Question

A highly pure dilute solution of sodium in liquid ammonia:

• A. Shows blue colour
• B. Exhibits electrical conductivity
• C. Produces hydrogen gas on standing
• D. All of the above

Answer 1

The correct option is D All of the above

Blue colour is due to the presence of solvated (ammoniated) electrons.
$Na\to N{a}^{+}+e^{-}N{a}^{+}+xN{H}_3\to \left[Na\right(N{H}_3{)}_x{]}^{+}{e}^{-}+yN{H}_3\to \left[e\right(N{H}_3{)}_y{]}^{-}\text{On adding}\overline{Na+(x+y)N{H}_3\to \left[Na\right(N{H}_3{)}_x{]}^{+}+\left[e\right(N{H}_3{)}_y{]}^{-}}\text{Ammoniated sodium}\text{Ammoniated electron}$
Note: Sodium in liquid ammonia forms $NaN{H}_2$ only in the presence of a catalyst like Pt black, iron oxide, etc.

$\left[e\right(N{H}_3{)}_y{]}^{-}\stackrel{F{e}_2{O}_3}{⟶}N{H}_2^{-}+(y-1)N{H}_3$
On standing, it slowly liberates hydrogen resulting in the formation of amide.

$N{a}^{+}\left(am\right)+e^{-}+N{H}_3\left(l\right)\to NaN{H}_2\left(am\right)+\frac{1}{2}{H}_2\left(g\right)$
where, ‘am’ denotes the solution in ammonia.
The electrical conductivity of the solution is due to both ammoniated cations and ammoniated electrons.