The reaction of acidified \(KMnO_4\) and \(NO_2^-\) gives:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(Mn^{2+}\) and \(NO_3^-\)
In acidic solution permanganate is a strong oxidising agent and is reduced to Mn2+; nitrite (N +3) is oxidised to nitrate (N +5). The two half-reactions are:
\( \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \) (× 2)
\( \ce{NO2^- + H2O -> NO3^- + 2H^+ + 2e^-} \) (× 5)
With 10 electrons on each side, adding them gives:
\( \ce{5NO2^- + 2MnO4^- + 6H^+ -> 2Mn^2+ + 5NO3^- + 3H2O} \)
Why the other options are incorrect
Option (1) \(Mn^{2+}\) and \(NO\) — Incorrect. NO (N +2) would mean that nitrite is reduced, but nitrite is the reducing agent here and must be oxidised.
Option (2) \(MnO_2\) and \(NO_3^-\) — Incorrect. \( \ce{MnO2} \) is the product of permanganate in neutral or faintly alkaline solution; in acid it goes to \( \ce{Mn^2+} \).
Option (4) \(Mn^{4+}\) and \(NO\) — Incorrect. \( \ce{Mn^4+} \) does not exist as a free ion in water; permanganate stops at the +4 state only as \( \ce{MnO2} \), in neutral or faintly alkaline solution. NO would again mean reduction of nitrite.
NCERT (2026-27 edition):
Potassium Permanganate – “… 1. In acid solutions: … (f) Nitrite is oxidised to nitrate:”
\( \ce{5NO2^- + 2MnO4^- + 6H^+ -> 2Mn^2+ + 5NO3^- + 3H2O} \)