When $H_{2} O_{2}$ is shaken with an acidified solution of $K_{2} C r_{2} O_{7}$ in the presence of ether, the ethereal layer turns blue due to the formation of :

C r_{2} O_{3}

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C r O_{4}^{2 -}

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

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C r O_{5}

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Solution

The correct option is D $C r O_{5}$
$K_{2} {C r}_{2} O_{7} + 4 H_{2} O_{2} + H_{2} S O_{4} ⟶ 2 C r O_{5} + K_{2} S O_{4} + 5 H_{2} O$

Blue layer is formed due to formation of $C r O_{5}$ .

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H_{2} O_{2}

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H_{2} O_{2}

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K_{2} C r_{2} O_{7}

K_{2} C r_{2} O_{7}

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