When $S O_{2}$ is passed through a solution of $K_{2} C r_{2} O_{7}$ turns green. The change in oxidation state of chromium and sulphur are ___________________ respectively.

+3 to +6 and +4 to +2

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+6 to +3 and +4 to +6

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+4 to +6 and +4 to -2

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no change in 0 states

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Solution

The correct option is B +6 to +3 and +4 to +6
$K_{2} C r_{2} O_{7} + S O_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O 4) 3 + H_{2} O$
Oxidation number of Cr in $K_{2} C r_{2} O_{7}$ is +6 and in $C r_{2} (S O_{4})_{3}$ is +3.
Oxidation number of S in $S O_{2}$ is +4 and in $(S O_{4})^{2 -}$ ion is +6.

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