Ionization Energy Trends

Ionization energies increase for a given atom as successive electrons are removed. This is because it becomes more difficult to remove a negatively charged particle from a positively charged particle. Additionally if an electron is being removed from a lower energy level the increased attraction the electron experiences to the nucleus is another important factor. Ionization energies generally increase going across a period. This is best understood in terms of the effective nuclear charge experienced by the valence electron. However, that in an of itself does not explain the trend as we proceed across the 2nd period.



















However, there are two variations on this basic trend. It is easier to remove an electron from a p orbital compared to an s orbital of the same quantum number. The p orbital is higher in energy, so the electron is further from the nucleus, easier to remove. Resulting in a lower ionization energy. (The Z* for the alkaline earth metals is Be (1.95), Mg (2.85), Ca (2.85), Sr (2.85), Ba (2.85)).

In going from nitrogen to oxygen there is a drop in the first ionization energy. This occurs because of the increased electron-electron repulsions that occur in the single filled 2p orbital in oxygen. So it is slightly easier to remove the electron in oxygen compared to nitrogen (or what you would extrapolate from nitrogen).

Why does sodium lose only one electron? why not two or more? While the exact details of the answer require a complete Born-Haber cycle treatment, one of the terms is the enrgy cost associated with ionization of the metal. For sodium the 1st ionization energy is 496 kJ·mol-1. The electron configuration is (1s2)(2s2,2p6)(3s1). The Z* for the electron in the 3s orbital is 11 - (2 · 1.00 + 8 · 0.85) = 2.2. However to remove the next electron requires much more energy since the Z* is much higher (4560 kJ·mol–1). 11 – (2 · 0.85 + 7 · 0.35) = 6.85 Over three times higher. The only place in the Born-Haber cycle to recover the energy is in the hydration of the ion and enough energy is not recovered in that step for the Na2+ ion.

What about removing electrons from the transition metal elements? Adding electrons we fill the 4s before the 3d. However when removing electrons from a transition metal we must take the first two electrons from the 4s than from the 3d. There is an apparent reversal of the relative energies. In general once we begin adding electrons to the 3d level the energy of the subshell drops. Lets compare the Z* for a 4s electron and a 3d electron in Fe.

(1s2)(2s2,2p6)(3s2,3p6) (3d6) (4s2)

Z*(4s) = 26 – (10 · 1.00 + 14 · 0.85 + 1 · 0.35) = 3.75

Z*(3d) = 26 – (18 · 1.00 + 5 · 0.35) = 6.25