Question

A metal ‘M’ reacts with nitrogen gas to give ‘\[{{M}_{3}}N\]’. \[{{M}_{3}}N\]on heating at high temperature gives back ‘M’ and on reaction with water produces gas ‘B’. Gas ‘B’ reacts with aqueous solutions of \[CuS{{O}_{4}}\] to form a deep blue color compound. ‘M’ and ‘B’ 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

Hint: Copper sulphate generally reacts with a gas having nitrogen and hydrogen and gives a deep blue colour complex. Then the gas B which is produced may be a gas with a pungent smell and it turns nessler's reagent brown. The metal M belongs to the second period and group 1 of the periodic table. It has an atomic number of 3.

Complete step by step answer:
In the question it is given that a metal ‘M’ reacts with nitrogen gas and gives ‘\[{{M}_{3}}N\]’ as a product.
\[\]\[M+{{N}_{2}}\to {{M}_{3}}N\]

In the compound\[{{M}_{3}}N\], Nitrogen has a valence of 3, metal ‘M’ has valence of 1.
Means metal M should have a valence of 1.
Coming to given options Na and Li only have a valence of 1. Barium has valence of 2 and aluminum has a valence of 3. Therefore option C and option D are wrong.
Later the formed \[{{M}_{3}}N\] reacts with water and produces a gas B.
\[{{M}_{3}}N+3{{H}_{2}}O\to 3MOH+N{{H}_{3}}\]

Here the formed gas is ammonia (\[N{{H}_{3}}\]).
Ammonia reacts with aqueous Copper sulphate and forms a deep blue color compound as a product.
\[N{{H}_{3}}+CuS{{O}_{4}}\to \underset{blue\text{ }color\text{ }complex}{\mathop{[Cu{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}}}\,\]
In the question it is mentioned that again on heating \[{{M}_{3}}N\] at high temperature gives back M.
Sodium also forms a metal nitride. But the problem is sodium nitride is unstable at room temperature. But in the question it is given that the\[{{M}_{3}}N\]at high temperature gives back metal (M).
So, option A is also wrong.

Therefore Li has all the properties mentioned in the question. We can confirm it by the following reactions.
\[\begin{align}
  & Li+{{N}_{2}}\to L{{i}_{3}}N \\
 & L{{i}_{3}}N+3{{H}_{2}}O\to 3LiOH+N{{H}_{3}} \\
 & 4N{{H}_{3}}+CuS{{O}_{4}}\to \underset{blue\text{ }color\text{ }complex}{\mathop{[Cu{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}}}\, \\
\end{align}\]
Here M is Li, B is\[N{{H}_{3}}\].
So, the correct answer is “Option B”.

Note: The nitride formed in the above reaction is lithium nitride (\[L{{i}_{3}}N\]). It is the only stable alkali metal nitride. All remaining metal nitrides of alkali metal are unstable at room temperature. \[L{{i}_{3}}N\]solid has a reddish-pink color.