Question

(a) Why is LiF least soluble in water among the fluorides of alkali metals ? Justify the given
(b) Justify the given order of mobilites of the alkali mteal cations in aqueous solution:
`Li lt Na^(+) lt K^(+) lt Rb^(+) lt Cs^(+)`
(c) Lithium is the only alkali metal which forms a nitride directly. Explain .
(d) `E^(@)` for `M^(2+)(aq)toM(s)`(where M=Ca,Sr or Ba) is nearly constant. Discuss.

Answer 1

(a) Lithium fluoride (LiF) is of covalent nature because of the high polarising power of `Li^(+)` ion due to its very small size and high effective nuclear charge. It destorts the electron cloud of the `F^(-)` ion the maximum as compared to the cations of other alkali metals. It is therefore, least soluble in water. On the other hand, the fluorides of other alkali metals are generally ionic and are water soluble.
(b) This is attributed to the hydration of the cation in water. As a result,size of the cation increases and its mobility decreases. Due to the smallest size, `Li^(+)` ion is hydrated to the maximum and exists as `Li^(+)` (aq) and has least mobility. `Cs^(+)` ion due to least hydration exists as `Cs^(+)` (aq) has maximum mobility.
(c) Lithium is a very strong redcuing agent. As a result, it directly exists as `Cs^(+)` (aq) combines with nitrogen to form its nitride `(Li_(3)N)`.
`3Li+N_(2) overset("Heat")toLi_(3)N_(2)`
(d) The overall magnitude of reduction potential `(E^(@))` depends upon three factors. These are (i) sublimation enthalpy (ii) ionisation enthalpy and (iii) hydration enthalpy. In case of the metals listed, the overall magnitude of `E^(@)` values remain almost the same. Therefore, these metals have almost same reducing strength.