Question

When $Na$ reacts with liquid $N{H_3}$ the following substance is formed:
A.${\left[ {Na{{\left( {N{H_3}} \right)}_x}} \right]^ - }$
B.${\left[ {e{{\left( {N{H_3}} \right)}_y}} \right]^ - }$
C.$NaN{H_2}$
D.$N{a_x}N{H_{3y}}$

Answer 1

To answer this question, you must recall the properties of alkali metals and their solutions in liquid ammonia. Alkali metals when added to liquid ammonia form a deep blue colored solution.

Complete step by step answer:
We know that liquid ammonia is a good polar solvent like water. It is obtained by the condensation of ammonia gas. Just like water, liquid ammonia undergoes self- ionization to form an ammonium ion and amide ion.
When sodium metal is added to the liquid ammonia in small amounts, that is, when a dilute solution of sodium in liquid ammonia is formed, we obtain a blue coloured solution. Sodium metal in a polar solvent loses an electron to give sodium cation. The sodium cation and the electron are both solvated by liquid ammonia to form solvated ions, ${\left[ {Na{{\left( {N{H_3}} \right)}_x}} \right]^ + }$ and ${\left[ {e{{\left( {N{H_3}} \right)}_y}} \right]^ - }$.
The solution is highly conducting due to the presence of these ammoniated ions and ammoniated electrons. The ammoniated electron formed is also responsible for the blue colour of the solution as it absorbs energy in the visible region of the electromagnetic spectrum. The solution is paramagnetic.
In concentrated solutions of concentration above 3 molar, the solution is copper-bronze in colour and has a metallic lustre because of formation of metal ion clusters. This solution is diamagnetic in nature.

Thus, the correct option is B.

Note:
If the liquid ammonia solution is impure, that is, if it contains some impurities like iron metal, then a different reaction takes place. The iron impurities present act as catalysts for the reaction occurring between the ammoniated metal ion and ammonia. Sodium reacts with ammonia and forms sodium amide, that is, $NaN{H_2}$ and hydrogen gas is released.