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

1 Answer
anor277
Apr 19, 2017

Well consider the electronic structure of aluminum versus magnesium.........Data are from here. #"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?

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