Question

Arrange the following ions in the increasing order of their radii:
\[ O^{2-}, \; N^{3-}, \; F^{-}, \; Mg^{2+} \]

• A. IV < II < I < III
• B. III < IV < II < I
• C. IV < III < I < II
• D. IV < II < I < IV

Hint:

For isoelectronic species, ionic radius decreases with increasing atomic number (Z).

Answer 1

The Correct Option is C

Step 1: Recognise these ions as an isoelectronic series.
Count the electrons on each ion: $N^{3-}$ has $7+3=10$, $O^{2-}$ has $8+2=10$, $F^{-}$ has $9+1=10$, and $Mg^{2+}$ has $12-2=10$. All four carry exactly 10 electrons, the neon configuration, so they form an isoelectronic series.

Step 2: Compare the number of protons pulling on the same 10 electrons.
\[ N: 7, \quad O: 8, \quad F: 9, \quad Mg: 12 \]
Since every ion has the same electron cloud, the one with more protons pulls that cloud in harder and ends up smaller.

Step 3: Rank by proton count directly.
More protons means a smaller ion, so from most protons (smallest) to fewest protons (largest):
\[ Mg^{2+}(12) \lt F^{-}(9) \lt O^{2-}(8) \lt N^{3-}(7) \]

Step 4: Translate protons into radius order.
Reading this the other way round, in increasing order of radius:
\[ Mg^{2+} \lt F^{-} \lt O^{2-} \lt N^{3-} \]

Step 5: Match to the roman numerals given.
\[ IV \lt III \lt I \lt II \]

Step 6: Why the other orders are wrong.
Any order that puts $N^{3-}$ before $Mg^{2+}$, such as options (1), (2), and (4), forgets that this ion has the fewest protons pulling on the same 10 electrons, which makes it the largest, not the smallest.

Final Answer:
\[ \boxed{IV \lt III \lt I \lt II} \]