A metal `X` on heating in nitrogen gas gives `Y,Y` on treatment with `H_(2)O` gives a colourless gas which when passed through `CuSO_(4)` solution gives a blue colour. `Y` is:

Options

A. `Mg (NO_(3))_(2)`
B. `Mg_(3)N_(2)`
C. `NH_(3)`
D. `MgO`

Explanation

To solve the question, we will follow a systematic approach to identify the metal `X` and the compound `Y` it forms, and then determine the properties of `Y` when treated with water.

### Step-by-Step Solution:

1. **Identify the Metal `X`:**
The question states that a metal `X` reacts with nitrogen gas to form `Y`. A common metal that reacts with nitrogen to form a nitride is magnesium (Mg).

**Reaction:**
\[ 3 \text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2 \]

Therefore, `X` is magnesium (Mg) and `Y` is magnesium nitride (Mg₃N₂).

2. **Hydrolysis of `Y`:**
The next part of the question states that `Y` (magnesium nitride) is treated with water. The hydrolysis of magnesium nitride can be represented as follows:

**Reaction:**
\[ \text{Mg}_3\text{N}_2 + 6 \text{H}_2\text{O} \rightarrow 3 \text{Mg(OH)}_2 + 2 \text{NH}_3 \]

This reaction produces magnesium hydroxide (Mg(OH)₂) and ammonia (NH₃).

3. **Identify the Colorless Gas:**
The colorless gas produced during this reaction is ammonia (NH₃).

4. **Ammonia and Copper Sulfate:**
The question states that the colorless gas (ammonia) when passed through a copper sulfate (CuSO₄) solution gives a blue color. This is due to the formation of a complex between copper ions and ammonia.

**Reaction:**
\[ \text{Cu}^{2+} + 4 \text{NH}_3 \rightarrow \text{[Cu(NH}_3\text{)}_4]^{2+} \]
This complex ion gives the solution a blue color.

5. **Conclusion:**
Thus, we conclude that `Y` is magnesium nitride (Mg₃N₂).

### Final Answer:
`Y` is magnesium nitride (Mg₃N₂).

---