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The correct order of the decreasing ionic radii among the following isoelectronic species is
A
$K^+ > Ca^{2+} > Cl^- > S^{2-}$
B
$Ca^{2+} > K^+ > S^{2-} > Cl^-$
C
$Cl^- > S^{2-} > Ca^{2+} > K^+$
D
$S^{2-} > Cl^- > K^+ > Ca^{2+}$
Detailed Solution
All four ions have 18 electrons (isoelectronic), but different nuclear charges: $S^{2-}$ (Z = 16), $Cl^-$ (Z = 17), $K^+$ (Z = 19), $Ca^{2+}$ (Z = 20).
For isoelectronic species, the greater the nuclear charge, the more strongly the same number of electrons is pulled in and the smaller the ion.
So the radius decreases as Z increases: $S^{2-}$ (184 pm) > $Cl^-$ (181 pm) > $K^+$ (138 pm) > $Ca^{2+}$ (100 pm).
Decreasing order of ionic radii: $S^{2-} > Cl^- > K^+ > Ca^{2+}$
For isoelectronic species, the greater the nuclear charge, the more strongly the same number of electrons is pulled in and the smaller the ion.
So the radius decreases as Z increases: $S^{2-}$ (184 pm) > $Cl^-$ (181 pm) > $K^+$ (138 pm) > $Ca^{2+}$ (100 pm).
Decreasing order of ionic radii: $S^{2-} > Cl^- > K^+ > Ca^{2+}$