# Why does aluminum have a lower first ionization energy than magnesium?

I used to use the explanation that $s$ orbitals penetrate better than $p$ orbitals, however, could the explanation be that $3s$ is shielding the $3p$ in aluminum?

The easiest way to explain it is that $\ce{Al}$ has one unpaired electron in it's highest energy orbital ($\mathrm{3p}$), and $\ce{Mg}$'s highest energy orbital ($\mathrm{3s}$) the electrons are paired. It is energetically favorable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Here's what their orbital pairings look like:

and if you look at the general trend you can see that this occurs whenever all the valence electrons are paired:

• Electrons repel each other so pairing them raises potential energy. Note that O is lower than N for this reason. – Brinn Belyea Oct 12 '14 at 23:47
• I've edited my answer to reflect that discrepancy. In the cases where the entire orbital level has electrons of one spin, adding a lone electron of opposite spin is unfavorable because of repulsion. = – John Snow Oct 13 '14 at 0:07
• Note that in the trend chart that i posted, the noble gasses have the highest 1st ionization energies. This is because all of their valence electrons are paired, and it is energetically unfavorable to unpair them. – John Snow Oct 13 '14 at 0:15

could the explanation be that 3s is shielding the 3p in aluminum?

Yes, one could say that for 2p, 3p and 4p orbitals; it becomes more confusing for 5p and 6p. One can also say that a half filled p orbital (3 electrons) shields the nucleus against the 4th electron.

Removal of an electron will disrupt the stable filled 3s subshell of magnesium. Also the 3p electron of aluminium is further from the nucleus compared to the 3s electrons of magnesium