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(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
Asked by Sumitha Last Modified
Answer (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 KI3to find the oxidation states.
In a KI3molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.
Hence, in a KI3molecule, the O.N. of the two I atoms forming the I2molecule 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) 
2 (x) + 6 (+1) + 1 (-2) = 0
or, 2x + 4 = 0
or, x = -2
Hence, the O.N. of C is –2.
(e) 
2 (x) + 4 (+1) + 2 (-2) = 0
or, 2x = 0
or, 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.
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