The radius of $Na^{+}$ cation is less than that of $Na$ atom. Give reason.

(N/A) For the formation of a cation, there is a loss of an electron, which means its radius decreases. Due to the decrease in the number of electrons, the nuclear pull per electron increases.
As a result, the radius of the cation decreases as the effective nuclear charge increases. For example, the ionic radius of $Na^{+}$ is smaller than that of the parent atom $Na$.
$Na \longrightarrow Na^{+} + 1e^{-}$
Electrons: $11, 10$
Nuclear charge: $11, 11$
Ionic size: $186 \text{ pm}, 95 \text{ pm}$
$(1 \text{ pm} = 10^{-12} \text{ m})$