Answer
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Hint: When Group-1 (alkali metals) and Group-2 (alkaline earth metals) are added to liquid ammonia, they dissolve to form metal cations and solvated electrons. Alkaline earth metals form metal-complex with ammonia i.e. ammoniates.
Complete step-by-step answer:
The alkaline earth metals are the metals of group 2 of the modern periodic table. Their general electronic configuration is \[n{s^2}\]. They readily lose two electrons to form cations of charge +2. Thus, the most common oxidation state shown by these metals is +2.
Alkaline earth metals form ammonia solvated cation and electrons when treated with ammonia. The solution formed is electrically conductive, paramagnetic in nature and reductive as well. The solvated electrons get absorbed in the visible region and thus the solution becomes blue in colour.
But the concentrated solution is brown coloured. On standing it for long, it decomposes into amide and ammonia with evolution of hydrogen gas.
\[M + (x + y)N{H_3} \to {[M{(N{H_3})_X}]^ + }{[M{(N{H_3})_Y}]^ - } \to M{(N{H_3})_6} + \dfrac{1}{2}{H_2}\]
The solvated electrons act as a reducing agent and get stabilized by hydrogen bonding and hexammoniates, \[M{(N{H_3})_6}\] is the precipitate formed as it is not soluble in liquid ammonia but when dissolved, it produces very concentrated solutions of alkaline earth metals.
Hence, the correct option is (A).
Note: Beryllium do not form hexammoniates on reacting with liquid ammonia rather that it forms tetraammine beryllium salts and hydrogen gas is evolved when excess of beryllium is used. Thus, it is an exception to this reaction.
Complete step-by-step answer:
The alkaline earth metals are the metals of group 2 of the modern periodic table. Their general electronic configuration is \[n{s^2}\]. They readily lose two electrons to form cations of charge +2. Thus, the most common oxidation state shown by these metals is +2.
Alkaline earth metals form ammonia solvated cation and electrons when treated with ammonia. The solution formed is electrically conductive, paramagnetic in nature and reductive as well. The solvated electrons get absorbed in the visible region and thus the solution becomes blue in colour.
But the concentrated solution is brown coloured. On standing it for long, it decomposes into amide and ammonia with evolution of hydrogen gas.
\[M + (x + y)N{H_3} \to {[M{(N{H_3})_X}]^ + }{[M{(N{H_3})_Y}]^ - } \to M{(N{H_3})_6} + \dfrac{1}{2}{H_2}\]
The solvated electrons act as a reducing agent and get stabilized by hydrogen bonding and hexammoniates, \[M{(N{H_3})_6}\] is the precipitate formed as it is not soluble in liquid ammonia but when dissolved, it produces very concentrated solutions of alkaline earth metals.
Hence, the correct option is (A).
Note: Beryllium do not form hexammoniates on reacting with liquid ammonia rather that it forms tetraammine beryllium salts and hydrogen gas is evolved when excess of beryllium is used. Thus, it is an exception to this reaction.
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