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The oxide that gives `H_(2)O_(2)` on treatment with dilute `H_(2)S0_(4)` isA. `PbO_(2)`B. `BaO_(2).8H_(2)O + O_(2)`C. `MnO_(2)`D. `TiO_(2)` |
Answer» Correct Answer - B Oxides such as `BaO_(2), Na_(2)O_(2)` etc, which contain peroxides (i.e., `-O-O or O_(2)^(2-)`) on treatment with dilute `H_(2)SO_(4)` give `H_(2)O_(2)` but dioxides (O= M= O, where M is the metal atom) such as `PbO_(2), MnO_(2), TiO_(2)` do not gives `H_(2)O_(2)` on treatment with dilute `H_(2)SO_(4)` `underset("peroxide")underset("Hydrated barium")(BaO_(2). 8H_(2)O(S) + H_(2)SO_(4)(aq)) to BaSO_(4)(S)+ underset("peroxide")underset("Hydrogen")(H_(2)O_(2)(aq)+ 8H_(2)O(l))` |
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