Question

Why is the electron configuration of `"Fe"^(2+)` NOT `[Ar] 4s^2 3d^4` but `[Ar] 3d^6`?

Answer 1

I can see why this can seem counterintuitive at first. The last orbital you fill when making neutral `"Fe"` is the `3d`.

`"Fe"` (neutral) is `[Ar]4s^2 3d^6`, where one `3d` orbital is doubly-occupied. Since one of them has two electrons, it adds a bit of repulsion that would help in removing an electron. Additionally, one might be tempted to remove electrons from the `3d` orbitals first, simply because they were filled last.

But there's a problem with that assumption: the `4s` orbital is doubly-occupied too.

It is important to realize that the electrons easiest to remove are removed first, and those in the highest-energy orbital tend to belong to that category. In this case, the `4s` orbitals are slightly higher in energy, so those get removed first. As a result... we get:

`color(blue)([Ar]3d^6)`

NOT:

`color(red)([Ar]4s^2 3d^4)`

As a result, this is paramagnetic (i.e. it has at least one unpaired electron in its orbital configuration).