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Question: An alkali earth metal \(\left( {\rm{M}} \right)\) dissolves in liquid ammonia forming a deep blue-bl...

An alkali earth metal (M)\left( {\rm{M}} \right) dissolves in liquid ammonia forming a deep blue-black solution. The reaction involved is:
(A) M+2NH3→M(NH2)2+H2{\rm{M + 2N}}{{\rm{H}}_{\rm{3}}} \to {\rm{M(N}}{{\rm{H}}_{\rm{2}}}{{\rm{)}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}
(A) M+xNH3→[M(NH3)x]{\rm{M + xN}}{{\rm{H}}_{\rm{3}}} \to \left[ {{\rm{M(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{x}}}} \right]
(A) M+2NH4OH→M(OH)2+2NH3{\rm{M + 2N}}{{\rm{H}}_4}{\rm{OH}} \to {\rm{M(OH}}{{\rm{)}}_{\rm{2}}}{\rm{ + 2N}}{{\rm{H}}_3}
(D) M+(x+2y)NH3→[M(NH3)x]2++2[e(NH3)y]−{\rm{M + }}\left( {{\rm{x + 2y}}} \right){\rm{N}}{{\rm{H}}_{\rm{3}}} \to {\left[ {{\rm{M(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{x}}}} \right]^{{\rm{2 + }}}}{\rm{ + 2}}{\left[ {{\rm{e(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{y}}}} \right]^{\rm{ - }}}(e−{{\rm{e}}^ - }represents electron)

Explanation

Solution

As we know that, the alkaline metals are the second family of the periodic table. These have two electrons in its outermost shell to which they can lose. As we go down the group in this family, the reducing character of the alkaline metal increases.

Complete step by step answer:
In solution, the alkaline metal atom readily loses its valence electron. Both the cation and the electron combine with ammonia to form ammoniated cation [M(NH3)x]2+{\left[ {{\rm{M(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_x}} \right]^{{\rm{2 + }}}}and ammoniated electron [e(NH3)y]−{\left[ {{\rm{e(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{y}}}} \right]^{\rm{ - }}}as shown below.
M+(x+2y)NH3→[M(NH3)x]2++2[e(NH3)y]−{\rm{M + }}\left( {{\rm{x + 2y}}} \right){\rm{N}}{{\rm{H}}_{\rm{3}}} \to {\left[ {{\rm{M(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{x}}}} \right]^{{\rm{2 + }}}}{\rm{ + 2}}{\left[ {{\rm{e(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{y}}}} \right]^{\rm{ - }}}
The ammoniated electron is responsible for the blue colour of the solution. The electrical conductivity of the solution is due to the ammoniated cation as well as ammoniated electron.
The dilute solutions are paramagnetic in nature because they contain free ammoniated electrons. As the concentration increases the ammoniated metal gets bound by free electrons and thus the solution becomes bronze colour. The decrease in paramagnetic character suggests that in concentrated solution the ammoniated electrons associate to form electron pairs.
2e−(NH3)y→[e(NH3)y]2{\rm{2}}{{\rm{e}}^{\rm{ - }}}{{\rm{(N}}{{\rm{H}}_{\rm{3}}}{\rm{)}}_y} \to {\left[ {{\rm{e(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_y}} \right]_{\rm{2}}}
Therefore, the correct option is option (D).
Note:
The blue coloured solutions (less concentrated) are paramagnetic while bronze coloured solutions (high concentrated) are diamagnetic.