Comment on each of the following observation:
a. The mobilities of the alkali metal ions in aqueous solution are `Li^(o+)ltNa^(o+)ltK^(o+)ltRb^(o+)ltCs^(o+)`.
b. Lithium is the only alkali metal to form a nitride directly.
c. `E^(ө)` for `M_(aq)^(2+)+2e^(-)to M_((s))` (where M=Ca,Sr or Ba) is nearly constant.

Explanation

(a) On moving down the alkali group, the ionic and atom sizes the metals increase The given alkali metal ions can be arranged in the increasing order of their ionic sizes as:
`Li^(+) lt Na^(+) lt K^(+) lt Rb^(+) lt Cs^(+)`
Smaller the size of an ion, the more highly is it hydrated. Since `Li^(+)` is the smallest, it gets heavily hydrated in an aqeous solution. On the other hand, `Cs^(+)` is the largest and so it the least hydrated. The given alkali metal ions can be arranged in the decreasing order of their hydrations as:
`Li^(+) gt Na^(+) gt K^(+) gt Rb^(+) gt Cs^(+)`
Greater the mass of a hydrated ion, the lower is its ionic mobility therefore, hydrated `Li^(+)` is the least mobile and hydrated `Cs^(+)` is the most mobile. Thus, the given alkali metal ions can be arranged in the increasing order of their mobilities as:
`Li^(+) lt Na^(+) lt K^(+) lt Rb^(+) lt Cs^(+)`
(b) Unlike the other elements of group 1, Li reacts directly with nitrogen to form lithium nitride. This is because `Li^(+)` is very small in size and so its size is the most compatible with the `N^(3-)` ion. Hence, the lattice energy released is very high. This energy also overcomes the high amount of energy required for the formation of the `N^(3-)` ion.
(c) Electrode potential `(E^(o))` and any `M^(2+)//M` electrode upon three factors:
(i) Ionisation enthalpy
(ii) Enthalpy of hydration
(iii) Enthalpy of vaporisation
The combined effect of these factors is approximately the same for Ca, Sr, and Ba Hence. Their electrode potentials and nearly constant.