Choose the smallest ion from the following.

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(A) For an isoelectronic series,the ionic radius decreases as the nuclear charge (atomic number) increases.All the given ions ($Mg^{2+}$,$Na^{+}$,$Cl^

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$Mg^{2+}$

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(A) For an isoelectronic series,the ionic radius decreases as the nuclear charge (atomic number) increases.All the given ions ($Mg^{2+}$,$Na^{+}$,$Cl^{-}$,$S^{2-}$) are not isoelectronic. Let us check their electron configurations:$Mg^{2+}$: $1s^2 2s^2 2p^6$ ($10$ electrons)$Na^{+}$: $1s^2 2s^2 2p^6$ ($10$ electrons)$Cl^{-}$: $1s^2 2s^2 2p^6 3s^2 3p^6$ ($18$ electrons)$S^{2-}$: $1s^2 2s^2 2p^6 3s^2 3p^6$ ($18$ electrons)Comparing the isoelectronic pairs:Among $Mg^{2+}$ and $Na^{+}$,$Mg^{2+}$ has a higher nuclear charge $(Z=12)$ compared to $Na^{+}$ $(Z=11)$,so $Mg^{2+}$ is smaller.Among $Cl^{-}$ and $S^{2-}$,$Cl^{-}$ has a higher nuclear charge $(Z=17)$ compared to $S^{2-}$ $(Z=16)$,so $Cl^{-}$ is smaller.Comparing the two groups,the ions with $10$ electrons are significantly smaller than those with $18$ electrons due to fewer shells.Thus,the overall order of ionic radii is $Mg^{2+} The smallest ion is $Mg^{2+}$.

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