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(D) $1$. For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases. In $O^{2-}$,$F^{-}$,and $Na^{+}$,all ha

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$Ni > Cu > Zn$

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(D) $1$. For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases. In $O^{2-}$,$F^{-}$,and $Na^{+}$,all have $10$ electrons. The order $O^{2-} (Z=8) > F^{-} (Z=9) > Na^{+} (Z=11)$ is correct.$2$. For $P^{3-} (Z=15)$,$S^{2-} (Z=16)$,and $K^{+} (Z=19)$,these are not all isoelectronic. However,$P^{3-}$ and $S^{2-}$ have $18$ electrons,and $K^{+}$ has $18$ electrons. The order $P^{3-} > S^{2-} > K^{+}$ is correct based on decreasing nuclear charge.$3$. For $Br^{-} > S^{2-} > Cl^{-} > F^{-}$,the order is incorrect. The correct order based on the periodic table (group and period trends) is $Br^{-} > S^{2-} > Cl^{-} > F^{-}$. Wait,let's re-evaluate: $Br^{-}$ (period $4$) is larger than $Cl^{-}$ (period $3$). $S^{2-}$ (period $3$) is larger than $Cl^{-}$ (period $3$) because $S^{2-}$ has a lower nuclear charge. So $Br^{-} > S^{2-} > Cl^{-} > F^{-}$ is actually correct.$4$. In the $3d$ transition series,the atomic radius decreases from $Sc$ to $Ni$ and then increases slightly due to increased electron-electron repulsion. The order $Ni > Cu > Zn$ is incorrect because the atomic radius of $Zn$ is larger than $Ni$ and $Cu$ due to the shielding effect of the fully filled $3d^{10}$ subshell.

Which of the following represents the incorrect order of ionic or atomic radii?

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