The total change in the oxidation state of manganese involved in...

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

${KMnO}_{4} + KI + {H}^{+} ⟶ {Mn}^{+ 2} + {I}_{2} + {H}_{2}O$

In ${KMnO}_{4}$, the oxidation state of manganese is $+ 7$, which is the highest possible oxidation state for manganese in this compound.

In the final product, ${MnSO}_{4}$, 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 $O . S$. of $Mn$ is \left(5\right).