Chapter 2 - Solutions

1.

3.a.

b.

c.

d.

  1. a.

Z / Configuration / No. of unpaired e / Element / Charge / Energy state
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

b.

Z / Configuration / No. of unpaired e- / Element / Charge / Energy state
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

7.

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.

  1. 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.

The oxidation number of the circled carbons in ascorbic acid is 4 – 3 = +1 and for the circled carbons in dehydroxyascorbic acid is 4 – 2 = +2. Therefore, the two circled carbon atoms in ascorbic acid are oxidized.