalent. MgBr2 should be
ionic. BaBr2 should be ionic
12.12
The two Grignard compounds C H MgBr
and 2,4,6-(CH ) C H MgBr dissolve in THF.
What differences would be expected in the
structures of the species formed in these
solutions?
2
3
3
6
Write and justify balanced equations for
plausible reactions between (a) BCl3 and
ethanol, (b) BCl3 and pyridine in
hydrocarbon solution, (c) BBr3 and
F3BN(CH3)3?
(b) BCl3 and pyridine in hydrocarbon
solution?
Predict structures for BeTe and BaTe.
2
5
Suggest a reaction or series of reactions for
the preparation of N, N’, N’’-trimethylB,B’,B’’-trimethylborazine starting with
methylamine and boron trichloride?
Cl3BNCH3 + 3CH3MgBr
(CH3)3B3N3(CH3)3 + 3Mg(Br, Cl)2
2
S13.5
How many framework electron pairs are
present in B H and to what structural
category does it belong? Sketch its
structure?
4
C2H5MgBr will be tetrahedral with two
molecules of solvent coordinated to the
magnesium. The bulky organic group in
2,4,6-(CH3)3C6 H2MgBr leads to a
coordination number of two.
12.13
BH3NH3 + LiCl + H2
BCl3(g) + 3 EtOH(l) → B(OEt)3(l) + 3 HCl(g)
BeTe, close to ZnS-like structure. BaTe, close
to CsCl-like structure.
12.11
THF
(a) BCl3 and ethanol?
Salts of divalent ions have low solubility.
12.10
11
S13.1
10
7, arachno species.
The structure of B4H10:
Predict the products of the following
reactions?
(a) MgCl2 + 2LiC2H5 → 2LiCl +
Mg(C2H5)2
(b) Mg + (C2H5)2Hg → Mg(C2H5)2 + Hg
(c) Mg + C2H5HgCl → C2H5MgCl + Hg
S13.6
CHAPTER 13
Self-tests
Propose a plausible product for the
reaction between Li[B H ] and Al (CH ) ?
10
–
[B10H11 (AlCH3)]
13
2
3
6
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 21
S13.7
Propose a synthesis for the polymer
precursor 1,7-B C H (Si(CH ) Cl) from 1,2B C H and other reagents of your choice?
10
2
10
3
2
2
13.6
1,2 - B10C2H12 ⎯
⎯→ 1,7-B10C2H12 (90%) + 1,12B10C2H12 (10%)
10
2
12
Δ
→
1,7-B10C2H10Li2
+
2Si(CH3)2Cl2
B10C2H10(Si(CH3)2 + 2LiCl
S13.8
1,7 -
Propose, with reasons, the chemical
equation (or indicate no reaction) for
reactions between (a) (CH3)2SAlCl3 and
GaBr3?
No, it explodes in air.
B2H6 + 3 O2 → B2O3 + 3 H2O2
13.7
Predict how many different boron
environments would be present in the
proton-decoupled 11B-NMR of a) B5H11, b)
B4H10? a) 3 b) 2.
13.8
Predict
the
products
from
the
hydroboration of (a) (CH3)2C=CH2, (b)
CH CH?
(Me)2SalCl3 + GaBr3 → Me2SGaBr3 + AlCl3.
(b) TlCl and formaldehyde (HCHO) in
acidic aqueous solution? No reaction.
Exercises
13.1
Give a balanced chemical equation and
conditions for the recovery of boron?
B2O3 + 3Mg → 2B + 3MgO
ΔH < 0
13.2
Describe the bonding in (a) BF3? Covalent.
(b) AlCl3? In the solid state, a layered
structure. At melting point, dimers. (c) B2H6?
Electron-deficient dimer.
13.3
Arrange the following in order of
increasing Lewis acidity: BF3, BCl3, AlCl3.
In the light of this order, write balanced
chemical reactions (or no reaction) for (a)
BF3N(CH3)3 + BCl3 →, (b) BH3CO +
BBr3→?
13.4
BCl3 > BF3 > AlCl3
(a) BF3N(CH3)3 + BCl3
BCl3N(CH3)3 + BF3
(BCl3 > BF3)
NR
(b) BH3CO + BBr3
Thallium tribromide (1.11 g) reacts
quantitatively with 0.257 g of NaBr to form
a product A. Deduce the formula of A.
Identify the cation and anion?
13.9
Diborane is extremely toxic, and the boron
containing product of combustion is a solid,
B2O3.
13.10
13.11
LiAlH4
(a) BCl3 + 3C2H5MgCl
B(C2H5)3 + 3MgCl2
(b)
H2O
13.12
C
(a) A = B2H6
(b) B = B(OH)3
heat
B
Given NaBH4, a hydrocarbon of your
choice, and appropriate ancillary reagents
and solvents, give formulas and conditions
for the synthesis of (a) B(C2H5)3, (b)
Et3NBH3?
Heat, Ether
A
CaF2
Using BCl3 as a starting material and other
reagents of your choice, devise a synthesis
for the Lewis acid chelating agent, F2B–
C2H4–BF2?
2BCl3 + 2Hg Æ B2Cl4 + 2HgCl2
B2Cl4 + 4AgF Æ B2F4 + 4AgCl
B2F4 + C2H4 Æ F2CH2CH2BF2
Identify compounds A, B, and C?
BF3
(a) BH3 +
(CH3)2C=CH2
B[CH2-CH(CH3)2]3
CH
(b) BH3 +
CH
B(CH=CH2)3
Diborane has been used as a rocket
propellant. Calculate the energy released
from 1.00 kg of diborane given the
following values of ΔfHө/kJ mol-1: B2H6 =
31, H2O = -242, B2O3 = -1264. The
combustion reaction is B2H6 (g) + 3 O2(g)
→ 3 H2O (g) + B2O3 (s). What would be the
problem with diborane as a fuel?
-73,172 kJ.
TlBr3 + NaBr → NaTlBr4
13.5
(c) C = B2O3
Does B2H6 survive in air? If not, write the
equation for the reaction?
[HN(C2H5)3]Cl + NaBH4
H2 + H3BN(C2H5)3 + NaCl
Draw the B12 unit that is a common motif
of boron structures; take a viewpoint along
a C2 axis?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
22
+ C2H2
(2) B10H12(SEt2)2
B10C2H12 + 2SEt2 + H2
(3) 2 B10C2H12
+ 2EtO–
+ 4EtOH
2 B9C2H12– + 2B(OEt)3 + 2H2
(4) Na[B9C2H12]
+
NaH
Na2[B9C2H11] + H2
THF
(5) 2Na2[B9C2H11] + FeCl2
2NaCl + Na2[Fe(B9C2H11)2]
2-
H
B
H
H
B
H
H
13.13
Which boron hydride would you expect to
be more thermally stable, B6H10 or B6H12?
Give a generalization by which the thermal
stability of a borane can be judged?
13.15
How many skeletal electrons are present in
B5H9? 14
(a) Give a balanced chemical equation
(including the state of each reactant and
product) for the air oxidation of
pentaborane(9). (b) Describe the probable
disadvantages, other than cost, for the use
of pentaborane as a fuel for an internal
combustion engine?
(b) The boron containing product of
combustion is a solid, B2O3.
(a) From its formula, classify B10H14 as
closo, nido, or arachno. (b) Use Wade’s
rules to determine the number of
framework
electron
pairs
for
decaborane(14). (c) Verify by detailed
accounting of valence electrons that the
number of cluster valence electrons of
B10H14 is the same as that determined in
(b)?
(a) nido .
(b) 12.
(c) The total number of valence elections is
(10x3)+(14x1)=44; the number of cluster
valence is the remainder of 44-20=24.
Starting with B10H14 and other reagents of
your choice, give the equations for the
synthesis of [Fe(nido-B9C2H11)2]2-, and
sketch the structure of this species?
(1) B10H14
+ 2SEt2
B10H12(SEt2)2 + H2
B
Fe
B H
H
H
C
B
C
B
B
B
B
H
H
H
H
B
B
H
H
B
H
13.18
Heat
13.17
H
H
H
B
H
(a) 2B5H9 (l) + 12O2 (g)
5B2O3 (s) + 9H2O (l)
13.16
H
B
B
B
B
C
C
H
B6H10
13.14
H
B
(a) What are the similarities and
differences in structure of layered BN and
graphite (Section 13.9)? (b) Contrast their
reactivity with Na and Br2. (c) Suggest a
rationalization for the differences in
structure and reactivity.
(a) Their structures? Both of these
substances have layered structures.
(b) Their reactivity with Na and
Br2? Graphite reacts, boron nitride is
unreactive.
(c) Explain the differences? The large
HOMO–LUMO gap in BN means it is more
difficult to remove an electron from it than
from the HOMO of graphite.
13.19
Devise a synthesis for the borazines (a)
Ph3N3B3Cl3 and (b) Me3N3B3H3, starting
with BCl3 and other reagents of your
choice. Draw the structures of the
products?
(a) Ph3N3B3Cl3?
3 PhNH3+Cl− + 3 BCl3 → Ph3N3B3Cl3 + 9 HCl
(b) Me3N3B3H3?
3 MeNH3+Cl− + 3 BCl3 → Me3N3B3Cl3 + 9 HCl
Me3N3B3Cl3 + 3 LiH → Me3N3B3H3 + 3 LiCl
The structures:
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 23
Exercises
13.20
14.1
Silicon forms the chlorofluorides SiCl3F,
SiCl2F2, and SiClF3. Sketch the structures
of these molecules?
14.2
Explain why CH4 burns in air whereas CF4
does not. The enthalpy of combustion of
CH4 is −888 kJ mol−1 and the C–H and C–F
bond enthalpies are −413 and −489 kJ
mol−1 respectively?
The bond enthalpy of a C–F bond is higher
than the bond enthalpy of a C–H bond.
14.3
SiF4 reacts with (CH3)4NF to form
[(CH3)4N][SiF5]. (a) Use the VSEPR rules
to determine the shape of the cation and
anion in the product; (b) Account for the
fact that the 19F NMR spectrum shows two
fluorine environments?
(a) The cation is [(CH3)4N]+
Give the structural type and describe the
structures of B4H10, B5H9, and 1,2B10C2H12?
B4H10, is an arachno borane.
B5H9, is a nido borane.
1,2-B10C2H12 is a closo carborane
13.21
Arrange the following boron hydrides in
order of increasing Brønsted acidity, and
draw a structure for the probable structure
of the deprotonated form of one of them:
B2H6, B10H14, B5H9? In a series of boranes,
the acidity increases as the size of the borane
increases.
CHAPTER 14
+
Self-tests
S14.1
CH3
Describe how the electronic structure of
graphite is altered when it reacts with (a)
potassium, (b) bromine?
N
(a) With potassium? Potassium results in a
material with a higher conductivity.
(b) With bromine? Bromine can remove
electrons from the π-symmetry HOMOs of
graphite. This also results in a material with a
higher conductivity.
S14.2
Use the bond enthalpy data in Table 14.2
and above to calculate the standard
enthalpy of formation of CH4 and SiH4?
C
H3
The anion is SiF5–
S14.3
CO(g) + 2MnO2(s) → CO2(g) + Mn2O3(s)
13
13
2Li(s) + D2 → 2LiD(s)
13
CO2(g) + LiD(et) → Li+D13CO2−(et)
-
F
F
CH4: ΔfH= –61kJ mol–1
SiH4: ΔfH= +39 kJ mol–1
Propose a synthesis of D13CO2– starting
from 13CO?
CH3
H3C
Si
F
F
F
(b) There are two different fluorine
environments.
14.4
Draw the structure and determine the
charge on the cyclic anion [Si4O12]ν฀?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
24
14.11
8O
SiO2(s) + C(s) → Si(s) + CO2(g)
ΔH < 0
GeO2(s) + 2H2(g) → Ge(s) + 2H2O(g)
ΔH < 0
Si
O
O
O
Si
Si
O
Give balanced chemical equations and
conditions for the recovery of silicon and
germanium from their ores?
O
O
Si
14.12
Sn-NMR
(a) Describe the trend in band gap energy,
Eg, for the elements carbon (diamond) to
tin (grey). (b) Does the electrical
conductivity of silicon increase or decrease
when its temperature is changed from 20˚C
to 40˚C?
14.6
Predict the appearance of the 1H-NMR
spectrum of Sn(CH3)4? Doublet
(a) There is a decrease in band gap energy
from carbon (diamond) to grey tin. (b)
Increase.
O
Charge = –8.
14.5
Predict the appearance of the
spectrum of Sn(CH3)4? Doublet
119
14.7
Use the data in Table 14.2 and the
additional bond enthalpy data given here
to calculate the enthalpy of hydrolysis of
CCl4 and CBr4. Bond enthalpies/(kJ mol–1):
O–H = 463, H–Cl = 431, H–Br = 366?
ΔhH˚CCl4 = 110 kJmol–1
ΔhH˚CBr4 = 86 kJmol–1
14.8
Identify the compounds A to F:?
(A) SiCl4 (B) SiRCl3 (C) RSi(OH)3. (D)
RSiOSiR +H2O (E) SiR4 (F)? SiO2
YYYYY14.9
(a) Summarize the trends in
relative stabilities of the oxidation states of
the elements of Group 14, and indicate the
elements that display the inert pair effect.
(b) With this information in mind, write
balanced chemical reactions or NR (for no
reaction) for the following combinations,
and explain how the answer fits the trends.
(i) Sn2+(aq) + PbO2(s) (excess) → (air
excluded) (ii) Sn2+(aq) + O2(air) →?
(a) +4 is the most stable oxidation state for
the lighter elements, but +2 is the most stable
oxidation state of Pb. Pb therefore displays
the inert-pair effect.
(b) (i) Sn2+ + PbO2 + 4 H+ → Sn4+ + Pb2+ +
2H2O, (ii) 2Sn2+ + O2 + 4H+ → 2Sn4+ +
2H2O.
14.10
Use data from Resource section 3 to
determine the standard potential for each
of the reactions in Exercise 14.5 (b). In
each case, comment on the agreement or
disagreement
with
the
qualitative
assessment you gave for the reactions?
(i) V = +1.31 V. (ii) V= 1.08 V. Both
reactions agree with the predictions made in
Excercise 14.5.
14.13
Preferably without consulting reference
material, draw a periodic table and
indicate the elements that
form saline,
metallic, and metalloid carbides?
Ionic(silane) Metallic
Metalloid
carbides
carbides
carbides
Group I
Li, Na, K,
elements
Rb, Cs
Group II
Be, Mg, Ca,
elements
Sr, Ba
Group 13
Al
B
elements
Group 14
Si
elements
3d-Block
Sc, Ti, V,
elements
Cr, Mn, Fe,
Co, Ni
4d-Block
Zr, Nb, Mo,
elements
Tc, Ru
5d-Block
La, Hf, Ta,
elements
W, Re, Os
6d-Block
Ac
elements
Lanthanides
Ce, Pr, Nd,
Pm, Sm,
Eu, Gd, Tb,
Dy, Ho, Er,
Tm, Yb, Lu
14.14 Describe the preparation, structure and
classification of (a) KC8, (b) CaC2, (c) K3C60?
(a) KC8? Formed by heating graphite with
potassium vapor or by treating graphite with a
solution of potassium in liquid ammonia.
There is a layered structure of alternating sp2
carbon atoms and potassium ions, a saline
carbide.
(b) CaC2?
Ca(l) + 2C(s) → CaC2(s) or
CaO(s) + 3C(s) → CaC2(s) + CO(g)
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 25
Calcium carbide contains discrete C22– ions.
(c) K3C60? A solution of C60 can be treated
with elemental potassium. It is ionic.
NH3 + ClO− + H+ → [H3N—Cl-O]− + H+ →
H2NCl + H2O
H2NCl + NH3 → H2NNH2 + HCl
Both can be thought of as redox reactions.
14.15
Write balanced chemical equations for the
reactions of K2CO3 with HCl(aq) and of
Na4SiO4 with aqueous acid?
K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) +
H2O(l)
S15.5
There are (30.4 cm3)/(7.6 cm3) = 4 strongly
acidic OH groups per molecule. A molecule
with 2 terminal OH groups and four further
OH groups is a tetrapolyphosphate.
Na4SiO4(aq) + 4HCl(aq) → 4NaCl(aq) + SiO2(s) +
2H2O(l)
14.16
14.17
Describe in general terms the nature of the
[SiO3]n2ν฀ ion in jadeite and the silicaalumina framework in kaolinite?
The structures of jadeite and kaolinite consist
of extended one- and two-dimensional
structures, respectively. The [SiO32–]n ions in
jadeite are a linear polymer of SiO4
tetrahedra.
The
two-dimensional
aluminosilicate layers in kaolinite represent
another way of connecting SiO4 tetrahedra.
Exercises
15.1
P
As
Sb
Bi
O
S
Se
Te
(b) Eight sodalite cages are linked together to
form the large cage of zeolite A.
Self-tests
S15.1
Consider the Lewis structure of a segment
of the structure of bismuth shown in Fig.
15.2. Is this puckered structure consistent
with the VSEPR model? Yes.
S15.2
Refined hydrocarbons and liquid hydrogen
are also used as rocket fuel. What are the
advantages of dimethylhydrazine over
these fuels?
Dimehylhydrazine ignites spontaneously, and
produces less CO2.
S15.3
S15.4
From trends in the periodic table, decide
whether phosphorus or sulphur is likely to
be the stronger oxidizing agent? Sulphur.
Summarize the reactions that are used for
the
synthesis
of
hydrazine
and
hydroxylamine. Are these reactions best
described as electron-transfer processes or
nucleophilic displacements?
List the elements in Groups 15 and indicate
the ones that are (a) diatomic gases, (b)
nonmetals, (c) metalloids, (d) true metals.
Indicate those elements that display the
inert-pair effect?
N
(a) How many bridging O atoms are in the
framework of a single sodalite cage? (b)
Describe the (supercage) polyhedron at the
centre of the Zeolite A structure in Fig.
14.3?
(a) 48.
CHAPTER 15
When titrated against base a sample of
polyphosphate gave end points at 30.4 and
45.6 cm3. What is the chain length?
15.2
Type of
element
Nonmetal
Nonmetal
nonmetal
metalloid
metalloid
nonmetal
nonmetal
nonmetal
nonmetal
Diatomic
gas?
yes
Inert pair
effect?
No
no
No
no
no
no
yes
no
no
no
No
No
Yes
No
No
No
No
(a) Give complete and balanced chemical
equations for each step in the synthesis of
H PO from hydroxyapatite to yield (a)
high-purity phosphoric acid and (b)
fertilizer-grade phosphoric acid. (c)
Account for the large difference in costs
between these two methods?
3
4
(a) high-purity phosphoric acid?
2Ca3(PO4)2 + 10C + 6SiO2 →
P4− + 10CO + 6CaSiO3
P4 (pure) + 5O2 → P4O10
P4O10 + 6H2O → 4H3PO4 (pure)
(b) Fertilizer grade H3PO4?
Ca5(PO4)3OH + 5H2SO4 →
3H3PO4 (impure) + 5CaSO4 + H2O
(c) Account for the difference in cost?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
26
P4O10 + 6H2O → 4H3PO4
Fertilizer-grade phosphoric acid involves a
single synthetic step for a product that
requires little or no purification.
15.3
Ammonia can be prepared by (a) the
hydrolysis of Li N or (b) the hightemperature, high-pressure reduction of N
by H . Give balanced chemical equations
for each method starting with N , Li, and
H , as appropriate. (c) Account for the
lower cost of the second method?
(c) Reaction of the product from part (b)
with CaCl2?
2H3PO4(l) + 3CaCl2(aq) →
3
Ca3(PO4)2(s) + 6HCl(aq)
2
2
15.9
2
2
Starting with NH3(g) and other reagents of
your choice, give the chemical equations
and conditions for the synthesis of (a)
HNO , (b) NO2– , (c) NH OH, (d) N3–?
3
(a) Hydrolysis of Li3N?
(a) HNO3?
6Li + N2 → 2Li3N
4NH3(aq) + 7O2(g) → 6H2O(g) + 4NO2(g)
2Li3N + 3H2O → 2NH3 + 3Li2O
High temperature
(b) Reduction of N2 by H2?
(b) NO2–?
2NO2(aq) + 2OH−(aq) →
N2 + 3H2 → 2NH3
NO2−(aq) + NO3−(aq) + H2O(l)
(c) Account for the difference in cost? The
second process is considerably cheaper than
the first, because lithium is very expensive.
15.4
(c) NH2OH? cold aqueous acidic solution
NO2−(aq) + 2HSO3−(aq) + H2O(l) →
NH3OH+(aq) + 2SO42−(aq)
Show with an equation why aqueous
solutions of NH NO are acidic?
4
3
NH4NO3(s) + H2O Æ NH4+ + NO3-(aq)
15.5
2
(d) N3–?
Carbon monoxide is a good ligand and is
toxic. Why is the isoelectronic N molecule
not toxic?
at elevated temperatures:
3NaNH2(l) + NaNO3 →
2
NaN3 + 3NaOH + NH3(g)
2NaNH2(l) + N2O → NaN3 + NaOH +
NH3
N2 itself, with a triple bond between the two
atoms, is strikingly unreactive.
15.6
Compare and contrast the formulas and
stabilities of the oxidation states of the
common nitrogen chlorides with the
phosphorus chlorides?
15.10
4
The only isolable nitrogen chloride is NCl3,
and it is thermodynamically unstable. Both
PCl3 and PCl5 are stable.
15.7
Use the VSEPR model to predict the
probable shapes of (a) PCl4+, (b) PCl4–, (c)
AsCl ?
5
(a)
PCl4+?
tetrahedron.
(b)
PCl4–?
A see-saw.
(c) AsCl5? A trigonal bipyramid.
15.8
Give balanced chemical equations for each
of the following reactions: (a) oxidation of
P with excess oxygen, (b) reaction of the
product from part (a) with excess water, (c)
reaction of the product from part (b) with
a solution of CaCl and name the product?
4
2
(a) Oxidation of P4 with excess O2?
P4 + 5O2 → P4O10
(b) Reaction of the product from part (a)
with excess H2O?
Write the balanced chemical equation
corresponding to the standard enthalpy of
formation of P O (s). Specify the structure,
physical state (s, l, or g), and allotrope of
the reactants. Do either of the reactants
differ from the usual practice of taking as
reference state the most stable form of an
element?
10
P4(s) + 5O2(g) → P4O10(s)
15.11
Without reference to the text, sketch the
general form of the Frost diagrams for
phosphorus (oxidation states 0 to +5) and
bismuth (0 to +5) in acidic solution and
discuss the relative stabilities of the +3 and
+5 oxidation states of both elements?
(i) Bi(III) is much more stable than Bi(V), and
(ii) P(III) and P(V) are both about equally
stable.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 27
and H2PO22− useful
reducing agents?
as
oxidizing
or
Eº = 0.839 V.
HPO32– and H2PO2– ions will be much better
reducing agents than oxidizing agents.
15.16
Identify the compounds A, B, C, and D?
D
LiAlH4
15.12
Cl2
Are reactions of NO2– as an oxidizing agent
generally faster or slower when pH is
lowered? Give a mechanistic explanation
for the pH dependence of NO2– oxidations?
AS
When equal volumes of nitric oxide (NO)
and air are mixed at atmospheric pressure
a rapid reaction occurs, to form NO and
N O . However, nitric oxide from an
automobile exhaust, which is present in the
parts per million concentration range,
reacts slowly with air. Give an explanation
for this observation in terms of the rate law
and the probable mechanism?
C
(a) A = AsCl3
(b) B = AsCl5
(c) C = AsR3
(d) D = AsH3
2
2
4
15.17
Sketch the two possible geometric isomers
of the octahedral [AsF Cl ]– and explain
how they could be distinguished by 19F–
NMR?
4
The rate law must be more than first order in
NO concentration.
15.14
B
3RMgBr
The rates of reactions in which nitrite ion is
reduced are increased as the pH is lowered.
15.13
Cl2/hv
A
2
The cis isomer gives two 19F signals and the
trans isomer gives one signal.
Give balanced chemical equations for the
reactions of the following reagents with
PCl and indicate the structures of the
products: (a) water (1:1), (b) water in
excess, (c) AlCl , (d) NH Cl?
5
3
4
(a) H2O? tetrahedral POCl3.
PCl5 + H2O → POCl3 + 2HCl
(b) H2O in excess?
2PCl5(g) + 8H2O(l) → 2H3PO4(aq) +
10HCl(aq)
(c) AlCl3?
tetrahedral.
PCl5 + AlCl3 → [PCl4]+[AlCl4]−
(d) NH4Cl? Cyclic molecules or linear
chain polymers
nPCl5 + nNH4Cl →
−[(N = P(Cl)2)n]− + 4nHCl
15.18
Identify the nitrogen compounds A, B, C,
D, and E?
(a) A = NO2
(b) B = HNO3; C = NO
15.15
Use standard potentials (Resource section
3) to calculate the standard potential of the
reaction of H3PO2 with Cu2+. Are HPO22−
(c) D = N2O4
(d) E= NO2
(e) F = NH4+
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
28
15.19
Use the Latimer diagrams in Resource
section 3 to determine which species of N
and P disproportionate in acid conditions?
The species of N and P that disproportionate
are N2O4, NO, N2O, NH3OH+, H4P2O6, and P.
16.3
Which hydrogen bond would be stronger:
S—H . . . O or O—H . . . S?
O–H hydrogen bonds are stronger.
16.4
Which of the solvents ethylenediamine
(which is basic and reducing) or SO (which
is acidic and oxidizing) might not react
with (a) Na S , (b) K Te ?
2
2
CHAPTER 16
16.5
Determine whether the decomposition of
H O is spontaneous in the presence of
either Br or Cl ?
2
2
2
16.6
In Presence of Cl–
The
decomposition
is
of
H2O2
thermodynamically favored in presence of Cl–
.
16.7
Probable structures of SO2F- and
(CH3) 3NSO2, and predict reactions with
OH-.
Both are trigonal pyramidal. OH- displaces
F— or (CH3) 3N—.
3
2
2
16.8
3
CO2, SO3, P2O5, and Al2O3 are amphoteric.
CO is neutral; MgO and K2O are acidic.
16.2
(a) Use standard potentials (Resource
section 3) to calculate the standard
potential of the disproportionation of H O
in acid solution. (b) Is Cr a likely catalyst
for the disproportionation of H O ? (c)
Given the Latimer diagram
2
16.9
2
2+
2
-0.13
1.51
HO2-
O2
2
H2O2
16.10
in acidic solution, calculate ΔrG for the
disproportionation of hydrogen superoxide
(HO ) into O and H O , and compare the
result
with
its
value
for
the
disproportionation of H O ?
2
2
2
16.11
(a) The disproportionation of H2O2 and
HO2? 1.068 V.
(b) Catalysis by Cr2+?
of decomposing H2O2.
Predict whether any of the following will be
reduced by thiosulfate ions,S2O32–, in acidic
conditions: VO2+, Fe3+, Cu+, Co3+?
SF reacts with BF to form [SF ][BF ]. Use
VSEPR theory to predict the shapes of the
cation and anion?
4
3
3
4
SF4+ trigonal pyramidal, BF4− tetrahedral.
Cr2+ is not capable
(c) The disproportionation of HO2? ΔrG° =
157 kJ. For the disproportionation of H2O2
(part (a)), ΔrG° = 103 kJ.
S2O62−and S2O32−.
Use the standard potential data in
Resource section 3 to predict whether
SeO32– is more stable in acidic or basic
solution?
Fe3+ and Co3+ will be reduced.
2–
2
Use the standard potential data in
Resource section 3 to predict which
oxoanions of sulfur will disproportionate in
acidic conditions?
The SeO32– is marginally more stable in acid
solutions.
ө
2
(a) Give the formula for Te(VI) in acidic
aqueous solution and contrast it with the
formula for S(VI). (b) Offer a plausible
explanation for this difference?
(b) An explanation? Tellurium is a larger
element than sulfur and can increase its
coordination number.
2
5
Predict which oxidation states of Mn will
be reduced by sulfite ions in basic
conditions?
(a) Formulas? H5TeO6– and HSO4–
State whether the following oxides are
acidic, basic, neutral, or amphoteric: CO ,
P O , SO , MgO, K O, Al O , CO?
2
Rank the following species from the
strongest reducing agent to the strongest
oxidizing agent: SO42–,SO32–,O3SO2SO32–?
Mn (+VII, +VI, +V, +IV, +III) will be
reduced by sulfite ions in basic solution.
Exercises
16.1
3
S2O82– > SO42– > SO32–
2
The
decomposition
of
H2O2
is
thermodynamically favored in presence of Br–
S16.2
2
ethylenediamine is a better solvent than sulfur
dioxide.
Self-tests
S16.1
4
16.12
Tetramethylammonium fluoride (0.70 g)
reacts with SF (0.81 g) to form an ionic
product. (a) Write a balanced equation for
4
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 29
the reaction and (b) sketch the structure of
the anion. (c) How many lines would be
observed in the F-NMR spectrum of the
anion?
19
16.13
(a) SF4 + (CH3)4NF → [(CH3)4N]+[SF5]−; (b)
square pyramidal structure; (c) two F
environments.
Identify the sulfur-containing compounds
A, B, C, D, E, and F?
A = S2Cl2, B = S4N4, C = S2N2, D = K2S2O3,
E = S2O62−, F = SO2.
CHAPTER 17
Self-tests
S17.1
One source of iodine is sodium iodate,
NaIO . Which of the reducing agents
SO (aq) or Sn (aq) would seem practical
from the standpoints of thermodynamic
feasibility and plausible judgements about
cost? Standard potentials are given in
Resource section 3?
B
r2
liqu
id
lower
soft
er
I2
soli
d
lowest
soft
est
H
e
gas
N
e
A
r
K
r
X
e
gas
gas
gas
gas
gree
n
dark
redbro
wn
dark
viol
et
Col
orle
ss
colo
rless
colo
rless
colo
rless
colo
rless
3
2+
2
17.2
Both. SO2 is cheaper.
S17.2
Predict the 19F-NMR pattern for IF7?
For F:
2 resonances.
S17.3
F2 + H2SO4 → CaSO4 + 2 HF
From the perspective of structure and
bonding, indicate several polyhalides that
are analogous to [py–I–py]+, and describe
their bonding?
2 HF + 2 KF → 2 K+HF2−
Examples include I3–, IBr2–, ICl2–, and IF2–.
The three centers contribute four electrons to
three molecular orbitals, one bonding, one
nonbonding, and one antibonding.
For Cl:
2 K+HF2− + electricity → F2 + H2 + 2 KF
2 Cl− + 2 H2O + electricity → Cl2 + H2 +
2 OH−
For Br and I:
Exercises
17.1
Describe how the halogens are recovered
from their naturally occurring halides and
rationalize the approach in terms of
standard
potentials.
Give
balanced
chemical equations and conditions where
appropriate?
Preferably without consulting reference
material, write out the halogens and noble
gases as they appear in the periodic table,
and indicate the trends in (a) physical state
(s, l, or g) at room temperature and
pressure, (b) electronegativity, (c) hardness
of the halide ion, (d) color?
F
Phy
sica
l
stat
e
Electron
egativit
y
gas
highest
(4.0)
2
C
l2
gas
lower
Har
dnes
s of
hali
de
ion
hard
est
soft
er
Col
or
2 X− + Cl2 → X2 + 2 Cl−
I−)
17.3
(X− = Br−,
Sketch a choralkali cell. Show the half-cell
reactions and indicate the direction of
diffusion of the ions. Give the chemical
equation for the unwanted reaction that
would occur if OH migrated through the
membrane
and
into
the
anode
compartment?
–
A drawing of the cell is shown in Figure 17.3.
anode: 2 Cl−(aq) → Cl2(g) + 2 e−
cathode: 2 H2O(l) + 2 e− → 2 OH−(aq) + H2(g)
light
yell
ow
yell
ow-
-
unwanted reaction:
2 OH−(aq) + Cl2(aq) → ClO−(aq) +
Cl−(aq) + H2O(l)
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
30
17.4
shapes of IF and the cation and anion in X,
(c) predict how many 19F-NMR signals
would be observed in IF and X?
Sketch the form of the vacant s* orbital of
a dihalogen molecule and describe its role
in the Lewis acidity of the dihalogens?
3
3
a).
b)
IF3:
X = IF4N(CH3)4
Different possible arrangements of
Since the 2σu* antibonding orbital is the
LUMO for a X2 molecule, it is the orbital that
accepts the pair of electrons from a Lewis
base.
17.5
Which dihalogens are thermodynamically
capable of oxidizing H O to O ?
2
17.6
The shape of anion IF4– is Square planar.
The shape of cation (CH3)4N+ is
Tetrahedral.
2
Cl2 and F2.
Nitrogen trifluoride, NF , boils at –129˚C
and is devoid of Lewis basicity. By
contrast, the lower molar mass compound
NH boils at –33˚C and is well known as a
Lewis base. (a) Describe the origins of this
very large difference in volatility. (b)
Describe the probable origins of the
difference in basicity?
3
c).IF3, two. IF4–, one.
3
(a) Difference in volatility? Ammonia
exhibits hydrogen bonding.
(b) Explain the difference in basicity? The
strong electron-withdrawing effect in NF3
reduces the basicity.
17.9
Use the VSEPR model to predict the shapes
of SbCl , FClO , and [ClF ]+?
5
Based on the analogy between halogens
and pseudohalogens write: (a) the balanced
equation for the probable reaction of
cyanogen, (CN) , with aqueous sodium
hydroxide, (b) the equation for the
probable reaction of excess thiocyanate
with the oxidizing agent MnO (s) in acidic
aqueous solution, (c) a plausible structure
for trimethylsilyl cyanide?
2
2
(a) The reaction of NCCN with NaOH?
NCCN(aq) + 2 OH−(aq) → CN−(aq) +
NCO−(aq) + H2O(l)
(b) The reaction of SCN– with MnO2 in
aqueous acid?
2SCN−(aq) + MnO2(s) + 4 H+(aq) →
(SCN)2(aq) + Mn2+(aq) + 2 H2O(l)
(c) The structure of trimethylsilyl
cyanide? Trimethylsilyl cyanide contains
an Si–CN single bond.
17.8
Given that 1.84 g of IF reacts with 0.93 g of
[(CH ) N]F to form a product X, (a) identify
X, (b) use the VSEPR model to predict the
3
3
4
6
SbCl5 is trigonal bipyramidal, FClO3 is
pyramidal, and ClF6 is octahedral.
17.10
Indicate the product of the reaction
between ClF and SbF ?
5
+
5
–
[ClF4] [SbF6] .
17.11
Sketch all the isomers of the complexes
MCl F and MCl F . Indicate how many
fluorine environments would be indicated
in the 19F-NMR spectrum of each isomer?
4
17.7
3
2
3
3
MCl4F2, the cis isomer, 1; the trans isomer, 2.
MCl3F3, the fac isomer, 1 the mer isomer, 2.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 31
17.12
(a) Use the VSEPR model to predict the
probable shapes of [IF ] and IF . (b) Give a
plausible chemical equation for the
preparation of [IF ][SbF ]?
17.19
(a) The structures of [IF6]+ and IF7?
octahedral; IF7, pentagonal bipyramid.
17.20
+
6
6
Explain why CsI (s) is stable with respect to
the elements but NaI (s) is not?
3
3
7
Large cations stabilize large, unstable anions.
6
IF6+,
Write plausible Lewis structures for (a)
ClO and (b) I O and predict their shapes
and the associated point group?
2
(b) The preparation of [IF6] [SbF6]?
2
6
(a).
IF7 + SbF5 → [IF6][SbF6]
17.13
Predict the shape of the doubly chlorine–
bridged I Cl molecule by using the VSEPR
model and assign the point group?
2
(b) I2O5?
6
A planar dimer (I2Cl6).
17.14
Predict the structure and identify the point
group of ClO F? trigonal pyramidal, Cs
symmetry.
2
17.15
Predict whether each of the following
solutes is likely to make liquid BrF a
stronger Lewis acid or a stronger Lewis
base: (a) SbF , (b) SF , (c) CsF?
17.21
3
5
6
(a) The formulas are HBrO4 and H5IO6. The
difference lies in iodine’s ability to expand its
coordination shell.
(a) SbF5? increases the acidity of BrF3.
(b) SF6? No effect on the acidity or basicity
of BrF3.
(b) Relative stabilities? Periodic acid is
thermodynamically more stable.
(c) CsF? Increases the basicity of BrF3.
17.16
Predict the appearance of the F-NMR
spectrum of IF ? Two resonances.
19
17.22
+
5
17.17
.
Predict whether each of the following
compounds is likely to be dangerously
explosive in contact with BrF and explain
your answer: (a) SbF , (b) CH OH, (c) F ,
(d) S Cl ?
3
5
2
3
(a) Describe the expected trend in the
standard potential of an oxoanion in a
solution with decreasing pH. (b)
Demonstrate
this
phenomenon
by
calculating the reduction potential of ClO4–
at pH = 7 and comparing it with the
tabulated value at pH = 0?
2
(a) The expected trend? E decreases as the
pH increases.
2
(a) SbF5? Since SbF5 cannot be oxidized, it
will not form an explosive mixture with BrF3.
(b) E at pH 0 and pH 7 for ClO4–?
Eº = 1.201 V (see Appendix 2).
(b) CH3OH? Methanol, being an organic
compound, is readily oxidized by strong
oxidants.
At pH 7, V = 0.788 V
17.23
(c) F2? No.
(d) S2Cl2? S2Cl2 will be oxidized to higher
valent sulfur fluorides.
17.18
(a) Give the formulas and the probable
relative acidities of perbromic acid and
periodic acid. (b) Which is the more stable?
The
formation
of
Br from
a
tetraalkylammonium bromide and Br is
only slightly exoergic. Write an equation
(or NR for no reaction) for the interaction
of [NR ][Br ] with I in CH Cl solution and
give your reasoning?
With regard to the general influence of pH
on the standard potentials of oxoanions,
explain why the disproportionation of an
oxoanion is often promoted by low pH?
Low pH results in a kinetic promotion:
protonation of an oxo group aids oxygen–
halogen bond scission.
–
3
2
4
3
2
Br3− + I2 → 2 IBr + Br−
2
2
17.24
Which oxidizing agent reacts more readily
in dilute aqueous solution, perchloric acid
or periodic acid? Give a mechanistic
explanation for the difference?
Periodic acid.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
32
17.25
(a) For which of the following anions is
disproportionation
thermodynamically
favourable in acidic solution: OCl , ClO2– ,
ClO2– , and ClO4– ? (If you do not know the
properties of these ions, determine them
from a table of standard potentials.) (b)
For which of the favourable cases is the
reaction very slow at room temperature?
18.2
2
The rates of disproportionation are probably
HClO > HClO2 > ClO3–.
17.26
(b) An electric discharge light source
requiring a safe gas with the lowest
ionization energy? Xenon.
(c)
The
atmosphere?
18.3
Which of the following compounds present
an explosion hazard? (a) NH ClO , (b)
Mg(ClO ) , (c) NaClO , (d) [Fe(H O) ][ClO ] .
Explain your reasoning?
4
4
4
2
2
4
6
4
2
least
Argon.
expensive
inert
By means of balanced chemical equations
and a statement of conditions, describe a
suitable synthesis of (a) XeF2? Xe and F2
at 400ºC, or photolyze Xe and F2 in glass:
Xe(g) + F2(g) → XeF2(s)
(a) NH4ClO4? Ammonium perchlorate is a
dangerous compound, since the N atom of the
NH4+ ion is in its lowest oxidation state and
can be oxidized.
(b) XeF6? High temperature, but have a large
excess of F2:
(b) Mg(ClO4)2? Not an explosion hazard.
(c) XeO3?
Xe(g) + 3 F2(g) → XeF6(s)
XeF6(s) + 3 H2O(l) → XeO3(s) + 6 HF(g)
(c) NaClO4? Not an explosion hazard.
(d) [Fe(H2O)6] [ClO4]2? An explosion
hazard, since Fe(II) can be oxidized to Fe(III).
17.27
Which of the noble gases would you choose
as
(a)
The
lowest-temperature
refrigerant? Helium.
Draw the Lewis structures of (a) XeOF4?
18.4
(b) XeO2F2? (c) XeO64−?
Use standard potentials to predict which of
the following will be oxidized by ClO– ions
in acidic conditions: (a) Cr3+, (b) V3+, (c)
Fe2+, (d) Co2+?
(a) Cr3+? No. (b) V3+? Yes. (c) Fe2+? Yes. (d)
Co2+? No.
18.5
CHAPTER 18
Self-tests
S18.1
2
(b) IBr2–? Linear geometry. Isostructural
with XeF2.
Write a balanced equation for the
decomposition of xenate ions in basic
solution for the production of perxenate
ions, xenon, and oxygen
HXeO4−(aq)
−
+ 2 OH (aq) →
+ O2(g) + 2 H2O(l)
XeO64−(aq)
+ Xe(g)
Give the formula and describe the
structure of a noble gas species that is
isostructural with (a) ICl4−? XeF4
(c) BrO3–? Trigonal pyramidal geometry.
Isostructural with XeO3.
(d) ClF? Isostructural with the cation XeF+.
18.6
(a) Give a Lewis structure for XeF7–?
Exercises
18.1
Explain why helium is present in low
concentration in the atmosphere even
though it is the second most abundant
element in the universe?
Helium present in today’s atmosphere is the
product of ongoing radioactive decay.
(b) Speculate on its possible structures by
using the VSEPR model and analogy with
other xenon fluoride anions? Pentagonal
bipyramid.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 33
18.7
Use molecular orbital theory to calculate
the bond order of the diatomic species E2+
with E=He and Ne? He22+ = 0.5, Ne22+ = 0.5.
18.8
Identify the xenon compounds A, B, C, D,
and E?
A = XeF2(g)
−
B = [XeF]+[MeBF3]
C = XeF6
D = XeO3
E = XeF4 (g).
18.9
the group oxidation number is not achieved
by N?
19.2
Explain why the enthalpy of sublimation of
Re(s) is significantly greater than that of
Mn(s)?
As atoms become heavier, more energy is
needed to vaporize them.
19.3
State the trend in the stability of the group
oxidation state on descending a group of
metallic elements in the d block. Illustrate
the trend using standard potentials in
acidic solution for Groups 5 and 6.
The group oxidation number increases in
stability as you descend a group.
19.4
For each part, give balanced chemical
equations or NR (for no reaction) and
rationalize your answer in terms of trends
in oxidation states.
Predict the appearance of the 129Xe-NMR
spectrum of XeOF3+.
1:3:3:1 quartet
18.10 Predict the appearance of the 19F-NMR
spectrum of XeOF4.
Strong central line, two other lines
symmetrically distributed around the central
line.
CHAPTER 19
Self-tests
S19.1
Refer to the appropriate Latimer diagram
in Resource section 3 and identify the
oxidation state and formula of the species
that is thermodynamically favoured when
an acidic aqueous solution of V2+ is exposed
to oxygen?
+4; VO2
S19.2
S19.3
(a) Cr2+ (aq) + Fe3+ (aq) →?
Cr2+(aq) + Fe3+(aq) → Cr3+(aq) + Fe2+(aq)
(b) CrO42– (aq) + MoO2 (s) →?
2 CrO42− + 3 MoO2(s) + 10 H+ →
+
Suggest a use for molybdenum(IV) sulfide
that makes use of its solid-state structure.
Rationalize your suggestion? MoS2 used as
a lubricant. The slipperiness of MoS2 is
because of the ease with which one layer can
glide over another.
2 Cr3+ + 3 H2MoO4 + 2 H2O
(c) MnO4– (aq) + Cr3+ (aq) →?
6 MnO4− + 10Cr3+ + 11H2O →
6Mn2+ + 5Cr2O72− + 22 H+
19.5
Describe the probable structure of the
compound formed when Re3Cl9 is dissolved
in a solvent containing PPh3?
2+
More likely for Ni2+ to form a sulphide.
Hardness decreases from left to right in the d
block.
Exercises
Without reference to a periodic table,
sketch the first series of the d block,
including the symbols of the elements.
Indicate those elements for which the
group oxidation number is common by C,
those for which the group oxidation
number can be reached but is a powerful
oxidizing agent by O, and those for which
2+
2
Discrete molecular species such as Re3Cl123–
or Re3Cl9(PPh3)3 are formed.
19.1
(a) Which ion, Ni (aq) or Mn (aq), is more
likely to form a sulfide in the presence of
H S? (b) Rationalize your answer with the
trends in hard and soft character across
Period 4. (c) Give a balanced chemical
equation for the reaction.
Ni2+(aq) + H2S(aq) → NiS(s) + 2 H+(aq)
19.6
Preferably without reference to the text (a)
write out the d block of the periodic table,
(b) indicate the metals that form
difluorides with the rutile or fluorite
structures, and (c) indicate the region of
the periodic table in which metal-metal
bonded halide compounds are formed,
giving one example?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
34
Metal–metal bonded halide compounds are
found within bold border. Example: Sc5Cl6.
19.7
Write a balanced chemical equation for the
reaction
that
occurs
when
cis[RuLCl(OH2)]+ (see Fig. 19.9) in acidic
solution at +0.2 V is made strongly basic at
the same potential. Write a balanced
equation for each of the successive
reactions when this same complex at pH =
6 and +0.2 V is exposed to progressively
more oxidizing environments up to +1.0 V.
Give other examples and a reason for the
redox state of the metal center affecting the
extent of protonation of coordinated
oxygen.
19.9
(a) [Re(O)2 (py)4]+, (b) [V (O)2 (ox)2]3–, (c)
[Mo(O)2(CN)4]4–, (d) [VOCl4]2–? The first
three are dioxo complexes and will have cisdioxometal structural units. (d) VOCl42–, has a
square-pyramidal structure with an apical oxo
ligand.
19.10
Which of the following are likely to have
structures that are typical of (a)
predominantly ionic, (b) significantly
covalent,
(c)
metal-metal
bonded
compounds: NiI , NbCl , FeF , PtS, and
WCl ? Rationalize the differences and
speculate on the structures? (a) NiI2? ionic
compound with significant degree of covalent
character
2
4
2
2
(b) NbCl4? significantly covalent.
(c) FeF2? ionic.
−
cis-[Ru LCl(OH2)] + Ox + OH → cis[RuIIILCl(OH)]+ + Red + H2O
(d) PtS ? significant amount of covalent
character.
cis-[RuIILCl(OH2)]+ + H2O → cis[RuIIILCl(OH)]+ + H3O+ + e−
(e) WCl2 ? metal–metal bonding.
II
+
cis-[RuIIILCl(OH2)]+ + H2O → cis[RuIVLCl(OH)]+ + H3O+ + e−
19.11
(b) [Cr2(O2CC2H5)4]? σ2π4δ2,chromium–
chromium quadruple bond.
Give plausible balanced chemical reactions
(or NR for no reaction) for the following
combinations, and state the basis for your
answer: (a) MoO42– (aq) plus Fe2+ (aq) in
acidic solution? No reaction.
(b) The preparation of [Mo6O19]2−(aq) from
K2MoO4(s)? 6 MoO42−(aq) + 10 H+(aq) →
[Mo6O19]2−(aq) + 5 H2O(l)
(c)
[Cu2(O2CCH3)4]?
σ2π4δ2δ*2π*4σ*2
configuration, no metal–metal bond in this
molecule.
19.12
Explain the differences in the following
redox couples, measured at 25oC?
Higher oxidation states become more stable
on descending a group.
19.13
Addition of sodium ethanoate to aqueous
solutions of Cr(II) gives a red diamagnetic
product. Draw the structure of the
product, noting any features of interest?
(c) ReCl5 (s) plus KMnO4(aq)?
5 ReCl5(s) + 2 MnO4−(aq) + 12 H2O(l) →
5 ReO4−(aq) + 2 Mn2+(aq) + 25 Cl−(aq) +
24 H+(aq)
(d) MoCl2(s) plus warm HBr(aq)?
6 MoCl2(s) + 2 Br−(aq) → [Mo6Cl12]2−(aq)
+ Br2(aq)
(e) TiO(s) with HCl(aq) under an inert
2TiO (s) + 6H+ (aq) Æ
atmosphere?
3+
2Ti + H2(g) + 2H2O(l) E˚ = +0.37 V.
(f) Cd(s) added to Hg2+(aq)?
Hg2+ (aq) + Cd(s) Æ Hg(l) + Cd2+ (aq)
Indicate the probable occupancy of s, p,
and d bonding and antibonding orbitals,
and the bond order for the following
tetragonal prismatic complexes?
(a) [Mo2(O2CCH3)4]? σ2π4δ2 configuration,
molybdenum-molybdenum quadruple bond.
As oxidation state of the metal increases,
ability to accept electron density from an OH–
or O2– ligand increases.
19.8
Speculate on the structures of the following
species and present bonding models to
justify your answers:
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 35
differences in the 6–8 eV region can be
attributed to the lack of d electrons for Mg(II).
O
What terms arise from a p1d1
configuration? 1F and 3F terms are possible.
S20.6
Identifying ground terms (Hint: Because d9
is one electron short of a closed shell with L
= 0 and S = 0, treat it on the same footing
as a d1 configuration.) (a) 2p2? 3P (called a
“triplet P” term).
(b) 3d9? 2D (called a “doublet D” term).
S20.7
What terms in a d2 complex of
symmetry correlate with the 3F and
terms of the free atom? An F terms
3
T1g, 3T2g, and 3A2g. Similarly, D terms
1
T2g and 1Eg.
S20.8
Use the same Tanabe-Sugano diagram to
predict the energy of the first two spinallowed quartet bands in the spectrum of
[Cr(OH2)6]3+ for which Δo = 17 600 cm-1
and B = 700 cm-1.
17500 cm–1 and 22400 cm–1.
S20.9
The spectrum of [Cr(NCS)6]3- has a very
weak band near 16 000 cm-1, a band at 17
700 cm-1 with εmax= 5160 dm3 mol-1 cm-1, a
band at 23 800 cm-1 with εmax= 5130 dm3
mol-1 cm-1, and a very strong band at 32
400 cm-1. Assign these transitions using the
d3 Tanabe-Sugano diagram and selection
rule considerations. (Hint: NCS- has lowlying π* orbitals.)
O
O
Cr
O
S20.5
Cr
O
O
O
O
19.14
Consider the two ruthenium complexes in
Table 19.8. Using the bonding scheme
depicted in Figure 19.19, confirm the
bonding orders and electron configurations
given in the table.
[Ru2Cl2(ClCO2)4]- :
2.5 +, mixed valence at the Ru center, in line
with the energy level scheme [0.5 (8-3)].
[Ru2 (CH3COCH3)2(ClCO2)4]: 2+, the bond
order of 2.0 should also be expected [0.5x(84)].
CHAPTER 20
Oh
1
D
are
are
Self-tests
S20.1
(i) The very low intensity of the band at 16,000
cm–1 is a clue that it is a spin-forbidden
transition, probably 2Eg ← 4A2g.
What is the LFSE for both high- and lowspin d7 configurations? A high-spin d7
configuration is t25geg2. High spin, 0.8 Δo.
(ii) Spin-allowed but Laporte-forbidden bands
typically have ε ~ 100 M–1 cm–1, so it is
likely that the bands at 17,700 cm–1 and
23,800 cm–1 are of this type.
S20.2 The magnetic moment of the complex
[Mn(NCS)6]4– is 6.06μB. What is its electron
3 2
configuration? t2g eg
S20.3
S20.4
Account for the variation in lattice
enthalpy of the solid fluorides in which
each metal ion is surrounded by an
octahedral array of F– ions: MnF– (2780 kJ
mol–1), FeF2 (2926 kJ mol–1), CoF2 (2976 kJ
mol–1), NiF2 (3060 kJ mol–1), and ZnF2
(2985 kJ mol–1).
If it were not for ligand field stabilization
energy (LFSE), MF2 lattice enthalpies would
increase from Mn(II) to Zn(II).
Suggest an interpretation of the
photoelectron spectra of [Fe(C5H5)2] and
[Mg(C5H5)2] shown in Fig. 20.18?
In the spectrum of [Mo(CO)6], the ionization
energy around 8 eV was attributed to the t2g
electrons that are largely metal-based. The
(iii) The band at 32,400 cm–1 is probably a charge
transfer band, since its intensity is too
high to be a ligand field (d–d) band.
Exercises
20.1
Determine the configuration (in the form
t2g x egy or ext2y, as appropriate), the
number of unpaired electrons, and the
ligand-field stabilization energy in terms of
ΔO or ΔT and P for each of the following
complexes using the spectrochemical series
to decide, where relevant, which are likely
to be high-spin and which low-spin.
(a) [Co(NH3)6]3+? d6, low spin, no unpaired
electrons, LFSE = 2.4Δ0.
(b) [Fe(OH2)6]2+? d6, high spin, four
unpaired electrons, LFSE = 0.4Δ0.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
36
(c) [Fe(CN)6]3–? d5, low spin, one unpaired
electron. LFSE is 2.0Δ0.
(d) [Cr(NH3)6]3+?
d3, three unpaired
electrons, low spin, LFSE = 1.2Δ0.
(e) [W(CO)6]? d6, low spin, no unpaired
electrons, LFSE = 2.4Δ0.
(c) [Fe(CN)6]3–
(d) The ruthenium complex.
(e) The Co2+ complex.
20.6
Interpret the variation, including the overall
trend across the 3d series, of the following
values of oxide lattice enthalpies (in kJ
mol–1). All the compounds have the rocksalt structure: CaO (3460), TiO (3878), VO
(3913), MnO (3810), FeO (3921), CoO
(3988), NiO (4071)? There are two factors
that lead to the values: (i) decreasing ionic
radius from left to right across the d block,
and (ii) LFSE.
.20.7
A neutral macrocyclic ligand with four
donor atoms produces a red diamagnetic
low-spin d8 complex of Ni(II) if the anion is
the weakly coordinating perchlorate ion.
When perchlorate is replaced by two
thiocyanate ions, SCN–, the complex turns
violet and is high-spin with two unpaired
electrons. Interpret the change in terms of
structure? Shift from square planar to
tetragonal complex.
20.8
Bearing in mind the Jahn-Teller theorem,
Predict
the
structure
of
[Cr(OH2)6]2+? Elongated octahedron.
20.9
The spectrum of d1 Ti3+(aq) is attributed to
a single electronic transition eg ← t2g. The
band shown in Fig. 20.3 is not symmetrical
and suggests that more than one state is
involved. Suggest how to explain this
observation
using
the
Jahn-Teller
theorem? The electronic excited state of
Ti(OH2)63+ has the configuration t2g0eg1, and
so the excited state possesses eg degeneracy.
Therefore, the “single” electronic transition is
really the superposition of two transitions, one
from an Oh ground-state ion to an Oh excitedstate ion, and a lower energy transition from
an Oh ground-state ion to a lower energy
distorted excited-state ion.
20.10
Write the Russell–Saunders term symbols
for states with the angular momentum
quantum numbers (L,S):
(f) Tetrahedral [FeCl4]2−? d6, high spin,
four unpaired electrons, LFSE = 0.6 ΔT.
(g) Tetrahedral [Ni(CO)4]? d10, 0 unpaired
electrons, LFSE = 0.
20.2
Both H- and P(C6H5)3 are ligands of similar
field strength, high in the spectrochemical
series. Recalling that phosphines act as π
acceptors, is π-acceptor character required
for strong-field behaviour? What orbital
factors account for the strength of each
ligand?
No.
Thus there are two ways for a complex to
develop a large value of Δ0, by possessing
ligands that are π-acids or by possessing
ligands that are strong σ-bases (or both).
20.3
20.4
20.5
Estimate the spin-only contribution to the
magnetic moment for each complex in
Exercise 20.1. Spin-only contributions are:
complex
N
μSO = [(N)(N + 2)]1/2
3+
[Co(NH3)6]
0
0
[Fe(OH2)6]2+
4
4.9
[Fe(CN)6]3–
1
1.7
[Cr(NH3)6]3+
3
3.9
[W(CO)6]
0
0
[FeCl4]2–
4
4.9
[Ni(CO)4]
0
0
Solutions of the complexes [Co(NH3)6]2+,
[Co(OH2)6]2+ (both Oh), and [CoCl4]2- are
colored. One is pink, another is yellow, and
the third is blue. Considering the
spectrochemical series and the relative
magnitudes of ΔT and ΔO, assign each color
to one of the complexes. [CoCl4]2– is blue,
[Co(NH3)6]2+ is yellow, [Co(OH2)6]2+ is pink.
For each of the following pairs of
complexes, identify the one that has the
larger LFSE:
(a) [Cr(OH2)6]2+ or [Mn(OH2)6]2+
(b) [Mn(OH2)6]2+ or [Fe(OH2)6]3+
(c) [Fe(OH2)6]3+ or [Fe(CN)6]3(d) [Fe(CN)6]3– or [Ru(CN)6]
(e) tetrahedral [FeCl4]2–or tetrahedral
[CoCl4]2–
(a) The chromium complex.
3+
(b) The Fe complex.
(a) L = 0, S = 5/2? 6S
(b) L = 3, S = 3/2? 4F
(c) L = 2, S = 1/2? 2D
(d) L = 1, S = 1? 3P
20.11
Identify the ground term from each set of
terms: (a) 3F, 3P, 1P, 1G? 3F.
(b) 5D, 3H, 3P, 1G, 1I? 5D
(c) 6S, 4G, 4P, 2I? 6S
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 37
20.12
(b) 3p2?
20.13
εmax52 dm3 mol-1 cm-1, and a strong band at
higher energy with εmax= 2 x 104 dm3 mol–1
cm–1. What do you suggest for the origins
of these transitions? First, the intense band
at relatively high energy is undoubtedly a
spin-allowed charge-transfer transition. The
two bands with εmax = 60 and 80 M–1 cm–1 are
probably spin-allowed ligand field transitions.
The very weak peak is most likely a spinforbidden ligand field transition.
Give the Russell-Saunders terms of the
configurations and identify the ground
term? (a) 4 s1? 2S.
1
D, 3P, 1S. The ground term is 3P.
The gas-phase ion V3+ has a 3F ground term.
The 1D and 3P terms lie, respectively, 10
642 and 12 920 cm–1 above it. The energies
of the terms are given in terms of Racah
parameters as E(3F) = A - 8B, E(3P) = A +
7B, E(1D) = A - 3B + 2C. Calculate the
values of B and C for V3+. B = (12920 cm–
1
)/(15) = 861.33 cm–1 and C = 3167.7 cm–1.
20.20
Ordinary bottle glass appears nearly
colorless when viewed through the wall of
the bottle but green when viewed from the
end so that the light has a long path
through the glass. The color is associated
with the presence of Fe3+ in the silicate
matrix. Suggest which transitions are
responsible for the color? The faint green
color, which is only observed when looking
through a long pathlength of bottle glass, is
caused by spin-forbidden ligand field
transitions.
20.21
Solutions of [Cr(OH2)6]3+ ions are pale blue–
green but the chromate ion, CrO42–, is an
intense yellow. Characterize the origins of
the transitions and explain the relative
intensities. The blue-green color of the Cr3+
ions in [Cr(H2O)6]3+ is caused by spinallowed but Laporte-forbidden ligand field
transitions. The relatively low molar
absorption coefficient is the reason that the
intensity of the color is weak. The oxidation
state of chromium in dichromate dianion is
Cr(VI); the intense yellow color is due to
LMCT transitions.
20.22
Classify the symmetry type of the d orbital
in a tetragonal C4v symmetry complex, such
as [CoCl(NH3)5] –, where the Cl lies on the
z-axis. (a) Which orbitals will be displaced
from their position in the octahedral
molecular
orbital
diagram
by
π
interactions with the lone pairs of the Cl–
ligand? (b) Which orbital will move
because the Cl– ligand is not as strong a
base as NH3? (c) Sketch the qualitative
molecular orbital diagram for the C4v
complex. The Cl atom lone pairs of electrons
can form π molecular orbitals with dxz and dyz.
These metal atomic orbitals are π-antibonding
MOs, and so they will be raised in energy.
20.14 Write the d-orbital configurations and use
the Tanabe–Sugano diagrams (Resource
section 6) to identify the ground term of (a)
Low-spin [Rh(NH3)6]3+? 1A1g.
(b) [Ti(H2O)6]3+? 2T2g.
(c) High-spin [Fe(H2O)6]3+? 6A1g.
20.15
Using the Tanabe-Sugano diagrams in
Resource section 6, estimate ΔO and B for
(a) [Ni(H2O)6]2+ (absorptions at 8500, 15400
and 26000 cm-1) ? Δ0 = 8500 cm–1 and B ≈
770 cm–1.
(b) [Ni(NH3)6]2+ (absorptions at 10750,
17500 and 28200 cm-1)? Δ0 = 10,750 cm–1
and B ≈ 720 cm–1.
20.16
The spectrum of [Co(NH3)6]3+ has a very
weak band in the red and two moderate
intensity bands in the visible to near-UV.
How should these transitions be assigned?
The first two transitions listed above
correspond to two low-spin bands. The very
weak band in the red corresponds to a spinforbidden transition.
20.17
Explain why [FeF6]3- is colorless whereas
[CoF6]3– is colored but exhibits only a
single band in the visible. No spin-allowed
transitions are possible for the Fe3+; the
complex is expected to be colorless. The d6
Co3+ ion in [CoF6]3– is also high spin, but in
this case a single spin-allowed transition
makes the complex colored and gives it a oneband spectrum.
20.18
The Racah parameter B is 460 cm–1 in
[Co(CN)6]3- and 615 cm-1 in [Co(NH3)6]3+.
Consider the nature of bonding with the
two ligands and explain the difference in
nephelauxetic effect? Ammonia and cyanide
ion are both σ-bases, but cyanide is also a πacid.
20.19
An approximately ‘octahedral’ complex of
Co(III) with ammine and chloro ligands
gives two bands with εmax between 60 and
80 dm3 mol–1 cm–1, one weak peak with
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
38
20.23
Consider the molecular orbital diagram for
a tetrahedral complex (based on Fig. 20.7)
and the relevant d-orbital configuration
and show that the purple color of MnO4ions cannot arise from a ligand-field
transition. Given that the wavenumbers of
the two transitions in MnO4- are 18 500 and
32 200 cm–1, explain how to estimate ΔT
from an assignment of the two chargetransfer transitions, even though ΔT cannot
be observed directly. The oxidation state of
manganese in permanganate anion is
Mn(VII), which is d0. Therefore, no ligand
field transitions are possible.
The difference in energy between the two
transitions, E(t2) – E(e) = 13700 cm–1, is just
equal to ΔT.
20.24
The lowest energy band in the spectrum of
[Fe(OH2)6]3+ (in 1M HClO4) occurs at
lower energy than the equivalent transition
in the spectrum of [Mn(OH2)6]2+. Explain
why this is.
The extra charge of the iron complexes keeps
the eg and t2g levels close
Exercises
21.1
The rate constants for the formation of
[CoX(NH3)5]2+ from [Co(NH3)5OH2]3+ for X
= Cl2, Br2, N3–, and SCN¯ differ by no more
than a factor of two. What is the
mechanism of the substitution? Dissociative.
21.2
If a substitution process is associative, why
may it be difficult to characterize an aqua
ion as labile or inert? The rate of an
associative process depends on the identity of
the entering ligand and, therefore, it is not an
inherent property of [M(OH2)6]n+.
21.3
The reactions of Ni(CO)4 in which
phosphines or phosphites replace CO to
give Ni(CO)3L all occur at the same rate
regardless of which phosphine or phosphite
is being used. Is the reaction d or a? d.
21.4
Write the rate law for formation of
[MnX(OH2)5]1 from the aqua ion and X–.
How would you undertake to determine if
the reaction is d or a?
rate = (kKE[Mn(OH2)62+][X−])/(1 + KE[X−])
21.5
CHAPTER 21
Self-tests
S21.1
Calculate the second-order rate constant
for the reaction of trans-[PtCl(CH3)
(PEt3)2] with NO2– in MeOH, for which npt
= 3.22 ? k2(NO2−) = 100.71 = 5.1 M−1s−1
S21.2
Given the reactants PPh , NH , and [PtCl ] ,
Propose efficient routes to both cis- and
trans- [PtCl2(NH3)(PPh3)]?
S21.3
Metal centers with high oxidation numbers
have stronger bonds to ligands than metal
centers with low oxidation numbers.
Furthermore, period 5 and 6 d-block metals
have stronger metal ligand bonds.
2-
3
3
4
Use the data in Table 21.8 to estimate an
appropriate value for KE and calculate kr2
for the reactions of V(II) with Cl- if the
observed second-order rate constant is
1.2x102 dm3 mol–1 s–1. KE = 1 M–1, k = 1.2 ×
102 s–1.
Octahedral complexes of metal centers with
high oxidation numbers or of d metals of
the second and third series are less labile
than those of low oxidation number and d
metals of the first series of the block.
Account for this observation on the basis of
a dissociative rate-determining step.
21.6
A
Pt(II)
complex
of
tetramethyldiethylenetriamine is attacked
by Cl- 105 times less rapidly than the
diethylenetriamine analogue. Explain this
observation in terms of an associative ratedetermining step. The greater steric
hindrance.
21.7
The rate of loss of chlorobenzene, PhCl,
from [W(CO)4L(PhCl)] increases with
increase in the cone angle of L. What does
this observation suggest about the
mechanism
is
mechanism? The
dissociative.
21.8
The pressure dependence of the
replacement of chlorobenzene (PhCl) by
piperidine in the complex
[W(CO)4(PPh3)(PhCl)] has been studied.
The volume of activation is found to be
111.3 cm3 mol–1. What does this value
suggest about the mechanism? Mechanism
must be dissociative.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 39
21.9
21.10
Does the fact that [Ni(CN)5]3– can be
isolated help to explain why substitution
reactions of [Ni(CN)4]2– are very rapid?
For a detectable amount of [Ni(CN)5]3– to
build up in solution, the forward rate constant
kf must be numerically close to or greater than
the reverse rate constant kr.
(iii) The implication is that a complex without
protic ligands will not undergo anomalously
fast OH– ion substitution.
21.14
Reactions of [Pt(Ph) 2(SMe2) 2] with the
bidentate ligand 1,10-phenanthroline
(phen) give [Pt(Ph) 2phen]. There is a
kinetic
pathway
with
activation
parameters Δ‡ H = 1101 kJ mol–1 and Δ ‡S
= 142 J K –1 mol –1. Propose a mechanism.
A possible mechanism is loss of one
dimethyl sulfide ligand, followed by the
coordination of 1,10-phenanthroline.
21.11
(ii) The anomalously high rate of substitution
by OH– signals an alternate path, that of base
hydrolysis.
Predict the products of the following
reactions: (a) [Pt(PR3)4]2+ + 2Cl–? cis[PtCl2(PR3)2].
(b) [PtCl4]2– + 2PR3? trans-[PtCl2(PR3)2].
(c) cis-[Pt(NH3)2(py)2]2+ + 2Cl–? trans[PtCl2(NH3)(py)].
21.15
A two-step synthesis for cis- and trans[PtCl2(NO2) (NH3)] – start with [PtCl4]2–?
Put in order of increasing rate of
substitution by H2O the complexes:
[Co(NH3)6]3+, [Rh(NH3)6]3+, [Ir(NH3)6]3+,
[Mn(OH2)6]2+, [Ni(OH2)6]2+?
The order of increasing rate is [Ir(NH3)6]3+ <
[Rh(NH3)6]3+ < [Co(NH3)6]3+ < [Ni(OH2)6]2+
< [Mn(OH2)6]2+.
21.16
State the effect on the rate of dissociatively
activated reactions of Rh(III) complexes of
each of (a) an increase in the positive
charge on the complex? decreased rate.
(b) changing the leaving group from NO3–
to Cl–? Decreased rate.
(c) changing the entering group from Cl– to
I–? This change will have little or no effect
on the rate.
21.12
How does each of the following affect the
rate
of
square-planar
substitution
reactions? (a) Changing a trans ligand
from H to Cl? Rate decreases.
(b) Changing the leaving group from Cl– to
I–? The rate decreases.
(c) Adding a bulky substituent to a cis
ligand? The rate decreases.
(d) Increasing the positive charge on the
complex? Rate increase.
21.13
The rate of attack on Co(III) by an entering
group Y is nearly independent of Y with
the spectacular exception of the rapid
reaction with OH–. Explain the anomaly.
What is the implication of your explanation
for the behaviour of a complex lacking
Brønsted acidity on the ligands? (i) The
general trend: octahedral Co(III) complexes
undergo dissociatively activated ligand
substitution.
(d) changing the cis ligands from NH3 to
H2O? Decreased rate.
21.17
Write out the inner- and outer-sphere
pathways for reduction of
azidopentaamminecobalt(III) ion with
V2+(aq). What experimental data might be
used to distinguish between the two
pathways?
The inner-sphere pathway:
[Co(N3)(NH3)5]2+ + [V(OH2)6]2+ →
[[Co(N3)(NH3)5]2+, [V(OH2)6]2+}
{[Co(N3)(NH3)5]2+, [V(OH2)6]2+} →
{[Co(N3)(NH3)5]2+, [V(OH2)5]2+, H2O}
{[Co(N3)(NH3)5]2+, [V(OH2)5]2+, H2O} →
[(NH3)5Co–N=N=N–V(OH2)5]4+
[(NH3)5Co–N=N=N–V(OH2)5]4+ → [(NH3)5Co–
N=N=N–V(OH2)5]4+
[(NH3)5Co–N=N=N–V(OH2)5]4+ → [Co(OH2)6]2+ +
[V(N3)(OH2)5]2+
The outer sphere pathway:
[Co(N3)(NH3)5]2+ + [V(OH2)6]2+ →
{[Co(N3)(NH3)5]2+ [V(OH2)6]2+}
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
40
Using these values gives K12 = 1.81 × 1028.
Substitution of these values in the MarcusCross relationship gives k12 = 8.51 × 1015
The
dm3mol−1s−1.21.21
photochemical
substitution
of
[W(CO)5(py)] (py = pyridine) with
triphenylphosphine
gives
W(CO)5(P(C6H5)3). In the presence of
excess phosphine, the quantum yield is
approximately 0.4. A flash photolysis study
reveals a spectrum that can be assigned to
the intermediate W(CO)5. What product
and quantum yield do you predict for
substitution of [W(CO)5(py)] in the
presence of excess triethylamine? Is this
reaction expected to be initiated from the
ligand field or MLCT excited state of the
The
product
will
be
complex?
[W(CO)5(NEt3)], and the quantum yield will
be 0.4.
{[Co(N3)(NH3)5]2+,[V(OH2)6]2+} → {[Co(N3)(NH3)5]+ ,
[V(OH2)6]3+}
{[Co(N3)(NH3)5] +,[V(OH2)6]3+} → {[Co(OH2)6]2+ ,
[V(OH2)6]3+
21.18
The compound [Fe(SCN)(OH2)5]2+ can be
detected
in
the
reaction
of
[Co(NCS)(NH3)5]2+ with Fe2+(aq) to give
Fe3+(aq) and Co2+(aq). What does this
observation
suggest
about
the
mechanism? Appears to be an inner-sphere
electron transfer reaction.
21.19
Calculate the rate constants for electron
transfer in the oxidation of [V(OH2)6]2+ (Eσ
(V3+/V2+) = –0.255 V) and the oxidants (a)
[Ru(NH3)6]3+ (EO(Ru3+/Ru2+) = + 0.07 V),
(b) [Co(NH3)6]3+ (EO(Co3+/Co2+) = +0.10 V).
Comment on the relative sizes of the rate
constants.
(a) [Ru(NH3)6]3+ k = 4.53 × 103 dm3mol−1s−1
(b) [Co(NH3)6]
3+
k = 1.41 × 10 dm mol s
−2
3
−1 −1
Relative sizes? The reduction of the Ru
complex is more thermodynamically favoured
and faster.
21.20
Calculate the rate constants for electron
transfer in the oxidation of [Cr(OH2)6]2+
(EO–(Cr3+/Cr2+) = –0.41 V) and each of the
oxidants [Ru(NH3)6]3+ (EO(Ru3+/Ru2+) =
+0.07 V), [Fe(OH2)6]3+ (EO(Fe3+/Fe2+) =
+0.77 V) and [Ru(bpy)3]3+ (EO(Ru3+/Ru2+) =
+1.26 V). Comment on the relative sizes of
the rate constants
(a) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22
(Ru3+/Ru2+ for the hexamine complex) = 6.6 ×
o
103 dm3mol−1s−1; f12 = 1; K12 = e [nF ε /RT]
where εo = 0.07 V – (–0.41V) = 0.48 V; n = 1;
F = 96485 C; R = 8.31 Jmol−1K−1 and T = 298
K. Using these values gives K12 = 1.32 × 108.
Substitution of these values in the MarcusCross relationship gives k12 = 2.95 × 103
dm3mol−1s−1.
(b) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1; k22
(Fe3+/Fe2+ for the aqua complex) = 1.1
o
dm3mol−1s−1; f12 = 1; K12 = e[nF ε /RT] where εo
= 0.77 V – (−0.41V) = 1.18 V; n = 1; F =
96485 C; R = 8.31 Jmol−1K−1 and T = 298 K.
Using these values we get K12 = 9.26 × 1019.
Substitution of these values in the MarcusCross relationship gives k12 = 3.19 × 107
dm3mol−1s−1.
(c) k11 (Cr3+/Cr2+) = 1 × 10−5 dm3mol−1s−1;
k22 (Ru3+/Ru2+ for the bipy complex) = 4 × 108
o
dm3mol−1s−1; f12 = 1; K12 = e[nF ε /RT] where εo
= 1.26 V – (–0.41V) = 1.67 V; n = 1; F =
96485 C; R = 8.31 Jmol−1K−1 and T = 298 K.
21.22 From the spectrum of [CrCl(NH3)s]2+ shown
in Fig. 20.32, propose a wavelength for
photoinitiation of reduction of Cr(III) to
Cr(II) accompanied by oxidation of a
ligand. ~250 nm.
CHAPTER 22
Self-tests
S22.1
Is Mo(CO)7 likely to be stable?
S22.2
What is the electron count for and
oxidation number of platinum in the anion
of Zeise’s salt, [PtCl3(C2H4)]–? Treat
CH =CH as a neutral two-electron donor.
Electron count, 16; oxidation number, +4.
2
No.
2
S22.3
What
is
the
formal
name
of
[Ir(Br)2(CH3)(CO)(PPh3)2]?
Dibromocarbonylmethylbis(triphenylphosphine)iridium(III).
S22.4
Which of the two iron compounds Fe(CO)5
and [Fe(CO)4(PEt3)] will have the higher
CO stretching frequency? Which will have
the longer M–C bond? Fe(CO)5
S22.5
Show that both are 18-electron species.
(a) [(η6-C7H8)Mo(CO)3] (49)?
η6-C7H8 = 6 electrons, Mo atom = 6.
carbonyl = 2 electrons, total = 18.
(b) [(η7-C7H7)Mo(CO)3]+(51)? η7-C7H7+ = 6
electrons, Mo atom = 6, 3 COs = 6, total = 18.
S22.6
Propose
a
synthesis
Mn(CO)4(PPh3)(COCH3) starting
[Mn2(CO)10], PPh3, Na and CH3I.
for
with
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 41
Mn2(CO)10 + 2 Na → 2Na[Mn(CO)5]
Exercises
Na[Mn(CO)5] + CH3I → Mn(CH3)(CO)5 + NaI
22.1
Mn(CH3)(CO)5 + PPh3 → Mn(CO)4(PPh3)(COCH3)
S22.7
S22.8
S 22.9
The IR spectrum of [Ni2(η5-Cp)2(CO)2] has
a pair of CO stretching bands at 1857 cm–1
(strong) and 1897 cm–1 (weak). Does this
complex contain bridging or terminal CO
ligands, or both? (Substitution of η5-C5H5
ligands for CO ligands leads to small shifts
in the CO stretching frequencies for a
terminal CO ligand.) Bridging
(a) Fe(CO)5? Pentacarbonyliron(0), 18e–;
(b) Mn2(CO)10?
decacarbonyldimanganese(0), 18e–;
(c) V(CO)6? hexcarbonylvanadium(0), 17e–
(d) [Fe(CO)4]2–? tetracarbonylferrate(–2),
18e–;
By using the same molecular orbital
diagram, comment on whether the removal
of an electron from [Fe(η5-Cp)2] to
produce [Fe(η5-Cp)2]+ should produce a
substantial change in M–C bond length
relative to neutral ferrocene. It will not.
(e) La(η5-Cp*)3?
tris(pentamethylcyclopentadienyl)lanthanum(I
II), 18e–;
(f) Fe(η3-allyl)(CO)3Cl?
allyltricarbonylchloroiron(II), 18e–;
The compound [Fe4(Cp)4(CO)4] is a darkgreen solid. Its IR spectrum shows a single
CO stretch at 1640 cm–1. The 1H NMR
spectrum is a single line even at low
temperatures. From this spectroscopic
information and the CVE, propose a
structure
for
[Fe4(Cp)4(CO)4]. TA
tetrahedron with 4 terminal Cp rings and four
capping COs.
(g) Fe(CO)4(PEt3)?
tetracarbonyltriethylphosphineiron(0);
(h) Rh(CO)2(Me)(PPh3)?
dicarbonylmethyltriphenylphosphinerhodium(
I),16e(i) Pd(Cl)(Me)(PPh3)2?
chloromethylbis(triphenylphosphine)
palladium(II), 16e–;
S22.10 If Mo(CO)3L3 is desired, which of the
ligands P(CH3)3 or P(t-Bu)3 would be
preferred? Give reasons for your choice.
PMe3 would be preferred.
(j) Co(η5-C5H5)(η4-C4Ph4)?
cyclopentadienyltetraphenylcylcobutadinecob
alt(I), 18e–;
(k) [Fe(η5-C5H5)(CO2)]–?
dicarbonylcyclopentadienylferrate(0), 18e–;
S22.11 Assess
the
relative
substitutional
reactivities of indenyl and fluorenyl (86)
compounds? Fluorenyl compounds are more
reactive than indenyl.
(l) Cr(η6-C6H6)(η6-C7H8)?
benzenecycloheptatrienechromium(0), 18e–;
S22.12 Show that the reaction is an example of
reductive elimination?
(m) Ta(η5-C5H5)2Cl3?
trichlorobiscyclopentadineyltanatalum(V),
18e–;
Cl
Ph3P
Cl
Cl
Ph
PPh3
Ph3P
(n) Ni(η5-C5H5)NO?
cyclopentadineylnitrosylnickel(0),18e–.
PPh3
Pd
Pd
+
PhCl
Cl
Cl
The decrease in both coordination number and
oxidation number by 2.
Name the species, draw the structures of,
and give valence electron counts to the
metal atoms? Do any of the complexes
deviate from the 18-electron rule? If so,
how is this reflected in their structure or
chemical properties?
22.2
Sketch an η2 the interactions of 1,4butadiene with a metal atom and (b) do the
same for an η4 interaction.
22.3
What hapticities are possible for the
interaction of each of the following ligands
S22.13 Explain why [Pt(PEt3)2(Et)(Cl)] readily
decomposes, whereas [Pt(PEt3)2 (Me)(Cl)]
The
ethyl
group
in
does
not?
[Pt(PEt3)2(Et)(Cl)] is prone to β-hydride
elimination.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
42
with a single d-block metal atom such as
cobalt? (a) C2H4? η2
(b) Cyclopentadienyl? Can be η5, η3, or η1.
(c) C6H6? η6, η4, and η2.
(b) Co(η4-C4H4)(η5-C5H5)? 18.
22.8
Provide plausible reasons for the
differences in IR wavenumbers between each of the
following pairs:
(a) Mo(PF3)3(CO)3 2040, 1991 cm–1 versus
Mo(PMe3)3(CO)3 1945, 1851 cm–1? CO
bands of the trimethylphosphine complex are
100 cm–1 or more lower in frequency. PMe3 is
primarily a σ-donor ligand. PF3 is primarily
a π-acid ligand.
(b) MnCp(CO)3 2023, 1939 cm–1 vs.
MnCp*(CO)3 2017, 1928 cm–1? CO bands
of the Cp* complex are lower in frequency
than the corresponding bands of the Cp
complex. Cp* is a stronger donor ligand than
Cp.
(c) Co(η3-C3H5)(CO)3? 18.
22.9
(d) Cyclooctadiene? η2 and η4 .
(e) Cyclooctatetraene?η8, η6, η4, η2.
22.4
spectrum? Check your answer and give the
number of expected bands for each by
consulting Table 22.7.
Cs symmetry complex, 3
Draw plausible structures and give the
electron count of (a) Ni(η3-C3H5)2 (b)
(c)
Co(η3Co(η4-C4H4)(η5-C5H5)
C3H5)(CO)3. If the electron count deviates
from 18, is the deviation explicable in
terms of periodic terms.
(a) Ni(η3-C3H5)2 16, very common for group
9 and group 10 elements.
The compound Ni3(C5H5)3(CO)2 has a
single CO stretching absorption at 1761
cm–1. The IR data indicate that all C5H5
ligands are pentahapto and probably in
identical environments. (a) On the basis of
these data, propose a structure.
22.5
State the two common methods for the
preparation of simple metal carbonyls and
illustrate your answer with chemical
equations. Is the selection of method based
on thermodynamic or kinetic
considerations?
(1) Mo(s) + 6 CO(g) → Mo(CO)6(s)
(high
temperature and pressure required)
(2) 2 CoCO3(s) + 2 H2(g) + 8 CO(g)
Co2(CO)8(s) + 2 CO2 + 2 H2O
(b) Does the electron count for each metal
in your structure agree with the 18electron rule? If not, is nickel in a
region of the periodic table where
deviations from the 18-electron rule
are common? No. Deviations from the
rule are common for cyclopentadienyl
complexes to the right of the d block.
→
The reason that the second method is
preferred is kinetic.
22.6
Suggest a sequence of reactions for the
preparation of Fe(CO)3(dppe), given iron
metal, CO, dppe (Ph2PCH2CH2PPh2), and
other reagents of your choice.
Fe(s) + 5 CO(g) → Fe(CO)5(l) (high temperature
and pressure required)
Fe(CO)5(l) + diphos(s) → Fe(CO)3(diphos)(s) + 2
CO(g)
22.7
Suppose that you are given a series of
metal tricarbonyl compounds having the
respective symmetries C2v, D3h, and Cs.
Without consulting reference material,
which of these should display the greatest
number of CO stretching bands in the IR
22.10
Decide which of the two complexes W(CO)6
or IrCl(CO)(PPh3)2 should undergo the
faster exchange with 13CO. Justify your
answer. The Ir complex. It has an A
mechanism.
22.11
Which metal carbonyl in each of (a)
[Fe(CO)4]2– or [Co(CO)4]– (b) [Mn(CO)5]–
or [Re(CO)5]–should be the most basic
toward a proton? What are the trends on
which your answer is based ?
(a) [Fe(CO)4]2– The trend involved is the
greater affinity for a cation that a species with
a higher negative charge has.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 43
(b) The rhenium complex. The trend involved
is the greater M–H bond enthalpy for a period
6 metal ion relative to a period 4 metal ion in
the same group.
22.12
Using the 18-electron rule as a guide,
indicate the probable number of carbonyl
ligands in (a) W(η6-C6H6)(CO)n, (b) Rh(η5C5H5)(CO)n, and (c) Ru3(CO)n.
(a) 3 (b) 2 (c) 12
22.13
Propose two syntheses for MnMe(CO)5,
both starting with Mn2(CO)10, with one
using Na and one using Br2? You may use
other reagents of your choice. (i) Reduce
Mn2(CO)10 with Na to give Mn(CO)5–; react
with MeI to give MnMe(CO)5. (ii), Oxidize
with Br2. to give MnBr(CO)5; displace the
bromide with MeLi to give MnMe(CO)5.
22.14
Give the probable structure of the product
obtained when Mo(CO)6 is allowed to react
first with LiPh and then with the strong
carbocation reagent, CH3OSO2CF3.
Mo(CO)5(C(OCH3)Ph)
22.15
Na[W(η5-C5H5)(CO)3] reacts with 3chloroprop-1-ene to give a solid, A, which
has the molecular formula
W(C3H5)(C5H5)(CO)3. Compound A loses
carbon monoxide on exposure to light and
forms compound B, which has the formula
W(C3H5)(C5H5)(CO)2. Treating compound
A with hydrogen chloride and then
potassium hexafluorophosphate, K+PF6–,
results in the formation of a salt, C.
Compound C has the molecular formula
[W(C3H6)(C5H5)(CO)3]PF6. Use this
information and the 18-electron rule to
identify the compounds A, B, and C.
Sketch a structure for each, paying
particular attention to the hapticity of the
hydrocarbon.
22.17
(a) Reflux Mo(CO)6 with cycloheptatriene to
give [Mo(η6-C7H8)(CO)3]; treat with the trityl
tetrafluoroborate to give [Mo(η7C7H7)(CO)3]BF4.
(b) React [IrCl(CO)(PPh3)2] with MeCl, then
expose to CO
atmosphere to give
[IrCl2(COMe)(CO)(PPh3)2].
22.18
W
CO
CO
CO
η1
22.16
W CO
CO C:
B:
η3
-2 C O
-C O
-H
Fe
OC
Fe
CO
CO
A
OC
H
CO
B
[π-C5H5Fe(CO)2]2
C
The compound B shows two 1H NMR
resonances due to Fe-H proton and the
aromatic Cp ring. C shows a single 1H NMR
resonance because of aromatic Cp ring.
22.19
W CO
CO
CO
η2
Treatment of TiCl4 at low temperature
with EtMgBr gives a compound that is
unstable above 270ºC. However, treatment
of TiCl4 at low temperature with MeLi or
LiCH2SiMe3 gives compounds that are
stable at room temperature. Rationalize
these observations.
The compounds formed are TiR4; ethyl has
the low energy β-hydride elimination
decomposition.
When
Fe(CO)5
is
refluxed
with
cyclopentadiene compound A is formed
which has the empirical formula C8H6O3Fe
and a complicated 1H NMR spectrum.
Compound A readily loses CO to give
compound B with two 1H-NMR resonances,
one at negative chemical shift (relative
intensity one) and one at around 5ppm
(relative intensity 5). Subsequent heating of
B results in the loss of H2 and the
formation of compound C. Compound C
has a single 1H-NMR resonance and the
empirical formula C7H5O2Fe. Compounds
A, B, and C all have 18 valence electrons:
identify them and explain the observed
spectroscopic data.
+ F e (C O ) 5
+
A:
Suggest syntheses of (a) [Mo(η7C7H7)(CO)3]BF4 from Mo(CO)6 and (b)
from
[IrCl2(COMe)(CO)(PPh3)2]
[IrCl(CO)(PPh3)2]?
When Mo(CO)6 is refluxed with
cyclopentadiene compound D is formed
which has the empirical formula
C8H5O3Mo and an absorption in the IR
spectrum at 1960 cm–1. Compound D can
be treated with bromine to yield E or with
Na/Hg to give compound F. There are
absorptions in the IR spectra of E and F at
2090 and 1860 cm–1, respectively.
Compounds D, E, and F all have 18 valence
electrons: identify them and explain the
observed spectroscopic data.
D = C5H5Mo(CO)3. E = C5H5Mo(CO)3Br.
F = C5H5Mo(CO)3Na.
22.20
Which compound would you expect to be
more stable,RhCp2 or RuCp2? Give a
plausible explanation for the difference in
terms of simple bonding concepts
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
44
(b) Can these CVE values be derived from
the 18-electron rule? No.
RuCp2 has 18 electrons.
Give the equation for a workable reaction
for the conversion of Fe(η5-C5H5)2 to Fe(η5C5H5) (η5-C5H4COCH3) and (b) Fe(η5C5H5) (η5-C5H4CO2H)
22.21
(c) Determine the probable geometry of
and
[Co6(C)(CO)15]2[Fe6(C)(CO)16]2–
? The iron complex probably contains an
octahedral Fe6 array. The cobalt complex
probably contains a trigonal-prismatic Co6
array.
(a)
Fe(η5 – C5H5)2 + CH3COCl →
5
Fe(η – C5H5)(η5 – C5H4COCH3) + HCl
22.27
(b)
Cl
Fe
22.22
22.23
22.24
+
O
O
Cl
A lC l 3
C H 2C l2
Cl
Fe
C O 2H
H 2 O , t-B u O K
Fe
DME
Sketch the a1’ symmetry-adapted orbitals
for the two eclipsed C5H5 ligands stacked
together with D5h symmetry. Identify the s,
p, and d orbitals of a metal atom lying
between the rings that may have nonzero
overlap, and state how many a1’ molecular
orbitals may be formed.
The symmetry-adapted orbitals of the two
eclipsed C5H5 rings in a metallocene are
shown in Resource Section 5, the dz2 orbital
on the metal has a1′ symmetry, as does s.
Three a1′ MOs will be formed.
22.28 Ligand substitution reactions on metal
clusters are often found to occur by associative
mechanisms, and it is postulated that these occur by
initial breaking of an M-M bond, thereby providing
an open coordination site for the incoming ligand. If
the proposed mechanism is applicable, which would
you expect to undergo the fastest exchange with
added13CO, Co4(CO)12 or Ir4(CO)12? Suggest an
explanation. Co. This is because metal–metal bond
strengths increase down a group in the d block.
The compound Ni(η5-C5H5)2 readily adds
one molecule of HF to yield [Ni(η5C5H5)(η4-C5H6)]+ whereas Fe(η5-C5H5)2
reacts with strong acid to yield [Fe(η5C5H5)2H]+. In the latter compound the H
atom is attached to the Fe atom. Provide a
reasonable explanation for this difference.
Protonation of FeCp2 at iron does not change
its number of valence electrons.
CHAPTER 23
Self-tests
S23.1
Derive the ground state of the Tm3+ ion
3
H6.
S23.2
The product of the reaction above is in fact
a hydride bridged dimer (9). Suggest a
strategy to ensure that the hydride is
monomeric. A simple approach would be to
use the Cp rings substituted by bulky groups
and examine rate of formation.
S23.3
Use the Frost diagrams and data in
Resource section 2 to determine the most
stable uranium ion in acid aqueous solution
in the presence of air and give its
formula. UO22+ if sufficient oxygen is
present.
Write a plausible mechanism, giving your
reasoning,
for
the
reactions:
(a)
[Mn(CO)5(CF2)]+ + H2O → [Mn(CO)6]+ +
2HF?
(i) The F atoms render the C atom subject to
nucleophilic attack
(ii)two equivalents of HF are eliminated
(b)
Rh(C2H5)
(CO)
RhH(CO)(PR3)2 + C2H4?
elimination reaction.
22.25
22.26
Based on isolobal analogies, choose the
groups that might replace the group in
boldface in (a) Co3(CO)9CH→ OCH3,
N(CH3)2, or SiCH3? SiCH3.
(b) (OC)5MnMn(CO)5 → I, CH2, or
CCH3? I.
→
(PR3)2
A β-hydrogen
Given mechanism of CO insertion, what
rate constant can be extracted from rate
data? Rate = ka[RMn(CO)5]
(a) What cluster valence electron (CVE)
count is characteristic of octahedral and
trigonal prismatic
complexes?
octahedral M6, 86; trigonal prismatic M6, 90.
Exercises
23.1
(a) Give a balanced equation for the
reaction of any of the lanthanoids with
aqueous acid. (b) Justify your answer with
redox potentials and with a generalization
on the most stable positive oxidation states
for the lanthanoids. (c) Name the two
lanthanoids that have the greatest tendency
to deviate from the usual positive oxidation
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 45
state and correlate this deviation with
electronic structure. ? (a) Balanced
equation 2 Ln(s) + 6 H3O+(aq) →
23.10
Predict a structure type for BkN based on
the ionic radii r(Bk3+) = 96 pm and r(N3−) =
146 pm.
The rock-salt structure.
23.11
Describe the general nature of the
distribution of the elements formed in the
thermal neutron fission of 235U, and
decide which of the following highly
radioactive nuclides are likely to present
the greatest radiation hazard in the spent
fuel from nuclear power reactors: (a) 39Ar,
(b) 228Th, (c) 90Sr, (d) 144Ce. 90Sr and 144Ce.
2 Ln3+(aq) + 3 H2(g) + 6 H2O(l)
(b) Redox potentials The potentials for the
Ln0/Ln3+ oxidations in acid solution range
from a low of 1.99 V for europium to 2.38 V.
(c) Two unusual lanthanides Ce4+, Eu2+.
23.2
Explain the variation in the ionic radii
between La3+ and Lu3+.
The lanthanide contraction.
23.3
From a knowledge of their chemical
properties, speculate on why Ce and Eu
were the easiest lanthanoids to isolate
before the development of ion-exchange
chromatography. Ce4+ and Eu2+, unusual
oxidation states were used in separation
procedures.
CHAPTER 24
S24.1
Synthesis for Sr2MoO4? Sealed tube, high
temperature, 6 SrO(s) + Mo(s) + 2 MoO3(s)
Æ 3 Sr2MoO4.
How would you expect the first and second
ionization energies of the lanthanoids to
vary across the series? Sketch the graph
that you would get if you plotted the third
ionization energy of the lanthanoids versus
atomic number? Identify elements at any
peaks or troughs and suggest a reason for
their occurrence? First and second IEs
would show general increase across the
lanthanoids. With the third, anomalies arise.
S24.2
Why does increased pressure reduce the
conductivity of K+ more than that of Na+ in
β-alumina? Because K+ is larger than Na+.
S24.3
Rationalize the observation that FeCr2O4 is
a normal spinel? The A2+ ions (Fe2+ in this
example) occupy tetrahedral sites and the B3+
ions (Cr3+) occupy octahedral sites.
23.4
7
23.5
Derive the ground state of the Tb3+ ion. F6
23.6
Predict the magnetic moment of a
compound containing the Tb3+ ion.
μ= 9.72 μB
23.7
23.8
23.9
Self-tests
Explain why stable and readily isolable
carbonyl complexes are unknown for the
lanthanoids? Carbonyl compounds need
back-bonding from metal orbitals of the
appropriate symmetry.
Suggest a synthesis of neptunocene from
NpCl4?
Np(COT)2 was prepared by the reaction of
K2COT
with
NpCl4
under
inert
atmosphere.
Account for the similar electronic spectra
of Eu3+ complexes with various ligands and
the variation of the electronic spectra of
Am3+ complexes as the ligand is varied.
The 5f orbitals of the actinide ions interact
strongly with ligand orbitals, and the splitting
of the 5f subshell, as well as the color of the
complex, varies as a function of ligand.
Exercises
24.1
NiO doped with Li2O? The electronic
conductivity of the solid increases owing to
formation of Ni1–xLix)O.
24.2
What is a crystallographic shear plane?
Both.
24.3
How might you distinguish between a solid
solution and a series of discrete
crystallographic shear plane structures? A
solid solution would contain a random
collection of crystallographic shear planes.
24.4
Wurtzite crystal structure and bottleneck?
The wurtzite structure is shown in Figure
3.35. The normal sites for cations in this
structure are the tetrahedral holes. The
bottleneck involves the space formed by three
close packed anions.
24.5
Synthesis of? (a) MgCr2O4 - heat
(NH4)2Mg(CrO4)2 gradually to 1100-1200°C;
(b) SrFeO3Cl - heat SrO + SrCl2 + Fe2O3 in a
sealed tube; (c) Ta3N5 - heat Ta2O5 under
NH3 at 700°C.
24.6
Products of reactions? (a) LiNiO2, (b)
Sr2WMnO6.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
46
24.7
Where might intercalated Na+ ions reside
in the ReO3 structure? The unit cell for
ReO3 is shown in Figure 24.16(a), the
structure is very open.
24.8
Antiferromagnet ordering?
In an
antiferromagnetic substance, the spins on
different metal centers are coupled into an
antiparallel alignment. As the temperature
approaches 0 K, the net magnetic moment
goes to zero.
24.9
Magnetic measurements on ferrite? inverse
spinel.
24.10
Using LFSEs determine site preference for
A= Ni(II) and B = Fe(III)? Better ligand
field stabilization and a strong preference for
inverse spinel.
24.11
High-temperature superconductors? All
except Gd2Ba2Ti2Cu2O11, which contains only
Cu2+, are superconductors.
24.12
Classify oxides as glass-forming or
nonglass-forming? BeO, B2O3, and to some
extent GeO2, glass forming. Transition metal
and rare earth oxides are typically nonglassforming.
24.13
Which metal sulfides might be glass
forming? Metalloid and nonmetal sulfides.
24.14
Examples of spinels containing? (a)
(b) fluoride?
sulfide? Zn(II)Cr(III)2S4
Li2NiF4
24.15
Synthesis of LiTaS2? Direct reaction of TaS2
with BuLi, or through electrochemical
insertion of Li ions.
24.16
Oxotetrahedral species in
structures? Be, Ga, Zn, and P.
24.17
Formulas for structures isomorphous with
SiO2 containing Al, P, B, and Zn replacing
Si? (AlP)O4, (BP)O4, (ZnP2)O6.
24.18
Mass percent of hydrogen in NaBH4 and
hydrogen storage? 10.7 %, it is a good
candidate to consider.
24.19
Formula for this lithium aluminium
magnesium dihydride and structures?
Order of band gaps? BN > C(diamond) >
AlP > InSb.
24.22
Fulleride structures? In Na2C60, all of the
tetrahedral holes are filled with sodium
cations. In Na3C60, all of the tetrahedral holes
and all of the octahedral holes are filled with
sodium cations.
CHAPTER 25
Self-tests
S25.1
Synthesis of core-shell nanoparticles? The
thermodynamic driving force is adjusted to a
level that allows for heterogeneous nucleation
of the shell material on the core but prevents
homogeneous nucleation of the shell material.
S25.2
Hemispherical imprint for 2 nm
nanoparticle? The radius is 4.21 nm and the
diameter is 8.42 nm.
S25.3
Host for QDs? MCM-41.
Exercises
25.1
Color intensity differences in Egyptian blue
pale versue blue-green spinel? The Cu site
in Egyptian blue is square planar. In copper
aluminate spinel blue, the site is tetrahedral.
(a) Surface areas? 3.14 × 102 nm2 versus
3.14 × 106 nm2 (a factor of 104). (b)
Nanoparticles based on size? The 10 nm
particle. (c) Nanoparticles based on
properties? A nanoparticle should exhibit
properties different from those of a molecule
or an extended solid.
25.2
Electron
length
and
quantum
confinement? A characteristic length is the
exciton Bohr radius.
25.3
Why are QDs better for bioimaging? One
light source can be used to excite different
quantum dots.
25.4
Band energies for QD versus bulk
semiconductor? The energies of the band
edges for a QD nanocrystal are more widely
separated.
25.5
(a) Top-down versus bottom-up? The “topdown” approach requires one to “carve out”
nanoscale features from a larger object. The
“bottom-up” approach requires one to “build
up” nanoscale features from smaller entities.
(b) Advangtages and disadvantages? Topdown methods allow for precise control over
the spatial relationships. Bottom-up methods
framework
Mg1-x(M)xH2y (M=Al, Li), The Li and Al will
be incorporated as metal hydride solid
solutions.
24.20
24.21
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 47
allow for the precise spatial control over
atoms and molecules.
25.6
25.7
25.8
(a) What are SPMs? A method to image the
microscale features by scanning a very small
probe over the surface and measuring some
physical interaction between the tip and the
material. (b) SPM and a specific
nanomaterial? Local magnetic domains of
magnetic nanomaterials, such as nanorods of
iron oxide, can be imaged using magnetic
force microscopy.
SEM versus TEM? SEM, an electron beam
is scanned over a material, and an image is
generated by recording the intensity of
secondary or back-scattered electrons. TEM,
an electron beam is transmitted through the
materials, and the image is the spatial
variation in the number of transmitted
electrons.
SEM sample preparation, ensure the material
is conductive. In TEM, the sample needs to be
made transparent to the electron beam..
(a) Steps in solutions synthesis of
nanoparticles? (i) solvation; (ii) stable nuclei
of nanometer dimensions formed; (iii) growth
of particles to desired size occurs.
(b) Why should the last two steps occur
independently? So that nucleation fixes the
total number of particles and growth leads to
controlled size and narrow size distribution.
(c) Stabilizers? Prevent surface oxidation and
aggregation, and they limit traps for the holes
and electrons, and improve quantum yields
and luminescence.
25.9
Vapor-phase
versus
solution-based
techniques? (a) Vapor-phase, large sizes. (b)
Vapor-phase, more agglomeration.
25.10
(a) What is a core-shell nanoparticle?
25.11
(a) Homogeneous versus heterogeneous?
Homogeneous nucleation leads to solid
formation throughout the vapour phase. (b)
thin film? Heterogeneous nucleation is
preferred
in
thin-film
growth.
(c)
nanoparticles? Homogeneous nucleation is
generally preferred for nanoparticle synthesis.
25.12
PVD versus CVD? In CVD, the atomic
species of interest are bound chemically to
other species. Also, their thermal energies are
typically rather low. In PVD, the atomic
species of interest are typically atomic.
25.13
(a) Purpose of QD layers? Multiple layers of
quantum dots can increase the intensity of any
optical absorption or emission. They can also
be used to form quantum cascade lasers. (b)
Limitations? The limitations come from the
requirements on how coherent the interface
between the two materials must be.
25.14
(a) Applications of quatum wells? Lasers
and optical sensors. (b) Why are they used
over other materials? They exhibit
properties that are not observed in molecular
or traditional solid state materials. (c) How
are they made? Molecular beam epitaxy.
25.15
Superlattices and improved properties?
AlN and TiN - different elastic constants,
large number of interfaces spaced on the
nanoscale.
Results are much improved hardness values.
25.16
(a) Self-assembly? (a) Offers methods to
bridge bottom-up and top-down approaches to
synthesis. (b) In nanotechnology? Offers a
route to assemble nanosized particles into
macroscopic structures.
25.17
Common features of self-assembly?
(i) molecular or nanoscale subunits; (ii)
spontaneous assembly of the subunits; (iii)
noncovalent interactions between the
assembled subunits; (iv) longer-range
structures arising from the assembly process.
25.18
Static versus dynamic self-assembly? Static
self-assembly is when a system selfassembles to a stable state, eg. a liquid crystal.
Dynamic self-assembly is when the system is
oscillating between states and is dissipating
energy in the process, eg. an oscillating
chemical reaction.
Shell
Core
(b) How are they made? Nucleation in one
solution, then grow the particle in another. (c)
Purpose? In biosensing, the dielectric
property of the shell can control the surface
plasmon of the core. In drug delivery, the
shell could react with a specific location and
the core could be used as a treatment.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
48
25.19
25.20
25.21
25.22
25.23
25.24.
25.25
Compare SAMs and cell membranes? SAM
can structurally resemble a phospholipid
bilayer.
What is morphosynthesis? Control of
nanoarchitectures in inorganic materials
through changes in synthesis parameters.
(a) What are the two classes of inorganicorganic nanocomposites? Class I, hybrid
materials where no covalent or ionic bonds
are present. Class II, at least some of the
components are linked through chemical
bonds. (b) Examples? Class I, block
copolymers.
Class
II,
polymer/clay
nanocomposites.
(a) Why is dispersion important in
nanocomposites? They lead to increased
exposed surface areas. (b) Why is dispersion
difficult? The often nonpolar organic
polymers do not have strong interactions with
the polar or ionic inorganic components.
S26.4
Without R groups attached to the Zr center
there is no preference for specific binding
during polymerization.
Exercises
26.1
(c) The production of Li3N and its reaction
with H2O? Not catalysis.
26.2
(c) Catalyst? A substance that increases the
rate of a reaction but is not itself consumed.
(d) Catalytic cycle? Sequence of chemical
reactions involving the catalyst that transform
the reactants into products.
(a) Biomimetics? Designing nanomaterials
that mimic biological systems. (b) example of
biomimetics? Cellulose fibers in paper have
been used to template the growth of titanium
oxide nanotubes.
(e) Catalyst support? Generally a ceramic
like γ-alumina or silica gel.
26.3
(c) The conversion of D-glucose to a
mixture by HCl? Homogeneous.
Self-tests
S26.3
Classify the following as homogeneous or
heterogeneous catalysis? (a) The increased
rate of SO2 oxidation in the presence of
NO? Homogeneous.
(b) The hydrogenation of oil using a finely
divided Ni catalyst? Heterogeneous.
26.4
S26.2
Define the following terms? (a) Turnover
frequency? The amount of product formed
per unit time per unit amount of catalyst.
(b) Selectivity? How much of the desired
product is formed relative to by-products.
CHAPTER 26
S26.1
Which of the following constitute catalysis?
(a) H2 and C2H4 in contact with Pt? An
example of genuine catalysis.
(b) H2 plus O2 plus an electrical arc? Not
catalysis..
A bionanomaterial and its application?
PPF/PPF-DA is an injectable bionanomaterial
used for bone-tissue engineering.
Bionanocomposites
and
improved
mechanical strength? Biomimetics need
trabecular and cortical bone tissues, which
combine compressibility and tensile strength.
Hybrid alumoxane nanoparticles dispersed in
PPF/PPF-DA shows enhanced strength.
Demonstate that the polymerization of
propene with a simple Cp2ZrCl2 catalyst
would give rise to atactic polypropene?
D,L
Which of the following processes would be
worth investigating? (a) The splitting of
H2O into H2 and O2? Not worthwhile.
The effect of added phosphine on the
catalytic activity of RhH(CO)(PPh3)3?
(b) The decomposition of CO2 into C and
O2? A waste of time.
Added phosphine will result in a lower
concentration of the catalytically active 16electron complex.
(c) The combination of N2 with H2 to
produce NH3? Very worthwhile reaction to
try to catalyze efficiently at 80ºC.
γ-Alumina heated to 900ºC, cooled, and
exposed to pyridine vapor? complete
dehydroxylation.
Would a pure silica analog of ZSM-5 be an
active catalyst for benzene alkylation (see
Figure 26.24)? No.
(d) The hydrogenation of double bonds in
vegetable oil? The process can be readily
set up with existing technology.
26.5
Why does the addition of PPh3 to
RhCl(PPh3)3 reduce the hydrogenation
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 49
turnover frequency? The catalytic species
that enters the cycle is RhCl(PPh3)2(Sol) (Sol
= a solvent molecule).
26.6
26.7
Explain the trend in rates of H2 absorption
by
various
olefins
catalyzed
by
RhCl(PPh3)3? In both cases, the alkene that
is hydrogenated more slowly has a greater
degree of substitution and so is sterically
more demanding.
Hydroformylation catalysis with and
without added P(n-Bu)3? The
transformation of (E) into CoH(CO)4 must be
the rate-determining step in the absence of
added P(n-Bu)3. In the presence of added
P(n-Bu)3, the formation of either (A) or (E) is
the rate-determining step.
exchange of the methyl group in the
chemisorbed –CHR(CH3) group (R =
C(CH3)2(C2H5)).
26.14
Why does CO decrease the effectiveness of
Pt in catalyzing the reaction 2H+(aq) + 2e–
→ H2(g)? Platinum has a strong tendency to
chemisorb CO.
26.15
Describe the role of an electrocatalyst?
Platinum is the most efficient electrocatalyst
for accelerating oxygen reduction at the fuel
cell cathode, but is expensive.
CHAPTER 27
Self-tests
26.8
26.9
26.10
26.11
How does starting with MeCOOMe instead
of MeOH lead to ethanoic anhydride
instead of ethanoic acid using the
Monsanto acetic acid process? The reaction
of the ethanoate ion with the acetyl iodide
leads to ethanoic anhydride.
S27.1
S27.2
S27.3
Why does saline contain NaCl? Osmotic
balance.
S27.4
Explain the significance of the Calcium ion
pumps activation by calmodulin? The
binding of calmodulin is a signal informing
the pump that the cytoplasmic Ca2+ level has
risen above a certain level.
S27.5
Why are iron-porphyrin complexes unable
to bind O2 reversibly? The Fe(II) gets
oxidized to Fe(III), yielding an oxo-Iron(III)
porphyrin complex.
S27.6
What is the nature of binding at Cu blue
centers as indicated by the EPR spectrum?
There is greater covalence in blue Cu centers
than in simple Cu(II) compounds.
Why is the platinum-rhodium in
automobile catalytic converters dispersed
on the surface of a ceramic rather than
used in the form of thin foil? A thin foil of
platinum-rhodium will not have as much
surface area as an equal amount of small
particles finely dispersed on the surface of a
ceramic support.
S27.7
What is the nature of an active site with
Copper (III)? Diamagnetic with squareplanar geometry.
Why mercury is so toxic because of the
action of enzymes containing cobalamin?
Cobalamins are very active methyl transfer
reactions, which can methylate anything in
the cell.
Devise a plausible mechanism to explain
the deuteration of 3,3-dimethylpentane?
(i) The mechanism of deuterium exchange is
probably related to the reverse of the last two
reactions in Figure 26.20.
S27.9
Suggest a reason why? (a) Ring opening
alkene metathesis polymerization (ROMP)
proceeds? ROMP can result in reduced
steric strain. (b) Ring-closing metathesis
(RCM) reaction proceeds? RCM results in
the loss of ethane.
(a) Attack by dissolved hydroxide?
Structure C in Figure 26.11 (b) Attack by
coordinated hydroxide? Structure E given in
Figure 26.11 (c) Can one differentiate the
stereochemistry? Yes.
(a) Enhanced acidity? When Al3+ replaces
Si4+ on lattice site charge is balanced by
H3O+.
(b) Three other ions? Ga3+, Co3+, and Fe3+.
26.12
26.13
Is Iron (II) expected to be present in the
cell as uncomplexed ions? No.
Unusual coordination of Mg? The protein’s
3D structure can place any particular atom in
a suitable position for axial coordination.
(ii) The second observation can be explained
by invoking a mechanism for rapid deuterium
S27.8
Suggest experiments that could establish
the structure of the MoFe cofactor? EPR,
single-crystal X-ray diffraction, and EXAFS.
S27.10 Why might Cu sensors be designed to bind
Cu(I) rather than Cu(II)? Cu(I) has an
almost unique ability to undergo linear
coordination by sulphur-containing ligands.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
50
Exercises
27.1
Lanthanides versus calcium? They are hard
Lewis acids and prefer coordination by hard
bases. Gd3+ has excellent fluorescence.
27.2
Substituting Co2+ for Zn2+?
Co(II)
commonly adopts distorted tetrahedral and
five-coordinate geometries typical of Zn(II) in
enzymes. Co(II) is paramagnetic, enabling Zn
enzymes to be studied by EPR.
27.3
Compare the acid/base catalytic activities
Acid
of Zn(II), Fe(III), and Mg(II)?
strengths: Fe(III) > Zn(II) > Mg(II). Ligand
binding rates are Mg(II) > Zn(II) > Fe(III).
Propose a physical method for the
EPR or
determination of Fe(V)?
Mossbauer.
27.4
27.5
Interpret the Mossbauer spectra of
ferredoxin? In the spectrum, the oxidized
spectrum is consistent with the iron atoms
having the same valence, intermediate
between the 3+ and the 2+ states, with one
pair spin-up and the other pair spin-down.
27.6
Explain the differences in the structures of
the oxidized and reduced forms of the Pcluster in nitrogenase? The change in
structure suggest that the coupling of proton
and electron transfer can also occur at the Pcluster, by controlling protonation of the
exchangeable ligands.
What metals are involved in the synthesis
of acetyl groups? Co(II), Fe(II) and Cu(I).
27.7