Question

The correct order of the ionic radii of
\(\mathrm{O^{2-}}\), \(\mathrm{N^{3-}}\), \(\mathrm{F^{-}}\), \(\mathrm{Mg^{2+}}\), \(\mathrm{Na^{+}}\) and \(\mathrm{Al^{3+}}\) is :

• A. \(\mathrm{N^{3-} \lt O^{2-} \lt F^{-} \lt Na^{+} \lt Mg^{2+} \lt Al^{3+}}\)
• B. \(\mathrm{N^{3-} \lt F^{-} \lt O^{2-} \lt Mg^{2+} \lt Na^{+} \lt Al^{3+}}\)
• C. \(\mathrm{Al^{3+} \lt Na^{+} \lt Mg^{2+} \lt O^{2-} \lt F^{-} \lt N^{3-}}\)
• D. \(\mathrm{Al^{3+} \lt Mg^{2+} \lt Na^{+} \lt F^{-} \lt O^{2-} \lt N^{3-}}\)

Hint:

All six ions are isoelectronic (10 electrons), so the ionic radius decreases as the nuclear charge (atomic number) increases.

Answer 1

The Correct Option is D

All the given ions have 10 electrons (isoelectronic with Ne), so size depends on nuclear charge: more protons pull the same electron cloud closer, giving a smaller radius.
Atomic numbers: N = 7, O = 8, F = 9, Na = 11, Mg = 12, Al = 13.
Nuclear charge order: \(\mathrm{N^{3-}} \lt \mathrm{O^{2-}} \lt \mathrm{F^{-}} \lt \mathrm{Na^{+}} \lt \mathrm{Mg^{2+}} \lt \mathrm{Al^{3+}}\).
Hence the radius order is the reverse: \(\mathrm{Al^{3+} \lt Mg^{2+} \lt Na^{+} \lt F^{-} \lt O^{2-} \lt N^{3-}}\), which is option (4).
Answer: (4) \(\mathrm{Al^{3+} \lt Mg^{2+} \lt Na^{+} \lt F^{-} \lt O^{2-} \lt N^{3-}}\)