Question

The correct order of reducing character of alkali metal is:
\[
  A)Rb < K < Na < Li \\
  B)Li < Na < K < Rb \\
  C)Na < K < Rb < Li \\
  D)Rb < Na < K < Li \\
\]

Answer 1

Hint: The alkali metals have low values of reduction potential and therefore have a strong tendency to lose electrons and act as good reducing agents. The reducing character increases from sodium to caesium. However lithium is the strongest reducing agent.

Complete answer: Among all the alkali metals, $Li$ is the strongest and $Na$ is the weakest reducing agent. $Li$, although, has the highest Ionization energy, yet it's the strongest reducing agent. The tendency of an element to lose an electron is measured by its oxidation potential. More the value of ${E^ \circ }$ of an element stronger will be it’s reducing character.
The energy change, measured in volts, required to add or remove electrons to or from an element or compound. The oxidation potential of other species are determined relatively by measuring the potential difference between a half-cell containing an aqueous solution of the oxidized and reduced forms of the test substance, and the standard hydrogen half-cell.
The alkali metals have a low value of ionization energy which decreases down the group and so can easily lose their valence electrons and thus act as good reducing agents.
So, the correct option is $B)Li < Na < K < Rb$

Note:
The excess energy required for ionization of the small lithium atom is more than compensated by the release of excess hydration energy. This accounts for the higher oxidation potential of lithium, the greater ease with which the following overall change occurs, Thus, in net shell the greater reducing power of lithium is due to its large heat of hydration which is due to its smaller size.