The number of electrons involved in the conversion of $M n O_{4}^{-} t o M n O_{2}$ is:

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Solution

The correct option is A 3
Given reaction:
$M n O_{4}^{-} \to M n O_{2}$
Calculate the oxidation state of reactant and product as
Let the oxidation number of Mn be 'x'.
Oxidation number of $O = - 2$ .
in $M n O_{4}^{-}$ : $x + 4 \times (- 2) = - 1$
$x = - 1 + 8$ or $x = + 7$
in $M n O_{2}$ : $x + 2 \times (- 2) = 0$
$x - 4 = 0$ or $x = + 4$
Therefore Change in oxidation number = 7-4=3
The number of electrons involved in the conversion of $M n O_{4}^{-} t o M n O_{2}$ is 3.

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