Number of electrons transferred in each case when KMnO_4, acts as oxidizing agent and convert into MnO₂, Mn²⁺, Mn(OH)₃
and MnO₄²⁻ are respectively

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3, 5, 4 and 1

5, 4, 3 and 1

1, 3, 4 and 5

4, 3, 1 and 5

answer is A.

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Detailed Solution

In KMnO₄ the O.N. of Mn is +7

in MnO₄²⁻+6

in MnO₂+4

in Mn₂O₃+3

and in Mn²⁺ is +2.

K $\overset{+ 7}{Mn}$ O₄ $\to$ $\overset{+ 2}{Mn}$ O₂ the number of electrons transfered =3

K $\overset{+ 7}{Mn}$ O₄ $\to$ $\overset{+ 2}{Mn}$ the number of electrons transfered =5

K $\overset{+ 7}{Mn}$ O₄ $\to$ $\overset{+ 3}{Mn}$ (OH)₃the number of electrons transfered = 4

K $\overset{+ 7}{Mn}$ O₄ $\to$ $\overset{+ 6}{Mn}$ O₄²⁻ the number of electrons transfered =1

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