Which of the following leaves a black residue on the addition of `NH_(3)` ?

Options

A. `AgCl`
B. `PbCl_(2)`
C. `Hg_(2)Cl_(2)`
D. `HgCl_(2)`

Answer

To determine which of the given compounds leaves a black residue upon the addition of ammonia (NH₃), we will analyze each compound one by one.

### Step 1: Analyze AgCl (Silver Chloride)
- When AgCl is treated with aqueous ammonia, it forms a complex ion:
\[ \text{AgCl} + 2 \text{NH}_3 \rightarrow \text{[Ag(NH}_3\text{)}_2]^+ + \text{Cl}^- \]
- The product is a colorless complex, meaning no black residue is formed.

**Hint:** Check if the compound forms a colored complex with ammonia.

### Step 2: Analyze PbCl₂ (Lead(II) Chloride)
- When PbCl₂ is treated with ammonia, it forms lead(II) hydroxide:
\[ \text{PbCl}_2 + 2 \text{NH}_3 + 2 \text{H}_2\text{O} \rightarrow \text{Pb(OH)}_2 \downarrow + 2 \text{NH}_4^+ + 2 \text{Cl}^- \]
- The precipitate formed is white (Pb(OH)₂), indicating no black residue.

**Hint:** Look for precipitates formed in the reaction.

### Step 3: Analyze Hg₂Cl₂ (Mercury(I) Chloride)
- When Hg₂Cl₂ is treated with ammonia, it forms:
\[ \text{Hg}_2\text{Cl}_2 + 2 \text{NH}_3 \rightarrow \text{HgNH}_2\text{Cl} \downarrow + \text{Hg} \downarrow + \text{Cl}^- \]
- Here, HgNH₂Cl is a white precipitate, and elemental mercury (Hg) is produced, which appears as a black residue.

**Hint:** Identify if elemental mercury is produced in the reaction.

### Step 4: Analyze HgCl₂ (Mercury(II) Chloride)
- When HgCl₂ is treated with ammonia, it forms:
\[ \text{HgCl}_2 + 2 \text{NH}_3 \rightarrow \text{HgNH}_2\text{Cl} \downarrow + \text{Cl}^- \]
- The product is a white precipitate (HgNH₂Cl) with no black residue.

**Hint:** Consider the oxidation state of mercury in the compound.

### Conclusion
After analyzing all the compounds, we find that only Hg₂Cl₂ (Mercury(I) Chloride) produces a black residue (elemental mercury) upon the addition of ammonia.

**Final Answer:** The correct choice is **Hg₂Cl₂**.