When $S O_{2}$ is passed through acidified solution of potassium dichromate, then chromium sulphate is formed. The change in oxidation number of chromium is?

+ 4 t o + 2

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+ 5 t o + 3

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+ 6 t o + 3

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+ 7 t o + 2

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Solution

The correct option is D $+ 6 t o + 3$
The formation of $C r_{2} {(S O_{4})}_{3}$ is given by the equation:
$K_{2} C r_{2} O_{7} + 2 H_{2} S O_{4} + 3 S O_{2} ⟶ 2 C r_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + H_{2} O$

In $K_{2} C r_{2} O_{7}$ ,
Oxidation no. of $C r = + 6$
$2 (1) + 2 x + 2 (- 7) = 0 x = + 6.$

In $C r_{2} {(S O_{4})}_{3}$ ,
Oxidation number of $C r = + 3$
$2 x + 3 (- 2) = 0 x = + 3$
$∴$ change in oxidation state is $+ 6$ to $+ 3$

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