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

Apr 19, 2017

Well consider the electronic structure of aluminum versus magnesium.........Data are from here. $\text{A priori}$, we would expect that the ionization enthalpies of aluminum to be higher.

#### Explanation:

$M g , Z = 12 : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2}$

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

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

${\text{3rd ionization energy, 7730 kJ"*"mol}}^{-} 1$.

$A l , Z = 13 : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{1}$

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

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

${\text{3rd ionization energy, 2881 kJ"*"mol}}^{-} 1$

${\text{4th ionization energy, 11600 kJ"*"mol}}^{-} 1$

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 $M g$, and the fourth ionization energy for $A l$, so disproportionately high?