Electrolysis of X gives Y at anode. Vacuum distillation of Y gives `H_(2)O_(2)`. The number of peroxy (O-O_ bonds present in X and Y respectively, are

Options

A. 1,1
B. 1,2
C. 0,1
D. 0,0

Answer

C

Explanation

Electrolysis of `H_(2)SO_(4)` is as follows:
`underset((X))(2H_(2)SO_(4))+2H^(+)+2HSO_(4)^(-)`
`2HSO_(4)^(-)+underset((Y))(H_(2)S_(2)O_(8))+2e^(-)`(At anode)
`H_(2)S_(2)O_(8)+2H_(2)O to 2H_(2)SO_(4)+H_(2)O_(2)`
`therfore`Number of peroxy `(O-O)` in X and Y are 0 and 1 respectively.