Mercuric chloride reacts with aqueous ammonia to form a white precipitate of

Options

A. `Hg (NH_(3))_(2)CI_(2)`
B. `Hg (NH_(2))_(2)CI_(2)`
C. `Hg (NH_(3))_(2)CI`
D. `HG (NH_(2))CI`

Explanation

To solve the question regarding the reaction of mercuric chloride (HgCl2) with aqueous ammonia (NH3) and to identify the white precipitate formed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are mercuric chloride (HgCl2) and aqueous ammonia (NH3).

2. **Write the Reaction**:
- When mercuric chloride reacts with aqueous ammonia, the reaction can be represented as:
\[ \text{HgCl}_2 + 2 \text{NH}_3 \rightarrow \text{Hg(NH}_2\text{Cl)} + \text{NH}_4\text{Cl} \]
- Here, mercuric chloride reacts with two moles of ammonia to form mercuric amidochloride (Hg(NH2Cl)) and ammonium chloride (NH4Cl).

3. **Identify the Precipitate**:
- The compound Hg(NH2Cl) formed in the reaction is known to be a white precipitate.

4. **Conclusion**:
- Therefore, the white precipitate formed when mercuric chloride reacts with aqueous ammonia is **mercuric amidochloride (Hg(NH2Cl))**.

### Final Answer:
The white precipitate formed is **Hg(NH2Cl)**.