`H_(2)` gas can not be prepared by :

Options

A. Be + NaOH
B. Na + NaOH
C. Mg + NaOH
D. By (2 & 3 ) method

Explanation

To determine which method does not produce hydrogen gas (H₂), let's analyze the options provided in the question step by step.

### Step 1: Analyze the first method
- **Method**: Reaction of beryllium with sodium hydroxide.
- **Reaction**: Beryllium (Be) reacts with sodium hydroxide (NaOH) to produce beryllium hydroxide (Be(OH)₂) and hydrogen gas (H₂).
- **Balanced Reaction**:
\[ 2 \text{NaOH} + \text{Be} \rightarrow \text{Na}_2\text{Be(OH)}_4 + \text{H}_2 \]
- **Conclusion**: This method produces hydrogen gas.

### Step 2: Analyze the second method
- **Method**: Reaction of sodium with excess sodium hydroxide.
- **Reaction**: Sodium (Na) reacts with sodium hydroxide (NaOH) to produce sodium oxide (Na₂O) and hydrogen gas (H₂).
- **Balanced Reaction**:
\[ 2 \text{Na} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{O} + \text{H}_2 \]
- **Conclusion**: This method also produces hydrogen gas.

### Step 3: Analyze the third method
- **Method**: Reaction of magnesium with sodium hydroxide.
- **Reaction**: Magnesium (Mg) reacts with sodium hydroxide (NaOH) to produce magnesium hydroxide (Mg(OH)₂).
- **Balanced Reaction**:
\[ \text{Mg} + 2 \text{NaOH} \rightarrow \text{Mg(OH)}_2 + \text{H}_2 \]
- **Conclusion**: This method does not produce hydrogen gas.

### Final Conclusion
Based on the analysis of the three methods:
- The first method (beryllium with sodium hydroxide) produces hydrogen gas.
- The second method (sodium with sodium hydroxide) produces hydrogen gas.
- The third method (magnesium with sodium hydroxide) does not produce hydrogen gas.

Thus, the correct answer to the question "H₂ gas cannot be prepared by:" is **option C** (magnesium with sodium hydroxide).

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