$K M n O_{4}$ acts as an oxidising agent and ultimately forms $[M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3}]$ and $M n^{+ 2} .$ Then the number of electrons transferred in each case respectively is:

4, 3, 1, 5

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1, 5, 3, 7

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1, 3, 4, 5

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3, 5, 7, 1

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Solution

The correct option is C 1, 3, 4, 5
The oxidation number of $M n$ in $K M n O_{4}, M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3}$ and $M n^{2 +}$ is 7, 6, 4, 3 and 2 respectively.

The number of electrons transferred corresponds to the change in the oxidation number.
So, when $K M n O_{4}$ acts as an oxidising agent and ultimately forms $M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3}$ and $M n^{+ 2},$ then the number of electrons transferred in each case are:

7-6=1,
7-4= 3,
7-3= 4,
7-2= 5 respectively.

Hence, (c) is the correct answer.

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