A metal $'M'$ reacts with $N_{2}$ to give a compound $'A^{'}\left( M_{3}N \right).\ \ 'A'$ on heating at high temperature gives back $'M^{'}and\ 'A'$ on reacting with $H_{2}O$ gives a gas $B.\ \ 'B'$ turns $CuSO_{4}$ solution blue on passing through it. $M\ and\ B$ can be

Question

Dr. Abubacker · Accepted Answer

(b)
The formula of $'A^{'}is\ M_{3}N.$ It suggests that $M$ is a monovalent metal.
$\ \ \ \ \ \ \ \ \ 6Li + N_{2}\  \rightarrow 2Li_{3}N\ \overset{\ \ \ \ \ \mathrm{\Delta}\ \ \ \ \ }{\rightarrow}\ 6\ Li + N_{2}$
$\ \ \ \ \ \ \ \ \ (M)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (A)$
lithium nitride
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ Li_{3}N + 3H_{2}O\ \  \rightarrow 3LiOH + NH_{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (B)$
$\ \ \ \ \ \ \ \ \ \ \ \ CuSO_{4} + 4NH_{3}\  \rightarrow \left\lbrack Cu\left( H_{3} \right)_{4} \right\rbrack SO_{4}$
blue solution
Hence, $M\ and\ B$ are Li and $NH_{3}$ respectively.

Q. A metal $'M'$ reacts with $N_{2}$ to give a compound $'A^{'}\left( M_{3}N \right).\ \ 'A'$ on heating at high temperature gives back $'M^{'}and\ 'A'$ on reacting with $H_{2}O$ gives a gas $B.\ \ 'B'$ turns $CuSO_{4}$ solution blue on passing through it. $M\ and\ B$ can be

Explanation

Reviewed by Dr. Abubacker

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