The radii of F,F^{-},O and O^{2-} are in the | vedclass

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(A) $1$. In a period,atomic radius decreases from left to right. Therefore,the atomic radius of $O > F$.$2$. $O^{2-}$ and $F^{-}$ are isoelectronic sp

Vedclass · Accepted Answer

$O^{2-} > F^{-} > O > F$

Vedclass · Answer

(A) $1$. In a period,atomic radius decreases from left to right. Therefore,the atomic radius of $O > F$.$2$. $O^{2-}$ and $F^{-}$ are isoelectronic species,both having $10$ electrons.$3$. For isoelectronic species,the ionic radius increases as the nuclear charge (atomic number) decreases.$4$. Since the atomic number of $O$ $(Z=8)$ is less than $F$ $(Z=9)$,the ionic radius of $O^{2-} > F^{-}$.$5$. Combining these,the overall order is $O^{2-} > F^{-} > O > F$.

The radii of $F$,$F^{-}$,$O$ and $O^{2-}$ are in the order of:

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