A solution of a sodium salt gives yellow precipitate with both `Pb^(2+)` and `Ag^(+)` ions. Moreover, the acidified solution of sodium salt with `KNO_(2)` liberates a coloured gas which turns starch paper blue. The anion is `:`

Options

A. `I^(-)`
B. `Br^(-)`
C. `NO_(3)^(-)`
D. `SO_(3)^(2-)`

Explanation

To determine the anion present in the sodium salt solution, we will analyze the information given step by step.

### Step 1: Identify the reaction with Pb²⁺ and Ag⁺
The problem states that the sodium salt gives a yellow precipitate with both Pb²⁺ and Ag⁺ ions.

- **Reaction with Pb²⁺:**
\[ \text{Pb}^{2+} + 2 \text{I}^- \rightarrow \text{PbI}_2 \downarrow \quad (\text{Yellow precipitate}) \]

- **Reaction with Ag⁺:**
\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \downarrow \quad (\text{Yellow precipitate}) \]

From these reactions, we can conclude that the anion must be iodide (I⁻) since both lead(II) iodide (PbI₂) and silver iodide (AgI) form yellow precipitates.

### Step 2: Analyze the reaction with KNO₂
The problem also mentions that when the acidified solution of the sodium salt is treated with KNO₂, a colored gas is liberated which turns starch paper blue.

- KNO₂ can be dissociated into K⁺ and NO₂⁻. The NO₂⁻ ion can react with I⁻ to produce iodine (I₂):
\[ 2 \text{I}^- + \text{NO}_2^- \rightarrow \text{I}_2 + \text{NO}_3^- \]

- The liberated iodine (I₂) is a colored gas. When iodine gas comes into contact with starch paper, it turns blue due to the formation of a starch-iodine complex.

### Conclusion
Based on the reactions observed, the anion present in the sodium salt is iodide (I⁻).

### Final Answer
The anion is: **I⁻ (iodide)**.

---