The ionic radii of Na^{+},F^{-},O^{2-},and N^ | vedclass

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(C) The given species $Na^{+}$,$F^{-}$,$O^{2-}$,and $N^{3-}$ are isoelectronic,meaning they all contain $10$ electrons.For isoelectronic species,the i

Vedclass · Accepted Answer

$N^{3-} > O^{2-} > F^{-} > Na^{+}$

Vedclass · Answer

(C) The given species $Na^{+}$,$F^{-}$,$O^{2-}$,and $N^{3-}$ are isoelectronic,meaning they all contain $10$ electrons.For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.$N^{3-}$ has an atomic number of $7$,$O^{2-}$ has $8$,$F^{-}$ has $9$,and $Na^{+}$ has $11$.As the nuclear charge increases from $7$ to $11$,the attraction between the nucleus and the electrons increases,causing the ionic radius to decrease.Therefore,the correct order of ionic radii is $N^{3-} > O^{2-} > F^{-} > Na^{+}$.

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

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