A salt of `NaX overset(MgCl_(2))rarr" white ppt."` on boiling. Thus, anion X is :

Options

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

Answer

To solve the question, we need to determine the anion \( X \) in the salt \( NaX \) that forms a white precipitate when boiled in the presence of \( MgCl_2 \).

### Step-by-Step Solution:

1. **Identify the Reaction**: When \( NaX \) is mixed with \( MgCl_2 \), a reaction occurs that leads to the formation of a white precipitate upon boiling.

2. **Assume the Anion**: Let's assume that the anion \( X \) is \( HCO_3^- \) (bicarbonate ion).

3. **Write the Initial Reaction**: The initial reaction between sodium bicarbonate (\( NaHCO_3 \)) and magnesium chloride (\( MgCl_2 \)) can be represented as:
\[ MgCl_2 + 2 NaHCO_3 \rightarrow Mg(HCO_3)_2 + 2 NaCl \]
Here, \( Mg(HCO_3)_2 \) is soluble in water.

4. **Heating the Solution**: Upon heating, \( Mg(HCO_3)_2 \) decomposes to form magnesium carbonate (\( MgCO_3 \)), water (\( H_2O \)), and carbon dioxide (\( CO_2 \)):
\[ Mg(HCO_3)_2 \rightarrow MgCO_3 + H_2O + CO_2 \]

5. **Formation of Precipitate**: The magnesium carbonate (\( MgCO_3 \)) formed is insoluble in water and precipitates out as a white solid:
\[ MgCO_3 \text{ (s)} \text{ (white precipitate)} \]

6. **Conclusion**: Since the formation of the white precipitate is due to the presence of the bicarbonate ion, we conclude that the anion \( X \) in the salt \( NaX \) is \( HCO_3^- \).

### Final Answer:
The anion \( X \) is \( HCO_3^- \) (bicarbonate ion).

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