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

1 Answer

Answer:

#[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#