(a) KI₃

Let assume oxidation number of l is x.

In KI₃, the oxidation number (O.N.) of K is +1.

1(+1) + 3(x) = 0

⇒ +1 +3x = 0

⇒ 3x = -1

⇒ x = -1/3

Hence, the average oxidation number of I is - 1/3

However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI₃ to find the oxidation states. In a KI₃ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI₃ molecule, the O.N. of the two I atoms forming the I₂ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

b) H₂S₄O₆

Let assume oxidation number of S is x.

The oxidation number (O.N.) of H is +1.

The oxidation number (O.N.) of O is -2.

2(+1) + 4(x) + 6(-2) = 0

⇒ 2 + 4x - 12 = 0

⇒ 4x -10 = 0

⇒ 4x  =  +10

⇒ x  = +10/4

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule. 

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0. 

(c) Fe₃O₄ 

Let assume oxidation number of Fe is x.

The oxidation number (O.N.) of O is -2.

3(x) + 4(-2) = 0

⇒ 3x  - 8 = 0

⇒ 3x  = 8

⇒ x  = 8/3

However, O.N. cannot be fractional. 

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3. 

(d) CH₃CH₂OH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

x + 3(+1) + x + 2(+1) + 1(-2) + 1(+1) = 0

⇒ x +3 + x +2 - 2  + 1 = 0

⇒ 2x  + 4 = 0

⇒ 2x  = -4

⇒ x  = -2

Hence, the oxidation number of C is -2.

(e) CH₃COOH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

x + 3(+1) + x + (-2) + (-2) + 1(+1) = 0

⇒ 2x + 3 - 2 -  2  +  1 = 0

⇒ 2x + 0 = 0

⇒ x = 0

However, 0 is average O.N. of C.

The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH₃COOH.