Electrolysis of X gives Y at anode. Vacuum distillation of Y gives ${\mathrm{H}}_2{\mathrm{O}}_2$. The number of peroxy (O –O) bonds present in X and Y respectively are

Hydrogen peroxide is manufactured by electrolysis of 50 % sulphuric acid followed by vacuum distillation. The distillate is 30 % hydrogen [peroxide. The first product of electrolysis is peroxy disulphuric acid which is obtained by the electrolytic oxidation of sulphuric acid.

$$\begin{gathered} 2H_{2}S{O}_4 \stackrel{ }{\to } 2H^{+}+ 2HS{O_4}^{-} \\ 2HS{O_4}^{-} \stackrel{ }{\to } H_2{S}_2{O}_8+2e^{-} ( at anode) \end{gathered}$$

Peroxy- disulphuric acid H_2​S₂​O₈​ reacts with water during distillation to form hydrogen peroxide.

$$\begin{gathered} H2S2O8\left(aq\right)+2H2O_{\left(l\right)} \stackrel{ }{\to } 2H_{2}S{O}_4{}_{\left(aq\right)}+H_2{O}_2{}_{\left(aq\right)} \\ 2H^{+}+2e^{-}\to H_2{ }_{\left(g\right)}(at cathode) \end{gathered}$$
 

 Hence X is H₂​SO_4​ and Y is  H_2​S₂​O₈​ ​.

In H₂​SO_4​ no peroxy bond i present . 

In H_2​S₂​O₈ one peroxy bond is present.