A sodium salt of unknown anion when treated with `MgCl_(2)` gives a white `ppt`. On boiling. The anion is

Options

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

Explanation

To solve the problem of identifying the unknown anion in the sodium salt that produces a white precipitate when treated with MgCl₂, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
We start with a sodium salt of an unknown anion (let's denote it as NaX, where X is the unknown anion). When this salt is treated with magnesium chloride (MgCl₂), we need to determine what happens.

2. **Formation of Precipitate**:
The reaction between NaX and MgCl₂ can produce a precipitate. The formation of a white precipitate indicates that the anion X likely forms an insoluble compound with magnesium.

3. **Consider Possible Anions**:
Common anions that form precipitates with magnesium include carbonate (CO₃²⁻) and bicarbonate (HCO₃⁻). However, bicarbonate is soluble in water, and it can decompose upon heating to form carbonate.

4. **Reaction of Bicarbonate with Magnesium Chloride**:
If we assume the anion is bicarbonate (HCO₃⁻), the reaction can be written as:
\[ 2 \text{NaHCO}_3 + \text{MgCl}_2 \rightarrow \text{Mg(HCO}_3)_2 + 2 \text{NaCl} \]
Here, magnesium bicarbonate is formed, which is soluble.

5. **Heating the Bicarbonate**:
Upon heating magnesium bicarbonate, it decomposes to form magnesium carbonate, water, and carbon dioxide:
\[ \text{Mg(HCO}_3)_2 \xrightarrow{\text{heat}} \text{MgCO}_3 + \text{H}_2\text{O} + \text{CO}_2 \]
Magnesium carbonate (MgCO₃) is insoluble in water and will precipitate out as a white solid.

6. **Conclusion**:
Since the white precipitate formed upon boiling is magnesium carbonate, we conclude that the unknown anion in the sodium salt is bicarbonate (HCO₃⁻).

### Final Answer:
The anion is bicarbonate (HCO₃⁻).