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

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

The solution turns blue

The solution is decolourised

${SO}_{2}$ is reduced

answer is D.

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Detailed Solution

Whenever ${SO}_{2}$ is acidified as it passes $K_{2} {Cr}_{2} O_{7}$ green solution of ${Cr}_{2} {({SO}_{4})}_{3}$ is created.

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

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