In the isoelectronic species,the ionic radii | vedclass

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(C) For isoelectronic species,the ionic radius decreases as the effective nuclear charge $(Z_{eff})$ increases.$N^{3-}$ $(Z=7)$,$O^{2-}$ $(Z=8)$,and $

Vedclass · Accepted Answer

$1.71, 1.40, 1.36$

Vedclass · Answer

(C) For isoelectronic species,the ionic radius decreases as the effective nuclear charge $(Z_{eff})$ increases.$N^{3-}$ $(Z=7)$,$O^{2-}$ $(Z=8)$,and $F^{-}$ $(Z=9)$ all have $10$ electrons.As the atomic number $(Z)$ increases from $7$ to $9$,the attraction of the nucleus for the electrons increases,causing the ionic radius to decrease.The ionic radii are:$N^{3-} = 1.71 \ \mathring{A}$$O^{2-} = 1.40 \ \mathring{A}$$F^{-} = 1.36 \ \mathring{A}$Therefore,the correct order 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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