Question

A metal ${}^{\prime }{M}^{\prime }$ reacts with nitrogen gas to give ${}^{\prime }{M}_3{N}^{\prime }$. ${}^{\prime }{M}_3{N}^{\prime }$ on heating at high temperature gives back ${}^{\prime }{M}^{\prime }$ and on reaction with water produces gas ${}^{\prime }{B}^{\prime }$. Gas ${}^{\prime }{B}^{\prime }$ reacts with aqueous solutions of $CuS{O}_4$ to form a deep blue compound. ${}^{\prime }{M}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ respectively are:

• A. $Na$ and $N{H}_3$
• B. $Li$ and $N{H}_3$
• C. $Ba$ and $N_2$
• D. $Al$ and $N_2$

Answer 1

Answer: (A), (B)

$Li$ and $N{H}_3$
$Na$ and $N{H}_3$

$Li$ and $N{H}_3$
The metal ‘M’ forms nitride $M_{3}N$. Since N has valency 3, metal ‘M’ has valence 1. Hence, ‘M’ can be either $Li$ or $Na$. So we can rule out $Ba$ and $Al$.
$M_{3}N$ reacts with water to produce gas ‘B’. This is given by $Li$
$L{i}_{3}N+3H_{2}O\to 3LiOH+N{H}_3$
Thus, gas ‘B’ is $ammonia$

$Ammonia$ reacts with aqueous $CuS{O}_4$ solution to form deep blue compound $\left[Cu\right(N{H}_3{)}_4]S{O}_4$
$N{H}_3+CuS{O}_4\to \left[Cu\right(N{H}_3{)}_4]S{O}_4$