Q13 / 75 · JEE Main 2026 (21 January Shift 1) · Chemistry
$\mathrm{MnO}_{4}{ }^{2-}$, in acidic medium, disproportionates to :
- Correct answerA$\mathrm{MnO}_{4}{ }^{-}$and $\mathrm{MnO}_{2}$
- B$\mathrm{Mn}_{2} \mathrm{O}_{7}$ and $\mathrm{MnO}_{2}$
- C$\mathrm{Mn}_{2} \mathrm{O}_{7}$ and MnO
- D$\mathrm{MnO}_{4}{ }^{-}$ and MnO
Solution
$MnO_4^{2-}$ (manganate) has Mn in +6 oxidation state.
In acidic medium, it undergoes disproportionation where Mn is simultaneously oxidized and reduced:
Oxidation: $+6 \rightarrow +7$ (forms $MnO_4^-$, permanganate)
Reduction: $+6 \rightarrow +4$ (forms $MnO_2$)
Reaction: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$
In acidic medium, it undergoes disproportionation where Mn is simultaneously oxidized and reduced:
Oxidation: $+6 \rightarrow +7$ (forms $MnO_4^-$, permanganate)
Reduction: $+6 \rightarrow +4$ (forms $MnO_2$)
Reaction: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$
Hence, the correct answer is option: 'A'.