A salt, when warmed with zinc powder and an NaOH solution, gives a gas that turns a filter paper soaked with an alkaline solution of `K_(2)[HgI_(4)]` brown. The salt responds to the brown ring test when acetic acid is used in place of sulphuric acid. The anion present in the salt is:

Options

A. `NO_(3)^(-)`
B. `NO_(2)^(-)`
C. `Br^(-)`
D. None of these

Answer

To identify the anion present in the salt based on the information provided, we can follow these steps:

### Step 1: Analyze the reaction with zinc and NaOH
The salt, when warmed with zinc powder and an NaOH solution, produces a gas. This gas is likely hydrogen gas (H₂) which is produced when zinc reacts with an alkaline solution.

**Hint:** Remember that zinc reacts with bases to produce hydrogen gas.

### Step 2: Identify the gas reaction with K₂[HgI₄]
The gas produced turns a filter paper soaked with an alkaline solution of K₂[HgI₄] brown. This indicates the presence of iodide ions (I⁻) in the solution. The reaction of iodide with the alkaline solution of K₂[HgI₄] leads to the formation of brown-colored mercury(I) iodide (Hg₂I₂).

**Hint:** The color change indicates a specific reaction; think about what happens when iodide ions are present.

### Step 3: Consider the brown ring test
The question states that the salt responds to the brown ring test when acetic acid is used instead of sulfuric acid. The brown ring test is used to identify nitrite (NO₂⁻) or nitrate (NO₃⁻) ions. The use of acetic acid suggests that the anion is nitrite (NO₂⁻), as acetic acid is used to confirm the presence of nitrite ions.

**Hint:** Recall the conditions under which the brown ring test is performed and what each reagent indicates.

### Step 4: Conclusion
Based on the above steps, the anion present in the salt is confirmed to be nitrite (NO₂⁻).

**Final Answer:** The anion present in the salt is nitrite (NO₂⁻).