**(a)** KI_{3}

In KI_{3}, the oxidation number (O.N.) of K is +1. 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 KI_{3} to find the oxidation states. In a KI_{3} molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI_{3} molecule, the O.N. of the two I atoms forming the I_{2} molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

**(b) **H2S4O6

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)** Fe_{3}O_{4 }

On taking the O.N. of O as –2, the O.N. of Fe is found to be . ^{ +2}(2/3) 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.

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 CH_{3}COOH.