The species $S^{2 -}, K^{+}, C a^{2 +}, C l^{-}$ contain the same number of electrons.
The increasing order of their radii is:

S^{2 -} > C l^{-} > K^{+} > C a^{2 +}

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$S^{2 -} > C l^{-} > C a^{2 +} > K^{+}$

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$C l^{-} > K^{+} > C a^{2 +} > S^{2 -}$

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$S^{2 -} > K^{+} > C l^{-} > C a^{2 +}$

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Solution

The correct option is A $S^{2 -} > C l^{-} > K^{+} > C a^{2 +}$

We can see that all the given species are isoelectronic with 18 electrons. As the charge on the cation increases, size decreases. Also, as the charge on the anion increase, size increases.
Therefore, the correct order of radii is $S^{2 -} > F^{-} > K^{+} > C a^{2 +}$

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Similar questions

The correct order of ionic size of the species $K^{+}, C l^{-}, C a^{2 +}$ and $S^{2 -}$ is:

K^{+}, C l^{-}, S^{2 -}, C a^{2 +}

S^{2 -}, K^{+}, C a^{2 +}, C l^{-}

S^{2 -}, N a^{+}, C a^{2 +}, C l^{-}

F, L i, M g, K

{C a}^{2 +}, K^{+}, {C l}^{-}, S^{2 -}

F < L i < M g < K

{C a}^{2 +} < K^{+} < {C l}^{-} < S^{2 -}

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