How many of following elements will liberate `H_2` gas when treated with dil `HNO_3`?
Mg, Zn, Sn, Pb

Answer

To determine how many of the given elements (Mg, Zn, Sn, Pb) will liberate hydrogen gas when treated with dilute nitric acid (HNO3), we need to analyze the reaction of each element with dilute nitric acid.

### Step-by-Step Solution:

1. **Identify the Elements**: The elements provided are Magnesium (Mg), Zinc (Zn), Tin (Sn), and Lead (Pb).

2. **Reaction of Magnesium (Mg)**:
- When magnesium reacts with dilute nitric acid, the reaction is as follows:
\[ \text{Mg} + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2 \]
- **Conclusion**: Magnesium liberates hydrogen gas.

3. **Reaction of Zinc (Zn)**:
- When zinc reacts with dilute nitric acid, the reaction is:
\[ \text{Zn} + 2 \text{HNO}_3 \rightarrow \text{Zn(NO}_3)_2 + \text{NO} + \text{H}_2\text{O} \]
- **Conclusion**: Zinc does not liberate hydrogen gas; instead, it produces nitric oxide (NO).

4. **Reaction of Tin (Sn)**:
- When tin reacts with dilute nitric acid, the reaction is:
\[ \text{Sn} + 4 \text{HNO}_3 \rightarrow \text{Sn(NO}_3)_2 + 2 \text{NO} + 2 \text{H}_2\text{O} \]
- **Conclusion**: Tin does not liberate hydrogen gas; it produces nitric oxide (NO).

5. **Reaction of Lead (Pb)**:
- When lead reacts with dilute nitric acid, the reaction is:
\[ \text{Pb} + 4 \text{HNO}_3 \rightarrow \text{Pb(NO}_3)_2 + 2 \text{NO}_2 + 2 \text{H}_2\text{O} \]
- **Conclusion**: Lead does not liberate hydrogen gas; it produces nitrogen dioxide (NO2).

6. **Final Count**:
- From the analysis:
- Magnesium (Mg) liberates hydrogen gas.
- Zinc (Zn) does not.
- Tin (Sn) does not.
- Lead (Pb) does not.
- Therefore, only **1 element (Magnesium)** liberates hydrogen gas when treated with dilute nitric acid.

### Final Answer:
Only **1 element** liberates hydrogen gas when treated with dilute nitric acid.

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