The oxide that gives `H_2 O_2` on treatment with dilute `H_2 SO_4` is

Options

A. `PbO_2`
B. `BaO_(2).8H_(2)O+O_2`
C. `MnO_2`
D. `TiO_2`

Answer

B

Explanation

Oxides such as `BaO_2, Na_2O_2` etc, which contain peroxide linkage (i.e., `O - O ` or `O_(2)^(2-)` ) on treatment with dilute `H_2SO_4` give `H_2O_2` but dioxides (O = M = O, where M is the metal atom) such as `PbO_2, MnO_2, TiO_2` do not give `H_2O_2` on treatment with dilute `H_2SO_4` .
`underset("peroxide")underset("Hydrated barium")(BaO_2 .8H_2O(s))+H_2SO_4(aq)rarrBaSO_4(s)+underset("peroxide")underset("Hydrogen")(H_2O_2(aq))+8H_2O(l)`