Calculate the n-factor for ${\text{KMnO}}_4$ when it acts as an oxidizing agent in an acidic medium.

Solution

1. In ${\text{KMnO}}_4$, the oxidation state of $\text{Mn}$ is $+7$.
2. In acidic medium, ${\text{MnO}}_4^{-}$ is reduced to ${\text{Mn}}^{2+}$.
3. The change in oxidation state is $|7-2|=5$.
4. The n-factor is equal to the change in oxidation state per mole.
5. Hence the answer is (C).