Question

What are the oxidation number 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.