Ncert Exercises & Solutions

The oxide that gives $H_{2} O_{2}$ on treatment with dilute $H_{2} {SO}_{4}$ is

1. ${PbO}_{2}$

2. ${BaO}_{2} . 8 H_{2} O + O_{2}$

3. ${MnO}_{2}$

4. ${TiO}_{2}$

Hint: Those compound which contains peroxy linkage give $H_{2} O_{2}$ when reacts with $H_{2} {SO}_{4}$ .

Step 1:

Oxides such as BaO₂, Na₂O₂ etc; which contain peroxide linkage (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 PbO₂, MnO₂, TiO₂ do not give H₂O₂ on treatment with dilute H₂SO₄.

Step 2:

The reaction is as follows:

\underset{Hydrated barium}{{BaO}_{2} . 8 H_{2} O + O_{2}}

peroxide + H_{2} {SO}_{4} (aq) \to {BaSO}_{4} (g) + \underset{Hydrogen}{H_{2} O_{2} (aq)}

peroxide + 8 H_{2} O (l)

Hence, option 2 is correct.