CHEMISTRY 163
HWCH#18
10, 14, 18, 24, 28, 30, 32, 34, 38, 40, 42, 46, 52, 56, 58, 60, 64, 68, 72, 76, 80, 84.
18-10
No. Any substance that is a Brønsted–Lowry acid is a Lewis acid because H+ is an electron-pair
acceptor.
18-14
The Lewis acid is CO2 and the Lewis base is O2–.
18-18
The Lewis acid is SbF5 and the Lewis base is HF.
18-24
NO3–, nitrate
18-28
Ag+ forms a complex with NH3 in solution, but the overall reaction has a very low equilibrium
constant, so AgI does not dissolve as well as AgCl in aqueous ammonia:
AgI(s)  2 NH3 (aq)
Ag(NH3 )2 (aq)  I – (aq)
K  Ksp  Kf  1.4  10–9
18-30
Initial
Change
Equilibrium
[Cu2+]
[en]
[Cu(en)2]2+
3.00  10– 4
–(3.00  10– 4 – x)
x
1.00  10–3
–2(3.00  10– 4 – x)
4.0  10– 4 + 2x
0
+(3.00  10– 4 – x)
3.00  10– 4 – x
3.2 1019 
3.00 10 – 4 – x
3.00 10 – 4

–4
2
x(4.0 10  2 x)
x(4.0 10 – 4 ) 2
x  5.9 10 –17 M
18-32
First, we can assume that all of the Ag+ present is converted to Ag(CN)2– so [Ag(CN)2–] = 4.00  10–5
M. The [CN–] therefore initially is [(4.00  10– 4) – (2  4.00  10–5)] = 3.2  10– 4 M and the amount
of Br– in solution is 4.00  10– 4 M. Using an ICE table we can then calculate the amount of Ag + in
the solution at equilibrium. To solve this problem for x we will allow AgBr to form as a soluble
species, even though we know it to be insoluble.
[AgBr]
[CN–]
[Ag(CN)2–]
[Br–]
Initial
Change
Equilibrium
0
+x
+x
3.2  10– 4
+2x
4.00  10–5
–x
4.00  10– 4
–x
3.2  10– 4 + 2x
4.00  10–5 – x
4.00  10– 4 – x
 4.00 10 – x  4.00 10
x  3.20 10  2 x 
–5
5.4 108 
–4
–4
– x
2
 4.00 10  4.00 10 
x  3.20 10 
–5

–4
–4 2
x  2.9 10–10 M
This is a very small amount of AgBr
mol
187.77 g
2.9  10–10
 0.250 L 
 1.36  10–8 g AgBr
L
1 mol
This small amount of AgBr could not be seen in the solution. The contents of the flask appear clear.
18-34
(a) Diamminecopper(I)
(b) Tetraaquadihydroxotitanium(IV)
(c) Tetraamminediaquanickel(II)
18-38
(a) Ammonium hexacyanocobaltate(III)
(b) Chlorobis(ethylenediamine)cobalt(III) nitrate
(c) Tetraaquadihydroxoiron(III) chloride
18-40
Ti4+ has the greatest positive charge and is the smallest cation of the group, so we would predict that
an aqueous solution of TiCl4 would have the lowest pH.
18-42
(a) As Al3+ reacts with OH– it first forms a precipitate of Al(OH)3:
Al3+(aq) + 3 OH–(aq)
Al(OH)3(s)
Upon further reaction with OH–, the precipitate forms the soluble anion Al(OH) 4– and so the
precipitate dissolves to give a clear solution:
Al(OH)3(s) + OH–(aq)
Al(OH)4–(aq)
(b) No, addition of KOH to FeCl3 does not produce the same changes as the addition of KOH to
AlCl3. Fe3+ forms an insoluble precipitate with OH–, but does not react further to form a soluble
complex ion:
Fe3+(aq) + 3 OH–(aq)
Fe(OH)3(s)
18-46
No. The solubility of FeCl3 is highest in more acidic solutions because the insoluble Fe(OH) 3
compound formed by Fe3+ hydrolysis reacts with H+ to form Fe3+:
Fe(OH)3(s) + 3 H+(aq)
Fe3+(aq) + 3 OH–(aq)
Because all these acid solutions have different pH, each contains a different [Fe 3+].
18-52
18-56
18-58
Ti4+ has an electron configuration of [Ar]4s03d 0. Because Ti4+ has no electrons in its d orbitals, no d –
d transitions can occur, and the compounds of Ti4+ are colorless.
18-60
The repulsions between the electrons on the ligands and the electrons in the d orbitals of the
transition metal cations raise the d orbitals in energy.
18-64
6.626  10 –34 J  s  3.00  108 m/s
 2.95  10 –7 m or 295 nm
6.74  10 –19 J/ion
This wavelength is in the UV region, so the solution is colorless.

18-68
In order to bond to six water ligands the Fe3+ ion must use the 4s, 4p3, and 4d 2 orbitals because
electrons occupy the 3d orbitals. The hybridization is then “4s4p34d 2” or sp3d 2 for Fe(H2O)63–.
Because the d Z2 and d x2 – y 2 orbitals of Fe(CN)63– are empty, this complex can use the 3d orbitals for
bonding to CN–. The hybridization is then “3d 24s4p3” or d 2sp3 for Fe(CN)63–.
18-72
High-spin Fe3+ has 5 unpaired electrons.
Rh+ has 2 unpaired electrons.
V3+ has 2 unpaired electrons.
Low-spin Mn3+ has 2 unpaired electrons.
18-76
After filling each orbital diagram with 8 electrons, we see that Ni(CN)42– is diamagnetic because it
has a square-planar geometry and NiCl42– is paramagnetic because it has a tetrahedral geometry.
18-80
No, not all square-planar complexes with two different ligands have geometric isomers. For example,
A is a chelating ligand such as ethylenediamine, that ligand occupies only cis positions and
if A
gives only one isomer of MA2B2.
18-84
H2
N
CN
CN
Cl
Cl
Cl
Cl
Ni
N
H2
H2
N
Ni
CN
CN
Cl
Cl
CN
CN
Ni
N
H2
H2
N
CN
CN
Cl
Cl
CN
CN
Cl
Cl
CN
Cl
Superimposable mirror images
Not chiral
H2
N
N
H2
H2
N
Ni
CN
Cl
N
H2
Ni
Ni
N
H2
H2
N
N
H2
Superimposable mirror images
Not chiral
Nonsuperimposable mirror
images
Chiral