Why do elements in columns 6 and 11 assume 'abnormal' electron configurations? When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital. 

What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.

Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.
As I understand this, there are basically two effects at work here.

When you populate an $\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in $\mathrm{d}$-orbitals and putting one in an $\mathrm{s}$-orbital increases as you fill the $\mathrm{d}$-orbitals.

Additionally, pairing electrons in one orbital (so, adding the second $\mathrm{s}$ electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.

I'm assuming that the effect isn't strong enough to avert fluorine having a $\mathrm{2s^2}$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the $\mathrm{s}$ from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the $\mathrm{5d}$ instead of the $\mathrm{4f}$ orbitals.

Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $\mathrm{6s^2}$ occupation, but is for gold - more $\mathrm{d}$ electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an $\mathrm{s}$-orbital.