H2O2 oxidises acidified K2Cr2O7 solution to blue CrO5 in ether.
A
True
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B
False
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Solution
The correct option is A True
K2Cr2O7+H2SO4+4H2O2→2CrO5+K2SO4+5H2O.
CrO5 has butterfly structure and has two peroxy linkages. Hence Cr is in +6 oxidation state. In enter, it forms hexacoordinated complex with ether and get stabilized.