Pink colour of acidified `KMnO_(4)` is decolourised but there is no evolution of any gas.This may happen with the compound containing the following acid radical.

Options

A. `SO_(3)^(2-)`
B. `NO_(2)^(-)`
C. `S^(2-)`
D. All of these

Explanation

To solve the question regarding the decolorization of pink acidified KMnO4 without the evolution of any gas, we need to analyze the given options and their reactions with KMnO4 in an acidic medium.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- KMnO4 (potassium permanganate) is a strong oxidizing agent. In an acidic medium, it can react with various anions (acid radicals) and get reduced, leading to the decolorization of its pink color.

2. **Analyzing the Options**:
- We have the following acid radicals to consider:
- **Option A: SO3^2- (Sulfite)**
- **Option B: NO2^- (Nitrite)**
- **Option C: S^2- (Sulfide)**
- **Option D: All of these**

3. **Reaction with SO3^2- (Sulfite)**:
- The reaction can be represented as:
\[ \text{SO}_3^{2-} + \text{MnO}_4^{-} \rightarrow \text{SO}_4^{2-} + \text{Mn}^{2+} + \text{H}_2\text{O} \]
- In this reaction, sulfite is oxidized to sulfate, and KMnO4 is reduced to Mn^2+. No gas is evolved.

4. **Reaction with NO2^- (Nitrite)**:
- The reaction can be represented as:
\[ \text{NO}_2^{-} + 2\text{MnO}_4^{-} \rightarrow \text{NO}_3^{-} + 2\text{Mn}^{2+} + \text{H}_2\text{O} \]
- Here, nitrite is oxidized to nitrate, and KMnO4 is again reduced to Mn^2+. No gas is evolved.

5. **Reaction with S^2- (Sulfide)**:
- The reaction can be represented as:
\[ \text{S}^{2-} + \text{MnO}_4^{-} \rightarrow \text{S}^{0} + \text{Mn}^{2+} + \text{H}_2\text{O} \]
- In this case, sulfide is oxidized to elemental sulfur, and KMnO4 is reduced to Mn^2+. No gas is evolved.

6. **Conclusion**:
- All three radicals (sulfite, nitrite, and sulfide) react with acidified KMnO4 without releasing any gas. Therefore, the correct answer is **Option D: All of these**.

### Final Answer:
**Option D: All of these**