A metal ‘M’ reacts with nitrogen gas to afford `'M_(3)N' . 'M_(3)N'` on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous solution of `CuSO_(4)` to form a deep blue compound. ‘M’ and ‘B’ respectively are :

Options

A. `Li " and "NH_(3)`
B. `Ba " and "N_(2)`
C. `Na " and "NH_(3)`
D. `Al " and "N_(2)`

Answer

A

Explanation

`Li(s)+N_(2)(g) underset("high temperature")(iff)2Li_(3)N(s)`
`Li_(3)N+3H_(2)O to 3LiOH+NH_(3)`
`{:(CuSO_(4)+4NH_(3)to [Cu(NH_3)_(4)]SO_(4)),(" deep blue compound "):}`