Which one of the following does not give a precipitate with excess of both NaOH solution and `NH_(3)` solution ?

Options

A. `Pb(NO_(3))_(2)`
B. `Zn(NO_(3))_(2)`
C. `Al(NO_(3))_(2)`
D. `Cu(NO_(3))_(2)`

Explanation

To solve the question "Which one of the following does not give a precipitate with excess of both NaOH solution and NH3 solution?", we will analyze the behavior of different metal ions with NaOH and NH3.

### Step-by-Step Solution:

1. **Identify the Metal Ions**: The first step is to identify the metal ions in the options provided (let's assume they are A, B, C, D). Common metal ions that we might consider include lead (Pb), aluminum (Al), copper (Cu), and zinc (Zn).

2. **Reaction with NaOH**:
- When metal ions react with NaOH, they can form metal hydroxides.
- For example:
- Pb²⁺ + 2OH⁻ → Pb(OH)₂ (precipitate)
- Al³⁺ + 3OH⁻ → Al(OH)₃ (precipitate)
- Cu²⁺ + 2OH⁻ → Cu(OH)₂ (precipitate)
- Zn²⁺ + 2OH⁻ → Zn(OH)₂ (precipitate)

3. **Excess NaOH**:
- When excess NaOH is added, some metal hydroxides can dissolve to form soluble complexes.
- For zinc:
- Zn(OH)₂ + 2OH⁻ → [Zn(OH)₄]²⁻ (soluble complex)
- This means that zinc does not form a precipitate with excess NaOH.

4. **Reaction with NH3**:
- Similarly, when metal ions react with NH3, they can form complex ions.
- For zinc:
- Zn²⁺ + 4NH₃ + 2H₂O → [Zn(NH₃)₄]²⁺ + 2OH⁻ (soluble complex)
- This indicates that zinc does not form a precipitate with excess NH3 either.

5. **Conclusion**:
- Based on the reactions, we find that while lead, aluminum, and copper form precipitates with both NaOH and NH3, zinc does not form a precipitate with excess of either reagent.
- Therefore, the correct answer to the question is **Zinc (Zn)**.

### Final Answer:
Zinc (Zn) does not give a precipitate with excess of both NaOH solution and NH3 solution.