The correct decreasing order of atomic size among the following species is :
$Ar, K^{+}, {Cl}^{-}, S^{2 -}, {Ca}^{2 +}$

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${Ca}^{2 +} > K^{+} > Ar > {Cl}^{-} > S^{2 -}$

$K^{+} > {Ca}^{2 +} > {Cl}^{-} > Ar > S^{2 -}$

$S^{2 -} > {Cl}^{-} > Ar > K^{+} > {Ca}^{2 +}$

$S^{2 -} > Ar > {Cl}^{-} > {Ca}^{2 +} > K^{+}$

answer is C.

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Detailed Solution

$Ar, K^{+}, {Cl}^{-}, S^{2 -}, {Ca}^{2 +}$ have the same number of electrons. Their radii would be different because of their different nuclear charges.

Due to the stronger attraction of the electrons to the nucleus, the cation with the larger positive charge will have a smaller radius.

The radius of an anion will be bigger the more negatively charged it is. In this scenario, the nuclear charge will be outweighed by the net repulsion of the electrons, causing the ion to grow in size.

Hence, the correct order will be : $S^{2 -} > {Cl}^{-} > Ar > K^{+} > {Ca}^{2 +}$ .