Question

Which ranks the following atoms (Li, Na, K, Rb) from the smallest to largest atomic radius?
(A) ${\text{Li}} < {\text{Na}} < {\text{K}} < {\text{Rb}}$
(B) ${\text{Rb}} < {\text{K}} < {\text{Na}} < {\text{Li}}$
(C) ${\text{K}} < {\text{Na}} < {\text{Li}} < {\text{Rb}}$
(D) ${\text{Na}} < {\text{Li}} < {\text{Rb}} < {\text{K}}$

Answer 1

Hint: Atomic radius can be defined as a distance from the nucleus of the atom to the outermost electron. Since the orbitals around the atoms are the probability distribution and do not have fixed boundaries. An average size for most atoms can be determined by comparing the bond lengths of the number of compounds of an element.

Complete step by step answer:
On moving down the group, number of shells increases in an atom and due to this increase in the number of shells, atomic size increases. Also on moving down the group, effective nuclear charge between the nucleus and the outermost shell starts decreasing due to which the atomic size increases.
Let us understand this by writing the electronic configuration of the elements given is as follows:
Li: $1{{\text{s}}^{\text{2}}}\;{\text{2}}{{\text{s}}^{\text{1}}}$
Na: $1{{\text{s}}^{\text{2}}}\;{\text{2}}{{\text{s}}^2}\;{\text{2}}{{\text{p}}^{\text{6}}}\;{\text{3}}{{\text{s}}^{\text{1}}}$
K: ${\text{1}}{{\text{s}}^{\text{2}}}\;{\text{2}}{{\text{s}}^{\text{2}}}\;{\text{2}}{{\text{p}}^{\text{6}}}\;{\text{3}}{{\text{s}}^{\text{2}}}\;{\text{3}}{{\text{p}}^{\text{6}}}\;{\text{4}}{{\text{s}}^{\text{1}}}$
Rb: ${\text{1}}{{\text{s}}^{\text{2}}}\;{\text{2}}{{\text{s}}^{\text{2}}}\;{\text{2}}{{\text{p}}^{\text{6}}}\;{\text{3}}{{\text{s}}^{\text{2}}}\;{\text{3}}{{\text{p}}^{\text{6}}}\;{\text{3}}{{\text{d}}^{{\text{10}}}}\;{\text{4}}{{\text{s}}^{\text{2}}}\;{\text{4}}{{\text{p}}^{\text{6}}}\;{\text{4}}{{\text{d}}^{{\text{10}}}}\;{\text{5}}{{\text{s}}^{\text{2}}}\;{\text{5}}{{\text{p}}^{\text{6}}}\;{\text{6}}{{\text{s}}^{\text{1}}}$
As we can see from the electronic configuration that the Li, Na, K and Cs belongs to the same group as the outer shell electron configurations are same i.e. ${\text{n}}{{\text{s}}^1}$. From the electronic configuration, we can conclude that the number of shells is increasing on moving down the group due to which atomic radius of the atom increases.
Therefore the correct order for the smallest to largest atomic radius will be ${\text{Li}} < {\text{Na}} < {\text{K}} < {\text{Rb}}$. So, the correct option is (A).

So, the correct answer is Option A.

Note: We need to remember that the atomic radius decreases on moving left to right in a period of periodic table because in period electrons are filled in the same shell. Therefore, they are not able to screen each other and thus the nuclear charge as well as the effective nuclear charge increases, which decrease the radius.