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 each option one by one.

### Step 1: Analyze the first compound (let's assume it's Boron, B)
1. **Reaction**: B + NaOH → Na₃BO₃ + H₂
2. **Balancing the equation**:
- We need to balance the reaction. The balanced reaction is:
- 6B + 6NaOH → 2Na₃BO₃ + 3H₂
3. **Conclusion**: This reaction liberates H₂.

### Step 2: Analyze the second compound (let's assume it's Aluminium, Al)
1. **Reaction**: Al + NaOH → NaAlO₂ + H₂
2. **Balancing the equation**:
- The balanced reaction is:
- 2Al + 6NaOH → 2NaAlO₂ + 3H₂
3. **Conclusion**: This reaction also liberates H₂.

### Step 3: Analyze the third compound (let's assume it's Zinc, Zn)
1. **Reaction**: Zn + NaOH → Na₂ZnO₂ + H₂
2. **Balancing the equation**:
- The balanced reaction is:
- Zn + 2NaOH → Na₂ZnO₂ + H₂
3. **Conclusion**: This reaction liberates H₂ as well.

### Final Conclusion
Since all three compounds (B, Al, and Zn) liberate hydrogen gas when reacted with sodium hydroxide, we can conclude that the correct option is D, which states that all of these compounds liberate H₂ with NaOH.

### Summary of the Solution
- B + NaOH liberates H₂.
- Al + NaOH liberates H₂.
- Zn + NaOH liberates H₂.
- Therefore, the answer is that all of the given options liberate H₂ with NaOH.