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When ${SO}_{2}$ is passed through acidified $K_{2} {Cr}_{2} O_{7}$ solution

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The solution turns blue

${SO}_{2}$ is reduced

The solution is decolourised

Green ${Cr}_{2} {({SO}_{4})}_{3}$ is formed

answer is D.

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Detailed Solution

$K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4} + 3 {SO}_{2} \to K_{2} {SO}_{4} + \underset{Green}{{Cr}_{2} {({SO}_{4})}_{3}} + 3 H_{2} O$

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