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.

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