• # question_answer 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)  $Rb<Na<K<Li$

The alkali metals are strong reducing agents, lithium the most and sodium the least powerful. The standard electrode potential $({{E}^{o}})$ which measures the reducing power represents the overall change $M(s)\xrightarrow{{}}M(g)$ Sublimation enthalpy $M(g)\xrightarrow{{}}{{M}^{+}}(g)+{{e}^{-}}$ ionization enthalpy ${{M}^{+}}(g)+{{H}_{2}}O\xrightarrow{{}}{{M}^{+}}(aq)$ 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.  Metal Li Na K Rb ${{E}^{\text{o}-}}/V$for $({{M}^{+}}/M)$ $-3.04$ $-2.714$ $-2.925$ $-2.930$
Hence, the order of reducing character of alkali metals is $Na<K<Rb<Li.$