The correct order of decreasing ionic radii among the following isoelectronic species is:

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

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

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

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

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Solution

The correct option is D $S^{2 -} > C l^{-} > K^{+} > C a^{2 +}$
Ionic radius of the cation is always smaller than that of the parent atom. Ionic radius of the anion is always greater than that of the parent atom. Greater the effective nuclear charge, smaller will be the ionic radii. Greater the negative charge, greater will be the size of the anion. Greater the postive charge, smaller will be the size of the cation.

Among the ions, $S^{2 -}, C l^{-}, K^{+}, C a^{2 +}$ the effective nuclear charge and the positive charge increases and the ionic radius decreases. Hence, the correct answer is D i.e,
$S^{2 -} > C l^{-} > K^{+} > C a^{2 +}$

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Similar questions

Which of the following sets contain only isoelectronic ions?
(i) $Z n^{2 +}, C a^{2 +}, G a^{3 +}, A l^{3 +}$

(ii) $K^{+}, C a^{2 +}, S c^{3 +}, C l^{-}$

(iii) $P^{3 -}, S^{2 -}, C l^{-}, K^{+}$

(iv) $T i^{4 +}, A r, C r^{3 +}, v^{5 +}$

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

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

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

C a^{2 +}

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

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