(a) In KI3, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = –1/3. In the structure, K+[I–I←I]–, a coordinate bond is formed between I2 molecule and I– ion. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine ion forming the coordinate bond is –1. Thus, the O.N. of three iodine atoms in KI3 are 0, 0 and –1 respectively.
(b) The structure of H2S4O6 is shown below :

The O.N. of each of the S atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.
(c)

x

Fe3

−2

O4

Let O.N. of Fe = x, then 3x + 4 (–2) = 0 or x = +

8

3

(average)
By stoichiometry Fe3O4 is

2+

Fe

2−

O

.

+3

Fe

2−

O3

. Thus Fe has O.N. of +2 and +3.
(d) H−

2

‌

H |

C

| H

−

1

‌

H |

C

| H

−OH
In this molecule, C-2 is attached to three H-atoms (less electronegative than carbon) and one CH2OH group (more electronegativity than carbon).
Therefore O.N. of C-2 = 3 (+1) + x + 1(–1) = 0 or x = –2.
C-1 is however, attached to one OH (charge = –1) and one CH3 (charge = +1), and two H-atoms, O.N. of +1.
Therefore, O.N. of C-1 = 1(+1) + 2( +1) + x + 1 (–1) = 0 or x = –2
(e) H−

2

‌

H |

C

| H

−

1

‌

O ||

C

−OH
In this molecule, C-2 is attached to three H-atoms (less electronegative than carbon) and one –COOH group (more electronegativity than carbon).
Therefore, O.N. of C-2 = 3 (+1) + x +1(–1) = 0 or x = –2.
C-1 is, however, attached to one oxygen atom by a double bond, one OH (charge = –1) and one CH3 (charge = +1) group, therefore, O.N. of C-1 = 1 (+1) + x + 1 (–2) + 1 (–1) = 0 or x = +2.