1. SO₂ 2. NO₂ 3. P₂O₅ 4. CO₂

Answer: 1. SO₂ Key idea: KMnO₄ is decolourised by a reducing gas. SO₂ is a reducing agent: 5SO₂ + 2MnO₄^- + 2H₂O → 5SO₄²⁻ + 2Mn²⁺ + 4H^+ . Purple MnO₄^- becomes colourless mathrmMn²⁺ . NO₂ (N is +4) can be oxidised too, but it is mainly an oxidiser and is not the expected answer. P₂O₅ and CO₂ are in their highest or stable oxidation states and cannot reduce KMnO₄ .

Topic: General trends, Group 15 and 16· p-Block Elements

Name the gas that can readily decolourise acidified KMnO_4 solution :

1. \(SO_2\)

2. \(NO_2\)

3. \(P_2O_5\)

4. \(CO_2\)

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

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Answer: 1. \(SO_2\)

Key idea: \(\mathrm{KMnO_4}\) is decolourised by a reducing gas.

\(\mathrm{SO_2}\) is a reducing agent: \(5SO_2 + 2MnO_4^- + 2H_2O \rightarrow 5SO_4^{2-} + 2Mn^{2+} + 4H^+\). Purple \(\mathrm{MnO_4^-}\) becomes colourless \(\mathrm{Mn^{2+}}\). \(\mathrm{NO_2}\) (N is +4) can be oxidised too, but it is mainly an oxidiser and is not the expected answer. \(\mathrm{P_2O_5}\) and \(\mathrm{CO_2}\) are in their highest or stable oxidation states and cannot reduce \(\mathrm{KMnO_4}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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