The ionic radii of $\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}$ and $\mathrm{Mg}^{2+}$ are in the order:

(D) $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}$

(A) $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Mg}^{2+}>\mathrm{Na}^{+}$

(B) $\mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-}$

(C) $\mathrm{F}^{-}>\mathrm{O}^{2-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Classification of Elements and Periodicity

The ionic radii of

O^{2 -}, F^{-}, Na^{+}

and

Mg^{2 +}

are in the order:

Answer: (D)

All four ions are isoelectronic, each having

10

electrons, so their sizes are set by the nuclear charge pulling on those electrons. The charges increase from

O X^{2 -}

(

Z = 8

) to

F X^{-}

(

Z = 9

) to

Na X^{+}

(

Z = 11

) to

Mg X^{2 +}

(

Z = 12

); the greater the nuclear charge, the tighter the pull and the smaller the ion. The order of ionic radii is therefore

O X^{2 -} > F X^{-} > Na X^{+} > Mg X^{2 +}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option