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

1 Answer
Dec 3, 2015

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).