JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Classification of Elements and Periodicity

${\mathrm{N}}^{3-}$, ${\mathrm{O}}^{2-}$ and ${\mathrm{F}}^{-}$ each carry ten electrons, so they form an isoelectronic series. Across such a series the same ten electrons are held by a nuclear charge that rises from $Z=7$ in nitrogen to $Z=8$ in oxygen and $Z=9$ in fluorine, so the electron cloud is pulled in progressively and the radius falls in the order ${\mathrm{N}}^{3-}>{\mathrm{O}}^{2-}>{\mathrm{F}}^{-}$. The covalent radius of neutral nitrogen, $0.74$ Å, is far smaller still, because adding three electrons to the same nucleus increases electron–electron repulsion and expands the ion. The ionic radius of ${\mathrm{N}}^{3-}$ is therefore larger than that of both ${\mathrm{O}}^{2-}$ and ${\mathrm{F}}^{-}$.