Question

The correct decreasing order of atomic size among the following species is:
$K^{+},C{l}^{-},S^{2-},C{a}^{2+}$

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

Answer 1

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

The given species are isoelectronic.
So in isoelectronic ions, the atomic size decreases as the $\frac{Z}{e}$ ratio increases.

$S^{2-}:\frac{Z}{e}=\frac{16}{18},C{l}^{-}:\frac{Z}{e}=\frac{17}{18}$

$K^{+}:\frac{2}{e}=\frac{19}{18},C{a}^{2+}:\frac{Z}{e}=\frac{20}{18}$