2005年第37届IChO预备试题答案英文7-12无机化学

Answer 11: Electrochemistry of Inorganic Compounds
2+
2+
11-1 For the concentration cell: Mn(s) | Mn (aq) (1M) || Mn (aq) / MnCO3 | Mn(s),
72
o
2+
2+
Ecell = E – (0.0592 / 2) log ([Mn ]right / [Mn ]left)
x
52 g
3 mol O2 2 mol Cr
= 1.5 mol O2
10-2-3 16 h
1 mol Cr
52 g Cr x
x
6F
96485 C x
x 1 sec x 1 min
x
1h
= 16 h
52 g
mol Cr 1 F
10 C 60 sec 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.
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.
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)
10-1-1
2 CrO42− (aq) + 2 H+ (aq)
Cr2O72− (aq) + H2O (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.
H
Adenine
H
H
NH
O
H
N
Cytosine
N
R
O
HN
HN H
N
H
N
R N
Guanine
70
Answer 9: Crystal Packing
9-1 Simple cubic: 6, body-centered cubic: 8 and face-centered cubic: 12
9-2
4 πr3
Answer 8: Intermolecular Forces
8-1-1 8-1-2
H2O H2O
OH2 OH2
Na+ OH2
OH2
69
O
O
O
Na+
O
O
O
8-2-1
F
H
H
F
F
H
H
F
F
H
H
F
8-2-2
OH O
CH3 C
C CH3
O HO
8-3 H
H3C
O
HN
N
H
Thymine
H
The decomposition of XeF2 in aqueous solution is favored in acidic solution.
мৰ: 2H2 мৰ: 00 мৰ: 2H
Answer 12: Metal Carbonyl Compounds
12-1 Compound A is anionic, the absorption bands attributed to CO stretching appear at lower 73
Cathode Anode Overall
Cr2O72− + 14 H+ + 12e− ( 2 H2O
Cr2O72− (aq) + 2 H+ (aq)
2 Cr + 7 H2O O2 + 4 H+ + 4e− ) x 3 H2O (l) + 2 Cr (s) + 3 O2 (g)
1 mol Cr
52 g Cr x
+
-o
2 H2O → O2 + 4H + 4e E = -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
9-3
2a = 4r , a = 2 2r = 407 pm
d
=
4
×
107.9 6.02 ×1023
= 10.6 g/cm3
(407 pm)3
9-4
2d sinθ = λ λ = 2 × 201× sin 17.34o = 229 pm
Answer 10: Applications of Transition Metals
-11
2+
2-
Ksp = 1.8×10 = [Mn ][CO3 ]
2+
-8
2+
o
[Mn ]right = 1.0×10 M and [Mn ]left= 1.0 M with E = 0.0 V (both are Mn)
-8
Ecell = 0.0 - (0.0592 / 2) log (1.0×10 M / 1.0 M) = 0.237 V
12-2
NaC5H5 W(CO)6
W Na OC
CO CO
A
FeSO4 Na/Hg
OC CO OC
WW OC
CO CO
B
HCCCH2Br
12-3
W OC
CO CO
C
W OC
CO CO
C
metal migration
W OC
CO CO
D
C CH2
metal migration D
W OC
CO CO
D
C CH2
For simple cubic, a = 2r,
fv =
3 a3
= 52.4%
For body-centered cubic,
2 × 4 πr3
3a = 4r , fv =
3 a3
= 68%
For face-centered cubic,
4 × 4 πr3
2a = 4r , fv =
3 a3
= 74%
o
For H2O2 → O2 + H2O E = 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.
11-2
Reduction of O2 to H2O is obtained as (0.70V+1.76V) / 2 = 1.23 V,
+
-
o
for O2 + 4H + 4e → 2 H2O E = 1.23V
o
The E 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).
Cr2O72− + 14 H+ + 12e− 2 H2O
Cr2O72− (aq) + 2 H+ (aq)
2 Cr + 7 H2O O2 + 4 H+ + 4e− H2O (l) + 2 Cr (s) + 3 O2 (g)
10-2-2 1.5 moles of oxygen gas will evolve.
10-1-4 Hydrogen ion concentration is the main factor to control the equilibrium position. 71
10-1-5
2−
O
Cr
O
O
O
O
O