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 \(-\frac{1}{3}\). 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 coordinated covalent bond with an iodine molecule.
K+[1-1←I]
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.

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(b) H₂S₄O₆
Now,2(+1) + 4(x) + 6(-2) = 0
2 + 4x -12 = 0
4x = 10
x =+2\(\frac{1}{2}\)
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.

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(c) Fe₃O₄
On taking the O.N. of O as -2, the O.N. of Fe is found to be +2\(\frac{2}{3}\) . 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.

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(d)CH₃CH₂OH
2 (x) + 6 (+1) + 1 (-2) = 0 
or, 2x + 4 = 0 
or, x = -2
Hence, the O.N. of C is -2.

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(e) CH₃COOH
2 (x) + 4 (+1) + 2 (-2) = 0 
or, 2x = 0 
or, x = 0 
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.