Question

The ionic radii of $N^{3-}$ and $F^{-}$ are $1.76\stackrel{o}{A}$ and $1.36\stackrel{o}{A}$ respectively. The ionic radius of $O^{2-}$ will be:

• A. $1.40\stackrel{o}{A}$
• B. $1.96\stackrel{o}{A}$
• C. $1.20\stackrel{o}{A}$
• D. $2.06\stackrel{o}{A}$

Answer 1

The correct option is A $1.40\stackrel{o}{A}$

Since radii $\alpha \frac{e^{-}}{P}\alpha \frac{1}{P}$ (if $e^{-}$ are same)

$\therefore$ radii $=N^{3-}>0^2>F^{-}$

$\downarrow$

$1.40A^0$