A metal `M` reacts with `N_(2)` to give a compound `'A'(M_(3)N)`. '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 `A` and `B` can be

Options

A. `Al and NH_(3)`
B. `Li and NH_(3)`
C. `Na and NH_(3)`
D. `Mg and NH_(3)`

Answer

B

Explanation

`underset((M))(6Li)+N_(2) to underset((A))(2Li_(3))N overset(Delta)to 6Li+N_(2) Li_(3) N + 3H_(2)Ooverset("Lithium")overset("nitride")to 3LiOH + underset((B))NH_(3) CuSO_(4) +4NH_(3) to [Cu underset("solution")underset("Blue")((NH_(3))_(4))]SO_(4)`
Hence , M and B are Li and `NH_(3)` respectively .