Select correct answers from the choices A, B, C, D which are given. Write down only the letter corresponding to the correct answer.
A chloride which forms a precipitate that is soluble in excess of ammonium hydroxide is

Options

A. calcium chloride
B. ferrous chloride
C. ferric chloride
D. copper chloride

Answer

To solve the question, we need to identify which chloride forms a precipitate that is soluble in excess ammonium hydroxide (NH4OH). Let's analyze each option step by step.

### Step 1: Analyze Calcium Chloride (CaCl2)
- When calcium chloride reacts with ammonium hydroxide, it forms calcium hydroxide (Ca(OH)2) and ammonium chloride (NH4Cl).
- The reaction is:
\[ \text{CaCl}_2 + 2 \text{NH}_4\text{OH} \rightarrow \text{Ca(OH)}_2 \downarrow + 2 \text{NH}_4\text{Cl} \]
- Calcium hydroxide is not soluble in excess ammonium hydroxide.

**Conclusion**: CaCl2 is not the answer.

### Step 2: Analyze Ferrous Chloride (FeCl2)
- Ferrous chloride reacts with ammonium hydroxide to form ferrous hydroxide (Fe(OH)2) and ammonium chloride.
- The reaction is:
\[ \text{FeCl}_2 + 2 \text{NH}_4\text{OH} \rightarrow \text{Fe(OH)}_2 \downarrow + 2 \text{NH}_4\text{Cl} \]
- Ferrous hydroxide is also not soluble in excess ammonium hydroxide.

**Conclusion**: FeCl2 is not the answer.

### Step 3: Analyze Ferric Chloride (FeCl3)
- Ferric chloride reacts with ammonium hydroxide to form ferric hydroxide (Fe(OH)3) and ammonium chloride.
- The reaction is:
\[ \text{FeCl}_3 + 3 \text{NH}_4\text{OH} \rightarrow \text{Fe(OH)}_3 \downarrow + 3 \text{NH}_4\text{Cl} \]
- Ferric hydroxide is not soluble in excess ammonium hydroxide.

**Conclusion**: FeCl3 is not the answer.

### Step 4: Analyze Copper Chloride (CuCl2)
- Copper chloride reacts with ammonium hydroxide to form copper hydroxide (Cu(OH)2) and ammonium chloride.
- The reaction is:
\[ \text{CuCl}_2 + 2 \text{NH}_4\text{OH} \rightarrow \text{Cu(OH)}_2 \downarrow + 2 \text{NH}_4\text{Cl} \]
- Copper hydroxide is soluble in excess ammonium hydroxide, forming a complex known as cupramine hydroxide:
\[ \text{Cu(OH)}_2 + 4 \text{NH}_4\text{OH} \rightarrow \text{Cu(NH}_3)_4\text{(OH)}_2 + 4 \text{H}_2\text{O} \]

**Conclusion**: CuCl2 is the answer.

### Final Answer
The chloride which forms a precipitate that is soluble in excess of ammonium hydroxide is **D (Copper Chloride)**.

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