Which of the following ionic/molecular species does not disproportionate in water at room temperature?

Options

A. `NO_(2)`
B. `Cu^(+)`
C. `MnO_(4)^(2-)`
D. `Ca(Ocl)Cl`

Explanation

To determine which of the following ionic/molecular species does not disproportionate in water at room temperature, we need to analyze the reactions of each species with water and check for any changes in oxidation states.

### Step-by-Step Solution:

1. **Understanding Disproportionation**:
Disproportionation is a type of redox reaction where a single species is simultaneously oxidized and reduced, resulting in two different products.

2. **Analyzing NO2**:
- When NO2 reacts with water, it forms HNO3 (oxidation state +5) and HNO2 (oxidation state +3).
- Here, nitrogen in NO2 (oxidation state +4) is oxidized to +5 and reduced to +3.
- Since both oxidation and reduction occur, NO2 undergoes disproportionation.

3. **Analyzing Cu+**:
- Cu+ in aqueous solution reacts with water to form Cu2+ (oxidation state +2) and Cu (oxidation state 0).
- Cu+ is oxidized to Cu2+ and reduced to Cu.
- This indicates that Cu+ also undergoes disproportionation.

4. **Analyzing MnO4^2-**:
- MnO4^2- reacts with water to form MnO4- (oxidation state +7) and MnO2 (oxidation state +4).
- Manganese in MnO4^2- (oxidation state +6) is oxidized to +7 and reduced to +4.
- Hence, MnO4^2- also undergoes disproportionation.

5. **Analyzing Ca(OCl)2**:
- When Ca(OCl)2 reacts with water, it forms Ca^2+ (oxidation state +2), Cl^- (oxidation state -1), and OCl^- (oxidation state -1).
- Here, the oxidation states of all species remain the same: Ca remains +2, Cl remains -1, and OCl remains -1.
- Since there are no changes in oxidation states, Ca(OCl)2 does not undergo disproportionation.

6. **Conclusion**:
Based on the analysis, the species that does not disproportionate in water at room temperature is **Ca(OCl)2** (option 4).

### Final Answer:
**Ca(OCl)2 does not undergo disproportionation in water at room temperature.**

---