A wire of an alkaline earth metal X, burnt in air and dipped in water, a gas 'Y' is evolved X and Y are respectively:-

Options

A. `Na,NO_(2)`
B. `Be,NO_(2)`
C. `Mg,CO_(2)`
D. `Mg,NH_(3)`

Explanation

To solve the problem, we need to identify the alkaline earth metal (X) and the gas (Y) produced when this metal is burnt in air and then dipped in water.

### Step-by-Step Solution:

1. **Identify Alkaline Earth Metals**:
- Alkaline earth metals belong to Group 2 of the periodic table. The elements in this group include Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), and Radium (Ra).

2. **Burning in Air**:
- When alkaline earth metals are burnt in air, they typically react with nitrogen and oxygen. For example, Magnesium (Mg) reacts with nitrogen (N2) in the air to form magnesium nitride (Mg3N2).
- The reaction can be represented as:
\[ 3Mg + N2 \rightarrow Mg3N2 \]

3. **Dipping in Water**:
- When magnesium nitride (Mg3N2) is dipped in water, it reacts to produce magnesium hydroxide (Mg(OH)2) and ammonia (NH3).
- The balanced chemical reaction is:
\[ Mg3N2 + 6H2O \rightarrow 3Mg(OH)2 + 2NH3 \]

4. **Identify X and Y**:
- From the above reactions, we can identify:
- X (the alkaline earth metal) is **Magnesium (Mg)**.
- Y (the gas evolved) is **Ammonia (NH3)**.

5. **Conclusion**:
- Therefore, the alkaline earth metal X and the gas Y are:
- X = Magnesium (Mg)
- Y = Ammonia (NH3)

### Final Answer:
- X = Magnesium (Mg)
- Y = Ammonia (NH3)