Question

Consider the isoelectronic ions:
$K^{+},S^{2-},C{l}^{-}$ and $C{a}^{2+}$
The radii of these ionic species follow the order:

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

Answer 1

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

As we move along a period, size decreases, and as we move down the group size increases.

Element	$Z$	Electron
$K^{+}$	19	18
$S^{2-}$	16	18
$C{l}^{-}$	17	18
$C{a}^{2+}$	20	18

Above are isoelectronic species, hence size $\propto \frac{1}{Z}$
So size order is $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$

Hence, option C is correct.