In the isoelectronic species,the ionic radii | vedclass

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(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases. The species $N^{3-}$,$O^{2-}$,and $F^{-}$ al

Vedclass · Accepted Answer

$1.71, 1.40, 1.36$

Vedclass · Answer

(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases. The species $N^{3-}$,$O^{2-}$,and $F^{-}$ all have $10$ electrons. The atomic numbers are $N (Z=7)$,$O (Z=8)$,and $F (Z=9)$. Since the nuclear charge increases in the order $N  O^{2-} > F^{-}$. The experimental values for the ionic radii are $N^{3-} = 1.71 \ \mathring{A}$,$O^{2-} = 1.40 \ \mathring{A}$,and $F^{-} = 1.36 \ \mathring{A}$. Therefore,the correct sequence is $1.71, 1.40, 1.36$.

In the isoelectronic species,the ionic radii in $\mathring{A}$ of $N^{3-}$,$O^{2-}$,and $F^{-}$ are respectively given by:

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