1. The solution turns blue. 2. The solution is decolourized. 3. SO₂ is reduced. 4. Green Cr₂(SO₄)₃ is formed.

Answer: 4. Green Cr₂(SO₄)₃ is formed. Key idea: SO₂ is a reducing agent and acidified dichromate is an oxidising agent. mathrmCr₂O₇²⁻+3SO₂+2H^+→ 2Cr³⁺+3SO₄²⁻+H₂O The orange mathrmCr₂O₇²⁻ turns green because green Cr₂(SO₄)₃ forms. The colour changes from orange to green, not blue; the solution is not decolourised; and SO₂ is oxidised, not reduced. So option 4.

Topic: K2Cr2O7 and KMnO4· d- and f-Block Elements

Which one of the following statements is correct when SO_2 is passed through acidified K_2Cr_2O_7 solution ?

1. The solution turns blue.

2. The solution is decolourized.

3. $SO_2$ is reduced.

4. Green $Cr_2(SO_4)_3$ is formed.

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

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Answer: 4. Green $Cr_2(SO_4)_3$ is formed.

Key idea: $\mathrm{SO_2}$ is a reducing agent and acidified dichromate is an oxidising agent.

$$\mathrm{Cr_2O_7^{2-}+3SO_2+2H^+\to 2Cr^{3+}+3SO_4^{2-}+H_2O}$$

The orange $\mathrm{Cr_2O_7^{2-}}$ turns green because green $\mathrm{Cr_2(SO_4)_3}$ forms. The colour changes from orange to green, not blue; the solution is not decolourised; and $\mathrm{SO_2}$ is oxidised, not reduced. So option 4.

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

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Which one of the following statements is correct when SO_2 is passed through acidified K_2Cr_2O_7 solution ?

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