Identify the species and electron count

The given ions are:

• $\mathrm{Cl^-}$
• $\mathrm{Ca^{2+}}$
• $\mathrm{K^+}$
• $\mathrm{S^{2-}}$

Let us check the number of electrons in each:

• $\mathrm{Cl^-}: 17 + 1 = 18$
• $\mathrm{Ca^{2+}}: 20 - 2 = 18$
• $\mathrm{K^+}: 19 - 1 = 18$
• $\mathrm{S^{2-}}: 16 + 2 = 18$

So all are isoelectronic species with $18$ electrons.

Rule for isoelectronic species

For isoelectronic species, the ionic radius decreases as nuclear charge increases.

This is because all species have the same number of electrons, but the ion with more protons attracts those electrons more strongly.

Compare nuclear charges

Atomic numbers are:

• $\mathrm{S} = 16$
• $\mathrm{Cl} = 17$
• $\mathrm{K} = 19$
• $\mathrm{Ca} = 20$

Since radius decreases with increasing atomic number for isoelectronic species:

$\mathrm{Ca^{2+}} < \mathrm{K^+} < \mathrm{Cl^-} < \mathrm{S^{2-}}$

Match with the options

This order corresponds to Option C.

Final answer

The increasing order of ionic radii is: $\boxed{\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}}$