Question

Find the correct order of the ionic radius of following isoelectronic species.

• A. $C{a}^{2+}<K^{+}<C{l}^{-}<S^{2-}$
• B. $C{a}^{2+}<K^{+}<S^{2-}<C{l}^{-}$
• C. $K^{+}<C{a}^{2+}<S^{2-}<C{l}^{-}$
• D. $K^{+}<C{a}^{2+}<C{l}^{-}<S^{2-}$

Answer 1

The correct option is A $C{a}^{2+}<K^{+}<C{l}^{-}<S^{2-}$

Isoelectronic species have same number of electrons in their electronic shells but their nuclear charge differs because of their difference in number of protons in the nucleus. With increase in number of protons in the nucleus the electrons are more attracted towards nucleus thereby causing the decrease in ionic radius.
$C{a}^{2+},K^{+},C{l}^{-1}$ and $S^{2-}$ have $18$ electrons. But $C{a}^{2+}$ have $20$ protons in its nucleus exerts higher effective nuclear charge than, $K^{+}$ having 19 protons than, $C{l}^{-}$ having 17 protons and than $S^{2-}$ having 16 protons. So, order of effective nuclear charge is $C{a}^{2+}>K^{+}>C{l}^{-}>S^{2-}$.
As size is inversely proportional to effective nuclear charge, therefore, order of size is: $C{a}^{2+}<K^{+}<C{l}^{-}<S^{2-}$