# What is the electron configuration of iron?

Jun 23, 2016

Iron is on the fourth row of the periodic table, sixth column of the transition metals, atomic number $26$.

What we have is:

• Its core orbitals are the $1 s$, $2 s$, $2 p$'s, $3 s$, and $3 p$'s.
• Its valence orbitals are the $4 s$ and $3 d$'s.

Writing the electron configuration, you really only need the valence orbitals, and you can omit the core orbitals by notating it via the noble gas shortcut.

So, $\left[A r\right]$ can be written instead of $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$.

Note that the total number of electrons in the neutral atom adds up to the atomic number, so $2 + 2 + 6 + 2 + 6 = 18$, which is the atomic number of $\text{Ar}$.

Finally, the $4 s$ orbitals are higher in energy than the $3 d$ orbitals by about $\text{3.75 eV}$ when they are partially filled like this, so we write the $3 d$ before the $4 s$.

Therefore, we get:

$\textcolor{b l u e}{\left[A r\right] 3 {d}^{6} 4 {s}^{2}}$

Or: (note, the original image was wrong. The $4 s$ is higher in energy, not lower, when the orbitals are filled. They are only lower in energy when they are empty.)

The long version is:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$

Indeed, $2 + 2 + 6 + 2 + 6 + 6 + 2 = 26$, the atomic number of $\text{Fe}$.