The ionic radii of ${\mathrm{F}}^{-}$ and ${\mathrm{O}}^{2-}$ respectively are $1.33 \stackrel{°}{\mathrm{A}}$ and $1.4 \stackrel{°}{A}$, while the covalent radius of $\mathrm{N}$ is $0.74 \stackrel{°}{A}$. The correct statement for the ionic radius of ${\mathrm{N}}^{3-}$ from the following is:

For isoelectronic species, as the charge on anion increases, the ionic radius increases ${\mathrm{F}}^{-}, {\mathrm{O}}^{2-}$ and ${\mathrm{N}}^{3-}$ has 10 electrons each. Their ionic radii are in decreasing order, as follows:

 ${\mathrm{N}}^{3-}>{\mathrm{O}}^{2-}>{\mathrm{F}}^{-}$

Now, Anions have more ionic radii than corresponding neutral atoms thus anion ${\mathrm{N}}^{3-}$ will have more ionic radii than covalent radii of $\mathrm{N}$.

$\therefore {\mathrm{N}}^{3-}$ is bigger than both ${\mathrm{O}}^{2-}$ and ${\mathrm{F}}^{-}$.

Hence, the required option is C.