Question

The ionic radii of N^3-, O^2-, F^- and Na⁺ follow the order:

• A. N^3- > O^2- > F^- > Na⁺
• B. N^3- > Na⁺ > O^2- > F^-
• C. Na⁺ > O^2- > N^3- > F^-
• D. O^2- > F^- > Na⁺ > N^3-

Hint:

These four ions are isoelectronic; more protons means smaller size.

Answer 1

The Correct Option is A

All four species, $N^{3-}$, $O^{2-}$, $F^{-}$ and $Na^{+}$, have exactly 10 electrons each. They are isoelectronic, so their electron count cannot explain any size difference between them.

• The only thing that differs is the number of protons in the nucleus: N has 7, O has 8, F has 9, Na has 11.
• A nucleus with more protons pulls harder on the same 10-electron cloud, so the cloud is dragged in tighter and the ion becomes smaller.
• Na⁺, with 11 protons pulling on only 10 electrons, is squeezed the most and is the smallest of the four.
• N^3-, with only 7 protons for 10 electrons, has the weakest pull and the largest cloud.

Ranking from weakest to strongest pull gives the size order $N^{3-} > O^{2-} > F^{-} > Na^{+}$, so option A is correct.