Amongst the following, the compound not soluble in excess of `NH_(4)OH` solution is:

Options

A. `CuSO_(4)`
B. `ZnSO_(4)`
C. `Fecl_(3)`
D. `Agcl`

Explanation

To determine which compound is not soluble in excess of `NH₄OH` (ammonium hydroxide), we will analyze the solubility of each compound mentioned in the question.

### Step-by-Step Solution:

1. **Identify the Compounds:**
The compounds we need to analyze are:
- Copper sulfate (`CuSO₄`)
- Zinc sulfate (`ZnSO₄`)
- Ferric chloride (`FeCl₃`)
- Silver chloride (`AgCl`)

2. **Reactions with `NH₄OH`:**
We will examine the reaction of each compound with `NH₄OH` to see if they form soluble complexes or precipitates.

3. **Copper Sulfate (`CuSO₄`):**
- When `CuSO₄` reacts with `NH₄OH`, it forms a precipitate of `Cu(OH)₂` (copper(II) hydroxide) and ammonium sulfate (`(NH₄)₂SO₄`).
- In excess `NH₄OH`, `Cu(OH)₂` dissolves to form a soluble complex: `Cu(NH₃)₄²⁺` (tetraamine copper(II)).
- **Conclusion:** `CuSO₄` is soluble in excess `NH₄OH`.

4. **Zinc Sulfate (`ZnSO₄`):**
- When `ZnSO₄` reacts with `NH₄OH`, it forms a precipitate of `Zn(OH)₂` (zinc hydroxide) and ammonium sulfate.
- In excess `NH₄OH`, `Zn(OH)₂` dissolves to form a soluble complex: `Zn(NH₃)₄²⁺` (tetraamine zinc).
- **Conclusion:** `ZnSO₄` is soluble in excess `NH₄OH`.

5. **Ferric Chloride (`FeCl₃`):**
- When `FeCl₃` reacts with `NH₄OH`, it forms a precipitate of `Fe(OH)₃` (ferric hydroxide) and ammonium chloride.
- `Fe(OH)₃` does not dissolve in excess `NH₄OH`.
- **Conclusion:** `FeCl₃` is **not** soluble in excess `NH₄OH`.

6. **Silver Chloride (`AgCl`):**
- When `AgCl` reacts with `NH₄OH`, it forms a soluble complex: `Ag(NH₃)₂Cl` (diamine silver(I) chloride).
- **Conclusion:** `AgCl` is soluble in excess `NH₄OH`.

### Final Answer:
The compound that is **not soluble** in excess of `NH₄OH` is **Ferric Chloride (`FeCl₃`)**.

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