Question

The correct order of the decreasing ionic radii among the following isoelectric species is :

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

Answer 1

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

In the formation of cation, the electrons are lost from the outer valence shell and the remaining electrons experience a greater force of attraction by the nucleus. In other words, nucleus hold the remaining electrons more tighty and this results is decreased radii.
In case of anion formation, the addition of electrons occurs in the same outer shell thus the hold of nucleus on the electron of outer shell decreases and results in increased ionic radii.
$\therefore$ Order of ionic radii is $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$
These are isoelectronic species with 18 electrons each. More than nucleus charge, smaller is the size of the ion. Nuclear charge are $S\left(16\right),Cl\left(17\right),,K\left(19\right),Ca\left(20\right)$