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

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(B) Lithium $(Li)$ reacts with nitrogen $(N_2)$ to form lithium nitride $(Li_3N)$.$6Li + N_2 \rightarrow 2Li_3N$ $(A)$Lithium nitride $(Li_3N)$ on hea

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$Li$ and $NH_3$

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(B) Lithium $(Li)$ reacts with nitrogen $(N_2)$ to form lithium nitride $(Li_3N)$.$6Li + N_2 \rightarrow 2Li_3N$ $(A)$Lithium nitride $(Li_3N)$ on heating at high temperature decomposes back to $Li$ and $N_2$.$Li_3N$ reacts with water $(H_2O)$ to produce ammonia gas ($NH_3$,which is $B$).$Li_3N + 3H_2O \rightarrow 3LiOH + NH_3(g)$Ammonia gas $(NH_3)$ reacts with $CuSO_4$ solution to form a deep blue colored complex,tetraamminecopper$(II)$ sulfate.$CuSO_4 + 4NH_3 \rightarrow [Cu(NH_3)_4]SO_4$Thus,$A$ is $Li_3N$ and $B$ is $NH_3$.

$A$ 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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