Which liberates `H_2` with NaOH

Options

A. B
B. Al
C. Zn
D. All

Answer

To determine which substance liberates hydrogen gas (H₂) when reacted with sodium hydroxide (NaOH), we will analyze the reactions of each given option step by step.

### Step 1: Reaction of Boron with NaOH
1. **Write the reaction**: When boron (B) reacts with sodium hydroxide (NaOH), the reaction can be represented as:
\[ B + 6 \, NaOH \rightarrow 2 \, Na_3BO_3 + 3 \, H_2 \]
2. **Identify the products**: The products of this reaction include sodium borate (Na₃BO₃) and hydrogen gas (H₂).
3. **Conclusion**: Boron liberates H₂ when reacted with NaOH.

### Step 2: Reaction of Aluminum with NaOH
1. **Write the reaction**: When aluminum (Al) reacts with sodium hydroxide (NaOH), the reaction can be represented as:
\[ 2 \, Al + 2 \, NaOH + 6 \, H_2O \rightarrow 2 \, NaAlO_2 + 3 \, H_2 \]
2. **Identify the products**: The products of this reaction include sodium aluminate (NaAlO₂) and hydrogen gas (H₂).
3. **Conclusion**: Aluminum also liberates H₂ when reacted with NaOH.

### Step 3: Reaction of Zinc with NaOH
1. **Write the reaction**: When zinc (Zn) reacts with sodium hydroxide (NaOH), the reaction can be represented as:
\[ Zn + 2 \, NaOH \rightarrow Na_2ZnO_2 + H_2 \]
2. **Identify the products**: The products of this reaction include sodium zincate (Na₂ZnO₂) and hydrogen gas (H₂).
3. **Conclusion**: Zinc liberates H₂ when reacted with NaOH.

### Final Conclusion
- All three substances (Boron, Aluminum, and Zinc) liberate hydrogen gas (H₂) when reacted with sodium hydroxide (NaOH). Therefore, the correct answer is:
**Option 4: ALL**