The ionic radii in mathring{A} of C^{4-} and | vedclass

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(B) $C^{4-}$,$N^{3-}$,and $O^{2-}$ are isoelectronic species,all having $10$ electrons.For isoelectronic species,the ionic radius decreases as the nuc

Vedclass · Accepted Answer

$1.71$

Vedclass · Answer

(B) $C^{4-}$,$N^{3-}$,and $O^{2-}$ are isoelectronic species,all having $10$ electrons.For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.The atomic numbers are: $C (Z=6)$,$N (Z=7)$,and $O (Z=8)$.Since the atomic number of $N$ $(Z=7)$ lies between $C$ $(Z=6)$ and $O$ $(Z=8)$,its ionic radius will lie between the ionic radii of $C^{4-}$ $(2.60 \ \mathring{A})$ and $O^{2-}$ $(1.40 \ \mathring{A})$.Among the given options,$1.71 \ \mathring{A}$ is the only value that falls between $1.40$ and $2.60$.

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

Similar Questions

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