JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Redox Reactions and Electrochemistry

A species disproportionates when it sits at an intermediate oxidation state and can simultaneously be oxidised and reduced to give products at higher and lower oxidation states. In ${\text{H}}_2{\text{O}}_2$, oxygen is $-1$, between 0 (in ${\text{O}}_2$) and $-2$ (in ${\text{H}}_2\text{O}$) — it disproportionates. In ${\text{ClO}}_3^{-}$, chlorine is $+5$, an intermediate state that can disproportionate to higher (${\text{ClO}}_4^{-}$) and lower (${\text{Cl}}^{-}$/${\text{Cl}}_2$) states. ${\text{P}}_4$ has phosphorus at 0, between $-3$ (phosphine) and $+3$/$+5$ (oxoacids) — it disproportionates on treatment with alkali. ${\text{Cl}}_2$, with chlorine at 0, disproportionates to ${\text{Cl}}^{-}$ and ${\text{OCl}}^{-}$/${\text{ClO}}_3^{-}$ in base. ${\text{Cu}}^{+1}$, intermediate between Cu(0) and ${\text{Cu}}^{2+}$, disproportionates in aqueous solution: $2{\text{Cu}}^{+}\to \text{Cu}+{\text{Cu}}^{2+}$. ${\text{NO}}_2$ has nitrogen at $+4$, between $+5$ (${\text{NO}}_3^{-}$) and $+3$/$+2$ states, and disproportionates in water to nitrate and nitrite/NO. $\text{Ag}$ (stable at 0 under these conditions), ${\text{F}}_2$ (fluorine has no accessible positive oxidation state) and ${\text{K}}^{+}$ (already at its only stable oxidation state) cannot disproportionate. Six species — ${\text{H}}_2{\text{O}}_2$, ${\text{ClO}}_3^{-}$, ${\text{P}}_4$, ${\text{Cl}}_2$, ${\text{Cu}}^{+1}$ and ${\text{NO}}_2$ — undergo disproportionation.