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

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.

& 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{aligned}$$ Here M is Li, B is$$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.