Ionic sizes increase in the order :-

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Question

(A) The species $Ca^{2+}$,$Cl^{-}$,$S^{2-}$,and $Ar$ are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic r

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$Ca^{2+} < Cl^{-} < S^{2-} < Ar$

Vedclass · Answer

(A) The species $Ca^{2+}$,$Cl^{-}$,$S^{2-}$,and $Ar$ are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $S$ $(Z=16)$,$Cl$ $(Z=17)$,$Ar$ $(Z=18)$,and $Ca$ $(Z=20)$.
Since the nuclear charge follows the order $S < Cl < Ar < Ca$,the ionic size follows the inverse order: $Ca^{2+} < Ar < Cl^{-} < S^{2-}$.
Comparing the given options with the standard ionic radii: $Ca^{2+} (0.99 \ \mathring{A}) < Ar (1.54 \ \mathring{A}) < Cl^{-} (1.81 \ \mathring{A}) < S^{2-} (1.84 \ \mathring{A})$.
None of the provided options are strictly correct based on the inclusion of $Ar$. However,considering the trend for the ions $Ca^{2+}$,$Cl^{-}$,and $S^{2-}$,the order is $Ca^{2+} < Cl^{-} < S^{2-}$. Option $A$ is the closest logical progression if $Ar$ is excluded or misplaced in the sequence.

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