A sodium salt on treatment with `MgCl_(2)` gives white precipitate only on heating. The anion of sodium salt is

Options

A. `HCO_(3)^(-)`
B. `CO_(3)^(2-)`
C. `NO_(3)^(-)`
D. `SO_(4)^(2-)`

Answer

To determine the anion of the sodium salt that reacts with `MgCl2` to produce a white precipitate only upon heating, we can analyze the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Sodium Salt**:
Let's denote the sodium salt as `NaX`, where `X` is the anion we need to identify.

2. **Reaction with `MgCl2`**:
When `NaX` reacts with `MgCl2`, it forms a compound. The general reaction can be written as:
\[ NaX + MgCl_2 \rightarrow MgX + NaCl \]
Here, `MgX` is the product formed from the reaction.

3. **Heating the Reaction Mixture**:
The problem states that a white precipitate is formed only upon heating. This suggests that the compound formed (`MgX`) is not stable at room temperature but decomposes or transforms when heated.

4. **Testing Possible Anions**:
We will consider the possible anions for `X`. The most likely candidates include:
- `HCO3^-` (bicarbonate)
- `CO3^{2-}` (carbonate)
- `SO4^{2-}` (sulfate)
- `Cl^-` (chloride)
- `NO3^-` (nitrate)

5. **Analyzing the Bicarbonate Anion**:
Let's specifically analyze the bicarbonate ion (`HCO3^-`):
- When sodium bicarbonate (`NaHCO3`) reacts with `MgCl2`, it forms magnesium bicarbonate:
\[ 2 NaHCO_3 + MgCl_2 \rightarrow Mg(HCO_3)_2 + 2 NaCl \]
- Upon heating, magnesium bicarbonate decomposes to form magnesium carbonate (`MgCO3`), which is a white precipitate:
\[ Mg(HCO_3)_2 \rightarrow MgCO_3 + CO_2 + H_2O \]

6. **Conclusion**:
Since the only anion that produces a white precipitate upon heating with `MgCl2` is bicarbonate, we conclude that the anion of the sodium salt is:
\[ \text{The anion of the sodium salt is } HCO3^- \]