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(A) The species given are $Mg^{2+}$,$Na^{+}$,$F^{-}$,and $Al$.$1$. $Al$ is a neutral atom,while $Mg^{2+}$,$Na^{+}$,and $F^{-}$ are ions.$2$. Comparing

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$Mg^{2+} < Na^{+} < F^{-} < Al$

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(A) The species given are $Mg^{2+}$,$Na^{+}$,$F^{-}$,and $Al$.$1$. $Al$ is a neutral atom,while $Mg^{2+}$,$Na^{+}$,and $F^{-}$ are ions.$2$. Comparing the ionic radii of isoelectronic species ($Mg^{2+}$,$Na^{+}$,$F^{-}$): These all have $10$ electrons. The ionic radius decreases as the nuclear charge $(Z)$ increases. The nuclear charges are $F^{-} (Z=9)$,$Na^{+} (Z=11)$,and $Mg^{2+} (Z=12)$. Thus,the order of size is $Mg^{2+} $3$. The neutral atom $Al$ $(Z=13)$ has a larger atomic radius than these cations. However,comparing $Al$ to the others,the correct order of size is $Mg^{2+} < Na^{+} < F^{-} < Al$.

The size of the following species increases in the order:

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