A colouless solution of a compound gives a precipitate with `AgNO_(3)` solution but no precipitate with a solution of `Na_(2)CO_(3)`. The action of concentrated `H_(2)SO_(4)` on the compound liberates a suffocating reddish brown gas.
The compound is `:`

Options

A. `Ba(CH_(3)COO)_(2)`
B. `CaCl_(2)`
C. `Nal`
D. `NaBr`

Explanation

To solve the problem step by step, let's analyze the information given in the question:

1. **Identify the Precipitate with AgNO₃**:
- The compound gives a precipitate when treated with silver nitrate (AgNO₃). This indicates that the compound likely contains a halide ion (Cl⁻, Br⁻, or I⁻) because AgNO₃ is known to form precipitates with halides (e.g., AgCl, AgBr, AgI).

2. **No Precipitate with Na₂CO₃**:
- The compound does not give a precipitate with sodium carbonate (Na₂CO₃). This suggests that the compound does not contain carbonate ions (CO₃²⁻) or does not produce a precipitate with Na₂CO₃, which typically indicates the absence of certain metal ions that form insoluble carbonates.

3. **Action of Concentrated H₂SO₄**:
- The action of concentrated sulfuric acid (H₂SO₄) on the compound liberates a suffocating reddish-brown gas. The reddish-brown gas is characteristic of bromine (Br₂) gas, which is produced when bromide ions (Br⁻) react with concentrated H₂SO₄.

4. **Conclusion**:
- Based on the above observations, the compound must be sodium bromide (NaBr). It gives a precipitate with AgNO₃ (forming AgBr) and does not precipitate with Na₂CO₃. Furthermore, when treated with concentrated H₂SO₄, it releases bromine gas, which is reddish-brown.

Thus, the compound is **sodium bromide (NaBr)**.