The blue colour produced on adding $H_{2} O_{2}$ to acidified $K_{2} C r_{2} O_{7}$ is due to the formation of:

C r O_{5}

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C r_{2} O_{3}

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C r O_{4}^{2 -}

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C r O_{3}

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Solution

The correct option is A $C r O_{5}$
$K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 4 H_{2} O_{2} \to 2 C r O_{5} + K_{2} S O_{4} + 5 H_{2} O$

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H_{2} O_{2}

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