The hydroxide which is soluble in excess of NaOH is .............

Options

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

Explanation

To solve the question "The hydroxide which is soluble in excess of NaOH is .............", we will follow these steps:

### Step 1: Understand the Concept of Amphoteric Hydroxides
Amphoteric hydroxides are those that can react with both acids and bases. This means they can dissolve in excess base (like NaOH) as well as in acids.

### Step 2: Identify Common Amphoteric Hydroxides
The most common amphoteric hydroxides are:
- Aluminum hydroxide (Al(OH)₃)
- Zinc hydroxide (Zn(OH)₂)

### Step 3: Analyze the Question
The question asks for the hydroxide that is soluble in excess NaOH. Since we know that amphoteric hydroxides can react with NaOH, we can conclude that both aluminum hydroxide and zinc hydroxide are soluble in excess NaOH.

### Step 4: Write the Reactions
1. **Aluminum Hydroxide Reaction:**
\[ \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + \text{H}_2\text{O} \]
This shows that aluminum hydroxide reacts with sodium hydroxide to form sodium aluminate and water.

2. **Zinc Hydroxide Reaction:**
\[ \text{Zn(OH)}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + 2\text{H}_2\text{O} \]
This shows that zinc hydroxide reacts with sodium hydroxide to form sodium zincate and water.

### Step 5: Conclusion
Based on the above analysis, the hydroxides that are soluble in excess NaOH are:
- Aluminum hydroxide (Al(OH)₃)
- Zinc hydroxide (Zn(OH)₂)

Thus, the answer to the question is **Aluminum hydroxide and Zinc hydroxide**.