Which of the following disproportionate (s) in heating with sodium hydroxide?

Options

A. `P_4`
B. `S_8`
C. `Cl_2`
D. `B`

Explanation

To determine which of the given substances undergo disproportionation when heated with sodium hydroxide (NaOH), we need to analyze the oxidation states of the elements involved in each reaction. Disproportionation reactions involve a single species being both oxidized and reduced simultaneously.

### Step-by-Step Solution:

1. **Identify the Compounds**: We are given four compounds: P4, S8, Cl2, and B.

2. **Reaction of P4 with NaOH**:
- **Reaction**: P4 + NaOH → PH3 + NaH2PO2
- **Oxidation States**:
- P in P4: 0
- P in PH3: -3 (reduction)
- P in NaH2PO2: +1 (oxidation)
- **Conclusion**: Since P4 shows both oxidation and reduction, it undergoes disproportionation.

3. **Reaction of S8 with NaOH**:
- **Reaction**: S8 + NaOH → Na2S + Na2SO3 + H2O
- **Oxidation States**:
- S in S8: 0
- S in Na2S: -2 (reduction)
- S in Na2SO3: +4 (oxidation)
- **Conclusion**: S8 also shows both oxidation and reduction, thus it undergoes disproportionation.

4. **Reaction of Cl2 with NaOH**:
- **Reaction**: Cl2 + NaOH → NaCl + HClO
- **Oxidation States**:
- Cl in Cl2: 0
- Cl in NaCl: -1 (reduction)
- Cl in HClO: +1 (oxidation)
- **Conclusion**: Cl2 undergoes both oxidation and reduction, indicating it undergoes disproportionation.

5. **Reaction of B with NaOH**:
- **Reaction**: B + NaOH → Na3BO3 + H2
- **Oxidation States**:
- B: 0
- B in Na3BO3: +3 (oxidation)
- **Conclusion**: There is no reduction occurring here, so B does not undergo disproportionation.

### Final Answer:
The substances that undergo disproportionation when heated with sodium hydroxide are **P4, S8, and Cl2**.