The ionic radii of N^{3-},O^{2-},F^{-},and Na | vedclass

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(A) The given ions $N^{3-}$,$O^{2-}$,$F^{-}$,and $Na^{+}$ are isoelectronic species,as each contains $10$ electrons.For isoelectronic species,the ioni

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given ions $N^{3-}$,$O^{2-}$,$F^{-}$,and $Na^{+}$ are isoelectronic species,as each contains $10$ electrons.For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases.The number of protons in these ions are: $N$ $(7)$,$O$ $(8)$,$F$ $(9)$,and $Na$ $(11)$.Since the nuclear charge increases from $N^{3-}$ to $Na^{+}$,the ionic radius decreases in the order: $N^{3-} > O^{2-} > F^{-} > Na^{+}$.

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

Similar Questions

Which has the highest value of ionic radius?

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The ionic radii $(\mathring{A})$ of $C^{4-}$ and $O^{2-}$ respectively are $2.60$ and $1.40$. The ionic radius of the isoelectronic ion $N^{3-}$ would be........$\mathring{A}$

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