# Why does F have a higher ionization energy than O?

The book I'm working from says F has a higher ionization energy than O because the nucleus has more protons so there is a higher attraction on the electrons. Isn't that balanced out by the higher number of electrons or is it because the atomic size is smaller?

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• Of course it is balanced to some extent, otherwise the ionization energy would depend monotonically on the atomic number, which in fact is not quite so. – Ivan Neretin Nov 1 '15 at 18:43

At first approximation, we need to consider the orbital we are extracting an electron from. Both oxygen and fluorine (and nitrogen, carbon, boron and neon) have a $\mathrm{2p}$ orbital as their highest occupied, so we expect to ionise from that. If you draw a trend of the ionisation energies, you observe that the energy rises from boron to nitrogen, is lower for oxygen and rises from oxygen to neon. Disregarding the slight decrease at oxygen at first approximation we can say that it becomes increasingly hard to ionise an electron from a $\mathrm{2p}$ orbital with higher element number.
This is because essentially the shape and size of the $\mathrm{2p}$ orbitals are (close enough to, at first approximation) equal. So the wavefunction of the electrons we are removing does not change significantly. However, in their vicinity we have a nucleus which changes charge. For boron, we have an effective nuclear charge of $+1$ or $+3$ (depending on whether you count the $\mathrm{2s}$ electrons as shielding or not). For every step we take and every move we make to the right, the effective nuclear charge ($n(\ce{protons}) - n(\ce{core~electrons})$) will increase by one. We need to counter that nuclear charge to remove an electron which, naturally, because increasingly harder the stronger the charge holding it back.