The element expected to form the largest ion to achieve the nearest noble gas configuration is:

Solution & Explanation

\textbf{Answer:} (C) $\mathrm{N}$ \textbf{Solution:} Each element gains or loses electrons to reach the nearest noble-gas ($\mathrm{Ne}$) configuration: $\mathrm{O}\to \mathrm{O}{}^{2-}$, $\mathrm{F}\to \mathrm{F}{}^{-}$, $\mathrm{N}\to \mathrm{N}{}^{3-}$, $\mathrm{Na}\to \mathrm{Na}{}^{+}$. All four ions are isoelectronic, each having $10$ electrons ($\mathrm{Ne}$ configuration). For an isoelectronic series, ionic size decreases as the nuclear charge $Z$ increases, because the greater effective nuclear charge pulls the electron cloud inward. The smallest $Z$ here is $\mathrm{N}$ ($Z=7$), so $\mathrm{N}{}^{3-}$ has the largest ionic radius.