Consider the isoelectronic species $K^{+}, S^{- 2}, {Cl}^{-}, {Ca}^{+ 2},$ the radii of the ions decrease as

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

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

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

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

answer is C.

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Detailed Solution

Isoelectronic species contain same no. of electrons with different charges. The size of ion depends on total no. of electrons to proton ratio.

For $S^{- 2} ion, e / p = 18 / 16 = 1.125; for {Cl}^{-} ion, e / p = 18 / 17 = 1.058$ $For K^{+} ion, e / p = 18 / 19 = 0.947; for {Ca}^{+ 2} ions, e / p = 18 / 20 = 0.900$

Hence, the decreasing order of radii of ions is

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

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