Which of the following is (are) soluble in excess of `NaOH`?

Options

A. `Cr(OH)_(3)`
B. `Fe(OH)_(3)`
C. `Al(OH)_(3)`
D. `Zn(OH)_(2)`

Answer

To determine which of the given hydroxides are soluble in excess sodium hydroxide (NaOH), we will analyze each compound one by one.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we are analyzing are:
- Barium hydroxide (Ba(OH)₂)
- Iron hydroxide (Fe(OH)₂ or Fe(OH)₃)
- Aluminum hydroxide (Al(OH)₃)
- Zinc hydroxide (Zn(OH)₂)

2. **Analyze Aluminum Hydroxide**:
- Aluminum hydroxide (Al(OH)₃) is amphoteric. This means it can react with both acids and bases.
- When aluminum hydroxide reacts with excess NaOH, it forms sodium aluminate (NaAlO₂) and water:
\[ \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + 2 \text{H}_2\text{O} \]
- **Conclusion**: Aluminum hydroxide is soluble in excess NaOH.

3. **Analyze Zinc Hydroxide**:
- Zinc hydroxide (Zn(OH)₂) is also amphoteric.
- When zinc hydroxide reacts with excess NaOH, it forms sodium zincate (Na₂ZnO₂) and water:
\[ \text{Zn(OH)}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + 2 \text{H}_2\text{O} \]
- **Conclusion**: Zinc hydroxide is soluble in excess NaOH.

4. **Analyze Iron Hydroxide**:
- Iron hydroxide (Fe(OH)₂ or Fe(OH)₃) is not amphoteric in the same way as aluminum and zinc hydroxides.
- It does not dissolve in excess NaOH and remains insoluble.
- **Conclusion**: Iron hydroxide is not soluble in excess NaOH.

5. **Analyze Barium Hydroxide**:
- Barium hydroxide (Ba(OH)₂) is a strong base and is soluble in water, but it does not form soluble complexes with excess NaOH.
- **Conclusion**: Barium hydroxide is not soluble in excess NaOH.

### Final Answer:
The compounds that are soluble in excess sodium hydroxide are:
- Aluminum hydroxide (Al(OH)₃)
- Zinc hydroxide (Zn(OH)₂)