Question

When $KMn{O}_4$ acts as an oxidising agent and ultimately forms $[Mn{O}_4{]}^{-1}$, $Mn{O}_2$, $M{n}_2{O}_3$, and $M{n}_2{O}^{+}$, then the number of electrons transferred in each case respectfully is

Answer 1

Total $e^{-}$ transferred are as follows:-

(Refer to Image)

Number of $e^{-}$ electron transferred are:-

$⟶[Mn{O}_4{]}^{-1}=1$

$⟶\left[Mn{O}_2\right]=3$

$⟶\left[M{n}_2{O}_3\right]=4$

$⟶\left[M{n}_2{O}^{+}\right]=5$