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Valency and Oxidation State

When KMnO4 ...

Question

When $K M n O_{4}$ acts as an oxidising agent and ultimately forms $[M n O_{4}]^{- 1}$ , $M n O_{2}$ , $M n_{2} O_{3}$ , and $M n_{2} O^{+}$ , then the number of electrons transferred in each case respectfully is

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Solution

Total $e^{-}$ transferred are as follows:-
(Refer to Image)
Number of $e^{-}$ electron transferred are:-
$⟶ [M n O_{4}]^{- 1} = 1$
$⟶ [M n O_{2}] = 3$
$⟶ [M n_{2} O_{3}] = 4$
$⟶ [M n_{2} O^{+}] = 5$

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