How do you determine the electron configuration of Fe?

Mar 18, 2018

The ground state electron configuration of Fe is:

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

Explanation:

For all but about 20 transition metals, the Aufbau diagram is a useful tool that helps to determine the ground state electron configuration of an element. Iron (Fe) is a transition metal that follows the Aufbau rule of the filling of atomic orbitals. The atomic number of Fe is 26, which means that its atoms contain 26 protons in their nuclei, and if neutral, 26 electrons in their electron clouds. The ground state electron configuration of Fe is:

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

If you look at the Aufbau diagram, you can see that the $\text{4s}$ sublevel fills before the $\text{3d}$ sublevel because it has lower energy. The electron configuration lists the sublevels of each energy level together, even though the $\text{4s}$ sublevel has lower energy than the $\text{3d}$ sublevel.

Mar 18, 2018

Refer to the Aufbau Principle

Explanation:

From the periodic table, iron has atomic number $26$, meaning that there are $26$ electrons in each ground state iron atom.

Numbers of electrons in each atomic orbital is twice the number of electron orbits in each orbital since two electrons of the opposite spin occupy one orbital:
$2 \cdot 1 = 2$ electron in each $s$ orbital
$2 \cdot 3 = 6$ electron in each $p$ orbital
$2 \cdot 5 = 10$ electron in each $d$ orbital,
and so on so forth.

Electron orbitals fill according to the Aufbau (Build-up) Principle. That is, each added electrons fill orbitals of lower energies before filling those of higher energies, so as to minimize the electrostatic potential energy within the atom.

Red arrows (from top to bottom) in the diagram below indicates the direction of increase in energy of each orbital. The $1 s$ orbital is filled the first, followed by $2 s$, $2 p$, $3 s$, $3 p$, $4 s$, $3 d$, until all the $26$ electrons end up in their place.

You might find something like this on your scratch paper.
$1 s = 2$
$2 s = 2 \text{, } 2 p = 6$
$3 s = 2 \text{, "3p=6", } 3 d = \textcolor{b l u e}{6}$
$4 s = 2$

Giving the electron configuration
(arranged by principle energy level, e.g. $3$ for $3 d$)
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$

Notice that the $4 s$ orbital (with principle energy level $n = 4$) is filled while the $3 d$ orbital ($n = 3$) remain partially empty to attain the most stable configuration.