# How do I generate the electron configuration of iron?

Mar 11, 2014

An atom in its ground state has the same number of electrons as protons. It is also know as a neutral atom or an uncharged atom.

So, looking at the periodic table, an atom of Iron $\left(\text{Fe}\right)$ in its ground state would have 26 electrons.

The electrons fill the orbitals closest to the nucleus first and then move out until all of the electrons are accounted for.

The first orbital filled is the $1 s$. It can hold a maximum of 2 electrons.

Next, the $2 s$, which can also hold a maximum of 2 electrons. Then the $2 p$, which can hold up to 6 electrons.

Then $3 s$ (max 2 electrons), $3 p$ (max 6 electrons), $4 s$ (max 2), $3 d$ (can hold a maximum of 10 electrons), then $4 p$ (max 6), $5 s$ (max 2), $4 d$ (max 10), $5 p$ (max 6), $6 s$ (max 2), $4 f$ (can hold a maximum of 14 electrons), $5 d$ (max 10), $6 p$ (max 6), $7 s$ (max 2), $5 f$ (max 14), $6 d$ (max 10), $7 p$ (max 6), 8s (max 2) and so on.

So, the electron configuration for Iron would then be:

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

If you add up the electron numbers (the numbers following the letters $s$ , $p$, and $d$, you will see that they add up to 26, which is the number of electrons that iron has in its ground state. You now know where each of the 26 electrons are located. I hope this helps.