Why is the electron configuration of #"Fe"^(2+)# NOT #[Ar] 4s^2 3d^4# but #[Ar] 3d^6#?
I can see why this can seem counterintuitive at first. The last orbital you fill when making neutral
But there's a problem with that assumption: the
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
As a result, this is paramagnetic (i.e. it has at least one unpaired electron in its orbital configuration).