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:

(B) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$

(A) $\mathrm{N}^{3-}<\mathrm{O}^{2-}<\mathrm{F}^{-}<\mathrm{Na}^{+}<\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}$

(C) $\mathrm{Al}^{3+}<\mathrm{Na}^{+}<\mathrm{Mg}^{2+}<\mathrm{O}^{2-}<\mathrm{F}^{-}<\mathrm{N}^{3-}$

(D) $\mathrm{N}^{3-}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{Al}^{3+}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Classification of Elements and Periodicity

The correct order of the ionic radii of

O^{2 -}, N^{3 -}, F^{-}, Mg^{2 +}, Na^{+}

and

Al^{3 +}

is:

(A) $N^{3 -} < O^{2 -} < F^{-} < Na^{+} < Mg^{2 +} < Al^{3 +}$
(B) $Al^{3 +} < Mg^{2 +} < Na^{+} < F^{-} < O^{2 -} < N^{3 -}$ ✓ Correct
(C) $Al^{3 +} < Na^{+} < Mg^{2 +} < O^{2 -} < F^{-} < N^{3 -}$
(D) $N^{3 -} < F^{-} < O^{2 -} < Mg^{2 +} < Na^{+} < Al^{3 +}$

Answer: (B)

Solution

All six ions are isoelectronic, each with ten electrons, so their sizes depend only on nuclear charge: the more protons pulling on the same ten electrons, the smaller the ion. Ordering from most protons to fewest,

Al X^{3 +}

is smallest and

N X^{3 -}

largest, giving

Al X^{3 +} < Mg X^{2 +} < Na X^{+} < F X^{-} < O X^{2 -} < N X^{3 -}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Classification of Elements and Periodicity

Solution

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