Question

The correct decreasing order of atomic size among the following species is:

$Ar$, $K^{+}$, $C{l}^{-}$, $S^{2-}$, $C{a}^{2+}$

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

Answer 1

Answer: (C)

$S^{2-}>C{l}^{-}>Ar>K^{+}>C{a}^{2+}$

In isoelectronic ions, the atomic size decreases as $\frac{z}{e}$ ratio increases.
$S^{2-}\left[\frac{z}{e}\right]=\frac{16}{18};C{l}^{-}\left[\frac{z}{e}\right]=\frac{17}{18};Ar\left[\frac{z}{e}\right]=\frac{18}{18};K^{+}\left[\frac{z}{e}\right]=\frac{19}{18};C{a}^{2+}\left[\frac{z}{e}\right]=\frac{20}{18}$