A solution when diluted with `H_(2)O` And boiled gives a white precipitate .On the addition of excess `NH_(4)Cl & NH_(4)OH` the volume of the precipitate decreases leaving behind a white gelationtious 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 step by step, we can follow this approach:

### Step 1: Understand the Problem
The problem states that a solution, when diluted with water and boiled, gives a white precipitate. We need to identify this precipitate, which dissolves in ammonium hydroxide (NH₄OH) and ammonium chloride (NH₄Cl).

### Step 2: Identify Possible Precipitates
The options given are:
- Zinc hydroxide (Zn(OH)₂)
- Aluminum hydroxide (Al(OH)₃)
- Magnesium hydroxide (Mg(OH)₂)
- Calcium hydroxide (Ca(OH)₂)

### Step 3: Analyze the Reaction
When a solution containing zinc ions (Zn²⁺) is diluted with water and boiled, it forms a white precipitate of zinc hydroxide (Zn(OH)₂). This is a characteristic reaction of zinc salts.

### Step 4: Addition of NH₄Cl and NH₄OH
When excess NH₄Cl and NH₄OH are added to the solution:
- NH₄OH acts as a reagent that can dissolve certain metal hydroxides.
- NH₄Cl is added to suppress the dissociation of NH₄OH.

In the case of zinc hydroxide, it reacts with NH₄OH to form a soluble complex, which decreases the volume of the precipitate.

### Step 5: Conclusion
Based on the reactions and the behavior of the precipitates with NH₄OH and NH₄Cl, we conclude that the white precipitate formed is zinc hydroxide (Zn(OH)₂), which dissolves in excess NH₄OH and NH₄Cl.

### Final Answer
The precipitate that dissolves in NH₄OH/NH₄Cl is **Zinc Hydroxide (Zn(OH)₂)**.

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