Chapter 2 - Solutions
1.
0
N
-1
0
+1
N
N
0
+1
N
N
3. a. H
N
-1
+1
N
N
+1
N
0
N
O
O
H
H
H
O
N
O
C
C
N
N
H
O
N
C
C
N
N
C
C
O
H
O
H
b.
H
H
N
C
N
C
C
C
O
H
H
H
H
N
H
c.
H
C
C
H
H
N
N
C
H
O
H
d.
H
H
H
N
C
H
C
C
C
H
5. a.
H
C
C
H
Configuration
No. of unpaired e
Element
Charge
26
[Ar]3d6
4
Fe
+2
ground state
52
[Kr]5s24d105p56s1
5
Te
2-
excited state
16
[Ne]3s23p3
1
S
+1
excited state
37
[Kr]4d1
1
Rb
0
excited state
30
[Ar]4s23d8
2
Zn
+2
excited state
Z
Energy state
b.
Configuration
No. of unpaired e- Element Charge
38
[Kr]5p1
1
Sr
+1
2nd excited state
45
[Kr]4d7
3
Rh
2+
ground state
43
[Kr]5s14d5
6
Tc
+1
ground state
8
[Ne]
0
O
2-
ground state
21
[Ar]4s13d1
2
Sc
+1
1st excited state
Z
Energy state
7.
H
N
H
-
H
N
N
+
N
N
H
0
N2H4
V (P.E.)
N3
N
N
9. Formal charges are not a method for calculating oxidation numbers. However, for the
purposes of this exercise, the formal charge is a hypothetical oxidation number of the
atom were the compound to have the actual Lewis structure indicated.
a. The oxidation numbers for H2O2, hydrogen peroxide, are by definition –1 or O
and +1 for H.
The oxidation numbers for H2O, water, are by definition –2 or O and +1 for H.
Therefore, hydrogen peroxide (H2O2) is being reduced as it is converted to water
because the oxygen in hydrogen peroxide has a (1) oxidation state in the
reactants and a (2) oxidation state as water in the products.
b. Because the carbons in ascorbic acid are not equivalent, and therefore have
different oxidation numbers, we need to look at a Lewis structure to see the
corresponding change in oxidation numbers between ascorbic acid and the
products. Lewis structures are drawn below and the C atoms that participate in
the redox reaction are circled.
H
H
H
H
O
C
H
O
H
C
H
H
C
H
O
H
H
C
O
H
C
O
The oxidation number of the circled carbons in ascorbic acid is 4 – 3 =O +1 and for the
circled carbonsHin dehydroxyascorbic
acid is 4 – 2 = +2. Therefore,
the two circled
C
C
O
C
O
C
carbon atoms in ascorbic
acid
are
oxidized.
C
C
O
O
C
O
H
O