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

Dec 3, 2015

I can see why this can seem counterintuitive at first. The last orbital you fill when making neutral $\text{Fe}$ is the $3 d$.

$\text{Fe}$ (neutral) is $\left[A r\right] 4 {s}^{2} 3 {d}^{6}$, where one $3 d$ 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 $3 d$ orbitals first, simply because they were filled last.

But there's a problem with that assumption: the $4 s$ 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 $4 s$ orbitals are slightly higher in energy, so those get removed first. As a result... we get:

$\textcolor{b l u e}{\left[A r\right] 3 {d}^{6}}$

NOT:

$\textcolor{red}{\left[A r\right] 4 {s}^{2} 3 {d}^{4}}$

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