When metals dissolve in liquid ammonia, they form solvated electrons. These solvated electrons are responsible for the characteristic blue-black color of the solution. Among the given options, calcium is the metal that forms solvated electrons in liquid ammonia, resulting in a blue-black solution. Other metals like aluminum, gallium, silicon, and germanium do not form such a solution in liquid ammonia.
The correct option is (C) : Calcium
The metal that dissolves in liquid ammonia to give a blue-black solution due to the formation of solvated electrons is calcium.
Explanation:
Alkali metals (like sodium, potassium, etc.) and alkaline earth metals (like calcium, strontium, barium) can dissolve in liquid ammonia. The dissolved metal atoms lose their valence electrons, which become solvated by ammonia molecules. These solvated electrons are responsible for the characteristic blue color of the solution. At higher concentrations, the solution becomes bronze-colored.
The general reaction can be represented as:
\(M \rightarrow M^+ + e^-\)
\(e^- + xNH_3 \rightarrow [e(NH_3)_x]^-\) (solvated electron)
The solvated electrons absorb light in the visible region, giving the solution its blue color. At higher concentrations, metal ions and solvated electrons associate to form metal clusters, leading to a bronze color.
Of the given options, calcium is the only alkaline earth metal, so it is the correct answer.
Therefore, the answer is calcium.
Given below are two statements. 
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In the light of the above statements, choose the correct answer from the options given below:
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is