Answer 7: Atomic Orbitals
7-1
1s: 0, 2s: 1 and 3s: 2.
7-2
2pz , 3pz
There is one angular node for 2pz; one angular node and one spherical node for 3pz.
7-3 (0, 2, 4, 1, 3)
Answer 8: Intermolecular Forces
8-1-1
Na
+OH2
OH2 OH2 H2O
OH2
H2O
8-1-2
70 O
O O
O O O
Na
+8-2-1
F H
F H F H F H F H F
H
8-2-2
CH
3C O O
C CH
3O
O H
H
8-3
N
N N
N N N
H O
O R
H
H3C H N
H
H
R
Thymine
HAdenine
N
N N
N N N H N
H
R O H
H
H O
N H
H
R H
Cytosine
Guanine
Answer 9: Crystal Packing
9-1 Simple cubic: 6, body-centered cubic: 8 and face-centered cubic: 12
9-2
For simple cubic, a = 2r,
52 . 4 % a
3 4
f
33
v
= π r =
For body-centered cubic,
3 a = 4 r
,68 % a
3 2 4
f
33
v
× =
= π r
For face-centered cubic,
2 a = 4 r
,74 % a
3 4 4
f
33
v
× =
= π r
9-3
2 a = 4 r
,a = 2 2 r = 407 pm
3 3
23
g/cm 6 . ) 10 407 (
10 02 . 6
9 . 4 107
× =
= × d pm
9-4
pm 229 17.34 sin 201 2 sin 2
o
=
×
×
=
= λ
λ θ d
Answer 10: Applications of Transition Metals
10-1-1
2 CrO
42−(aq) + 2 H
+(aq) Cr
2O
72−(aq) + H
2O (l)
10-1-2 CrO42−: + 6, Cr2O72−: +6.
10-1-3 This is not a redox reaction because the oxidation state in each metal center does not change.
10-1-4 Hydrogen ion concentration is the main factor to control the equilibrium position.
72 10-1-5
Cr O
O
Cr O
Cr
O O O
O
O
O
O
O
2− 2−
10-2-1
+ + 12e− +
+ + 4e−
O2
Cr2O7
2− (aq) + 2 H+ (aq) H2O (l) + 2 Cr (s) + 3 O2 (g) Cathode
Anode
Overall
Cr2O7
2− 14 H+ 2 Cr 7 H2O
2 H2O 4 H+
10-2-2 1.5 moles of oxygen gas will evolve.
+ + 12e− +
+ + 4e−
O2
Cr2O72− (aq) + 2 H+ (aq) H2O (l) + 2 Cr (s) + 3 O2 (g) Cathode
Anode
Overall
( ) x 3
52 g Cr x 1 mol Cr 52 g
x 3 mol O2
2 mol Cr = 1.5 mol O2
Cr2O72− 14 H+ 2 Cr 7 H2O
2 H2O 4 H+
10-2-3 16 h
52 g Cr x 1 mol Cr
52 g x 6 F
mol Cr x 96485 C = 16 h
1 F x 1 sec 10 C
1 min 60 sec
x x 1 h
60 min
10-2-4 Chromium readily forms a thin, adherent, transparent coating of Cr2O3 in air, making the metal extremely useful as an attractive protective coating on easily corroded metals.
Answer 11: Electrochemistry of Inorganic Compounds
11-1 For the concentration cell: Mn(s) | Mn2+(aq) (1M) || Mn2+(aq) / MnCO3 | Mn(s),
Ecell = E – (0.0592 / 2) log ([Mn ]right / [Mn ]left) Ksp = 1.8×10-11 = [Mn2+][CO3
2-]
[Mn2+]right = 1.0×10-8 M and [Mn2+]left= 1.0 M with Eo = 0.0 V (both are Mn) Ecell = 0.0 - (0.0592 / 2) log (1.0×10-8 M / 1.0 M) = 0.237 V
11-2
Reduction of O2 to H2O is obtained as (0.70V+1.76V) / 2 = 1.23 V, for O2 + 4H+ + 4e- → 2 H2O Eo = 1.23V
The Eo value could be obtained directly from the diagram by dividing the differences (2.46) of O2 and H2O by the differences of the oxidation number (2).
For H2O2 → O2 + H2O Eo = 1.06 > 0.0 The disproportionation reaction is spontaneous.
11-3 The number of electron pair should be 5 (trigonal bipyramidal) with three electron pairs in the equatorial plane, thus the molecular geometry of XeF2 is linear.
2 H2O → O2 + 4H+ + 4e- Eo = -1.23V
XeF2(aq) + 2H+(aq) + 2e- → Xe(g) + 2HF(aq) E° = 2.32V
2 XeF2(aq) + 2H+(aq) + 2 H2O → 2 Xe(g) + O2 + 4HF(aq) E° = 1.09 V
The decomposition of XeF2 in aqueous solution is favored in acidic solution.
Answer 12: Metal Carbonyl Compounds
12-1 Compound A is anionic, the absorption bands attributed to CO stretching appear at lower
刪除: 2H2
刪除: 00
刪除: 2H
74
frequency because of stronger back donation of the anionic charge to the anti bonding orbital of CO thus weakening the CO bond. For the neutral species B, absorption bands appear at the higher frequency.
12-2
W OC
CO CO W
OC CO OC
W OC
CO CO
C CH2 W
OC CO CO
W OC
CO CO W(CO)6 NaC5H5
Na
HCCCH2Br FeSO4
Na/Hg
metal migration
A B
C D
12-3
C CH2 W
OC CO CO
W OC
CO CO D
D
metal migration
