JEE Main 2023 — D And F Block Elements Question with Solution

The reaction of ${\mathrm{KMnO}}_4$ and potassium iodide ($\mathrm{KI}$) in acidic medium can be represented as follows:

$$\mathrm{Mn}{\mathrm{O}}_4^{-}+{\mathrm{I}}^{-}+{\mathrm{H}}^{+}\to \mathrm{M}{\mathrm{n}}^{2+}+{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}$$

In this reaction, ${\mathrm{KMnO}}_4$ acts as an oxidizing agent, while $\mathrm{KI}$ acts as a reducing agent. The oxidation state of manganese changes from +7 in $\mathrm{KMn}{\mathrm{O}}_4$ to +2 in $\mathrm{M}{\mathrm{n}}^{2+}$, which is a reduction of 5 units.

Therefore, the total change in the oxidation state of manganese involved in the reaction is $\boxed{5}$.