Question

What are the oxidation number of the underlined elements in
each of the following and how do you rationalise your results?
(a) Kl; (b) H.S.O. (c) Fe, O. (d) CH.CH,OH (e) CH,COOH​

Answer 1

Answer:

Explanation:

(a) KI3  In KI3, 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 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) Fe3O4  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 the 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.  

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