Which of the following combination of species undergo(es) comproportionation?

Options

A. `MnO_(4)^(-)(aq.)+Mn^(2+)(aq.)overset(ZnO//ZnSO_(4))to`
B. `S+conc.underset((excess))(H_(2)SO_(4)) overset(warm)to`
C. `PH_(3)+H_(3)PO_(4) to`
D. `NO(g)+NO_(2)(g) overset("cool")to`

Answer

To determine which combinations of species undergo comproportionation, we need to analyze the oxidation states of the elements involved in each reaction. Comproportionation occurs when two reactants containing the same element in different oxidation states react to form a single product with an intermediate oxidation state.

### Step-by-Step Solution:

1. **Understand Comproportionation**:
- Comproportionation is a redox reaction where two reactants with the same element in different oxidation states combine to form a single product with an intermediate oxidation state.

2. **Identify the Reactions**:
- We will analyze each given reaction to check if they fit the criteria for comproportionation.

3. **Analyze Each Reaction**:
- **Reaction 1**: \( \text{MnO}_4^- + \text{Mn}^{2+} \rightarrow \text{Mn}^{6+} \)
- Manganese in \( \text{MnO}_4^- \) has an oxidation state of +7, and in \( \text{Mn}^{2+} \) it is +2. The product \( \text{Mn}^{6+} \) indicates a comproportionation since +7 and +2 combine to form +6.

- **Reaction 2**: \( \text{S} + \text{H}_2\text{SO}_4 \rightarrow \text{SO}_2 \)
- Sulfur in elemental form is 0, in \( \text{H}_2\text{SO}_4 \) it is +6, and in \( \text{SO}_2 \) it is +4. This indicates a decrease from +6 to +4 and an increase from 0 to +4, thus it is also a comproportionation reaction.

- **Reaction 3**: \( \text{PH}_3 + \text{H}_3\text{PO}_4 \rightarrow \text{H}_4\text{PO}_3 \)
- Phosphorus in \( \text{PH}_3 \) is -3, in \( \text{H}_3\text{PO}_4 \) it is +5, and in \( \text{H}_4\text{PO}_3 \) it is +3. This shows an increase from -3 to +3 and a decrease from +5 to +3, confirming it as a comproportionation reaction.

- **Reaction 4**: \( \text{NO} + \text{NO}_2 \rightarrow \text{N}_2\text{O}_3 \)
- Nitrogen in \( \text{NO} \) is +2, in \( \text{NO}_2 \) it is +4, and in \( \text{N}_2\text{O}_3 \) it is +3. Here, nitrogen increases from +2 to +3 and decreases from +4 to +3, indicating it is also a comproportionation reaction.

4. **Conclusion**:
- All four reactions exhibit comproportionation as they involve the same element in different oxidation states reacting to form a product with an intermediate oxidation state.

### Final Answer:
All of the given combinations of species undergo comproportionation.