The ionic radii (in mathring{A}) of N^{3-},O^ | vedclass

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Question

(A) $N^{3-}$,$O^{2-}$,and $F^{-}$ are isoelectronic species,as each contains $10$ electrons.For isoelectronic species,the ionic radius decreases as th

Vedclass · Accepted Answer

$1.71, 1.40$ and $1.36$

Vedclass · Answer

(A) $N^{3-}$,$O^{2-}$,and $F^{-}$ are isoelectronic species,as each contains $10$ electrons.For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.The atomic numbers are $N (7)$,$O (8)$,and $F (9)$.Therefore,the ionic radii decrease in the order $N^{3-} > O^{2-} > F^{-}$.The values are $1.71 \ \mathring{A}$,$1.40 \ \mathring{A}$,and $1.36 \ \mathring{A}$ respectively.

The ionic radii (in $\mathring{A}$) of $N^{3-}$,$O^{2-}$ and $F^{-}$ are respectively :

Similar Questions

Consider the following species: $N^{3-}, O^{2-}, F^{-}, Na^{+}, Mg^{2+}, Al^{3+}$
$(a)$ What is common in them?
$(b)$ Arrange them in the order of increasing ionic radii.

The radius of $Na^{+}$ cation is less than that of $Na$ atom. Give reason.

Of the following,the one with the largest size is:

Arrange $Na, Na^+, F, F^-$ in the decreasing order of their radii (size).

For $Na^{+}$,$Mg^{2+}$,$F^{-}$,and $O^{2-}$; the correct order of increasing ionic radii is

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