Element A burns in nitrogen to give an ionic compound B. The compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. What is the nature of compound (D)?

The reactions proceed as follows:

$\mathrm{A}+ {\mathrm{N}}_2 \to \mathrm{B}$

$\mathrm{B} + {\mathrm{H}}_2\mathrm{O} \to \mathrm{C} + \mathrm{D}$

$\mathrm{C} + {\mathrm{CO}}_2 \to \mathrm{Milkiness}$

The production of a white precipitate of carbonate causes the milkiness to emerge.

It is known that when carbon dioxide is passed through lime water ($\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$), it produces milkiness due to the formation of Calcium carbonate (${\mathrm{CaCO}}_3$).

Thus, compound C is Calcium hydroxide ($\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$).

This suggests that the element A is Ca and the Cation of the ionic compound B is Calcium and anion comes from nitrogen, so the B should be calcium nitride (${\mathrm{Ca}}_3{\mathrm{N}}_2$).

Now, when calcium nitride reacts with water, then –

${\mathrm{Ca}}_3{\mathrm{N}}_2 + 6{\mathrm{H}}_2\mathrm{O} \to 3\mathrm{Ca}{\left(\mathrm{OH}\right)}_2 + 2{\mathrm{NH}}_3$

Hence, compound D is ammonia (${\mathrm{NH}}_3$) and it is basic in nature.

Hence, option B is correct.