JEE Main 2025 — Redox Reactions Question with Solution

$$\text{Oxidation States:}$$

In $$\mathrm{Cl}{\mathrm{O}}_4^{-}$$ (perchlorate), chlorine is in the +7 oxidation state.

In $$\mathrm{Cl}{\mathrm{O}}_3^{-}$$ (chlorate), chlorine is in the +5 oxidation state.

In $$\mathrm{Cl}{\mathrm{O}}_2^{-}$$ (chlorite), chlorine is in the +3 oxidation state.

In $$\mathrm{Cl}{\mathrm{O}}^{-}$$ (hypochlorite), chlorine is in the +1 oxidation state.

$$\text{Disproportionation Reaction:}$$

A disproportionation reaction is one in which a species simultaneously undergoes oxidation and reduction. For this to occur, the element must be in an intermediate oxidation state such that it can be oxidized to a higher state and reduced to a lower one.

In $$\mathrm{Cl}{\mathrm{O}}^{-}$$ and $$\mathrm{Cl}{\mathrm{O}}_2^{-}$$, chlorine is in lower oxidation states (+1 and +3, respectively), making them susceptible to disproportionation. For example, hypochlorite can disproportionate in basic solution as follows:

$$3\mathrm{Cl}{\mathrm{O}}^{-}\to 2\mathrm{C}{\mathrm{l}}^{-}+\mathrm{Cl}{\mathrm{O}}_3^{-}$$

In $$\mathrm{Cl}{\mathrm{O}}_3^{-}$$, chlorine is in an intermediate oxidation state (+5) that, under certain conditions, can undergo disproportionation.

However, in $$\mathrm{Cl}{\mathrm{O}}_4^{-}$$, chlorine is in its highest possible oxidation state (+7) and cannot be oxidized further. Since disproportionation requires one part of the species to be oxidized and the other reduced, $$\mathrm{Cl}{\mathrm{O}}_4^{-}$$ is thermodynamically stable and does not undergo disproportionation.

$$\boxed{\mathrm{Cl}{\mathrm{O}}_4^{-}\text{ (perchlorate ion) does not undergo disproportionation.}}$$