Question

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a) KI₃ (b) H₂S₄O₆ (c) Fe₃O₄ (d) CH₃CH₂OH (e) CH₃COOH

Answer 1

(a) KI₃

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

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)

On taking the O.N. of O as –2, the O.N. of Fe is found to be. 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)

Hence, the O.N. of C is –2.

(e)

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.