JEE Main · 2025 · 22 Jan 2025 · Shift 2 · p-Block Elements

Disproportionation requires the element to sit at an intermediate oxidation state, so that part of it can be oxidized while the rest is reduced.

Assigning oxidation states of chlorine (O taken as $-2$):
$\mathrm{Cl}{\mathrm{O}}_2^{-}$: $x-4=-1\Rightarrow x=+3$
$\mathrm{Cl}{\mathrm{O}}_4^{-}$: $x-8=-1\Rightarrow x=+7$
$\mathrm{Cl}{\mathrm{O}}^{-}$: $x-2=-1\Rightarrow x=+1$
$\mathrm{Cl}{\mathrm{O}}_3^{-}$: $x-6=-1\Rightarrow x=+5$

Chlorine's accessible oxidation states run from $-1$ to $+7$, so $+7$ is its ceiling. In $\mathrm{Cl}{\mathrm{O}}_4^{-}$, chlorine already sits at this maximum $+7$ state and has nowhere higher to go, so it cannot be oxidized further — disproportionation is impossible for perchlorate, while $\mathrm{Cl}{\mathrm{O}}_2^{-}$, $\mathrm{Cl}{\mathrm{O}}^{-}$, and $\mathrm{Cl}{\mathrm{O}}_3^{-}$ all sit at intermediate states and readily disproportionate.

$\mathrm{Cl}{\mathrm{O}}_4^{-}$ is the species that does not undergo disproportionation.