A metal (A) on heating in nitrogen gas gives compound B. B on treatment with $H_{2}O$ gives a colorless gas which when passed through $CuS{O}_4$ the solution gives a dark blue-violet-colored solution. A and B respectively, are:

The metal (A) should be magnesium, which it reacts with nitrogen to give compound (B) $$M{g}_3{N}_2$$. The obtained $$M{g}_3{N}_2$$ upon treatment with  $$H_{2}O$$ gives ammonia $$N{H}_3$$ which is a colorless gas. When this ammonia $$N{H}_3$$ passed through $$CuS{O}_4$$ solution it forms dark blue-violet complex 

$$\left[Cu(N{H}_3{)}_4\right]S{O}_4$$. ​     

$$3Mg+N_2\to M{g}_3{N}_2$$

$$M{g}_3{N}_2+6H_{2}O\to 3Mg(OH{)}_2+2N{H}_3$$     

$$CuS{O}_4+4N{H}_3\to \left[Cu(N{H}_3{)}_4\right]S{O}_4$$

Hence the correct option is (C).