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Shriver 5e Answers to Self Tests and Exercises
Inorganic Chemistry (Shriver & Atkins) 5th ed 的习题解答
Answers to self-tests and exercises
CHAPTER 1
Self-tests
S1.1 S1.2 S1.3 S1.4 S1.5 S1.6 S1.7 S1.8
8035Br
1.7
10
n8135Br
γ
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d orbitals, 5 orbitals 4 3p
The added p electron is in a different (p) orbital, so it is less shielded.
Ni:[Ar]3d84s2, Ni2 :[Ar]3d8
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 1.11
1524nm, 1.524 X 104 cm-1
Period 4, Group 2, s block
Going down a group the atomic radius
increases and the first ionization energy generally decreases.
? 11? =R?=1.0974X107m?http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a41 λ?
1∞?
1
? 11? =R?=1.0288X107m?1 ? λ? 14?
1
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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 Adding
another electron to C would result in
the stable half filled p subshell. S1.11 Cs
1.12 1.13 1.14
? 11?
=R?=9.7547X106m?1 λ? 13?
1
? 11?
=R?=8.2305X106m?1 λ? 12?
1
0 up to n-X n2
N l ml
Orbital
designation
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Number of
orbitals Exercises
144171
1.1 (a)
(b)
7N 2He→8O 1p γ
6C 1p→7N γ
1312
(c)
147N 0n→1H 6C
12113
1.2 1.3
24696
2571
Cm 126C→112Uub 0n
?1
3 2
2,
…, ?2
1, 3d 5
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4 0 0 4s 1 4 3
3http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4,
2, 4f 7
…, ?3
1.15 1.16
n=5, l = 3, and ml = -3,-2,-1,0,1,2,3 Li:
σ = Z – Zeff ; σ = 3-1.28 = 1.72
The higher value of I2 for Cr relative to Mn is a consequence of the special stability
of half-filled subshell configurations and the higher Zeff of a 3d electron verses
a 4s electron.
221094
Be:
C:
σ = Z – Zeff ; σ = 4-1.19 = 2.09
σ = Z – Zeff ; σ = 6-3.14 = 2.86
σ = Z – Zeff ; σ = 8-4.45 = 3.55
B:
N:
1.4 1.5 1.6
251
Ne 42He→12Mg 0n
941 Be 4Be→126C 2He 20n
σ = Z – Zeff ; σ = 7-3.83 = 3.17
F: σ = Z – Zeff ; σ = 9-5.10 = 3.90
The 1s electrons shield the positive charge
form the 2s electrons.
σ = Z – Zeff ; σ = 5-2.42 = 2.58
O:
1.17
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Since helium-4 is the basic building block,
most additional fusion processes will produce nuclei with even atomic numbers.
1
2 1.18 1.19 1http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4.20
See Figs 1.11 through 1.16 See Table 1.6 and discussion.
Table 1.6 shows SR > Ba
Anomalously high value for Cr is associated with the stability of a half filled d
shell.
(b) [He]2s22p5 (c) [Ar]4s2 (d) [Ar]3d10 (e) [Xe]4f5d6s6p (f) [Xe]4f145d106s2
1.23 (a) [Ar]3d14s2
(b) [Ar]3d (c) [Ar]3d (d) [Ar]3d4 (e) [Ar]3d6
(f) [Ar]
(g) [Ar]3d104s1 (h) [Xe]4f 7
1.24 (a) Xe]4f145d46s2
(b) [Kr]4d6 (c) [Xe]4f6 (d) [Xe]4f7 (e) [Ar] (f) [Kr]4d2
1.25 (a) S
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(b) Sr (c) V
(d) Tc (e) In (f) Sm
1.26 1.27
See Figure 1.4.
(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).
Radii of Period 4 and 5 d-metals are similar because of lanthanide contraction.
and 2p0
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par5214
10
2
3
ANSWERS TO SELF-TESTS AND EXERCISES 1.30
1.21
1.22 (a) [He]2s22p2
CHAPTER 2
2s2
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Self-tests
S2.1
S2.2 (a) Angular
(b) square planar
S2.3 S2.4
Linear
S22– : 1σg22σu23σg21πu42πg4 ;
Cl2– : 1σg22σu23σg21πu42πg44σu1.
S2.5
1σg22σu23σg21πu42πg4
S2.6
?[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 S2.9
If it contains 4 or fewer electrons. –21 kJ mol–1 (b)
S2.10 (a)
1/2
5
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Exercises
2.1 (a) angular
(b) tetrahedral
(c) tetrahedral 2.2 (a) trigonal planar
trigonal pyramidal
2.3 (a) T-shaped
(c) square pyramidal
1.28 1.29
ANSWERS TO SELF-TESTS AND EXERCISES
(b) square planar (c) linear
2.4
(b) one
(http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4a)
(c) none
(b) 1σg22σu21πu2
(d) two 2.15 (a) 1σg22σu2
3
(b)
2.16
(b)
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(c) 1σg22σu21πu43σg1 (d) 1σg22σu23σg21πu42πg3
The configuration for the neutral C2 would be
1σg21σ
u2 1πu4
.
The bond order would be ?[2-
2 4] = 2.
(b) 1
(c) There is no bond between the two atoms.
2.17 (a)
1
σg22σu23σg21πu42πg4
2.18 (a) 2
2.5 (a) tetrahedral
2.7
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(b) octahedral
(b) 217 pm (c) 221 pm
2.6 (a) 176 pm
1200
(b) 1 (c) 2
2.19 (a)
0.5
(b) –0.5 (c)
0.5
2.20
2(Si–O) = 932kJ > Si=O = 640kJ; therefore
two Si–O are preferred and SiO2 should (and does) have four single Si–O bonds.
Multiple
bonds
are
much
stronger
for
period
2
elements
elements://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par
2.21 (a-c)
2.8
than
heavier
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2.9 –483 kJ difference is smaller than expected because bond energies are not
accurate.
2.10 (a) 0
(b) 205 kJ mol-1 2.11
Difference in electronegativities are AB 0.5,
AD 2.5, BD 2.0, and AC 1.0. The increasing covalent character AD
2.12 (a) covalent
(b) ionic
(c) ionic 2.13 (a) sp2
(b) sp3
(c) sp3d or spd3
(d)
p2d2 or sp2d. 2.14 (a) one
(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.
HOMO exclusively F; LUMO mainly S.
2.22 2.23
2.24 (a) electron deficient
4
1
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(b) electron precise
ANSWERS TO SELF-TESTS AND EXERCISES 3.4
3.5
3.6
3.7
3.8
3.9
XA2 K3C60 281 pm 429 pm
CuAu. Primihttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tive 12 carat.
Zintl phase region
CHAPTER 3
Self-tests
S3.1
S3.2
S3.3
S3.4
S3.5
S3.6
S3.7
S3.8
S3.9
S3.10
S3.11
S3.12
S3.13 S3.14
S3.15
S3.16
S3.17
See Fig. 3.7 and Fig. 3.32. See Figure 3.35. 52%
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rh = ((3/2)1/2 – 1) r = 0.225 r 409 pm 401 pm FeCr3 X2A3.
Ti CN = 6 (thought these are as two slightly different distances it is often described
as 4
2) and O CN = 3. LaInO3 2421 kJ mol–1 Unlikely
MgSO4
Phosphorus and aluminium.
The dx2-y2 and dz2 have lobes pointing along the cell edges to the nearest neighbor
metals.
3.10 (a) 6:6 and 8:8
(b) CsCI
3.11 6
3.12
2B type and 4A type – in a distorted
octahedral arrangement.
3.13
(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
Nahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4Cl
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3.14
3.15
CsCl.
Lattice enthalpies for the di- and the trivalent
ions. Also, the bond energy and third electron gain enthalpy for nitrogen will be
large.
3.16
4 four times the NaCl value or 3144 kJmol–1.
(a) 10906 kJmol–1 (b) 1888 kJ mol–1 (c)
664 kJ mol–1
3.18 (a) MgSO4
(b) NaBF4
3.19 CsI
3.20 Ba2 ; solubilities decrease with increasing
radius of the cation.
S3.18 (a) n-type
3.17
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(b) p-type
Exercises
3.1
3.2
a≠b≠c
and α = 90°, β =
90°,
γ = 90° 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 (?,?,0), (?,1,?), (0,?,?) (?,?,1), (?,1,?), and (1,?,?).
3.21
(a) Schottky defects
(b) Frenkel defects
3.2http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a42
3.23
3.24
Solids have a greater number of defects as temperatures approaches their melting
points. The origin of the blue color involves electron transfer from cationic centres.
Vanandium carbide and manganese oxide.
3.3 (c) and (f) are not as they have neighbouring
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layers of the same position.
ANSWERS TO SELF-TESTS AND EXERCISES
3.25
3.26
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.
S4.9
Identify the acids and bases?
5
(a) FeCl3
(b) I–
Cl– → [FeCl4]–, acid is FeCl3, base is Cl–.
I2 → I3–, acid is I2, base is I–.
S4.10
The
difference
in
struhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4cture 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
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Draw the structure of BF3·OEt2?
3.27
CHAPTER 4
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. (c) NH3
H2S → NH4
HS–
Ammonia, base; NH4 , conjugate acid; hydrogen sulphide, acid; HS–, conjugate base.
S4.2 S4.3
What is the pH of a 0.10 M HF solution? pH= 2.24
Calculate the pH of a 0.20 M tartaric acid solution?
Which solvent?
dimethylsulfoxide (DMSO) and ammonia. A base.
Arrange in order of increasing acidity?
The order of increasing acidity
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4is [Na(H2O)6]
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Exercises
4.1
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
S4.6
S4.5 Is aKBrF4 an acid or a base in BrF3?
4.2
Identify the conjugate bases of the
following acids?
S4.7 Predict pKa values? (a) H3PO4
pKa ≈ 3. The
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actual value, given in Table 4.1, is 2.1.
(b) H2PO4–
pKa(2) ≈ 8. The actual value, given in Table 4.1, is 7.4.
(c) HPO42–
pKa(3) ≈ 13. The actual value, given in Table 4.1, is 12.7.
S4.8 What happens to Ti(IV) in aqueous
solution as the pH is raised?
Treatment
withhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
causes the precipitation of TiO2.
ammonia
Further treatment with NaOH causes the TiO2 to
redissolve.
(a) [Co(NH3)5(OH2)]3 , conjugate base is
[Co(NH3)5(OH)]2 .
(b) HSO4–?
The conjugate base is SO4–. (c) CH3OH?
CH3O–. (d) H2PO4–?
The conjugate base is HPO42–. (e) Si(OH)4?
base is SiO(OH)3–.
(f) HS??
The conjugate base is S2–.
4.3
Identify the conjugate acids of the following bases?
6
The conjugate base is
The conjugate
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(a) C5H5N (pyridine)?
The conjugate acid
is pyridinium ion, C5H6N .
(b) HPO4?
The conjugate acid is H2PO4. (c) O2–?
(d) CH3COOH?
The conjugate acid is OH–.
The conjugate acid is
CH3C(OH)2.
(e) [Co(CO)4]–?
(f) CN–?
The conjugate acid is HCo(CO)4.
The conjugate acid is HCN.
4.4
Calculate
the
[H3O
]
and
pH
of
a
0.10
M
sohttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4lution?
5.6 × 10–10 Ka =
butanoic
acid
pH=2.85 Kb =
5.6 × 10–6
Predict if F will behave as an acid or a base in water?
F will behave as a base in water.
4.8
What are the structures and the pKa values of chloric (HClO3) and chlorous (HClO2)
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acid?
--2–
2–
ANSWERS TO SELF-TESTS AND EXERCISES
ClO4–, cannot be studied in sulfuric acid.
4.9
Is the –CN group electron donating or withdrawing?
electron withdrawing
4.11
Is the pKa for HAsO42– consistent with Pauling’s rules?
No.
Pauling’s rules
are only approximate.
What is the order of increasing acid strength for HNO2, H2SO4, HBrO3, and HClO4?
the order is HClO4 > HBrO3 > H2SO4 > HNO2.
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
HSiO4http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a43–
Which of the following is the stronger acid? (a) [Fe(OH2)6]3
or [Fe(OH2)6]2 ?
The
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Fe(III) complex, [Fe(OH2)6]3 , is the stronger acid.
(b) [Al(OH2)6]3
or [Ga(OH2)6]3 ?
the aluminum-containing species is more
acidic.
4.12
4.13
4.5 What is the Kb of ethanoic acid?
4.7
4.6 What is the Ka for C5H5NH ?
4.14
(c) Si(OH)4 or Ge(OH)4?
Si(OH)4, is more acidic. (d) HClO3 or HClO4? HClO4 is
a stronger acid. (e) H2CrO4 or HMnO4?
H2SO4?
HMnO4 is the stronger acid. (f) H3PO4 or
H2SO4 is a stronger acid.
4.15
Arrange the following oxides in order of
increasing basicity? Al2O3, B2O3, BaO, CO2, Cl2O7, and SO3?
order of increasing basicity is Cl2O7
chloric acid
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4.8
chlorousacid
Chloric acid, the predicted pKa = –2; actual value = –1.
Chlorous
acid,http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
the
predicted pKa = 3; actual value = 2.
Which bases are too strong or too weak to be studied experimentally? (a) CO32– O2–,
ClO4–, and NO3– in water?
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.
(b) HSO4–, NO3–, and ClO4–, in H2SO4?
HSO4–, not too strong to be studied
experimentally.
NO3– is of directly measurable base strength in liquid H2SO4.
4.16 Arrange the following in order of
increasing acidity? HSO4?, H3O , H4SiO4, CH3GeH3, NH3, and HSO3F?
4.17
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increasing acidity is NH3
Which aqua ion is the stronger acid, Na
or Ag ?
Ag (aq).
ANSWERS TO SELF-TESTS AND EXERCISES
4.18
Which of the following elements form oxide polyanions or polycations? Al, As, Cu,
Mo, Si, B, Ti?
polhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ycations: Al, Cu, and
Ti.
polyoxoanions (oxide polyanions): Mo polyoxoanions: As, B, and Si.
The change in charge upon aqua ion polymerization?
4.26
7
(d) AsF3(g)
SbF5(g) → [AsF2][SbF6]?
the Lewis acid [AsF2]
The very strong Lewis acid SbF5 displaces
from the Lewis base F–. (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.
Select the compound with the named characteristic? (a) Strongest Lewis acid: BF3,
BCl3, or BBr3?
BBr3. BeCl2 or BCl3?
Boron trichloride. B(n-Bu)3 or B(t-Bu)3?
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B(n-Bu)3.
(b) More basic toward BMe3: NMe3 or NEt3?
2-Me-py or 4-Me-py?
NMe3.
4-Me-py.
4.27
Which of the following reactions have Keq > 1? (a) R3P–BBr3
R3N–BF3→ R3P–BF3
R3N–BBr3?
(b)
SO2
Ph3P–HOCMe3
→
Ph3P–SO2
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
(c) CH3HgI
HCl → CH3HgCl
HI?
HOCMe3?
> 1.
1
4.28
Choose between the two basic sites in Me2NPF2?
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
The hard nitrogen atom will bond more
strongly to the hard Lewis acid BF3.
4.19
Polycation formation reduces the average
positive charge per central M atom by
1 per M. 4.20
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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 ?
3–
4–
3
4PO4
8H3O →
P4O12
12H2O
[(H2O)4Fe(OH)2Fe(OH2)4
2H3O
More balanced equations?
CO2
CaCO3
Ca2
H3PO4
2[Fe(OH2)6] →
4.21
HPO4 ?? 2H2PO4 (b) CO2 and CaCO3?
H2O
2HCO3-
2--
(http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4a) H3PO4 and Na2HPO4?
4.22
4.23
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Give the equations for HF in H2SO4 and
HF in liquid NH3?
H2SO4
NH3
NH2-
HF ? H3SO4
F-
HF
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.
4.24 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.
Identifying acids and bases: (a) SO3
H2O → HSO4–
reaction are the Lewis acids SO3 and H
and the base is the Lewis base OH–.
(b) Me[B12]–
Hg2
→ [B12]
MeHg ?
acid [B12] from the Lewis base CH3–. (c) KCl
acid
SnCl2
H ?
4.25
The Lewis acid Hg2 displaces the Lewis
SnCl2 → K
displaces
[SnCl3]–?
the
achttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4id K
Cl–.
The acids in this
The Lewis
Lewis
from the Lewis base
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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. (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 for the
dissolution of SiO2 by HF?
8 ANSWERS TO SELF-TESTS AND EXERCISES
SiO2
6HF
or
SiO2
4HF
2H2O
H2SiF6
2H2O
SiF4
4.37 4.39
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(c) C6H5COOH?
C6H5
COOH
HF
C6H5COO-
H2F
The
dissolution
of
silicates
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
both
by
HF?
Calculate
the
enthalpy change for I2 with phenol?
Both a Br?nsted acid–base reaction and a Lewis acid–base reaction.
4.32
Write a balanced equation to explain the foul odor of damp Al2S3?
suggests H2S formation.
4.38 Are the f-block elements hard? yes.
ΔfH=-20.0kJ/mol
φ
The foul odor
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Al2S3
3H2O
Al2O3
3H2S
4.33 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:
CHAPTER 5
Self-tests
S5.1
Half-reactions and balanced reaction for
oxidation of zinc metal by permanganate ions?
2[MnO4? (aq)
8H
(aq)
5e–
→ Mn2 (aq)
4H2O(l)] reduction
oxidation http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 5 [ Zn(s)
Zn2 (aq)
2e– ]
→
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2MnO4? (aq)
5Zn2 (aq)
S5.2
5Zn(s)
16H (aq)
2Mn2 (aq)
→
8H2O(l)
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.
acid-Cl-
I-
acid-I-
Cl-
(b) Favor basicity of R3As over R3N?
Alcohols such as methanol or ethanol would
be suitable.
(c) Favor acidity of Ag
over Al3 ?
An example of a suitable solvent is diethyl
ether. Another suitable solvent is H2O.
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(d) Promote the reaction 2FeCl3
ZnCl2 →Zn2
2[FeCl4]??
A suitable solvent
is acetonitrhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ile, MeCN.
4.34
Propose a mechanism for the acylation of
An alumina surface, such as the benzene?
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 Br?nsted acid–base
reactions in
liquid HF?
(a) CH3CH2OH?
The balanced equation
is:
S5.6 Can Fe2
conditions?
disproportionate under standard
No. S5.7
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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
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4into
Pu(VI) and Pu(IV).
S5.10 Frost diagram for thallium in aqueous
acid?
CH3CH2
OH
HFNH3
H
F
CH3CH2OH2
F- NH
4
(b) NH3?
The equation is
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F-
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)
2H2O(l)
E? = 1.65 V (b) Fe2 ?
4Fe2 (aq)
O2(g)
4H (aq)
→
4Fe3 (aq)
2H2O(l)
E? = 0.46 V
S5.11
The oxidation number of manganese?
2
(c) Cl–?
no reaction. (d) HOCl?
No reaction. (e) Zn(s)?
Mn(aq)
S5.12 Compare the strength of NO3– as an
oxidizing agent in acidic and basic solution?
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2Zn(s)
O2(g)
4H
(aq)
→
2Zn2
(aq)
2H2O(l)
E?
=
1.99
V
A
competinghttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 reaction is:
Zn(s)
2H (aq)
→
Zn2 (aq)
H2(g) (E? = 0.763 V).
5.4
Balanced equations for redox reactions?
(a) Fe2 ?
Nitrate is a stronger oxidizing agent in acidic
solution than in basic solution.
S5.13 The possibility of finding Fe(OH)3 in a
waterlogged soil?
Fe(OH)3
is not stable. S5.14 The minimum temperature for
reduction
of MgO by carbon?
(i) Fe2
1800?C or above.
(iii) Fe2
will not oxidize water.
(ii) Fe2
will not reduce water.
will reduce O2 and in doing so will be oxidized to Fe3 .
(iv) disproportionation will not occur.
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(b) Ru2 ?
Ru2
will not oxidize or reduce water. Ru2
so will be oxidized to Ru3 . Ru2
will reduce O2 and in doing
will disproportionate in aqueous acid to Ru3
and
metallic ruthenium.
(c) HClO2? HClO2 will oxidize water, will not reduce water.
HClO2 will
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4reduce O2 and in doing so
will be oxidized to ClO3–.
HClO2 will disproportionate in aqueous acid to ClO3–
and HClO.
(d) Br2? Br2 will not oxidize or reduce water.
Br2 will not reduce O2.
Br2 will
not disproportionate in aqueous acid to Br– and HBrO.
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.
Exercises
5.1
Oxidation numbers?
2 NO(g)
2Mn3 (aq)
1
O2(g)
2H2O
→ 2 NO2(g)
→ MnO2
1 -2
Mn2
0
4H (aq)
4 -2
3
1 -2
4 -2
2
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→ LiC(s)
1
CoO2(s)
3 -2
Ca(s)
0 0
LiCoO2(s)
0
H2(g) →
1-1
C(s)
4 -2
CaH2(s)
2 -1 5.2
Suggest chemical rhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4eagents
for redox
transformations? (a) Oxidation of HCl to Cl2? S2O82–, H2O2, or α–PbO2 to oxidize
Cl– to Cl2.
(b) Reducing Cr3 (aq) to Cr2 (aq)?
(c) Reducing Ag (aq) to Ag(s)?
metallic manganese, metallic zinc, or NH3OH .
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?
10ANSWERS TO SELF-TESTS AND EXERCISES 5.13
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Calculate the equilibrium constant for Au (aq)
2CN–(aq)
→
[Au(CN)2]–(aq)?
K = 5.7 × 1038
Find the approximate potential of an aerated lake at pH = 6, and predict the
predominant species? (a) Fe? 0.5 – 0.6 V (b) Mn?
(c) S?
At pH 0, 0.387 V.
E = 0.55 V
At pH 14, SO42– would again predominate. HSO4– is the
prehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4dominant sulfur species
at pH 6.
5.15
Frost diagram and standard potential for the HSO4?/S8(s) couple? 0.387 V
2MnO4? (aq)
2Mn2 (aq)
5H2SO3(aq)
5HSO3? (aq)
H (aq)
→
3H2O(l)
The potential decreases as the pH increases.
5.7
Write the Nernst equation for (a) The reduction of O2?
Q = 1/(p(O2)[H ]4)
5.14
and
E = E?
– [(0.059V)/4][log(1/(p(O2)[H ]4)] (b) The reduction of Fe2O3(s)?
Q = 1/[H ]6
and
E = E?
– (RT/nF)(13.8 pH)
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5.8
Using Frost diagrams? (a) What happens when Cl2 is dissolved in aqueous basic solution?
Cl2 is thermodynamically susceptible to disproportionation to Cl– and ClO4– when
it is dissolved in aqueous base.
5.16
Equilibrium constant for the reaction
Pd2 (aq)
Reduction
4 Cl–(aq) ≡ [PdCl4]2–(aq) in 1 M HCl(aq)? K = 4.37 × 1010 5.17
potential
for
MnO4–
5.18
to
MnO2http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4(s) at pH = 9.00? E =
0.98 V
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)
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(b) Zinc metal is added to aqueous acidic sodium triiodide?
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.
5.19
Thermodynahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4mic
tendency of HO2 to
undergo disproportionation? E =
1.275 V. (is positive), HO2 will undergo
disproportionation.
5.20
Dissolved carbon dioxide corrosive towards
iron? Carbon dioxide and water generate carbonic acid which encourages the corrosion
process by lowering solution pH.
5.21 What is the maximum E for an anaerobic
environment rich in Fe2
and H2S?
–0.1 V. 5.22
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. 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 ].
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Under what conditions wilhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4l
Al reduce MgO?
Zn(s)
Above about 1400?C.
I3– (aq)
→
Zn2 (aq)
3I–(aq) (c) I2 is added to excess aqueous
acidic HClO3?
3I2(s)
5.10
5ClO3– (aq)
3H2O(l)
→
6IO3– (aq)
5Cl– (aq)
6H (aq)
5.11
Electrode potential for Ni2 /Ni couple at pH = 14? E =– 0.21 V
Will acid or base most favour the following half-reactions? (a) Mn2
(b) ClO4– → ClO3–?
(d) I2 → 2I–?
Acid
(c) H2O2 → O2?
Acid or base, no difference.
5.12 Determine the standard potential for the
reduction of ClO4– to Cl2?
1.392 V
5.23
5.24
ANSWERS TO SELF-TESTS AND EXERCISES 11
CHAPTER 6
Base
→ MnO4–?
Base
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Self-tests
Sketch the S4 axis of an NH4
ion. How many of these axes are there in the ion? Three
S4 axes.
S6.2 (a) BF3 point group? D3h. S6.1
(b) SO42– point group? Td.
S6.3
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4arSymmetry species of all five
d orbitals of
the central Xe atom in XeF4 (D4h, Fig. 6.3)?
dxz and dyz are Eg;
dx2-y2 is B1g;
dxy is B2g;
dz2 is A1g.
What is the maximum possible degeneracy for an Oh molecule? 3.
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.
Is the skew form of H2O2 chiral? Yes. Can the bending mode of N2O be Raman active?
Yes.
Confirm that the symmetric mode is Ag? D2h character table, which is the Ag symmetry
type.
SF6 has Oh symmetry. Analysis of the stretching vibrations leads to:
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Γstr
=
(Raman)
A1g
T1u
(Raman, polarized)
Eg
(IR).
SF5Cl has C4v symmetry.
Analysis of the
stretching vibrations leads to:
Γstr
=
3A1 (IR and Raman, polarized)
2B1 (Raman)
E (IR, Raman).
S6.14http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 Symmetries of all the
vibration modes of [PdCl4]2-? A1g
SALCs for sigma bonding in O? A1g
B1g
B2g
A2u
B2u
2Eu S6.15
Eg
S6.4 S6.5
Exercises
6.1
Symmetry elements? (a) a C3 axis and a σv plane in the NH3 molecule?
S6.6 S6.7 S6.8
T1u.
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H
C3
planar [PtCl4]2– ion?
σv
S6.9
Show that the four CO displacements in the square-planar (D4h) [Pt(CO)4]2
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 2σv
2σd
Γ3N 4 0 0 2
0 0 0 4 2 0
6.2
A1g
Reduces to
cation
A1g
B1g
Eu
B1g are Raman active. Eu is IR active.
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S6.10 Orbital symmetry for a tetrahedral array
of H atoms in mehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4thane? 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.
Predict how the IR and Raman spectra of SF5Cl differ from that of SF6?
6.3
(b) a C4 axis and a σh plane in the square-
C4
S4 or i? (a) CO2? i (b) C2H2? i. (c) BF3? neither.
(d) SO42–? three different S4.
σh
Assigning point groups: (a) NH2Cl? Cs
(b) CO32–? D3h
(c) SiF4? Td
(d) HCN? C∞v.
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(e) SiFClBrI? C1. (f) BrF4–? D4h.
S6.13
126.4
How many planes of symmetry does a benzene molecule possess? What chloro-substituted
benzene has exactly four planes of symmetry? 7,and C6H3Cl3.
The syhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4mmetry 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.
(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 45?
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
SO32–
ion?
(a)
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4group? C3v
Point
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(b) Degenerate MOs? 2
(c) Which s and p orbitals have the maximum degeneracy? 3px and 3py orbitals are doubly
degenerate.
6.7 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 6.9
AsCl5 Raman spectrum consistent with a trigonal bipyamidal geometry?
No.
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
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.
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ANSWERS
TO
SELF-TESTS
AND
EXERCISES
6.1http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a42 [AuCl4]? ion? Γ of
all 3N displacements
and irreducible representations?
A1g
A2g
B1g
B2g
Eg
2A2u
B2u
3Eu
6.13 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:
(Raman, polarized)
A1’
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)
B2u (IR)
2B2g (Raman)
B3g (Raman)
Au(inactive)
2B1u
(IR)
2B3u (IR).
For the 90o-twisted form of B2F4 (D2d)
The vibrations are:
3A1 (Raman, polarized)
B1 (Raman)
2B2 (IR and Raman )
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3E (IR and Raman).
6.14 (a) Take the 4 hydrogen 1s orbitals of CH4
and
determhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ine
transform under Td.
to A1
how
they
(b) Confirm that it is possible to reduce this representation
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.
6.5
6.15
Use the projection operator method to
construct the SALCs of A1
T2 symmetry that derive from the four H1s orbitals in
methane..
s = (1/2)(?1
?2
?3
(1/2)(?1 – ?2 – ?3
SALCs for σ-bonds
?3)
?3)
(= A1) px = (1/2)(?1 – ?2
(a) BF3?
(= T2) pz = (1/2)(?1
?3 – ?3)
?2 – ?3 – ?3)
(= T2) py =
(= T2)
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1
(/√3)(?1
?2
?3)
(= A1’)
(1/√6)(2?1 – ?2 – ?3)
and (1/√2)(?2 – ?3)
(=
E’)
(b) PF5?
(axial F atoms are ?4
(1/√2)(?4
?5)
?5)
(=
A1’)
(1/√2)(?4
?
?5)
(1/√3)(http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4?1
(=
?2
A2”)
?3)
(=
A1’)
(1/√6)(2?1 – ?2 – ?3) and (1/√2)(?2 – ?3)
(= E’)
ANSWERS TO SELF-TESTS AND EXERCISES 13
CHAPTER 7
Self-tests
S7.1
Give
formulas
corresponding
Cis-diaquadichloroplatinum(II)?
to
the
following
cis-[PtCl2(OH2)2],
trans-diaquadichloroplatinum(II), trans-[PtCl2(OH2)2].
(b) Diamminetetra(isothiocyanato) chromate(III)?
names?
(a)
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[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 (III)? [Rh(en)3]3 .
(d) Bromopentacarbonylmanganese (I)?
[MnBr (CO)5].
(e) Chlorotris(triphenylphosphine)rhodium (I)?
[RhCl(PPh3)3].
S7.2
What type of isomers are possible for [Cr(NO2)2?6H2O]?
linkage isomers of the NO2 group.
The hydrate isomers and
Also, [Cr(ONO)(H2O)5]NO2 ?H2O.
Exercises
7.1
Name
and
draw
thhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4e
structures of the complexes? (a) [Ni(CO)4]?
nickel(0).
(b) [Ni(CN)4]2–
?
2-
Tetracyanonickelate (II).
Nickel tetracarbonyl or tetracarbonyl
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(c) [CoCl4]
2–?
Tetrachlorocobaltate (II)
Cl
2-
S7.3 Identifying isomers?
Note that the two
phosphine ligands in the trans isomer are related, therefore, they exhibit the same
chemical shift. S7.4
Sketches of the mer and fac isomers of [Co(gly)3]?
(d) [Mn(NH3)6]2 ?
Hexaamminemanganesium (II)
H3
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H3H3
H3
H3H3
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]
(d) [Cr(NH3)5μ–OH–Cr(NH3)5]Cl5
7.3
Name
the
following
complexes?
(a)
cis-[CrCl2(NH3)4]
?
cis-tetra(amminhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4e)di(chlor
o)chromium(III)
(b)
trans-[Cr(NCS)4(NH3)2]?
trans-di(ammine)tetrakis(isothiocyanato)chromate (III)
-
S7.5
Which of the following are chiral? (a) cis-[CrCl2 (ox)2]3–?
Chiral .
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(b) trans-[CrCl2(ox)2]3?
Not chiral. (c) cis-[RhH(CO)(PR3)2]?
Not chiral.
S7.6
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.
–
(c)
[Co(C2O4)(en)2] ?
bis(ethylenediamine)oxalatocobalt(III).
14ANSWERS TO SELF-TESTS AND EXERCISES
7.4
Four-coordinate complexes? (a)
L
(b)
B
Isomers expected for MA2
Sketch the two observed structures?
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2
?
tetrahedral complex, no isomers, for a square-planar complex, two isomers, cis and
trans.
7.5://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4r
For five-coordinate complexes, sketch the two observed structures?
AA
E
A
tetrahedral
square planar
7.6
Trigonal BipyramidalA = axial ligandsE = equatorial ligands
Square based pyramid
A = axialB = basal
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[Mg(edta)(OH2
)]2–
7.11 7.12
7.13
What types of isomers are [RuBr(NH3)5]Cl and [RuCl(NH3)5]Br?
Ionization isomers.
Which complexes have isomers?
Six-coordinate complexes? (a) Sketch the two observed structures?
[CoBrClI(OH2)]
Which complexes have isomers?
(b) Which one of these is rare?
Trigonal prism.
(a) [Pt(ox)(NH3)2] no isomers
(b) [PdBrCl(PEt
3)2] has two isomers. (c) [IrHCO(PR3)2] has two isomers.
7.7 Explain the difference between
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(d)
[Pd(gly)2]
has
two
isomers.
monodentate,
bidhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4entate, and
quadridentate? A monodentate ligand can
7.14 How many isomers are possible for the bond to a metal atom only at a single atom,
a
following complexes?
bidentate ligand can bond through two atoms,
a quadridentate ligand can bond through four
(b) [IrCl3(PEt3)2]?
(a) [FeCl(OH2)5]2 ?
None. atoms.
2 7.8 What type of isomers do you get with
ambidentate ligands? linkage isomers.
(c) [Ru(biby)3]2 ?
2
7.9 Which ligand could act like a chelating (d) [CoCl2(en)(NH3)2] ?
ligand? (a) Triphenylphosphite, no/
(b) Bis(dimethyl)phosphino ethane (dmpe) (e)
yes.
[W(CO)4(py)2] 2
(c) Bipyridine (bipy), yes.
(d) Pyrazine, no.
4
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7.10 Draw structures of complexes that contain
the ligands (a) en, (b) ox, (c) phen, and (d)
edta?
.
ANSWERS TO SELF-TESTS AND EXERCISES 15
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ar7.15
Draw all possible isomers for [MA2BCDE]? Including optical isomers, 15 isomers are
possible!
.
identical reflections to those of rutile TiO2 but shifted to slightly higher
diffraction angles.
S8.2
S8.3 S8.4
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?
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S8.5
7.16
Which of the following complexes are chiral? (a) [Cr(ox)3]3–?
(b) cis-[PtCl2(en)]?
(c)
Chiral
Chiral (The en is not planar).
cis-[RhCl2(NH3)4]
?
not
chiral.
[Ru(bipy)http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a43]2 ?
(e) fac-[Co(NO2)3(dien)]?
Not chiral. (f) mer-[Co(NO2)3(dien)]?
(d)
chiral
Chiral (dien is
not planar).
7.17
Which isomer, Λ or Δ, is the complex Mn(acac)3, shown in the exercise? The Λ isomer.
Draw both isomers, Λ or Δ, of the complex [Ru(en)3] 2 ?
N
Δ isomer
14% of naturally occurring tungsten is 183W, which has I = ?. Thus, the signal is
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split into 2 lines.
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.
S8.6 S8.7
Exercises
8.1
How would you determine crystalline components in mineral sample? Powder X-ray
diffraction.
Why are there no diffraction maxima in borosilicate glass? Glass has no long-range
periodicity or order.
What
is
the
minimum
size
of
a
cubic
crystal
that
can
behttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 studied? 0.5 μm by 0.5
μm by 0.5 μm.
Wavelength of neutron at 2.20 km/s? 1.80 × 10–12 m or 180 pm.
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.
Wavenumber for O–O in O2 ? In the region of 1800 cm 1.
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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. Raman bands assignments? N(SiH3)3 is planar. N(CH3) 3 is pyramidal.
8.2
7.18 8.3
8.4
8.5
7.19
Λ isomer
Suggest a reason why Kf5 is so different?
Because of a change in coordination.
Compare these values with those of
ammonia
given
in
exercise
7.19
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4and suggest why they are
different?
8.6
7.20
8.7
The chelate effect.
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CHAPTER 8
Self-tests
S8.1
Main features of the CrO2 powder XRD pattern? XRD pattern for CrO2 will show
8.8
16
Single 13C peak in NMR? Chemically distinct carbonyls are exchanging position
sufficiently quickly.
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%.
1.56. Slower process, NMR or EPR?
NMR.
g-values? 1.94, 1.74, and
Differences in EPR spectrum for
d-mehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tal with one electron
in solution versus frozen? In aqueous solution at room temperature, molecular
tumbling removes the effect of the g-value anisotropy.
anisotropy can be observed.
In frozen solution, g-value
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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 M?ssbauer.
No quadrupole splitting in M?ssbauer 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
one-electron reduction with a reduction potential of 0.21 V. Above 0.720 V the complex
is oxidized.
Zeolite
of
composihttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tion
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.
ANSWERS TO SELF-TESTS AND EXERCISES 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.
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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.
8.9
8.10
S9.2
8.11
8.12
8.13
8.14
8.15
8.16
8.17
S9.3 Shape of XeO4 and identify the Z
compound
with
8
the
shttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tructure?
same
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A tetrahedral geometry. The same structure is
SmO4.
S9.4 Comment on ΔfH? 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
9.2
9.3
Maximum stable oxidation state? (a) Ba;
2, (b) As;
5, (c) P;
5, (d) Cl;
7.
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. Elements vary from metals through
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metalloids to non-metals; form halides in oxidation states
5 and
3 and toxic
gaseous hydrides? Elements in Group 15. Born–Haber cycle for the formation of the
hypotheticalhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
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. 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.
8.18
8.19
8.20
9.4
8.21
8.22
CHAPTER 9
Self-tests
9.5
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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.
Names of ores? (a) Mg; MgCO3 magnesite, (b) Al; Al2O3 bauxite, and (c) Pb; PbS galena.
Identify thhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4e Z
for P. Similarities? V (vanadium).
Calculate ΔfH? for SeF6?
8 element
ΔfH? = ?1397 kJ
mol-1.
10.4 10.3
Assign oxidation numbers to elements?
H = –1, Re =
7. (d) H2SO4?
H =
1,
(b) KH?
H = –1, K =
O = –2, S =
1. (c) [ReH9]2–?
6. (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 =
(a) H2S?
9.7
9.8
(i)
CH4(g)
H2(g)
H =
1. Preparation of hydrogen gas?
1, S = –2.
9.9
H2O
→ CO(g)
(1000°C) (iii)
CO(g)
3H2(g)
H2O
→
(1000°C) (ii)
CO2(g)
C(s)
H2 (g) 10.5
H2O
→
CO(g)
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Properties of hydrides of the elements? (a) Position in the periodic table?
See
Figure 10.2.
(b) Trends in ΔfG??
See Table 10.1.
(c) Different molecular hydrides?
Molecular hydrides are found in groups 13/III
through 17/VII.://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par
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.
Which molecule has the stronger hydrogen bonds?
S–H···O has a weaker hydrogen
bond than O–H···S.
Name and classify the following?
AsH3?
(b) SiH4?
silane. (c) NH3?
ammonia,
(d)
Arsine.
(e) PdH0.9?
palladium hydride. (f) HI?
hydrogen iodide.
10.9
Chemical characteristics of hydrides?
Br?nsted acidity?
basicity?
(a) Hydridic character?
Hydrogen iodide. (c) Variable composition?
Barium hydride (b)
PdH0.9 . (d) Lewis
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.
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Structures of H2Se, P2H4, and H3O ?
The Lewis structures of these three species
are:
CHAPTEhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4R 10
Self-tests
S10.1
S10.2
Which of the following CH4, SiH4,, or GeH4
-would best H
or H donor?
CH4, the strongest Bronsted acid. GeH4 would be the best hydride donor. Reactions
of hydrogen compounds?
(a) Ca(s)
H3N–BF3(g). (c) LiOH(s)
H2(g) → CaH2(s).
(b) NH3(g)
BF3(g) →
H2(g) → NR.
S10.3 A procedure for making Et3MeSn?
10.7
2Et3SnH
2Na
→
2Na Et3Sn–
NaBr
10.8
(a) BaH2?
barium hydride.
H2 Na Et3Sn–
CH3Br
→
Et3MeSn
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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.://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4r
(c) Hydrogen in group 14? There is no reason for hydrogen to be placed in this group.
Exercises
10.1
10.2
Low reactivity of hydrogen?
a high bond enthalpy.
Hydrogen exists as a diatomic molecule (H2).
It has
It also only has two electrons shared between two protons.
10.11
18ANSWERS TO SELF-TESTS AND EXERCISES 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.
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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.
(a) Increasing acidity?
acid?
10.14 Arrange H2O, H2S, and H2Se in order?
H2O
H2Se ://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
10.15 The synthesis of binary hydrogen
(i) direct combination of the compounds?
elements, (ii) protonation of a Br?nsted base, and (iii) metathesis using a compound
such as LiH, NaBH4, or LiAlH4.
10.16 Compare BH4–, AlH4–, and GaH4–?
AlH4– is more “hydride-like,” it is the strongest reducing agent.
Since
10.17 Compare
period 2 and period 3 hydrogen
compounds?
Period 2 compounds:
- except for B2H6, are all exoergic
- tend to be weaker Br?nsted acids and stronger Br?nsted bases
- bond angles in period 2 hydrogen compounds reflect a greater degree of sp3
hybridization
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- Several period 2 compounds exhibit strong hydrogen bonding.
- boiling points of HF, H2O, and NH3 are all higher than their respective period 3
homologues.
CHAPTER 11
Self-tests
S11.1 Change in cell parameter for CsCl? At 445
° C thttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4he CsCl structure
changes to rock-salt and assumes the face centered cubic.
S11.2 S11.3
Lattice enthalpies of formation? LiF is 625 kJ mol?1 and for NaF is 535 kJ mol?1.
Trend is stability of Group 1 ozonides? Group 1 ozonides are less stable compared
to the superoxides.
S11.4 Sketch the thermodynamic cycle of Group 1 carbonate.
M2CO3(s)
2M (g)
M2O(s)
CO32?(g)
CO2
2M (g)
O2?(g)
CO2(g)
S11.5 Explain the differences in temperature of
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decomposition of LiNO3 and KNO3?
KNO3 decomposes in two steps at two different
temperatures.
KNO3(s)
→
KNO2(s)
1/2O2(g)
K2O(s)
2NO2(g)
2KNO2(s)
→
LiNO3(s)
→ 1/2 Li2O(s)
1/2O2(g)
NO2(g)
LiNO3 decomposes in one step.
1/4O2(g)
7
S11.6 Predicted Li NMR of Li3N?
the
NMR
spectrum
at
low
Two peaks in
temperature.
Only
one
resonance
in
the
athttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 high temperature.
10.18
Suggest a method for the preparation of
BiH3?
The redistribution of methylbismuthine, BiH2Me.
3BiH2Me →
2BiH3
BiMe3
10.19 Describe the compound formed between
NMR
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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.
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.
ANSWERS TO SELF-TESTS AND EXERCISES 19
(b)
Why are group 1 metals poor complexing agents?
They are large, electropositive
metals and have little tendency to act as Lewis acids.
Trends
of
the
fluorides
and
chlorides
ohttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4f 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. Synthesis of group 1 alkyls?
Most alkyl
lithiums are made using elemental lithium with the corresponding alkyl chlorides.
Which is more likely to lead to the desired result? (a) Cs
complex?
Mg2 .
(b) Be or Sr, dissolve in liquid ammonia?
Strontium.
or Mg2 , form an acetate
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CHAPTER 12
Self-tests
S12.1 Predict whether (a) BeCl2 and (b) BaCl2
are predominantly ionic or covalent?
BeCl2 is covalent; BaCl2 is ionic.
S12.2 Calculate the lattice enthalpies for CaO
and CaO2 and check that the above trend is confirmed?
11.2
11.3
11.4
Calcium oxide and calcium peroxide are 3465 kJ mol–1 and 3040 kJ mol–1.
S12.3
Use
ionic
radii
to
predict
type://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
of BeSe?
(c) Li
or K , form a complex with C2.2.2?
11.5 Identify the compounds?
Potassium ion.
a
structure
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According to Table 3.6 should be close to
ZnS-like structure.
S12.4 Calculate the lattice enthalpy of MgF2and
comment on how it will affect the solubility compared to MgCl2?
NaOH ←
heat
H2O
Sodium metal
O2
→
Na2O2
→ Na2O
MgF2 is 2991 kJ mol–1, will reduce solubility compared to MgCl2.
NH3 ↓ NaNH2
11.6
Trends in solubility?
Exercises
Higher for LiF and CsI, lower for CsF and LiI.
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12.1
Why are compounds of beryllium covalent
whereas those of the other group 2 elements are predominantly ionic?
11.7 Thermal stability of hydrides versus
carbonates?
Hydrides
decompose
to
elements.
Carbonates
decompose
to
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4oxides.
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. 11.10 Predict the products of the following
reactions?
12.2
Be has large polarizing power and a high charge density due.
Why are the properties of beryllium more similar to aluminium and zinc than to
magnesium?
Because of a diagonal relationship between Be and Al.
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Identify the compounds A, B, C, and D of the group 2 element M?
12.3
(a) CH3Br
Li → Li(CH3)
(b) MgCl2
LiC2H5 → Mg(C2H5)Br
(c) C2H5Li
LiBr
C6H6 → LiC6H5
LiBr
C2H6
12.4
M
H2O
→
M(OH)2;
A
=
M(OH)2
M(OH)2
CO2
→
MCOhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a43 2MCO3
3CO2; C = MC2 MC2
M(OH)2
2H2O → M(OH)2
2HCl → MCl2
MCO3;
B
=
5C → 2MC2
C2H2
2H2O; D = MCl2.
Why does beryllium fluoride form a glass when cooled from a melt? BeF2 adopts SiO2
like arrangement.
Why is magnesium hydroxide a much more effective antacid than calcium or barium
hydroxide? Mg(OH)2 is sparingly soluble and mildly basic.
12.5
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2012.6
Explain why Group 1 hydroxides are much more corrosive to metals than Group 2
hydroxides?
ANSWERS TO SELF-TESTS AND EXERCISES S13.1
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.
11
12.7
12.8
Group 1 hydroxides are more soluble than group 2 hydroxides, and therefore have higher
OH? concentrations.
Which
of
the
salts
MgSeO
or
BaSeOwould
tohttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
be
be
more
expected
soluble
in
water?
4
4
S13.2 Write an equation for the reaction of
LiBH4 with propene in ether solvent and a 1:1 stoichiometry and another equation for
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its reaction with ammonium chloride in THF with the same stoichiometry?
MgSeO4
Which Group 2 salts are used as drying agents and why?
Simple alkenes are inert towards LiBH4.
LiBH4
NH4Cl
THF
BH3NH3
LiCl
H2
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?
S13.3 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?
(a) BCl3 and ethanol?
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BCl3(g)
3
EtOH(l)
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4B(OEt)3(l)
→
3 HCl(g)
(b) BCl3 and pyridine in hydrocarbon solution?
Salts of divalent ions have low solubility.
12.10
Predict structures for BeTe and BaTe.
BeTe, close to ZnS-like structure. BaTe, close to CsCl-like structure.
BCl3(g)
py(l) → Cl3B ? py(s) (c) BBr3 and F3BN(CH3)3?
12.11
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
2
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2
BBr3(l)
F3BN(CH3)3(s) → BF3(g)
Br3BN(CH3)3(s)
Suggest a reaction or series of reactions for the preparation of N, N’,
N’’-trimethyl-B,B’,B’’-trimethylborazine starting with methylamine and boron
trichloride?
S13.4
12.12 The two Grignard compounds CHMgBr
and
2,4,6-(CH)CHMgBr
dissolve
in
THF.
What
differences
ehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4xpected in the
structures of the species formed in these solutions?
2
5
3
3
would
be
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6
2
BeBr2 should be covalent.
Cl3BNCH3
3CH3MgBr
MgBr2 should be ionic.
(CH3)3B3N3(CH3)3
BaBr2 should be ionic
3Mg(Br, Cl)2
S13.5 How many framework electron pairs are
present in BH and to what structural category does it belong? Sketch its structure?
4
10
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
Predict the products of the following
reactions?
(a)
MgCl2
(b) Mg
2LiC2H5 → 2LiCl
(C2H5)2Hg → Mg(C2H5)2
Mg(C2H5)2
Hg
(c) Mg
C2H5HgCl → C2H5MgCl
Hg
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7, arachno species.
The structure of
B4H10:
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ar
S13.6 Propose a plausible product for the
reaction between Li[BH] and Al(CH)?
10
13
2
3
6
CHAPTER 13
Self-tests
[B10H11 (AlCH3)]–
ANSWERS TO SELF-TESTS AND EXERCISES 21
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S13.7 Propose a synthesis for the polymer
precursor 1,7-BCH(Si(CH)Cl) from 1,2-BCH and other reagents of your choice?
10
2
10
3
2
2
10
2
12
(c) C = B2O3 13.6 Does B2H6 survive in air? If not, write the
equation for the reaction?
1,2 - B10C2H12 ??→1,7-B10C2H12 (90%)
2Si(CH3)2Cl2
→
B10C2H10(Si(CH3)2
1,12-B10C2H12 (10%) 1,7-B10C2H10Li2
1,7 -
2LiCl S13.8 Propose, with reasons, the chemical
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equation (or indicate no reaction) for reactions between (a) (CH3)2SAlCl3 and GaBr3?
Δ
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par13.7
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) CHCH?
No, it explodes in air.
B2H6
3 O2 → B2O3
(Me)2SalCl3
3 H2O2
GaBr3 → Me2SGaBr3
AlCl3.
(b) TlCl and formaldehyde (HCHO) in acidic aqueous solution? No reaction.
Exercises
13.1
13.2
Give a balanced chemical equation and conditions for the recovery of boron? B2O3
3Mg
→
2B
3MgO ΔH
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(a) BH3
(CH3)2C=CH2
B[CH2-CH(CH3)2]3 (b) BH3
CHCH
B(CH=CH2)3
13.9 Diborane has been used as a rocket
propellant. Calculate the energy released from 1.00 kg of diborane given the following
values of ΔfH?/kJ mol-1: B2H6 = 31, H2O
is B2H6 (g)
H2O (g)
-242, B2O3
-1264. The combustion reaction
3 O2(g) http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4→ 3
B2O3 (s). What would be the problem with diborane as a fuel?
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→?
-73,172 kJ.
Diborane is extremely toxic, and the boron containing product of combustion is a solid,
B2O3.
BCl3 > BF3 > AlCl3
(a) BF3N(CH3)3
BCl3N(CH3)3
13.4
BCl3
BF3 (BCl3 > BF3) (b) BH3CO
Thallium tribromide (1.11 g) reacts
BBr3
NR
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quantitatively with 0.257 g of NaBr to form a product A. Deduce the formula of A.
Identify the cation and anion?
13.5
13.10 Using BCl3 as a starting material and other
reagents of your choice, devise a synthesis for the Lewis acid chelating agent,
F2B–C2H4–BF2?
TlBr3
NaBr → NaThttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4lBr4
Identify compounds A, B, and C?
BF
3
LiAlH4
2BCl3
2Hg ? B2Cl4
B2Cl4
4AgF
?
B2F4
2HgCl2
4AgCl
B2F4
C2H4
?
F2CH2CH2BF2
NaBH4, a hydrocarbon of your 13.11 Given
choice, and appropriate ancillary reagents and solvents, give formulas and conditions
for the synthesis of (a) B(C2H5)3, (b) Et3NBH3?
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(a) BCl3
3C2H5MgCl
B(C2H5)3
3MgCl2
Heat, Ether
A
CaFH2O
C
(a) A = B2H6 (b) B = B(OH)3
heat
B
13.12 Draw the B12 unit that is a common motif
of boron structures; take a viewpoint along a C2 axis?
(b) [HN(C2H5)3]Cl
NaBH4
H2
NaCl
H3BN(C2H5)3
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22ANSWERS TO SELF-TESTS AND EXERCISES
(2) B10H12(SEt2)2
B10C2H12
(3)
C2H2
2SEt2
2
H2
B10C2H12
2EtO–
4Ehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tOH
2 B9C2H12–
(4) Na[B9C2H12]
(5) 2Na2[B9C2H11]
2NaCl
2B(OEt)3
NaH
2H2
Na2[B9C2H11]
H2
FeCl2
Na2[Fe(B9C2H11)2]
2-
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?
B6H10
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13.14
How many skeletal electrons are present in
B5H9? 14 13.15
(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?
(a)
2B5H9 (l)
5B2O3 (s)
12O2
(g)
9H2O (l)
(b) The boron containing product of combustion is a solid, B2O3.
13.16
(ahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4)
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)?
13.18
(a) What are the similarities and
differences in structure of layered BN and graphite (Section 13.9)? (b) Contrast their
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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.
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4rDevise 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?
(b) Me3N3B3H3?
13.19
(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. 13.17
Starting with B10H14 and other reagents of
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your choice, give the equations for the synthesis of [Fe(nido-B9C2H11)2]2-, and
sketch the structure of this species?
3 PhNH3 Cl?
3 BCl3 → Ph3N3B3Cl3
HCl Me3N3B3Cl3
3 LiH → Me3N3B3H3
9 HCl 3 MeNH3 Cl?
3 BCl3 → Me3N3B3Cl3
9
3 LiCl
The structures:
(1) B10H14
2SEt2
B10H12(SEt2)2
H2
ANSWERS TO SELF-TESTS AND EXERCISES
23
Exercises
14.1
Silicon
forms
chlorofluoridhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4es
SiCl2F2, and SiClF3. Sketch the structures
of these molecules?
the
SiCl3F,
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13.20 Give the structural type and describe the
structures of B4H10, B5H9, and 1,2-B10C2H12?
13.21
Arrange the following boron hydrides in
order of increasing Br?nsted 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.
B4H10, is an arachno borane. B5H9, is a nido borane.
1,2-B10C2H12 is a closo carborane
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
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4(CH3)4NF to form
with
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[(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]
14.2
CHAPTER 14
Self-tests
S14.1 Describe how the electronic structure of
graphite is altered when it reacts with (a) potassium, (b) bromine?
H33
H3
Potassium results in a (a) With potassium?
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.
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The anion is SiF5
–
S14.2 Use the bond enthalpy data in Table 14.2
and above to calculate the standard enthalpy of formation of CH4 and SiH4?
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
CH4: ΔfH= –61kJ mol
–1
SiH4: ΔfH=
39 kJ mol–1
S14.3 Propose a synthesis of D13CO2– starting
from 13CO?
13
CO(g)
2Li(s)
13
2MnO2(s) → CO2(g)
D2 → 2LiD(s)
Mn2O3(s)
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13
(b) There are two different fluorine environments.
?
CO2(g)
LiD(et) → LiDCO2(et)
13
14.4 Draw the structure and determine the charge on the cyclic anion [Si4O12]ν???
24
ANSWERS TO SELF-TESTS AND EXERCISES
14.11 Give balanced chemical equations and
conditions for the recovery of silicon and germanium from their ores?
SiO2(s)
ΔH
C(s) → Si(s)
CO2(g)
ΔH
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14.12
(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
dhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ecrease
when
its
temperature is changed from 20?C to 40?C?
(a) There is a decrease in band gap energy
from carbon (diamond) to grey tin. (b) Increase.
14.5 Predict the appearance of the 119Sn-NMR
spectrum of Sn(CH3)4?
14.6 Predict the appearance of the 1H-NMR spectrum of Sn(CH3)4?
Doublet
Doublet
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
to F:?
(A) SiCl4 (B) SiRCl3 (C) RSi(OH)3. (D)
RSiOSiR
H2O (E) SiR4 (F)? SiO2
14.8 Identify the compounds A
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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
inerthttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 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)
(a)
PbO2(s) (excess) → (air excluded) (ii) Sn2 (aq)
O2(air) →?
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
2Sn4
PbO2
4 H
→
Sn4
Pb2
2H2O,
(ii) 2Sn2
O2
4H
→
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.
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14.13
Preferahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4bly
without
consulting reference material, draw a periodic table and indicate the elements that
form saline,
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.
2
There is a layered structure of alternating sp carbon atoms and potassium ions, a
saline carbide.
(b) CaC2?
CaO(s)
Ca(l)
2C(s) → CaC2(s) or
3C(s) → CaC2(s)
CO(g)
ANSWERS TO SELF-TESTS AND EXERCISES 25
Calcium carbide contains discrete C22– ions. (c) K3C60?
be treated with elemental potassium. It is ionic.
14.15 Write balanced chemical equations for the
A solution of C60 can
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reactions of K2CO3 with HCl(aq) and of Na4SiO4 with aqueous acid?
K2CO3(aq)
2HCl(aq)
→
2KCl(aq)
CO2http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4(g)
H2O(l)
NH3
ClO?
H2NCl
H2O
H2NCl
NH3
H
→
→
H2NNH2
[H3N—Cl-O]?
H
→
HCl Both can be thought of as redox reactions.
S15.5 When titrated against base a sample of
polyphosphate gave end points at 30.4 and 45.6 cm3. What is the chain length?
Na4SiO4(aq)
4HCl(aq) → 4NaCl(aq)
SiO2(s)
2H2O(l) 14.16
14.17
Describe in general terms the nature of the
Exercises [SiO3]n2ν?? ion in jadeite and the silica-alumina framework in kaolinite?
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15.1 List the elements in Groups 15 and indicate
The structures of jadeite and kaolinite consist the ones that are (a) diatomic gases,
(b) of extended one- and two-dimensional nonmetals, (c) metalloids, (d) true metals.
structures, respectively. The [SiO32–]n ions in Indicate those elements that display
the
jadeite
are
a
linear
polymer
inhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ert-pair
of
SiO4
effect?
tetrahedra. The two-dimensional
aluminosilicate layers in kaolinite represent
element gas? effect? another way of connecting SiO4 tetrahedra.
metal
(a) How many bridging O atoms are in the framework of a single sodalite cage? (b)
metal Describe the (supercage) polyhedron at the no No centre of the Zeolite A
structure in Fig.
no No
14.3?
Yes
(a) 48.
yes No
(b) Eight sodalite cages are linked together to no No form the large cage of zeolite
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A. no No
no No
15.2
(a) Give complete and balanced chemical
equations for each step in the synthesis of HPO 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
There
are
(30.4
cm3)/(7.6
cm3)
=
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4acidic
4
OH
strongly
groups
per
molecule. A molecule with 2 terminal OH groups and four further OH groups is a
tetrapolyphosphate.
CHAPTER 15
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
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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?
(a) high-purity phosphoric acid?
2Ca3(PO4)2
P4?
10C
10CO
P4 (pure)
6SiO2
→
6CaSiO3
5O2
→
P4O10 P4O10
6H2O
→
4H3PO4 (pure) (b) Fertilizer grade
H3PO4?
Dimehylhydrazine ignites spontaneously, and produces less CO2.
S15.3 From trends in the periodic table, decide
whether
phosphorus
or
sulphur
is
likely
to
be
the
ahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4gent?
stronger
oxidizing
Sulphur.
S15.4
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?
Ca5(PO4)3OH
5H2SO4
→
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3H3PO4 (impure)
5CaSO4
H2O
(c) Account for the difference in cost?
26ANSWERS TO SELF-TESTS AND EXERCISES
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 LiN or (b) the high-temperature,
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?
3
2
2
2
2
P4O10
(c)
6H2O
→
4H3PO4
Reaction
of
the
product
from
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parthttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 (b) with CaCl2?
2H3PO4(l)
3CaCl2(aq)
Ca3(PO4)2(s)
→
6HCl(aq)
15.9 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) NHOH, (d) N3–?
3
2
(a) Hydrolysis of Li3N?
6Li
N2
2Li3N
→
3H2O
2Li3N
→
2NH3
3Li2O (b) Reduction of N2 by H2?
(a) HNO3?
4NH3(aq)
7O2(g)
→
6H2O(g)
High temperature (b) NO2–?
N2
3H2
→
2NH3
4NO2(g)
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(c) Account for the difference in cost?
The
second process is considerably cheaper than the first, because lithium is very
expensive.
15.4 Show with an equation why aqueous
solutions of NHNO are acidic?
4
3
2NO2(aq)
2OH?(aq)
NO2?(aq)
NO3?(aq)
→
H2O(l)
(c) NH2OH? http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
cold aqueous
acidic solution
NO2?(aq)
2HSO3?(aq)
H2O(l)
→
(d) N3–?
at elevated temperatures:
NH3OH (aq)
2SO42?(aq)
15.5
NH4NO3(s)
H2O ? NH4
NO3-(aq) Carbon monoxide is a good ligand and is
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toxic. Why is the isoelectronic N molecule not toxic?
2
3NaNH2(l)
NaN3
NaNO3
3NaOH
2NaNH2(l)
→
NH3(g)
N2O
→
NaN3
NaOH
NH3
15.10 Write the balanced chemical equation
corresponding to the standard enthalpy of formation of PO(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?
4
10
N2 itself, with a triple bond between the two atoms, is strikingly unreactive.
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15.6 Compare and http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4contrast
the formulas and
stabilities of the oxidation states of the common nitrogen chlorides with the
phosphorus chlorides?
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
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
bismuth (0 to
and
5) in acidic solution and discuss the relative stabilities of the
5 oxidation states of both elements?
(a) PCl4 ?
tetrahedron.
(b) PCl4–?
A see-saw.
(c) AsCl5?
15.8
5) and
A trigonal bipyramid.
3
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Give balanced chemical equations for each
of
the
following
reactions:
(a)
oxidation
of
P
with
excess
oxygenhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4, (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
(i) Bi(III) is much more stable than Bi(V), and (ii) P(III) and P(V) are both about
equally stable.
(a) Oxidation of P4 with excess O2?
P4
5O2
→
P4O10
(b) Reaction of the product from part (a) with excess H2O?
ANSWERS TO SELF-TESTS AND EXERCISES
27
and H2PO22? useful as oxidizing or reducing agents?
E? = 0.839 V.
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HPO32– and H2PO2– ions will be much better reducing agents than oxidizing agents.
15.16
Identify the compounds A, B, C, and D?
D
15.12
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?
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4parAS
ClA
LiAlH4
Cl2/hv
B
The rates of reactions in which nitrite ion is reduced are increased as the pH is
lowered.
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C
3RMgBr
15.13 When equal volumes of nitric oxide (NO)
and air are mixed at atmospheric pressure a rapid reaction occurs, to form NO and
NO. 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?
2
2
4
(a) A = AsCl3 (b) B = AsCl5 (c) C = AsR3 (d) D = AsH3
The rate law must be more than first order in NO concentration.
15.17 Sketch the two possible geometric isomers
of the octahedral [AsFCl]– and explain how they could be distinguished by 19F–NMR?
4
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2://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par
15.14 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) NHCl?
5
3
4
The cis isomer gives two 19F signals and the trans isomer gives one signal.
(a) H2O?
PCl5
tetrahedral POCl3.
H2O
2PCl5(g)
→
POCl3
8H2O(l)
→
2HCl (b) H2O in excess?
2H3PO4(aq)
10HCl(aq)
(c) AlCl3?
PCl5
AlCl3
tetrahedral.
→
[PCl4] [AlCl4]?
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(d) NH4Cl?
Cyclic molecules or linear chain polymers nPCl5
nNH4Cl
→
15.18 Identify the nitrogen compounds A, B, C,
D, and E?
(a) A = NO2
(b) B = HNO3; C = NO
(c)
D = N2O4
(d)
?[(N = P(Cl)2)n]?
15.15
E= NO2
(e)
F = NH4
4nHCl
Use
standard
potehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ntials
(Resource
section
3) to calculate the standard potential of the reaction of H3PO2 with Cu2 . Are HPO22?
28
15.19 Use the Latimer diagrams in Resource
section 3 to determine which species of N and P disproportionate in acid conditions?
ANSWERS TO SELF-TESTS AND EXERCISES 16.3
16.4
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Which hydrogen bond would be stronger: S—H . . . O or O—H . . . S?
O–H hydrogen bonds are stronger.
Which of the solvents ethylenediamine (which is basic and reducing) or SO (which is
acidic and oxidizing) might not react with (a) NaS, (b) KTe?
2
2
4
2
3
The species of N and P that disproportionate are N2O4, NO, N2O, NH3OH , H4P2O6, and
P.
CHAPTER 16
Self-tests
S16.1 Determine whether the decomposition of
HO is spontaneous in the presence of either Br or Cl?
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2
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a42
2
2
ethylenediamine is a better solvent than sulfur dioxide.
16.5 Rank the following species from the
strongest
reducing
agent
to
the
strongest
oxidizing
agent:
SO42–,SO32–,O3SO2SO32–?
16.6
S2O82– > SO42– > SO32–
Predict which oxidation states of Mn will be reduced by sulfite ions in basic
conditions?
The decomposition of H2O2 is thermodynamically favored in presence of Br– In Presence
of Cl–
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The decomposition of H2O2 is thermodynamically favored in presence of Cl–.
16.7
Mn ( VII,
VI,
V,
IV,
III) will be reduced by sulfite ions in basic solution. (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
increhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ase its coordination
number.
S16.2 Probable structures of SO2F- and
(CH3) 3NSO2, and predict reactions with
displaces
OH-.
Both are trigonal pyramidal.
F— or (CH3) 3N—.
(a) Formulas?
H5TeO6– and HSO4–
Exercises
16.1
State whether the following oxides are
acidic, basic, neutral, or amphoteric: CO, PO, SO, MgO, KO, AlO, CO?
2
OH-
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2
5
3
2
2
3
CO2, SO3, P2O5, and Al2O3 are amphoteric. CO is neutral; MgO and K2O are acidic.
16.8 Use the standard potential data in
Resource section 3 to predict which oxoanions of sulfur will disproportionate in
acidic conditions?
16.2 (a) Use standard potentials (Resource
section 3) to calculate the standard potential of the disproportionation of HO in
acid
solution.
(b)
Is
Cr
a
likely
catalyst
for
the
disproportionatiohttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4n of HO?
(c) Given the Latimer diagram
2
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2
2
2
2
S2O62?and S2O32?.
16.9 Use the standard potential data in
Resource section 3 to predict whether SeO32– is more stable in acidic or basic
solution?
The SeO32– is marginally more stable in acid solutions.
16.10
Predict whether any of the following will be reduced by thiosulfate ions,S2O32–,
in acidic conditions: VO2 , Fe3 , Cu , Co3 ?
O2
-0.13
HO2-
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H2O2
in acidic solution, calculate ΔrG? for the disproportionation of hydrogen superoxide
(HO) into Oand HO, and compare the result with its value for the disproportionation
of HO?
2–
2
2
22
2
Fe3
4
3
3
4
and Co3
will be reduced.
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(a)
The
disproportionation
of
andhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 HO2?
(b) Catalysis by Cr2 ?
Cr2
H2O2
1.068 V.
is not capable of decomposing H2O2.
(c) The disproportionation of HO2?
ΔrG° = 157 kJ. For the disproportionation of
H2O2 (part (a)), ΔrG° = 103 kJ.
16.11 SF reacts with BF to form [SF][BF]. Use
VSEPR theory to predict the shapes of the cation and anion?
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
SF4
trigonal pyramidal, BF4? tetrahedral.
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
Bliqur2 id I2d e Ne Ar Kr e
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17.2
er est
(a) SF4
(CH3)4NF → [(CH3)4N] [SF5]?; (b) square pyramidal structure; (c) two F
environments.
16.13
Identify
thhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4e
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?
3
2
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2
green dark red-brown dark violet orless colorless colorless colorless rless
Both.
SO2 is cheaper. 2 resonances.
19the F-NMR pattern for IF7? S17.2 Predict
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?
For F:://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
F2
H2SO4
2 Cl?
2 H2O
2 X?
Cl2
→
CaSO4
2 HF 2 HF
electricity
→
X2
→
2 KF
Cl2
H2
→
2 K HF2? For Cl:
2 OH? For Br and I:
2 Cl? I?)
(X? = Br?,
S17.3 From the perspective of structure and
bonding, indicate several polyhalides that are analogous to [py–I–py] , and
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describe their bonding?
2 K HF2?
electricity
→
F2
H2
2 KF
Examples include I3–, IBr2–, ICl2–, and IF2–.
The three centers contribute four electrons to three molecular orbitals, one bonding,
one nonbonding, and one antibonding.
Exercises
17.1 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
ihttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4on, (d) color?
17.3
sical state F2 dness of halide ion
hard
est (4.0)
egativity
halide
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or
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.
e? cathode:
2 H2O(l)
2 e?
→
2 OH?(aq)
anode:
2 Cl?(aq)
→
Cl2(g)
2
H2(g)
l2 er
light
yellow ow-
- unwanted reaction:
2 OH?(aq)
Cl?(aq)
Cl2(aq)
→
ClO?(aq)
H2O(l)
3017.4
Sketch the form of the vacant s* orbital of a dihalogen molecule and describe its
role in the Lewis acidity of the dihalogens?
ANSWERS TO SELF-TESTS AND EXERCISES
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shapes
of
IF
and
the
cation
and
anhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ion in X, (c) predict how
many 19F-NMR signals would be observed in IF and X?
3
3
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 HO to O?
2
2
a).
X = IF4N(CH3)4 b) Different possible arrangements of IF3:
17.6
Cl2 and F2.
Nitrogen trifluoride, NF, boils at –129?C and is devoid of Lewis basicity. By
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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
3
The shape of anion IF4– is Square planar.
The shape of cation (CH3)4N
Tetrahedral.
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par
17.9
c).IF3, two. IF4–, one.
Use the VSEPR model to predict the shapes of SbCl, FClO, and [ClF] ?
5
3
6
SbCl5 is trigonal bipyramidal, FClO3 is pyramidal, and ClF6 is octahedral.
5
5
is
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(a) Difference in volatility?
Ammonia exhibits hydrogen bonding.
(b) Explain the difference in basicity?
The strong electron-withdrawing effect
in NF3 reduces the basicity.
17.10 Indicate the product of the reaction
between ClF and SbF?
[ClF4] [SbF6] –.
4
2
3
3
17.7
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,
phttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4lausible
trimethylsilyl cyanide?
(c)
structure
a
for
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2
2
17.11 Sketch all the isomers of the complexes
MClFand MClF. Indicate how many fluorine environments would be indicated in the
19F-NMR spectrum of each isomer?
MCl4F2, the cis isomer, 1; the trans isomer, 2. MCl3F3, the fac isomer, 1 the mer
isomer, 2.
(a) The reaction of NCCN with NaOH?
NCCN(aq)
2 OH?(aq)
NCO?(aq)
H2O(l) (b) The reaction of SCN– with MnO2 in aqueous acid?
2SCN?(aq)
MnO2(s)
→
CN?(aq)
4 H (aq)
→
(SCN)2(aq)
The structure of trimethylsilyl cyanide?
Si–CN single bond.
17.8
Mn2 (aq)
2 H2O(l) (c)
Trimethylsilyl cyanide contains an
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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
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4rANSWERS
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]?
6
7
6
6
17.19 Explain why CsI(s) is stable with respect to
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the elements but NaI(s) is not?
3
3
Large cations stabilize large, unstable anions.
2
2
6
(a) The structures of [IF6]
and IF7?
IF6 , octahedral; IF7, pentagonal bipyramid.
(b) The preparation of [IF6] [SbF6]?
17.20 Write plausible Lewis structures for (a)
ClO and (b) IO and predict their shapes and the associated point group?
(a).
IF7
SbF5
→
[IF6][SbF6]
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2
6
17.13 Predict the shape of the doubly chlorine–
bridged
IClmolecule
by
uhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4sing the VSEPR model and
assign the point group?
(b) I2O5?
17.14
A planar dimer (I2Cl6).
Predict the structure and identify the point group of ClOF? trigonal pyramidal, Cs
symmetry.
2
3
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17.21 (a) Give the formulas and the probable
relative acidities of perbromic acid and periodic acid. (b) Which is the more stable?
17.15 Predict whether each of the following
solutes is likely to make liquid BrFa stronger Lewis acid or a stronger Lewis base:
(a) SbF, (b) SF, (c) CsF?
5
6
SbF5?
increases the acidity of BrF3.
(b) SF6?
(a)
No effect on the acidity or basicity of BrF3.
(c) CsF? Increases the basicity of BrF3.
17.16 Predict the appearance of the F-NMR
spectrum of IF?
Two resonances.
19
5
(a)
The
formulas
are
HBrO4
and
H5IO6http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4. The difference lies
in iodine’s ability to expand its coordination shell.
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(b) Relative stabilities?
Periodic acid is thermodynamically more stable.
17.22
.
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) CHOH, (c) F, (d) SCl?
3
5
3
2
2
2
(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?
(a) The expected trend?
E decreases as the pH increases.
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(b) E at pH 0 and pH 7 for ClO4–?
SbF5?
E? = 1.201 V (see Appendix 2).
Since SbF5 cannot be oxidized, it
(a)
will not form an exphttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4losive
mixture with BrF3.
(b) CH3OH?
Methanol, being an organic compound, is readily oxidized by strong
oxidants. (c) F2?
(d) S2Cl2?
No.
S2Cl2 will be oxidized to higher valent sulfur fluorides.
17.18 The formation of Brfrom 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 CHCl solution and give
your reasoning?
3–
2
4
3
2
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2
2
At pH 7, V
=
0.788 V
17.23
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.
17.24 Which oxidizing agent reacts more readily
in dhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ilute aqueous solution,
perchloric acid or periodic acid? Give a mechanistic explanation for the difference?
Br3?
I2
→
2 IBr
Br?
Periodic acid.
32
17.25 (a) For which of the following anions is
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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?
2
ANSWERS TO SELF-TESTS AND EXERCISES 18.2
Which of the noble gases would you choose as (a) The lowest-temperature refrigerant?
Helium.
(b) An electric discharge light source requiring a safe gas with the lowest ionization
energy?
Xenon.
(c) The least expensive inert atmosphere?
Argon.
18.3
By
means
of
balanced
chemical
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4equations and a statement
of conditions, describe a suitable synthesis of (a) XeF2?
Xe and F2 at 400?C, or
photolyze Xe and F2 in glass:
17.26
The rates of disproportionation are probably HClO > HClO2 > ClO3–.
Which of the following compounds present an explosion hazard? (a) NHClO, (b) Mg(ClO),
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(c) NaClO, (d) [Fe(HO)][ClO]. Explain your reasoning?
4
4
4
2
4
2
6
4
2
Xe(g)
F2(g)
→
XeF2(s)
(b) XeF6? High temperature, but have a large excess of F2:
(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) Mg(ClO4)2?
Not an explosion hazard. (c) NaClO4?
Not an explosion hazard.
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(d) [Fe(H2O)6] [ClO4]2?
An explosion hazard, since Fe(II) can be oxidized to
Fe(III).
17.http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a427
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.
Xe(g)
3 F2(g)
→
XeF6(s)
(c) XeO3?
XeF6(s)
3 H2O(l)
→
XeO3(s)
6 HF(g)
18.4 Draw the Lewis structures of (a) XeOF4? (b) XeO2F2? (c) XeO64??
CHAPTER 18
Self-tests
S18.1
Write a balanced equation for the
decomposition of xenate ions in basic solution for the production of perxenate ions,
xenon, and oxygen
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18.5 Give the formula and describe the
structure of a noble gas species that is isostructural with (a) ICl4??
(b) IBr2–?
Linear geometry. Isostructural with XeF2.
(c) BrO3–?
Trigonal pyramidal geometry. Isostructural with XeO3.
(d) ClF?
XeF4
Isostrhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4uctural
with the cation XeF .
18.6
(a) Give a Lewis structure for XeF7–?
2 HXeO4?(aq)
O2(g)
2 OH?(aq)
→
XeO64?(aq)
Xe(g)
2 H2O(l)
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?
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(b) Speculate on its possible structures by using the VSEPR model and analogy with
other xenon fluoride anions? Pentagonal bipyramid.
Helium present in today’s atmosphere is the product of ongoing radioactive decay.
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.
Identify the xenon compounds A, B, C, D,
the
group
oxidation
number
is
not
achieved
N?://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4par
18.8 and E?
18.9
A = XeF2(g)
?
B = [XeF] [MeBF3] C = XeF6 D = XeO3 E = XeF4 (g).
Predict the appearance of the 129Xe-NMR spectrum of XeOF3 . 1:3:3:1 quartet
by
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19.2
Explain why the enthalpy of sublimation of
that of
Re(s) is significantly greater than
Mn(s)?
As atoms become heavier, more energy is needed to vaporize them.
State the trend in the stability of the group 19.3
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.
18.10
Predict
the
appearance
of
the
19F-NMR
spehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ctrum of XeOF4.
Strong central line, two other lines symmetrically distributed around the central
line.
CHAPTER 19
Self-tests
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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?
(a) Cr2
(aq)
Cr2 (aq)
Fe3
(aq) →?
Fe3 (aq)
(b) CrO42– (aq)
→
Cr3 (aq)
Fe2 (aq)
MoO2 (s) →?
4; VO2
S19.2
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. S19.3 Describe the probable structure of the
compound
formed
when
Re3Cl9
is
dissolved
in
a
solvent
containing
PPh3http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4?
2 CrO42?
Cr3
3 MoO2(s)
10 H
→
2 Cr3
3 H2MoO4
2 H2O (c) MnO4– (aq)
(aq) →?
6 MnO4?
10Cr3
11H2O
→
6Mn2
5Cr2O72?
22 H
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19.5
(a) Which ion, Ni(aq) or Mn(aq), is more likely to form a sulfide in the presence
of HS? (b) Rationalize your answer with the trends in hard and soft character across
Period 4. (c) Give a balanced chemical equation for the reaction.
2
2
2
Discrete molecular species such as Re3Cl123– or Re3Cl9(PPh3)3 are formed.
Exercises
19.1
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 th
ose http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4for which
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More likely for Ni2
to form a sulphide. Hardness decreases from left to right in
the d block.
Ni2 (aq)
H2S(aq)
→
NiS(s)
2 H (aq)
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?
19.6
34ANSWERS TO SELF-TESTS AND EXERCISES
Metal–metal bonded halide compounds are found within bold border. Example: Sc5Cl6.
19.9
Speculate on the structures of the following species and present bonding models to
justify your answers:
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
madehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4 strongly basic at the
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same potential. Write a balanced equation for each of the successive reactions when
this same complex at pH = 6 and
environments up to
0.2 V is exposed to progressively more oxidizing
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.
(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 cis-dioxometal 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?
significant
degree
of
character://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
2
4
2
2
(b) NbCl4? significantly covalent.
(d) PtS ?
(c) FeF2?
ionic.
significant amount of covalent character.
(e) WCl2 ? metal–metal bonding.
ionic compound with
covalent
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19.11 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.
(b) [Cr2(O2CC2H5)4]? σ2π4δ2,chromium–chromium quadruple bond.
(c) [Cu2(O2CCH3)4]? σ2π4δ2δ*2π*4σ*2 configuration, no metal–metal bond in
this molecule.
cis-[RuIILCl(OH2)]
Ox
cis-[RuIILCl(OH2)]
H2O
cis-[RuIIILCl(OH2)]
As
oxidation
H2O
state
OH?
→
→
of
→
cis-[RuIIILCl(OH)]
Red
cis-[RuIIILCl(OH)]
H3O
e?
cis-[RuIVLCl(OH)]
H3O
e?
the
metal
increases,
ahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ccept
H2O
ability
electron
to
density
from an OH– or O2– ligand increases.
19.8
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)?
→
[Mo6O19]2?(aq)
5 H2O(l)
6 MoO42?(aq)
10 H (aq)
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19.12
Explain the differences in the following
redox couples, measured at 25oC?
Higher oxidation states become more stable
(c) ReCl5 (s) plus KMnO4(aq)? on descending a group.
5 ReCl5(s)
19.13
2 MnO4?(aq)
12 H2O(l)
→
Addition of sodium ethanoate to aqueous
?2 ?
solutions of Cr(II) gives a red diamagnetic 5 ReO4(aq)
2 Mn(aq)
25 Cl(aq)
product. Draw the structure of the 24 H(aq)
product,
noting
any
feahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tures of interest?
(d) MoCl2(s) plus warm HBr(aq)?
[Mo6Cl12]2?(aq)
6 MoCl2(s)
2 Br?(aq)
→
Br2(aq)
(e) TiO(s) with HCl(aq) under an inert atmosphere? 2TiO (s)
6H
(aq) ? 2Ti3
H2(g)
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2H2O(l)
E? =
0.37 V.
(f) Cd(s) added to Hg2 (aq)?
Hg2
(aq)
Cd(s) ? Hg(l)
Cd2
(aq)
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).
OO
OS20.5 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,
footing as a d1 configuration.) (a) 2p2?
treat it on the same
3P (called a “triplet P” term).
(b) 3d9? 2D (called a “doublet D” term). S20.7 What terms in a d2 complex of Oh
symmetry
correlate
with
the
3F
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4and 1D terms of the free
atom?
An F terms are 3
T1g, 3T2g, and 3A2g. Similarly, D terms are 1
T2g and 1Eg. S20.8 Use the same Tanabe-Sugano diagram to
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predict the energy of the first two spin-allowed 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 low-lying π* orbitals.)
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
tablehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4.
[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
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order of 2.0 should also be expected [0.5x(8-4)].
CHAPTER 20
Self-tests
S20.1
What is the LFSE for both high- and low-spin d7 configurations? A high-spin
d7 configuration is t25geg2. High spin, 0.8 Δo.
S20.2 The magnetic moment of the
complex [Mn(NCS)6]4– is 6.06μB. What is its electron
32
configuration? t2geg
S20.3 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 lahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ttice
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?
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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 S20.4
(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.
(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.
(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
unphttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4aired
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.
d6, low spin, no unpaired
(a) [Co(NH3)6]3 ?
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electrons, LFSE = 2.4Δ0.
(b) [Fe(OH2)6]2 ?
d6, high spin, four unpaired electrons, LFSE = 0.4Δ0.
36
(c) [Fe(CN)6]3–?
d5, low spin, one unpaired electron. LFSE is 2.0Δ0.
(d) [Cr(NH3)6]?
d, three unpaired electrons, low spin, LFSE = 1.2Δ0.
(e) [W(CO)6]?
d6, low spin, no unpaired electrons, LFSE = 2.4Δ0.
(f) Tetrahedral [FeCl4]2??
(g) Tetrahedral [Ni(CO)4]?
d6, high spin, four unpaired electrons, LFSE = 0.6 ΔT.
d10, 0 unpaired electrons, LFSE = 0.
Both H- and P(C6H5)3 are ligands of similar field strength, high in the
spectrochemical
series.
Recalling
that
phosphines
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4act as π acceptors, is
π-acceptor character required for strong-field behaviour? What orbital factors
account for the strength of each ligand?
No.
20.2
3
3
ANSWERS TO SELF-TESTS AND EXERCISES
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(c) [Fe(CN)6]3–
(d) The ruthenium complex. (e) The Co2
20.6
complex.
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 rock-salt 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.
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
σ-baseshttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4
are
(or
strong
both).
Estimate the spin-only contribution to the magnetic moment for each complex in
Exercise 20.1.
[Co(NH3)6]3
Spin-only contributions are: complex N μSO = [(N)(N
0 0 [Fe(OH2)6]2
3
[Cr(NH3)6] 3 3.9 [W(CO)6] 0 0
2–
4 4.9 [Fe(CN)6]3– 1 1.7
2)]1/2
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[FeCl4] 4 4.9 [Ni(CO)4] 0 0
.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?
to tetragonal complex. 20.8
Shift from square planar
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
elehttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ctronic
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 excited-state 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
6S
(b) L = 3, S = 3/2? 4F
(c) L = 2, S = 1/2? 2D
20.11 Identify the ground term from each set of
terms: (a) 3F, 3P, 1P, 1G?
3F.
(L,S):
(a) L = 0, S = 5/2?
(d) L = 1, S = 1? 3P
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(b)
5D,
3H,
3P,
1G,
1I?
5D
(c)
6S,
4http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4G, 4P, 2I? 6S
20.3
20.4
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.
is blue, [Co(NH3)6]2
pairs of
is yellow, [Co(OH2)6]2
is pink. 20.5 For each of the following
complexes, identify the one that has the
(a) [Cr(OH2)6]2
or [Fe(CN)6]3-
or [Mn(OH2)6]2
[CoCl4]2–
(b) [Mn(OH2)6]2
larger LFSE:
or [Fe(OH2)6]3
(c) [Fe(OH2)6]3
(d) [Fe(CN)6]3– or [Ru(CN)6]
(e) tetrahedral [FeCl4]2–or tetrahedral [CoCl4]2–
(a)
The chromium complex.
(b) The Fe3
complex.
ANSWERS TO SELF-TESTS AND EXERCISES 37
20.12 Give the Russell-Saunders terms of the
configurations and identify the ground term? (a) 4 s1?
2S.
://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ar(b) 3p2?
1D, 3P, 1S.
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The ground term is 3P.
20.13
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
E(1D) = A - 3B
2C. Calculate the values of B and C for V3 .
)/(15) = 861.33 cm–1 and C = 3167.7 cm–1. 20.14
7B,
B = (12920 cm–1
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)?
Δhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a40 = 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
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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
spin-forbidden 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.
d6 Co3
The
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 one-band 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
nephelauxhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4etic
in
effect?
Ammonia and cyanide ion are both σ-bases, but cyanide is also a π-acid. 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
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 spin-forbidden ligand field transition.
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
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thhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4e presence of Fe3
silicate matrix. Suggest which transitions are responsible for the color?
in the
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
[Cr(OH2)6]3
Solutions of
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 spin-allowed but Laporte-forbidden ligand
field transitions. The relatively low molar absorption coefficient is the reason that
the intensity of the color is weak.
dianion is Cr(VI); the
The oxidation state of chromium in dichromate
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
thttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4he 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.
38
20.23
Consider the molecular orbital diagram for
a tetrahedral complex (based on Fig. 20.7) and the relevant d-orbital configuration
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and show that the purple color of MnO4- ions 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
charge-transfer
trhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4ansitions, 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.
ANSWERS TO SELF-TESTS AND EXERCISES
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 .
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
spectruhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4m of
in
the
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[Fe(OH2)6]3
(in 1M HClO4) occurs at lower energy than the equivalent transition in
the spectrum of [Mn(OH2)]2 . Explain why this is.
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?
The extra charge of the iron complexes keeps the eg and t2g levels close
rate = (kKE[Mn(OH2)62 ][X?])/(1
KE[X?]) 21.5
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
observahttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tion on the basis of
a dissociative rate-determining step.
CHAPTER 21
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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)]?
3
3
4
2-
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.
21.6
A Pt(II) complex of
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tetramethyldiethylenetriamine is attacked by Cl- 105 times less rapidly than the
diethylenetriamine analogue. Explain this observation in terms of an associative
rate-
The
greater
steric
determining
stephttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4.
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.
S21.3
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.
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
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dissociative.
ANSWERS TO SELF-TESTS AND EXERCISES 39
21.9
Does
the
fact
that
[Ni(Chttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4N)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. 21.10
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.
(ii) The anomalously high rate of substitution by OH– signals an alternate path,
that of base hydrolysis.
(iii) The implication is that a complex without protic ligands will not undergo
anomalously fast OH– ion substitution.
21.14 Predict the products of the following
reactions: (a) [Pt(PR3)4]2
2Cl–?
cis-[PtCl2(PR3)2].
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(b)
[PtCl4]2–
2PR3?
trans-[PtCl2(PR3)2].
(chttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4)
2Cl–?
cis-[Pt(NH3)2(py)2]2
trans-[PtCl2(NH3)(py)].
21.15
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 ?
A possible mechanism is loss of one dimethyl sulfide ligand, followed by the
coordination
of
1,10-phenanthroline.
A
two-step
synthesis
for
cis-
and
trans-[PtCl2(NO2) (NH3)] – start with [PtCl4]2–?
21.11
The order of increasing rate is [Ir(NH3)6]3
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?
rate. (b) changing the leaving group from NO3– to Cl–?
(c) changing the entering group from Cl– to I–?
or no effect on the rate.
21.16
decreased
Decreased rate.
This change will have little
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21.12
How
does
each
of
the
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4following 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?
21.13
Rate increase.
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 Br?nsted 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?
21.17
Decreased rate.
Write out the inner- and outer-sphere
pathways for reduchttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4tion 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
→
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[[Co(N3)(NH3)5]2 , [V(OH2)6]2
[Co(N3)(NH3)5]2 , [V(OH2)6]2
[V(OH2)5]2 , H2O [Co(N3)(NH3)5]2 , [V(OH2)5]2 , H2O
¡ú
[Co(N3)(NH3)5]2 ,
¡ú [(NH3)5Co¨CN=N=N¨CV(OH2)5]4
[(NH3)5Co¨CN=N=N¨CV(OH2)5]4 ¡ú [(NH3)5Co¨CN=N=N¨CV(OH2)5]4
[(NH3)5Co¨CN=N=N¨CV(OH2)5]4
¡ú [Co(OH2)6]2
[V(N3)(OH2)5]2
The outer sphere pathway:
[Co(N3)(NH3)5]2
[V(OH2)6]2
¡ú [Co(N3)(NH3)5]2 [V(OH2)6]2
40ANSWERS TO SELF-TESTS AND EXERCISES
[Co(N3)(NH3)5]2 ,[V(OH2)6]2 ¡ú [Co(N3)(NH3)5] , [V(OH2)6]3
[Co(N3)(NH3)5] ,[V(OH2)6]3 ¡ú [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
http://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4with Fe2 (aq) to give Fe3 (aq) and Co2 (aq). What
does this observation suggest about the mechanism?
21.19
Appears to be an inner-sphere electron transfer reaction.
Calculate the rate constants for electron
transfer in the oxidation of [V(OH2)6]2 (E¦Ò (V3 /V2 ) = ¨C0.255 V) and the oxidants (a) [Ru(NH3)6]3
(EO(Ru3 /Ru2 ) =
the rate constants.
0.07 V), (b) [Co(NH3)6]3 (EO(Co3 /Co2 ) = 0.10 V). Comment on the relative sizes of
(a) [Ru(NH3)6]3
k = 4.53 ¡Á 103 dm3mol?1s?1
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Using these values gives K12 = 1.81 ¡Á 1028.
Substitution of these values in the Marcus-Cross relationship gives k12 = 8.51 ¡Á 1015
The dm3mol?1s?1.21.21
photochemical substitution of
= pyridine) with [W(CO)5(py)] (py
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
thhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4at 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.
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.
(b) [Co(NH3)6]3 k = 1.41 ¡Á 10?2 dm3mol?1s?1 Relative sizes?
complex is more thermodynamically favoured and faster.
The reduction of the Ru
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21.20
Calculate the rate constants for electron
transfer in the oxidation of [Cr(OH2)6]2 (EO¨C(Cr3 /Cr2 ) = ¨C0.41 V) and each of the oxidants [Ru(NH3)6]3
(EO(Ru3 /Ru2 )
0.07 V), [Fe(OHhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a42)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
CHAPTER 22
Self-tests
S22.1 Is Mo(CO)7 likely to be stable?
No. S22.2
What is the electron count for and
oxidation number of platinum in the anion of Zeise¡¯s salt, [PtCl3(C2H4)]¨C? Treat CH=CH as a neutral
two-electron donor. Electron count, 16; oxidation number, 4.
2
2
(a) k11 (Cr3 /Cr2 ) = 1 ¡Á 10?5 dm3mol?1s?1; k22 (Ru3 /Ru2 for the hexamine complex) = 6.6 ¡Á
o
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103 dm3mol?1s?1; f12 = 1; K12 = e [nF ¦Å/RT] where ¦Åo = 0.07 V ¨C (¨C0.41V) = 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 ¡Á 108. Substitution of these values in
the Marcus-Cross relationship gives k12 = 2.95 ¡Á 103 dm3mol?1s?1.
(b)
k11
(Cr3
/Cr2
)
=
1
¡Á
10?5
dm3mol?1s?1;
k22
(Fe3
/Fhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4e2 for the aqua complex) = 1.1
o
dm3mol?1s?1; f12 = 1; K12 = e[nF ¦Å/RT] where ¦Åo = 0.77 V ¨C (?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 ¡Á 1019. Substitution of these values in the
Marcus-Cross relationship gives k12 = 3.19 ¡Á 107 dm3mol?1s?1.
(c)
k11 (Cr3 /Cr2 ) = 1 ¡Á 10?5 dm3mol?1s?1; k22 (Ru3 /Ru2 for the bipy complex) = 4 ¡Á 108
o
dm3mol?1s?1; f12 = 1; K12 = e[nF ¦Å/RT] where ¦Åo = 1.26 V ¨C (¨C0.41V) = 1.67 V; n = 1; F = 96485 C; R =
8.31 Jmol?1K?1 and T = 298 K.
S22.3 What is the formal name of
[Ir(Br)2(CH3)(CO)(PPh3)2]?
Dibromocarbonylmethylbis(triphenylphos-
phine)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¨CC bond?
Fe(CO)5
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S22.5 Show that bhttp://www.wendangwang.com/doc/9f5b37fc1fa388230700f6a4oth are 18-electron species.
(a) [(¦Ç6-C7H8)Mo(CO)3] (49)?
¦Ç6-C7H8 = 6 electrons, Mo atom = 6. carbonyl = 2 electrons, total = 18.
(b) [(¦Ç7-C7H7)Mo(CO)3] (51)? ¦Ç7-C7H7 = 6
electrons, Mo atom = 6, 3 COs = 6, total = 18. S22.6 Propose a synthesis for
Mn(CO)4(PPh3)(COCH3) starting with [Mn2(CO)10], PPh3, Na and CH3I.
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