Which of the following are not soluble in excess of `NaOH`?

Options

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

Answer

To determine which of the given hydroxides are not soluble in excess NaOH, we need to analyze the amphoteric nature of each hydroxide. Here's a step-by-step solution:

### Step 1: Identify the Hydroxides
The hydroxides given in the question are:
1. Al(OH)₃ (Aluminum hydroxide)
2. Cr(OH)₃ (Chromium(III) hydroxide)
3. Zn(OH)₂ (Zinc hydroxide)
4. Fe(OH)₃ (Iron(III) hydroxide)

### Step 2: Understand Amphoteric Behavior
Amphoteric hydroxides can react with both acids and bases. When they react with excess NaOH, they dissolve to form soluble complexes. Non-amphoteric hydroxides do not dissolve in excess NaOH.

### Step 3: Analyze Each Hydroxide
- **Al(OH)₃**: Amphoteric. It dissolves in excess NaOH to form [Al(OH)₄]⁻ (aluminate ion).
- **Cr(OH)₃**: Amphoteric. It dissolves in excess NaOH to form [Cr(OH)₄]⁻ (chromate ion).
- **Zn(OH)₂**: Amphoteric. It dissolves in excess NaOH to form [Zn(OH)₄]²⁻ (zincate ion).
- **Fe(OH)₃**: Non-amphoteric. It does not dissolve in excess NaOH and remains insoluble.

### Step 4: Conclusion
Among the given hydroxides, only **Fe(OH)₃** is not soluble in excess NaOH. Therefore, the answer to the question is:

**Fe(OH)₃ is not soluble in excess NaOH.**

### Final Answer
The correct answer is **Fe(OH)₃**.

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