White crystal (A) on treatment with `AgNO_(3)` gives white crystalline precipitate. (A) discharges the colour of `KMnO_(4)` solution but no gas is evolved. Probable radical present in (A) is:

Options

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

Answer

To solve the problem step by step, we need to analyze the information given and determine the probable radical present in the white crystal (A).

### Step 1: Analyze the reaction with AgNO3
- The question states that white crystal (A) gives a white crystalline precipitate when treated with AgNO3.
- This indicates that (A) contains an anion that forms an insoluble salt with silver ions (Ag+). Common anions that form white precipitates with AgNO3 include Cl⁻ (chloride), Br⁻ (bromide), and I⁻ (iodide).

### Step 2: Analyze the reaction with KMnO4
- The question also states that (A) discharges the color of KMnO4 solution but no gas is evolved.
- KMnO4 is a strong oxidizing agent, and it can react with certain anions. We need to identify which anions can react with KMnO4 without evolving gas.

### Step 3: Test possible anions
1. **Chloride ion (Cl⁻)**:
- When Cl⁻ reacts with KMnO4, it produces Cl2 gas. Therefore, this option is eliminated as gas is evolved.

2. **Bromide ion (Br⁻)**:
- When Br⁻ reacts with KMnO4, it produces Br2 gas. This option is also eliminated for the same reason.

3. **Nitrite ion (NO2⁻)**:
- When NO2⁻ reacts with KMnO4, it gets oxidized to NO3⁻ and does not evolve any gas. This option remains valid.

4. **Carbonate ion (CO3²⁻)**:
- When CO3²⁻ reacts with KMnO4, it produces CO2 gas. Thus, this option is eliminated as well.

### Step 4: Conclusion
- The only anion that fits both conditions (forms a white precipitate with AgNO3 and does not evolve gas when reacting with KMnO4) is the nitrite ion (NO2⁻).
- Therefore, the probable radical present in (A) is **NO2⁻**.

### Final Answer:
The probable radical present in (A) is **NO2⁻**.