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(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» <html><body><p></p>Solution :(a) Lithium fluoride (LiF) is of covalent nature because of the high polarising <a href="https://interviewquestions.tuteehub.com/tag/power-2475" style="font-weight:bold;" target="_blank" title="Click to know more about POWER">POWER</a> of `Li^(+)` ion due to its very small <a href="https://interviewquestions.tuteehub.com/tag/size-1211196" style="font-weight:bold;" target="_blank" title="Click to know more about SIZE">SIZE</a> 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 <a href="https://interviewquestions.tuteehub.com/tag/hand-1015171" style="font-weight:bold;" target="_blank" title="Click to know more about HAND">HAND</a>, the fluorides of other alkali metals are generally ionic and are water soluble. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) 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. <br/> (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)`. <br/> `3Li+N_(2) overset("Heat")toLi_(3)N_(2)` <br/> (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 <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a>.</body></html>


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