# What is the electron configuration of iron iii?

Jun 3, 2018

$1 {s}^{2} \setminus 2 {s}^{2} \setminus 2 {p}^{6} \setminus 3 {s}^{2} \setminus 3 {p}^{6} \setminus 3 {d}^{5}$

#### Explanation:

Iron(III) is the ion of iron in its $3 +$ oxidation state, meaning that it has lost $3$ electrons from its valence shell.

The standard electron configuration for an iron atom is:

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

Check to see if the total electrons add up to the proton number of the iron, which is $26$.

Now, for iron to lose three electrons, it'll lose its outermost $4 s$ electrons (both of them), and one of its $3 d$ electrons. This leaves a configuration of:

$1 {s}^{2} \setminus 2 {s}^{2} \setminus 2 {p}^{6} \setminus 3 {s}^{2} \setminus 3 {p}^{6} \setminus 3 {d}^{5}$