Question

An alkaline earth metal $\left(M\right)$ dissolves in liquid ammonia forming a deep blue-black solution. The reaction involved is:

• A. $M+2N{H}_3\to M(N{H}_2{)}_2+H_2$
• B. $M+xN{H}_3\to \left[M\right(N{H}_3{)}_x]$
• C. $M+2N{H}_{4}OH\to M(OH{)}_2+2N{H}_3$
• D. $M+(x+2y)N{H}_3\to \left[M\right(N{H}_3{)}_x{]}^{2+}+2\left[e\right(N{H}_3{)}_y{]}^{-}$. (e represents electron)

Answer 1

Answer: (D)

$M+(x+2y)N{H}_3\to \left[M\right(N{H}_3{)}_x{]}^{2+}+2\left[e\right(N{H}_3{)}_y{]}^{-}$. (e represents electron)

An alkaline earth metal $\left(M\right)$ dissolves in liquid ammonia forming a deep blue-black solution. The reaction involved is $M+(x+2y)N{H}_3\to \left[M\right(N{H}_3{)}_x{]}^{2+}+2\left[e\right(N{H}_3{)}_y{]}^{-}$ (e represents electron).
An alkaline earth metal $\left(M\right)$ will lose two electrons to form $M^{2+}$ ions.
$M\to M^{2+}+2e^{-}$
Both $M^{2+}$ ions and electrons are solvated (surrounded by ammonia molecules to form $\left[M\right(N{H}_3{)}_x{]}^{2+}+2\left[e\right(N{H}_3{)}_y{]}^{-}$).