Question

How do successive ionization energies of aluminum, and magnesium compare?

Answer 1

Well consider the electronic structure of aluminum versus magnesium.........Data are from [here.](https://en.wikipedia.org/wiki/Ionization_energy) `"A priori"`, we would expect that the ionization enthalpies of aluminum to be higher.

Explanation:

`Mg, Z= 12:1s^(2)2s^(2)2p^(6)3s^(2)`

`"1st ionization energy, 738 kJ"*"mol"^-1;`

`"2nd ionization energy, 1450 kJ"*"mol"^-1;`

`"3rd ionization energy, 7730 kJ"*"mol"^-1`.

`Al, Z= 13:1s^(2)2s^(2)2p^(6)3s^(2)3p^1`

`"1st ionization energy, 577 kJ"*"mol"^-1;`

`"2nd ionization energy, 1816 kJ"*"mol"^-1;`

`"3rd ionization energy, 2881 kJ"*"mol"^-1`

`"4th ionization energy, 11600 kJ"*"mol"^-1`

Now the electron removed that is first removed from aluminum is a p-orbital based electron. This has zero electron density at the nucleus, even though there is a greater nuclear charge. We would expect a priori that the p electron is easier to remove than an s electron, which of course, is the valence electron for magnesium. And thus the ionization energies reflect the electronic structure of the atom.

Why are the third ionization energy for `Mg`, and the fourth ionization energy for `Al`, so disproportionately high?