`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 solve the question regarding the reaction of lead(IV) oxide (Pb3O4) with dilute nitric acid (HNO3) to form lead(II) nitrate (Pb(NO3)2) and lead(IV) oxide (PbO2), we will analyze the oxidation states of the elements involved and determine the type of reaction.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- Reactants: Pb3O4 and HNO3 (dilute)
- Products: Pb(NO3)2 and PbO2

2. **Assign Oxidation States:**
- For Pb3O4:
- Let the oxidation state of Pb be \( x \).
- The equation for oxidation state: \( 3x + 4(-2) = 0 \)
- This simplifies to \( 3x - 8 = 0 \) → \( 3x = 8 \) → \( x = \frac{8}{3} \) (approximately +2.67)
- For HNO3:
- H is +1, O is -2, and the overall charge of NO3 is -1.
- For Pb(NO3)2:
- The oxidation state of Pb is +2.
- For PbO2:
- The oxidation state of Pb is +4 (since O is -2, and there are two O atoms).

3. **Determine Changes in Oxidation States:**
- In Pb3O4, Pb is in the +8/3 state (approximately +2.67).
- In Pb(NO3)2, Pb is in the +2 state.
- In PbO2, Pb is in the +4 state.
- Therefore, one Pb atom is reduced from +8/3 to +2 (reduction), and another Pb atom is oxidized from +8/3 to +4 (oxidation).

4. **Identify the Type of Reaction:**
- Since one species (lead) is undergoing both oxidation and reduction, this type of reaction is classified as a **disproportionation reaction**.

5. **Conclusion:**
- The reaction of Pb3O4 with dilute HNO3 results in the formation of Pb(NO3)2 and PbO2, and it is a disproportionation reaction.

### Final Answer:
The type of reaction depicted here is a **disproportionation reaction**.

---