Question

The correct order of reducing character of alkali metals is

• A. $ Rb < K < Na < Li $
• B. $ Li < Na < K < Rb $
• C. $ Na < K < Rb < Li $
• D. $ Rb < Na < K < Li $

Answer 1

The Correct Option is C

The alkali metals are strong reducing agents, lithium the most and sodium the least powerful. The standard electrode potential $\left(E^{o}\right)$ which measures the reducing power represents the overall change
$M(s) \rightarrow M(g)$ Sublimation enthalpy
$M(g) \rightarrow M^{+}(g)+e^{-}$ ionization enthalpy
$M^{+}(g)+H_{2} O \rightarrow M^{+}(a q)$ hydration enthalpy
With the small size of its ion, lithium has the highest hydration enthalpy which accounts for its high negative $E^{\circ}$ value and its high reducing power.
Hence, the order of reducing character of alkali metals is $ Na < K < Rb < Li$