Is #"CoCl"_4^(2-)# paramagnetic or diamagnetic?

1 Answer
Truong-Son N.
Sep 14, 2017

Cobalt in this case has an oxidation state of #+2# to add with the four #"Cl"^(-)# ligand charges and give an overall charge of #-2#. With a #+2# oxidation state, #"Co"# therefore is a #d^7# metal.

A four-coordinate complex with four #"Cl"^(-)# ligands (which are weak-field) generally is said, under crystal field theory, to have a small d-orbital splitting energy, making it high spin, since the #"Cl"^(-)#, treated as point charges, repel the metal #d# orbitals fairly little.

That makes the geometry easily tetrahedral (also favored since the metal is small). Thus, a #d^7# metal in a high spin four-coordinate complex would have a configuration of:

#ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "(t_2)#
#3d_(xy)color(white)(.....)3d_(xz)color(white)(.....)3d_(yz)#

#" "ul(uarr darr)" "ul(uarr darr)" "" "" "(e)#
#" "3d_(z^2)color(white)(.....)3d_(x^2-y^2)#

Clearly, there are unpaired electrons, which makes the complex paramagnetic.

Related questions
See all questions in Electron Configuration
Impact of this question
12951 views around the world
You can reuse this answer
Creative Commons License