The evolution of a red-brown gas on heating a salt with `K_(2)Cr_(2)O_(7)` and concentrated `H_(2)SO_(4)` can arise from:

Options

A. chlroride
B. bromide
C. nitrate
D. nitrite

Explanation

To solve the question regarding the evolution of a red-brown gas when heating a salt with potassium dichromate (K₂Cr₂O₇) and concentrated sulfuric acid (H₂SO₄), we will analyze the possible reactions with different anions: chloride (Cl⁻), bromide (Br⁻), nitrate (NO₃⁻), and nitrite (NO₂⁻).

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants are potassium dichromate (K₂Cr₂O₇) and concentrated sulfuric acid (H₂SO₄).

2. **Consider the Anions**: We will evaluate the reactions of K₂Cr₂O₇ with each of the following anions:
- Chloride (Cl⁻)
- Bromide (Br⁻)
- Nitrate (NO₃⁻)
- Nitrite (NO₂⁻)

3. **Reaction with Chloride (Cl⁻)**:
- When chloride ions react with K₂Cr₂O₇ and concentrated H₂SO₄, chromyl chloride (CrO₂Cl₂) is formed.
- Chromyl chloride is a red-colored gas.
- **Conclusion**: This reaction produces a red gas.

4. **Reaction with Bromide (Br⁻)**:
- The reaction of bromide ions with K₂Cr₂O₇ and concentrated H₂SO₄ produces bromine gas (Br₂).
- Bromine gas is red-brown in color.
- **Conclusion**: This reaction produces a red-brown gas.

5. **Reaction with Nitrate (NO₃⁻)**:
- Nitrate ions do not react with K₂Cr₂O₇ and concentrated H₂SO₄.
- **Conclusion**: No gas is produced.

6. **Reaction with Nitrite (NO₂⁻)**:
- Nitrite ions react with K₂Cr₂O₇ and concentrated H₂SO₄ to form nitrate ions (NO₃⁻) and chromium (III) ions (Cr³⁺).
- **Conclusion**: No gas is produced.

7. **Final Evaluation**:
- The only anion that produces a red-brown gas when reacted with K₂Cr₂O₇ and concentrated H₂SO₄ is bromide (Br⁻), which produces bromine gas.

### Conclusion:
The evolution of a red-brown gas on heating a salt with K₂Cr₂O₇ and concentrated H₂SO₄ can arise from bromide ions (Br⁻). Therefore, the correct answer is **bromide (option B)**.