Which of the following salts does not form any precipitate with excess of NaOH ?

Options

A. `ZnCI_(2)`
B. `FeCI_(3)`
C. `CrCI_(3)`
D. `CuSO_(4)`

Explanation

To determine which of the following salts does not form any precipitate with excess NaOH, we will analyze each salt provided in the options.

### Step-by-Step Solution:

1. **Identify the salts**: The question provides us with different salts. We need to analyze each one to see how they react with NaOH.

2. **Examine ZnCl2 (Zinc Chloride)**:
- When ZnCl2 is added to excess NaOH, it reacts to form sodium zincate (Na2ZnO2) and does not form any precipitate.
- Reaction:
\[ ZnCl_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2NaCl \]
- Conclusion: **No precipitate forms**.

3. **Examine FeCl3 (Ferric Chloride)**:
- When FeCl3 is added to excess NaOH, it reacts to form ferric hydroxide (Fe(OH)3), which is a brown precipitate.
- Reaction:
\[ FeCl_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaCl \]
- Conclusion: **Brown precipitate forms**.

4. **Examine CrCl3 (Chromium Chloride)**:
- When CrCl3 is added to excess NaOH, it reacts to form chromium hydroxide (Cr(OH)3), which is a green precipitate.
- Reaction:
\[ CrCl_3 + 3NaOH \rightarrow Cr(OH)_3 + 3NaCl \]
- Conclusion: **Green precipitate forms**.

5. **Examine CuSO4 (Copper Sulfate)**:
- When CuSO4 is added to excess NaOH, it reacts to form copper hydroxide (Cu(OH)2), which is a light blue precipitate.
- Reaction:
\[ CuSO_4 + 2NaOH \rightarrow Cu(OH)_2 + Na_2SO_4 \]
- Conclusion: **Light blue precipitate forms**.

6. **Final Conclusion**:
- The only salt that does not form any precipitate with excess NaOH is **ZnCl2**.

### Answer:
**ZnCl2 (Zinc Chloride)** does not form any precipitate with excess NaOH.