The ionic radii (. textin Å.) of N3 -, O2 - and F- are respectively:

Question

The ionic radii (. textin Å.) of N3 -, O2 - and F- are respectively: (A) 1.71,1.36 and 1.40 (B) 1.36,1.40 and 1.71 (C) 1.36,1.71 and 1.40 (D) 1.71,1.40 and 1.36

Tardigrade Admin · Accepted Answer

As (Z/ e ) uparrow ionic radius decreases for isoelectronic species. Z= numberofprotonse = numberofelectrons ( N )3-(( Z / e ))=(7/10) ( O )2-(( Z / e ))=(8/10) ( F )-(( Z / e ))=(9/10) Trend of ionic radius: N 3-> O 2-> F -

Q. The ionic radii $(in Å)$ of $N^{3 -}, O^{2 -}$ and $F^{-}$ are respectively:

$1.71, 1.36$ and $1.40$

$1.36, 1.40$ and $1.71$

$1.36, 1.71$ and $1.40$

$1.71, 1.40$ and $1.36$

As $\frac{Z}{e} ↑$ ionic radius decreases for isoelectronic species.
$Z =$ numberofprotonse $=$ numberofelectrons
$(N)^{3 -} (\frac{Z}{e}) = \frac{7}{10}$
$(O)^{2 -} (\frac{Z}{e}) = \frac{8}{10}$
$(F)^{-} (\frac{Z}{e}) = \frac{9}{10}$
Trend of ionic radius: $N^{3 -} > O^{2 -} > F^{-}$

Q. The ionic radii (inÅ) of N3−,O2− and F− are respectively:

1.71,1.36 and 1.40

1.36,1.40 and 1.71

1.36,1.71 and 1.40

1.71,1.40 and 1.36