Question

When $KMn{O}_4$ acts as an oxidizing agent and ultimately forms $Mn{O}_4^{2-},Mn{O}_2,M{n}_2{O}_3$ and $M{n}^{2+}$ then the number of electrons transferred in each case respectively are:

• A. 1, 3, 4, 5
• B. 1, 5, 3, 7
• C. 4, 3, 2, 1
• D. 3, 2, 1, 4

Answer 1

The correct option is A 1, 3, 4, 5

The oxidation state of $Mn$ in $Mn{O}_4^{-}$ is +7.
The oxidation state of $Mn$ in $Mn{O}_4^{2-}$ is +6.
The oxidation state of $Mn$ in $Mn{O}_2$ is +4.
The oxidation state of $Mn$ in $M{n}_2{O}_3$ is +3.
The oxidation state of $Mn$ in $M{n}^{2+}$ is +2.

The number of transfer of electron from $Mn{O}_4^{-}$ to $Mn{O}_4^{2-}$ is 1.
The number of transfer of electron from $Mn{O}_4^{-}$ to $Mn{O}_2$ is 3.
The number of transfer of electron from $Mn{O}_4^{-}$ to $M{n}_2{O}_3$ is 4.
The number of transfer of electron from $Mn{O}_4^{-}$ to $M{n}^{2+}$is 5.

Hence, option (a) is correct.