The ionic radii of F^{-} and O^{2-} respectiv | vedclass

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(C) $F^{-}$,$O^{2-}$,and $N^{3-}$ are all isoelectronic species with $10$ electrons each. In isoelectronic species,the ionic radius increases as the n

Vedclass · Accepted Answer

It is bigger than $O^{2-}$ and $F^{-}$

Vedclass · Answer

(C) $F^{-}$,$O^{2-}$,and $N^{3-}$ are all isoelectronic species with $10$ electrons each. In isoelectronic species,the ionic radius increases as the nuclear charge (number of protons) decreases. The number of protons in $N^{3-}$,$O^{2-}$,and $F^{-}$ are $7$,$8$,and $9$ respectively. Since $N^{3-}$ has the lowest nuclear charge,it experiences the least nuclear attraction on its valence electrons,resulting in the largest ionic radius. Therefore,the order of ionic radii is $N^{3-} > O^{2-} > F^{-}$. Thus,the ionic radius of $N^{3-}$ is bigger than both $O^{2-}$ and $F^{-}$.

The ionic radii of $F^{-}$ and $O^{2-}$ respectively are $1.33 \ \mathring{A}$ and $1.40 \ \mathring{A}$,while the covalent radius of $N$ is $0.74 \ \mathring{A}$. The correct statement for the ionic radius of $N^{3-}$ from the following is:

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