The oxide that gives H_{2}O_{2}  on treatment with dilute H_{2}SO_4 is —
(i) PbO_{2}
(ii) BaO_{2}. 8H_{2}O + O_{2}
(iii) MnO_{2}
(iv) TiO_{2}

Answers (1)

The answer is the option (ii) BaO_{2}. 8H_{2}O + O_{2}

This oxide has peroxide linkage (O^{2-}_{2}) when reacted with dilute H_{2}SO_4, it produces H_2O, however, dioxides do not produce the same products and the metal atom doesn’t give out water on treatment with dilute H_{2}SO_4.

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