A solution of $K_{2} C r_{2} O_{7}$ acidified with $H_{2} S O_{4}$ when treated with a substance 'X', gives a blue coloured solution of 'Y'. The substance 'X' and the oxidation state of central atom of compound 'Y' are respectively:

C r O_{5}, + 6

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H_{2} C r_{2} O_{7}, + 4

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H_{2} O_{2}, + 6

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H_{2} O_{2}, + 4

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Solution

The correct option is C $H_{2} O_{2}, + 6$
(I) $K_{2} C r_{2} O_{7} + H_{2} S O_{4} ⟶ K_{2} S O_{4} + H_{2} C r_{2} O_{7}$
(II) $[H_{2} O_{2} ⟶ H_{2} O + [O]] \times 4$
(III) $H_{2} C r_{2} O_{7} + 4 [O] ⟶ 2 C r O_{5} + H_{2} O$
Adding (I), (II) and (III) we get,
$K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 4 H_{2} O_{2} ⟶ 2 C r O_{5} + K_{2} S O_{4} + 5 H_{2} O$
Here 'X' is $H_{2} O_{2}$ and blue coloured solution is due to formation of $C r O_{5}$ which has chromium in +6 oxidation state.

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