Question

All alkali metals dissolve in anhydrous liquid ammonia to give a blue colour solution. It is the ammoniated electron which is responsible for the blue colour of the solution, and the electrical conductivity is due to ammoniated cation, ${[M{(N{H_3})_x}]^ + }$ as well as the ammoniated electron, ${[e{(N{H_3})_y}]^ - },$ values of $x$ and $y$ depends on the extent of solvation by $N{H_3}$ ​. Dilute solutions are paramagnetic due to free ammoniated electrons.
What happens if alkali metals are allowed to react with concentrated liquid ammonia?
(A) Paramagnetic character of solvated electrons is retained
(B) Solvated electrons associate to form electron pairs and paramagnetic character decreases
(C) Reducing character is increased
(D) Reducing character is not affected

Answer 1

To answer this question, you should recall the concept of properties of s-block elements and the effect of their addition in liquid ammonia. They form blue color solutions due to the presence of electrons.

Complete step by step solution:
The s block elements are known as alkali metals and have only one electron in their s-orbital. In liquid ammonia, you can generate solvated electrons. Here, when you dissolve sodium metal in very cold liquid ammonia solvent, the sodium metal loses its electron, which results in the ion becoming surrounded by the positive end as ammonia is polar of several ammonia molecules. The solvated electron is of special importance in radiation chemistry. Alkali metals show the property that when dissolved in liquid ammonia they give deep blue solutions which are conducting in nature. The resulting blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light. These solutions contain ammoniated cations and ammoniated electrons can be represented as: $M{\text{ }} + {\text{ }}\left( {{\text{ }}x{\text{ }} + {\text{ }}y{\text{ }}} \right){\text{ }}N{H_3}{\text{ }} \to {\text{ }}{M^ + }{\text{ }}{\left( {{\text{ }}N{H_3}} \right)_{x}} + {\text{ }}{e^ - }{\left( {N{H_3}} \right)_y}$
Dilute solutions of alkali metals in liquid ammonia are dark blue in colour but as the concentration increases above $3{\text{ M}}$ , the colour changes to copper-bronze and the solutions acquire metallic lustre due to formation of metal ion clusters.
Thus, the correct answer to this question is option B.

Note:
You should know about the preparation of liquid ammonia used in the above process. The liquid ammonia required for the production of ammoniated electrons requires Ammonia which is prepared using Haber’s process: ${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g) + 22.0{\text{kcal}}$
Make sure you remember the conditions which maximize the yield of ammonia:
Using Le Chatelier’s principle, as the above reaction is exothermic, low temperature will shift the equilibrium to the right leading to a greater yield of ammonia. A temperature of $450^\circ C$ will maximise the preparation of ammonia.
As there is a decrease in gaseous moles, high pressure on the reaction at equilibrium favours the shift of the equilibrium to the right. A pressure of ${\text{200 atm}}$ will maximize the preparation of ammonia.
To increase the rate of reaction, a catalyst is used and reaction quickly attains equilibrium.
The reactants nitrogen and hydrogen gas should be pure to increase the yield of ammonia.