Which one of the following statements is correct when $S O_{2}$ is passed through acidified $K_{2} C r_{2} O_{7}$ solution?

The solution turns blue

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The solution is decolourized

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S O_{2}

is reduced

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Green

C r_{2} (S O_{4})_{3}

is formed

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Solution

The correct option is D Green $C r_{2} (S O_{4})_{3}$ is formed
When $S O_{2}$ is passed through acidified $K_{2} C r_{2} O_{7}$ solution, green $C r_{2} (S O_{4})_{3}$ is formed.
$K_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$
Sulphur dioxide is oxidized to sulphuric acid. Potassium dichromate is reduced to chromic sulphate. Orange colour of solution turns green.

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