Question

Which of the following statements is not true about the dilute solutions of alkali metals in liquid ammonia ?

• A. They are deep blue coloured solutions
• B. They are highly conductive in nature
• C. They are diamagnetic in nature
• D. Ammoniated cation and solvated electrons are formed in the solution

Answer 1

The correct option is C They are diamagnetic in nature

$M+(x+y)N{H}_3\to \left[M\right(N{H}_3{)}_x{]}^{+}+\left[e\right(N{H}_3{)}_y{]}^{-}$.
where M is an alkali metal.
Dilute solutions of alkali metals in liquid ammonia are dark blue in colour and the main species present are ammoniated cation and solvated electrons.
These solution of metals in liquid ammonia conduct electricity better than any salt in any liquid. Here, the conductions is mainly due to the presence of solvated electrons.
This dilute solution of alkali metal in liquid ammonia is paramagnetic in nature due to the presence of the unpaired electrons.
Hence option (c) is a wrong statement.

Theory
General Characteristics:
Halides of alkali metals:
Other halides of lithium such as LiCl, LiBr, LiI, LiF are soluble in ethanol, acetone, and ethyl acetate. LiCl is soluble in pyridine too.
Solutions in liquid ammonia:
The alkali metals dissolve in liquid ammonia giving a deep blue solution that is conducting in nature.
$M+(x+y)N{H}_3\to \left[M\right(N{H}_3{)}_x{]}^{+}+\left[e\right(N{H}_3{)}_y{]}^{-}$
The solution is paramagnetic due to the presence of free or ammoniated electrons i.e $\left[e\right(N{H}_3{)}_y{]}^{-}$
The ammoniated electrons in the solution absorb energy in the visible region of light, Imparting blue color to the solution.
In a Concentrated solution (above 3 M), blue color changes to copper-bronze/bronze color with a metallic lustre.
Due to the formation of metal ion clusters, it becomes diamagnetic.
Ammoniated electrons with opposite electron spin associate to form diamagnetic electron pairs.
$m_s=+\frac{1}{2}$, $m_s=-\frac{1}{2}$
The conductivity of deep blue solutions is similar to that of pure metals.
Due to the presence of solvated electrons (also due to solvated cations), the conductivity decreases as the concentration increases.
Evaporation of the ammonia from the blue solutions of alkali metals yields the metal.
The blue solutions are moderately stable at room temperature, where ammonia is still a liquid.
But the reaction to give amide can occur photochemically and is catalyzed by transition metals (like Fe) and impurities.
On standing, it slowly liberates hydrogen resulting in the formation of amide.
$M_{am}^{+}+e^{-}+N{H}_3\left(l\right)\to MN{H}_{2\left(am\right)}+\frac{1}{2}{H}_{2\left(g\right)}$
where ‘am’ denotes solution in ammonia.