A solution when diluted with `H_(2)O` and boiled, it gives a white precipitate. On addition of excess `NH_(4)Cl//NH_(4)OH` the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in `NH_(4)OH//NH_(4)Cl`

Options

A. `Zn(OH)_(2)`
B. `Al(OH)_(3)`
C. `Mg(OH)_(2)`
D. `Ca(OH)_(2)`

Explanation

To solve the problem, we need to identify the precipitate formed when a solution is diluted with water and boiled, and then determine which precipitate dissolves in excess ammonium chloride (NH₄Cl) or ammonium hydroxide (NH₄OH).

### Step-by-Step Solution:

1. **Understanding the Initial Reaction**:
- When a solution is diluted with water and boiled, a white precipitate is formed. This indicates the presence of certain metal ions that form insoluble hydroxides upon heating.

2. **Identifying the Possible Metal Ions**:
- Common metal ions that form white precipitates upon heating include zinc (Zn²⁺), aluminum (Al³⁺), and lead (Pb²⁺). However, we need to focus on the ions that are part of group 3 cations, which typically include Zn²⁺, Al³⁺, and Fe³⁺.

3. **Analyzing the Precipitate**:
- The white precipitate formed upon dilution and boiling is likely zinc hydroxide (Zn(OH)₂), as it is known to form a white precipitate when zinc ions are present in the solution.

4. **Reaction with NH₄Cl/NH₄OH**:
- When excess NH₄Cl or NH₄OH is added, the volume of the precipitate decreases, indicating that some of the precipitate is dissolving. Zinc hydroxide (Zn(OH)₂) can dissolve in excess NH₄OH to form a soluble complex, which is [Zn(NH₃)₄]²⁺.

5. **Final Identification**:
- The precipitate that dissolves in NH₄OH/NH₄Cl is identified as zinc hydroxide (Zn(OH)₂), which forms a soluble complex when reacted with these reagents.

### Conclusion:
The precipitate that dissolves in NH₄OH or NH₄Cl is **Zinc Hydroxide (Zn(OH)₂)**.