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 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 D $1, 3, 4, 5$
The oxidation numbers of manganese in $K M n O_{4}, M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3}$ and $M n^{2 +}$ are $+ 7, + 6, + 4, + 3$ and $+ 2$ respectively.
Hence, the number of electrons transferred during conversion of $K M n O_{4}$ to $M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3}$ and $M n^{2 +}$ are $+ 1, + 3, + 4$ and + $5$ respectively.

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Similar questions

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

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[M n O_{4}^{2 -}, M n O_{2}, M n_{2} O_{3}]

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When KMnO4 acts as an oxidising agent and ultimately forms MnO−24, MnO2, Mn2O3 and Mn+2, then the number of electrons transferred in each case respectively, is :

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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 respectively, is :