When KMnO₄ acts as an oxidizing agent and ultimately forms MnO₄²⁻, MnO₂, Mn₂O₃ and Mn²⁺. The number of electrons transferred in each case respectively is

$$\begin{gathered} \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+{\mathrm{e}}^{-}\to \stackrel{+6}{\mathrm{Mn}}{\mathrm{O}}_4^{-2}; \mathrm{no}. \mathrm{of} {\mathrm{e}}^{-}\mathrm{s} \mathrm{transfered}=1 \\ \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+3{\mathrm{e}}^{-}\to \stackrel{+4}{\mathrm{Mn}}{\mathrm{O}}_2 ; \mathrm{no}. \mathrm{of} {\mathrm{e}}^{-}\mathrm{s} \mathrm{transfered}=3 \\ \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+4{\mathrm{e}}^{-}\to {\stackrel{+3}{\mathrm{Mn}}}_2{\mathrm{O}}_3 ; \mathrm{no}. \mathrm{of} {\mathrm{e}}^{-}\mathrm{s} \mathrm{transfered}=4 \\ \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{+2} ; \mathrm{no}. \mathrm{of} {\mathrm{e}}^{-}\mathrm{s} \mathrm{transfered}=5 \end{gathered}$$