A metal $$M$$ on heating in nitrogen gas gives $$Y$$. $$Y$$ on treatment with $$\text{H}_2\text{O}$$ gives a colourless gas which when passed through $$\text{CuSO}_4$$ solution gives a blue colour. $$Y$$ is

When a metal is heated in a current of nitrogen it generally forms the corresponding metal nitride. Thus

$$M + \dfrac{n}{2}\,N_2 \;\xrightarrow{\;\Delta\;}\; M\text{N}_n$$

The product $$Y$$ therefore should be a nitride of the given metal.

Nitrides react with water (or moist air) giving ammonia gas:

$$\text{Metal nitride} + 3\,H_2O \;\longrightarrow\; \text{Metal hydroxide} + NH_3$$

Ammonia, $$NH_3$$, is a colourless gas. When passed through an aqueous $$CuSO_4$$ solution it forms the deep-blue tetrammine complex $$[Cu(NH_3)_4]^{2+}$$, turning the solution blue: $$Cu^{2+} + 4\,NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$$.

Hence the unknown solid $$Y$$ must be a metal nitride that furnishes $$NH_3$$ on hydrolysis.

Examining the options:

A. $$NH_3$$ - already a gas, not formed by heating a metal in $$N_2$$.
B. $$Mg(NO_3)_2$$ - a nitrate, gives $$NO_2$$/$$O_2$$ upon heating, not $$NH_3$$ on hydrolysis.
C. $$Mg_3N_2$$ - magnesium nitride; reacts with water to give $$NH_3$$:
$$Mg_3N_2 + 6\,H_2O \rightarrow 3\,Mg(OH)_2 + 2\,NH_3$$
The liberated $$NH_3$$ produces the observed blue colour with $$CuSO_4$$.
D. $$MgO$$ - oxide, gives no gas on treatment with water.

Only Option C satisfies all the observations.

Therefore, $$Y = Mg_3N_2$$.

Option C which is: $$\textbf{Mg}_3\textbf{N}_2$$