Discussion
The ionic radii of N3-, O2-, F- and Na+ follows the order
Answer: A
The order is \(\mathrm{N^{3-} > O^{2-} > F^- > Na^+}\).
All four are isoelectronic, each having 10 electrons, so the size depends only on the nuclear charge pulling them in: \[\mathrm{N}\ (Z=7) < \mathrm{O}\ (8) < \mathrm{F}\ (9) < \mathrm{Na}\ (11)\] The larger the nuclear charge for the same electron count, the smaller the ion - so radius falls as \(Z\) rises.
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