An inorganic substance liberates `O_(2)` when heated, turns an acidified solution of Kl brown and reduces acidified `KMnO_(4)`. The substance is
(a)`SO_(2)`
(b)`H_(2)O_(2)`
(c)`KNO_(3)`
(d)`Pb(NO_(3))_(2)`

Options

A. `SO_(2)`
B. `H_(2)O_(2)`
C. `KNO_(3)`
D. `Pb(NO_(3))_(2)`

Answer

To solve the problem, we need to analyze the given conditions and determine which of the provided substances meets all of them.

### Step-by-Step Solution:

1. **Identify the Conditions:**
- The substance liberates \( O_2 \) when heated.
- It turns an acidified solution of potassium iodide (KI) brown.
- It reduces acidified potassium permanganate (KMnO4).

2. **Evaluate Each Option:**
- **(a) \( SO_2 \)**:
- When heated, \( SO_2 \) does not liberate \( O_2 \).
- Therefore, this option is not correct.

- **(b) \( H_2O_2 \)** (Hydrogen Peroxide):
- When heated, \( H_2O_2 \) decomposes to release \( O_2 \):
\[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]
- It reacts with acidified KI, producing iodine, which turns the solution brown:
\[ H_2O_2 + 2 KI + 2 H^+ \rightarrow I_2 + 2 K^+ + 2 H_2O \]
- It reduces acidified KMnO4, changing the oxidation state of manganese from +7 to +2:
\[ 2 KMnO_4 + 5 H_2O_2 + 6 H^+ \rightarrow 2 MnSO_4 + 5 O_2 + 8 H_2O \]
- All conditions are satisfied for \( H_2O_2 \).

- **(c) \( KNO_3 \)**:
- \( KNO_3 \) does not liberate \( O_2 \) upon heating under normal conditions.
- Therefore, this option is not correct.

- **(d) \( Pb(NO_3)_2 \)**:
- \( Pb(NO_3)_2 \) does not liberate \( O_2 \) when heated.
- Therefore, this option is not correct.

3. **Conclusion:**
- The only substance that satisfies all the conditions is **(b) \( H_2O_2 \)**.

### Final Answer:
The substance is \( H_2O_2 \) (Hydrogen Peroxide).

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