Question

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

Answer 1

`[Ar] 3d^4`

or

`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(4)`

Explanation:

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

The reason for this is because this configuration minimizes electron repulsion. Half filled orbitals for `"Cr"` in particular is its most stable configuration.

So the electron configuration for elemental Chromium is

`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5)`.

And the electrons in the `4s` 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 `Cr^(2+)` ion we remove 1 `4s` electron and 1 `3d` electron leaving us with:

`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(4)`

or

`[Ar] 3d^4`