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: 
![enter image description here](https://i.stack.imgur.com/e9aSc.jpg)


and if you look at the general trend you can see that this occurs whenever all the valence electrons are paired: 
![enter image description here](https://i.stack.imgur.com/i8l4C.gif)




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Sources:
<http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html>
<http://en.wikipedia.org/wiki/Electron_pair>


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


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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


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