A metal M reacts with N_2 to form a compound | vedclass

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(C) $1$. The metal $M$ that reacts with $N_2$ to form a nitride of the type $M_3N$ is Lithium $(Li)$.$2$. The reaction is: $6Li + N_2 \rightarrow 2Li_

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$Li_3N$ and $NH_3$

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(C) $1$. The metal $M$ that reacts with $N_2$ to form a nitride of the type $M_3N$ is Lithium $(Li)$.$2$. The reaction is: $6Li + N_2 \rightarrow 2Li_3N$ (Compound $A$).$3$. When $Li_3N$ is treated with water,it undergoes hydrolysis to produce ammonia gas $(B)$: $Li_3N + 3H_2O \rightarrow 3LiOH + NH_3(g)$.$4$. Ammonia $(NH_3)$ reacts with $CuSO_4$ solution to form a deep blue complex $[Cu(NH_3)_4]^{2+}$,which confirms the presence of $NH_3$.$5$. Therefore,$A$ is $Li_3N$ and $B$ is $NH_3$.

$A$ metal $M$ reacts with $N_2$ to form a compound $A$ $(M_3N)$. Heating $A$ at a high temperature gives back $M$,and passing $A$ through $H_2O$ results in the evolution of gas $B$. Gas $B$ turns $CuSO_4$ solution blue. What are $A$ and $B$ respectively?

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$A$ metal $M$ reacts with $N_2$ to give a compound '$A$' $(M_3N)$. '$A$' on heating at high temperature gives back '$M$' and '$A$' on reacting with $H_2O$ gives a gas '$B$'. '$B$' turns $CuSO_4$ solution blue on passing through it. $A$ and $B$ can be

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