# What is the electron configuration of "Cr"^(2+) ?

May 5, 2018

$\left[A r\right] 3 {d}^{4}$

or

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

#### Explanation:

Chromium and Copper are two special cases when it comes to their electron configurations- having only 1 electron in the $4 s$ orbital, as opposed to the other transition metals in the first row which has a filled $4 s$ orbital.

The reason for this is because this configuration minimizes electron repulsion. Half filled orbitals for $\text{Cr}$ in particular is its most stable configuration.

So the electron configuration for elemental Chromium is

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{1} 3 {d}^{5}$.

And the electrons in the $4 s$ orbital is removed first because this orbital lies further from the nucleus, making electrons easier to remove in ionization.

So if we remove 2 electrons to form the $C {r}^{2 +}$ ion we remove 1 $4 s$ electron and 1 $3 d$ electron leaving us with:

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

or

$\left[A r\right] 3 {d}^{4}$