To an acidic solution of an anion, a few drops of `Kmno_(4)` solution are added. Which of the following, if present, will not decolourise the `KMnO_(4)` solution?

Options

A. `CO_(3)^(2-)`
B. `NO_(2)^(ө)`
C. `S^(2-)`
D. `Cl^(ө)`

Explanation

To solve the question, we need to determine which anion will not decolorize the KMnO4 solution when added to an acidic solution. The key to solving this problem lies in understanding the oxidation states of the elements in the given anions and their ability to undergo oxidation.

### Step-by-Step Solution:

1. **Identify the Anions**: The question asks us to consider several anions. Let's denote them as follows:
- A: CO3^2- (Carbonate)
- B: NO2^- (Nitrite)
- C: S2- (Sulfide)
- D: Cl^- (Chloride)

2. **Determine the Oxidation States**:
- For CO3^2- (Carbonate), the oxidation state of carbon is +4.
- For NO2^- (Nitrite), the oxidation state of nitrogen is +3.
- For S2- (Sulfide), the oxidation state of sulfur is -2.
- For Cl^- (Chloride), the oxidation state of chlorine is -1.

3. **Assess Oxidation Potential**:
- KMnO4 is a strong oxidizing agent in acidic medium. It can oxidize various anions.
- Anions that have lower oxidation states can be oxidized, while those at their maximum oxidation state cannot.

4. **Analyze Each Anion**:
- **CO3^2-**: Carbon is at +4, which is its maximum oxidation state. It cannot be oxidized further, so it will not reduce KMnO4 and thus will not decolorize it.
- **NO2^-**: Nitrogen can be oxidized from +3 to +5, so it can reduce KMnO4 and will decolorize it.
- **S2-**: Sulfur can be oxidized from -2 to higher oxidation states (like +6), so it can also reduce KMnO4 and will decolorize it.
- **Cl^-**: Chlorine can be oxidized from -1 to higher oxidation states, so it can reduce KMnO4 and will decolorize it.

5. **Conclusion**: Since CO3^2- cannot be oxidized further and will not reduce KMnO4, it will not decolorize the KMnO4 solution. Therefore, the correct answer is:
- **Option A: CO3^2-**