Question

A metal $X$ on heating in nitrogen gas gives $Y$. $Y$ on treatment with $H_{2}O$ gives a colorless gas which when passed through $CuS{O}_4$ solution gives a blue color. $Y$ is:

• A. $Mg(N{O}_3{)}_2$
• B. $M{g}_3{N}_2$
• C. $N{H}_3$
• D. $MgO$

Answer 1

Answer: (B)

$M{g}_3{N}_2$

The metal $X$ is $Mg$. On heating in nitrogen gas gives $M{g}_3{N}_2$
$\underset{X}{3Mg}+N_2\stackrel{\Delta }{\to }\underset{Y}{M{g}_3{N}_2}$
$M{g}_3{N}_2$ on treatment with water gives colorless ammonia gas.
$M{g}_3{N}_2+6H_{2}O\to 3Mg(OH{)}_2+2N{H}_3$
Ammonia reacts with $CuS{O}_4$ to form blue coloured complex $\left[Cu\right(N{H}_3{)}_4]S{O}_4$.
$CuS{O}_4+4N{H}_3\to \left[Cu\right(N{H}_3{)}_4]S{O}_4$