JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Redox Reactions and Electrochemistry

Disproportionation needs the element in an intermediate oxidation state. In set (A): Mn(+6) in ${\mathrm{MnO}}_4^{2-}$ ($3{\mathrm{MnO}}_4^{2-}+4H^{+}\to 2{\mathrm{MnO}}_4^{-}+{\mathrm{MnO}}_2+2H_{2}O$); Cl(+3) in ${\mathrm{ClO}}_2^{-}$; Cl(0) in ${\mathrm{Cl}}_2$ ($\to C{l}^{-}+Cl{O}^{-}$); and Mn(+3) in ${\mathrm{Mn}}^{3+}$ ($\to {\mathrm{Mn}}^{2+}+{\mathrm{MnO}}_2$) — all intermediate, so all disproportionate. Every other set contains a species already in its highest oxidation state (${\mathrm{MnO}}_4^{-},{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-},{\mathrm{ClO}}_4^{-}$) or ${\mathrm{F}}_2$ (fluorine has no positive state), which cannot disproportionate.