# What is the electronic configuration of Fe(II) ion?

The electronic configuration of Fe is $\ce{[Ar] 3d^6 4s^2}$. So after removing two electrons the configuration becomes: $\ce{[Ar] 3d^6}$

But why can't the electrons rearrange themselves to give a more stable $\ce{[Ar] 3d^5 4s^1}$ configuration?

When researching the explanation for this phenomenon, I found this:

It is natural to question why one or two electrons are usually pushed into a higher energy orbital. The answer is because $\mathrm{3d}$ orbitals are more compact than $\mathrm{4s}$, and as a result any electrons entering $\mathrm{3d}$ orbitals will experience greater mutual repulsion. The slightly unsettling feature is that although the relevant $\mathrm{s}$ orbital can relieve such additional electron-electron repulsion, different atoms do not always make full use of this form of sheltering because the situation is more complicated than just described. One thing to consider is that nuclear charge increases as we move through the atoms, and there is a complicated set of interactions between the electrons and the nucleus as well as between the electrons themselves. This is what ultimately produces an electronic configuration and, contrary to what some educators may wish for, there is no simple qualitative rule of thumb that can cope with this complicated situation.

On a semi-related note, according to one source I found (the source sites many others as well), it is in fact the $\mathrm{3d}$ electrons that fill first, followed by the $\mathrm{4s}$ electrons. This more satisfactorily explains why the $\mathrm{s}$ electrons are first lost in the ionization of the $\mathrm{d}$-block transition metals.

• But wont this configuration explain why Fe2+ is more stable than Fe+ even when Fe+ has half filled orbital but Fe2+ does not? Mar 9 at 8:49

Your mixed up with the use of half orbital. When iron become to $\ce{Fe^2+}$, it needs to donate two valance electrons away which is at the highest energy (orbital $\ce{4s}$). So, that is why we write $\ce{Fe^2+}$ as $\ce{[Ar] 3d^6}$.

• But given that half-filled orbitals have a special stability, the OP is asking, why isn't a 4s1 3d5 configuration preferred over a 4s0 3d6 configuration.
– ron
May 1 '15 at 1:36