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

Sep 14, 2017

Cobalt in this case has an oxidation state of $+ 2$ to add with the four ${\text{Cl}}^{-}$ ligand charges and give an overall charge of $- 2$. With a $+ 2$ oxidation state, $\text{Co}$ therefore is a ${d}^{7}$ metal.

A four-coordinate complex with four ${\text{Cl}}^{-}$ 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 ${\text{Cl}}^{-}$, 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:

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" } \left({t}_{2}\right)$
$3 {d}_{x y} \textcolor{w h i t e}{\ldots . .} 3 {d}_{x z} \textcolor{w h i t e}{\ldots . .} 3 {d}_{y z}$

$\text{ "ul(uarr darr)" "ul(uarr darr)" "" "" } \left(e\right)$
$\text{ } 3 {d}_{{z}^{2}} \textcolor{w h i t e}{\ldots . .} 3 {d}_{{x}^{2} - {y}^{2}}$

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