A) IV < III < II < I
B) I < II < III < IV
C) II < III < IV < 1
D) IV < II < III < I
E) None of these
Correct Answer: C
Solution :
In aqueous solution the reducing character o given alkali metals follows the sequence. \[\underset{(II)}{\mathop{Na}}\,<\underset{(III)}{\mathop{K}}\,<\underset{(IV)}{\mathop{Rb}}\,<\underset{(I)}{\mathop{Li}}\,\] This trend is explained on the basis of electrode potentials more negative is the electrode potential, higher is the tendency of the element to lose electrons and nence, stronger is the reducing agent. \[E{}^\circ \]values of other alkali metals become more and more negative as we move down the group from Na to Rb. But, because electrode potential depends upon (i) enthalpy of sublimation (ii) ionisation enthalpy (iii) enthalpy of hydration ? Sublimation enthalpies of alkali metals are almost similar. ? Lithium has the smallest ionic size among all thus, its enthalpy of hydration is highest. ? Although ionisation enthalpy of lithium is the highest among alkali metals, it is more than compensated by large hydration enthalpy released, therefore \[Li\] has most negative \[E{}^\circ \], .i.e. \[-\,3.0\,4\,V.\]
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