The oxide that give `H_(2)O_(2)` on treatment with dilute `H_(2)SO_(4)` is

Options

A. `PbO_(2)`
B. `BaO_(2).8H_(2)O`
C. `MnO_(2)`
D. `TiO_(2)`

Explanation

To determine which oxide produces hydrogen peroxide (H₂O₂) when treated with dilute sulfuric acid (H₂SO₄), we will analyze the options provided:

1. **Identify the options**:
- A. PbO₂ (Lead(IV) oxide)
- B. BaO₂ (Barium peroxide)
- C. MnO₂ (Manganese(IV) oxide)
- D. TiO₂ (Titanium dioxide)

2. **Understanding the reaction**:
We need to find out which of these oxides reacts with dilute H₂SO₄ to yield H₂O₂.

3. **Analyzing each option**:
- **PbO₂**: Lead(IV) oxide does not produce H₂O₂ when treated with dilute H₂SO₄.
- **BaO₂**: Barium peroxide reacts with dilute H₂SO₄ to produce barium sulfate (BaSO₄), water (H₂O), and hydrogen peroxide (H₂O₂). The reaction can be represented as:
\[ BaO_2 \cdot 8H_2O + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 + 8H_2O \]
- **MnO₂**: Manganese(IV) oxide does not yield H₂O₂ with dilute H₂SO₄.
- **TiO₂**: Titanium dioxide does not produce H₂O₂ when treated with dilute H₂SO₄.

4. **Conclusion**:
The only oxide that gives H₂O₂ upon treatment with dilute H₂SO₄ is **Barium peroxide (BaO₂)**.

Thus, the answer is **B. BaO₂**.