The ions $\mathrm{S}^{2-}, \mathrm{Cl}^{-}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}$ are isoelectronic. Their ionic…
The ions $\mathrm{S}^{2-}, \mathrm{Cl}^{-}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}$ are isoelectronic. Their ionic radii show
a decrease from $\mathrm{S}^{2-}$ to $\mathrm{Cl}^{-}$and then increase from $\mathrm{K}^{+}$to $\mathrm{Ca}^{2+}$
an increase from $\mathrm{S}^{2-}$ to $\mathrm{Cl}^{-}$and then decrease from $\mathrm{K}^{+}$to $\mathrm{Ca}^{2+}$
a significant decrease from $\mathrm{S}^{2-}$ to $\mathrm{Ca}^{2+}$
a significant increase from $\mathrm{S}^{2-}$ to $\mathrm{Ca}^{2+}$
Solution
For isoelectronic species, the ionic radii decrease with increase in nuclear charge.
So, the cation with greater + ve charge i.e. $\mathrm{Ca}^{2+}$ will have smaller radius and the anion with greater - ve charge i.e. $\mathrm{S}^{2-}$ will have a larger radius.
$\because$ The ionic radii show a significant decrease from $\mathrm{S}^{2-}$ to $\mathrm{Ca}^{2+}$.