$K_{2} {C r}_{2} O_{7} + c o n c . H_{2} {S O}_{4} + H_{2} O_{2} + e t h e r ⟶$ blue perchromic anhydride (in ethereal layer). Blue colour is due to :

{C r O}_{3}

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H_{2} {C r O}_{4}

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H_{2} {C r}_{2} O_{3}

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{C r O}_{5}

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Solution

The correct option is D ${C r O}_{5}$
The blue colour is due to formation of chromic diperoxide $(C r O_{5})$
$2 H^{+} + C r_{2} O_{7}^{2 -} + 4 H_{2} O_{2} \to 2 C r O_{5} c h r o m i c d i p e r o x i d e (b l u e) + 5 H_{2} O$

Option D is correct.

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Similar questions

K_{2} C r_{2} O_{7} + c o n c . H_{2} S O_{4} + H_{2} O_{2} + e t h e r ⟶

H C H O + K_{2} {C r}_{2} O_{7} + H_{2} {S O}_{4} ⟶ ?

H_{2} O_{2}

K_{2} {C r}_{2} O_{7}

K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

K M n O_{4} + {N a}_{2} {S O}_{3} \to M n O_{2} + {N a}_{2} {S O}_{4} + K O H

C u + H {N O}_{3} \to C u {({N O}_{3})}_{2} + {N O}_{2} + H_{2} O

K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

[X] + H_{2} {S O}_{4} ⟶ [Y]

[Y] + K_{2} {C r}_{2} O_{7} + H_{2} {S O}_{4} ⟶

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