A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’.  Gas ‘B’ reacts with an aqueous solution of CuSO4 to form a deep blue compound.  ‘M’ and ‘B’ respectively are :

Answers (1)

Li and NH3?
The metal ‘M’ forms nitride M3?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.

M3?N reacts with water to produce gas ‘B’. This is given by Li

\begin{aligned} &\mathrm{Li}_{3} \mathrm{N}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{LiOH}+\mathrm{NH}_{3}\\ &\text { Thus, gas 'B' is ammonia } \end{aligned}

\begin{aligned} &\text { Ammonia reacts with aqueous } \mathrm{CuSO}_{4} \text { solution to form deep blue compound }\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}\\ &\mathrm{NH}_{3}+\mathrm{CuSO}_{4} \rightarrow\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \end{aligned}

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