The total change in the oxidation state of manganese involved in the reaction of KMnO 4 and potassium iodide…

In the reaction of ${\mathrm{KMnO}}_{4 }$and potassium iodide ($\mathrm{KI}$) in an acidic medium, the ${\mathrm{KMnO}}_{4 }$ acts as an oxidizing agent, and the iodide ions (${\mathrm{I}}^{-}$) in $\mathrm{KI}$ act as reducing agents.   

${\mathrm{KMnO}}_4+\mathrm{KI}+{\mathrm{H}}^{+}⟶{\mathrm{Mn}}^{+2}+{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}$

In ${\mathrm{KMnO}}_4$, the oxidation state of manganese is $+7$, which is the highest possible oxidation state for manganese in this compound.

In the final product, ${\mathrm{MnSO}}_4$, the oxidation state of manganese is $+2$.

Therefore, the total change in the oxidation state of manganese is:

$+7$ (initial)$- +2$ (final) $= 5$

Hence the change in $\mathrm{O}.\mathrm{S}$. of $\mathrm{Mn}$ is $\left(5\right)$.