Comment on each of the following observations:
$(a)$ The mobilities of the alkali metal ions in aqueous solution are
$Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$
$(b)$ Lithium is the only alkali metal to form a nitride directly.
$(c)$ $E^{\circ}$ for $M^{2+}_{(aq)} + 2e^{-} \longrightarrow M_{(s)}$ (where $M = Ca, Sr$ or $Ba$) is nearly constant.

(A) On moving down the alkali group,the ionic sizes of the metals increase as: $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$. Smaller ions have a higher charge density and are more extensively hydrated in aqueous solution. Since $Li^{+}$ is the smallest,it is the most heavily hydrated,while $Cs^{+}$ is the least hydrated. The mobility of an ion in water depends on the size of the hydrated ion; larger hydrated ions move more slowly. Thus,the order of mobility is $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$.
$(b)$ Unlike other group $1$ elements,$Li$ reacts directly with nitrogen to form lithium nitride $(Li_{3}N)$. This is because the small size of the $Li^{+}$ ion is highly compatible with the $N^{3-}$ ion,resulting in a very high lattice energy that compensates for the energy required to form the $N^{3-}$ ion.
$(c)$ The standard electrode potential $(E^{\circ})$ for the reduction $M^{2+} + 2e^{-} \longrightarrow M$ depends on the sum of the enthalpy of sublimation,ionization enthalpy,and hydration enthalpy. For $Ca, Sr$,and $Ba$,the combined effect of these factors remains approximately constant,leading to nearly identical $E^{\circ}$ values.