Question

The correct decreasing order of atomic size of 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 atoms and ions are isoelectronic.
In isoelectronic ions, the atomic size decreases as $\frac{z}{e}$ ratio increases.
$\text{For}{S}^{2-},\frac{z}{e}=\frac{16}{18}$
$C{l}^{-},\frac{z}{e}=\frac{17}{18}$
$K^{+},\frac{z}{e}=\frac{19}{18}$
$C{a}^{2+},\frac{z}{e}=\frac{20}{18}$
So the correct decreasing order of atomic size is
$S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$