`Pb_(3)O_(4)+HNO_(3)(dil.) overset(R.T.) to Pb(NO_(3))_(2)+PbO_(2)darr`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To determine the type of reaction taking place in the equation \( \text{Pb}_3\text{O}_4 + \text{HNO}_3 \, (\text{dil.}) \rightarrow \text{Pb(NO}_3)_2 + \text{PbO}_2 \), we will analyze the oxidation states of lead (Pb) in both the reactants and products.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- Reactants: \( \text{Pb}_3\text{O}_4 \) and \( \text{HNO}_3 \) (dilute)
- Products: \( \text{Pb(NO}_3)_2 \) and \( \text{PbO}_2 \)

2. **Calculate the Oxidation State of Lead in \( \text{Pb}_3\text{O}_4 \):**
- Let the oxidation state of Pb be \( x \).
- The formula for calculating the oxidation state is:
\[ 3x + 4(-2) = 0 \]
- This simplifies to:
\[ 3x - 8 = 0 \implies 3x = 8 \implies x = \frac{8}{3} \approx 2.67 \]
- Therefore, the oxidation state of Pb in \( \text{Pb}_3\text{O}_4 \) is approximately +2.67.

3. **Determine the Oxidation States of Lead in the Products:**
- In \( \text{Pb(NO}_3)_2 \), the oxidation state of Pb is +2.
- In \( \text{PbO}_2 \), the oxidation state of Pb is +4.

4. **Analyze the Changes in Oxidation State:**
- In the reaction, Pb in \( \text{Pb}_3\text{O}_4 \) is being oxidized to +4 in \( \text{PbO}_2 \) and reduced to +2 in \( \text{Pb(NO}_3)_2 \).
- This indicates that the same element (Pb) is undergoing both oxidation and reduction.

5. **Conclusion: Identify the Type of Reaction:**
- Since the same element (Pb) is both oxidized and reduced in the reaction, this is classified as a **disproportionation reaction**.

### Final Answer:
The reaction \( \text{Pb}_3\text{O}_4 + \text{HNO}_3 \, (\text{dil.}) \rightarrow \text{Pb(NO}_3)_2 + \text{PbO}_2 \) is a **disproportionation reaction**.

---