Arrange \[{\text{Li, Na, K, Rb, Cs}}\] in order of increasing ionization energy:
A. \[{\text{Cs < Rb < K < Na < Li}}\]
B. \[{\text{K < Rb < Cs < Na < Li}}\]
C. \[{\text{Cs < K < Rb < Na < Li}}\]
D. None of these

Hint: For solving this question we first need to understand the basic concepts of alkali metals. We know that alkali metals are highly reactive elements. The alkali metals are soft with low density and low melting and low boiling point. So, based on this we arrange the alkali metals.

Complete step by step answer:
As we know that alkali metals are responsible for giving a characteristic color to the flame. Thus, they possess a strong electro-positive nature. When these alkali metals react with water it liberates ${{\text{H}}_{\text{2}}}$ gas and reacts with acids by replacing hydrogen.
Now, when we will refer to the ionization enthalpies of alkali metals that are low and gradually decreases as we move down the group from \[{\text{Li}}\] to \[{\text{Cs}}\]. Also, the melting and boiling points of alkali metals are high.
So, according to the question when we move from top to bottom in a group then the ionization energy gradually decreases. This happens due to the atomic radius. As the atomic radius increases then the removal of electrons will be easy.
Hence, \[{\text{Li,Na, K, Rb, Cs}}\] are Group IA elements from \[{\text{Li}}\] to \[{\text{Rb}}\] the ionization energy decreases. Thus, the correct order of ionization energy is \[{\text{Cs < Rb < K < Na < Li}}\](option A).

$\therefore $ The option A is correct answer.

Note:
In case of alkaline earth metals, they have higher density, higher boiling point, and higher melting point. Thus, they possess a less electropositive nature and are less reactive. Also, alkaline earth metals react with water by liberating \[{{\text{H}}_{\text{2}}}\] gas. These metals react with acids in a less vigorous way than alkali metals and displace hydrogen.