The precipitate of which of the following compounds is soluble in excess of ammonia solution?

Options

A. Iron (II) chloride
B. Magnesium chloride
C. Copper (II) sulphate
D. Lead nitrate

Explanation

To solve the question regarding which compound's precipitate is soluble in excess ammonia solution, we will analyze each given compound step by step.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given in the question are:
- Iron(II) chloride (FeCl2)
- Magnesium chloride (MgCl2)
- Copper(II) sulfate (CuSO4)
- Lead(II) nitrate (Pb(NO3)2)

2. **Analyze Iron(II) Chloride (FeCl2)**:
- When FeCl2 reacts with ammonium hydroxide (NH4OH), it forms iron(II) hydroxide (Fe(OH)2), which is a red precipitate.
- Fe(OH)2 is insoluble in excess ammonia.
- **Conclusion**: FeCl2 does not yield a soluble precipitate in excess ammonia.

3. **Analyze Magnesium Chloride (MgCl2)**:
- When MgCl2 reacts with NH4OH, it forms magnesium hydroxide (Mg(OH)2), which is a white precipitate.
- Mg(OH)2 is also insoluble in excess ammonia.
- **Conclusion**: MgCl2 does not yield a soluble precipitate in excess ammonia.

4. **Analyze Copper(II) Sulfate (CuSO4)**:
- When CuSO4 reacts with NH4OH, it forms copper(II) hydroxide (Cu(OH)2), which is a blue precipitate.
- Cu(OH)2 is insoluble in excess ammonia initially, but it can react with excess NH4OH to form a soluble complex, specifically tetraamminecopper(II) complex [Cu(NH3)4]²⁺.
- **Conclusion**: CuSO4 yields a soluble precipitate in excess ammonia.

5. **Analyze Lead(II) Nitrate (Pb(NO3)2)**:
- When Pb(NO3)2 reacts with NH4OH, it forms lead(II) hydroxide (Pb(OH)2), which is a white precipitate.
- Pb(OH)2 is insoluble in excess ammonia.
- **Conclusion**: Pb(NO3)2 does not yield a soluble precipitate in excess ammonia.

### Final Answer:
The only compound whose precipitate is soluble in excess ammonia solution is **Copper(II) sulfate (CuSO4)**.

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