The species that undergoes diproportionation in an alkaline medium are:

Options

A. `Cl_(2)`
B. `MnO_(4)^(2-)`
C. `NO_(2)`
D. `ClO_(4)^(-)`

Explanation

To solve the question regarding which species undergoes disproportionation in an alkaline medium, we need to analyze each option based on the definition of disproportionation. Disproportionation is a reaction where a single species is both oxidized and reduced simultaneously.

### Step-by-Step Solution:

1. **Understand Disproportionation**:
- Disproportionation involves a single element undergoing both oxidation and reduction. This means that in the reaction, the element must change its oxidation state to a higher and a lower value.

2. **Analyze Option A: Cl2**:
- The oxidation state of chlorine (Cl) in Cl2 is 0.
- In an alkaline medium (reacting with NaOH), Cl2 can form NaCl (where Cl is -1) and NaOCl (where Cl is +1).
- Here, Cl is reduced from 0 to -1 and oxidized from 0 to +1.
- **Conclusion**: Cl2 undergoes disproportionation.

3. **Analyze Option B: MnO4^2-**:
- The oxidation state of manganese (Mn) in MnO4^2- is +6.
- In alkaline medium, it can form MnO4^- (where Mn is +7) and MnO2 (where Mn is +4).
- Here, Mn is oxidized from +6 to +7 and reduced from +6 to +4.
- **Conclusion**: MnO4^2- undergoes disproportionation.

4. **Analyze Option C: NO2**:
- The oxidation state of nitrogen (N) in NO2 is +4.
- In alkaline medium, it can form NaNO2 (where N is +3) and NaNO3 (where N is +5).
- Here, N is reduced from +4 to +3 and oxidized from +4 to +5.
- **Conclusion**: NO2 undergoes disproportionation.

5. **Analyze Option D: ClO4^-**:
- The oxidation state of chlorine (Cl) in ClO4^- is +7, which is the highest oxidation state for chlorine.
- Since it is already in the highest oxidation state, it cannot be oxidized further.
- **Conclusion**: ClO4^- does not undergo disproportionation.

### Final Answer:
The species that undergo disproportionation in an alkaline medium are **Cl2, MnO4^2-, and NO2**.