Which of the following anion (s) evolve(s) reddish brown gas with concentrated `H_(2)SO_(4)`?

Options

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

Explanation

To determine which anion(s) evolve reddish-brown gas when reacted with concentrated sulfuric acid (H₂SO₄), we can analyze each anion provided in the question.

### Step-by-Step Solution:

1. **Identify the Anions**: The question involves several anions, including Br⁻ (bromide), NO₃⁻ (nitrate), SO₃²⁻ (sulfite), and I⁻ (iodide).

2. **Reactions with Concentrated H₂SO₄**:
- **For Br⁻ (Bromide Ion)**:
- When Br⁻ reacts with concentrated H₂SO₄, it forms HBr.
- HBr can further react with H₂SO₄ to produce Br₂ (bromine gas) and SO₂ (sulfur dioxide).
- The balanced reaction is:
\[ 2 HBr + H₂SO₄ \rightarrow Br₂ + SO₂ + 2 H₂O \]
- Br₂ is reddish-brown in color.

- **For NO₃⁻ (Nitrate Ion)**:
- When NO₃⁻ reacts with concentrated H₂SO₄, it produces NO₂ (nitrogen dioxide), which is also reddish-brown.
- The balanced reaction is:
\[ 4 NO₃⁻ + 2 H₂SO₄ \rightarrow 4 NO₂ + 2 SO₄²⁻ + 2 H₂O \]
- NO₂ is reddish-brown in color.

- **For SO₃²⁻ (Sulfite Ion)**:
- When SO₃²⁻ reacts with H₂SO₄, it produces SO₂, which is colorless.
- The balanced reaction is:
\[ SO₃²⁻ + 2 H⁺ \rightarrow SO₂ + H₂O \]
- SO₂ is colorless, so this option is incorrect.

- **For I⁻ (Iodide Ion)**:
- When I⁻ reacts with concentrated H₂SO₄, it forms I₂ (iodine), which appears violet in color.
- The balanced reaction is:
\[ 2 I⁻ + H₂SO₄ \rightarrow I₂ + SO₄²⁻ + 2 H⁺ + H₂O \]
- I₂ is violet, so this option is also incorrect.

3. **Conclusion**:
- The anions that evolve reddish-brown gas when reacted with concentrated H₂SO₄ are **Br⁻ (bromide)** and **NO₃⁻ (nitrate)**.

### Final Answer:
The anions that evolve reddish-brown gas with concentrated H₂SO₄ are **Br⁻ and NO₃⁻**.