Is cobalt(II) in the hexaamminecobalt(II) complex high spin or low spin?

Question

I want to determine the electronic configuration of $\ce{Co}$ in $\ce{[Co(NH3)_6]Cl2}$ with Crystal Field Theory.

I know that since the complex is octahedral, there would be degeneracy breaking and the d-orbitals would undergo splitting to form a $\mathrm{t_{2g}}$ level having $3$ orbitals and an $\mathrm{e_g}$ level above that with $2$ orbitals.

As cobalt is in the $+2$ oxidation state, it would have a $\mathrm{d^7}$ configuration.

Thus, if the complex is high spin, the electronic configuration is $\mathrm{(t_{2g})^5}\ \mathrm{(e_g)^2}$ and $\mathrm{(t_{2g})^6}\ \mathrm{(e_g)^1}$ if it is low spin.

This is where I ran into a problem, what type of complex is this?

Ammonia is generally considered strong field ligand, but it is weaker than say, $\ce{CO}$ as it is a σ only donor.

Wikipedia does not give anything definitive and I get inconclusive/contradictory information elsewhere on the Internet, so can someone please clarify whether the electronic configuration of cobalt in $\ce{[Co(NH3)_6]Cl2}$ is high spin or low spin?

Answer 1

Theoretically, you cannot predict a priori whether a compound is high- or low-spin. For some reason, a lot of people seem to think that it depends only on the ligand and that it is possible to unambiguously use the position of the ligand in the spectrochemical series to figure out whether a complex is high- or low-spin. This is a very narrow viewpoint and leads to lots of mistakes: for example $\ce{[Co(H2O)6]^3+}$ is low-spin although $\ce{H2O}$ is fairly low on the spectrochemical series. In truth it depends on (at least) the ligand, the metal, as well as the oxidation state, and there is no magic formula or rule that allows you to combine all three factors.

However, as a general rule of thumb, most 3d metal complexes are high-spin. If you return to the fundamental criterion for high- vs low-spin, i.e. the relative magnitudes of $\Delta_\mathrm o$ and the pairing energy, you will find that for many 3d metals, $\Delta_\mathrm o$ is small due to the poor overlap of the 3d orbitals with ligand orbitals. $\ce{[Co(NH3)6]^2+}$ falls squarely into this category.

The main exceptions occur when you have ligands that are very high on the spectrochemical series. These are nearly always π-acceptors, such as $\ce{CN-}$, bpy, phen, etc. Even so, it should be noted that there are some 3d π-acceptor complexes that are still high-spin, such as $\ce{[Co(bpy)3]^2+}$, so this shouldn't be taken as a rule but rather a rough generalisation. The other big exception is when you have high oxidation states, mainly +3 or higher. So, for example, Co(III) is nearly always low-spin except in $\ce{[CoF6]^3-}$. But there are plenty of high-spin Mn(III) and Fe(III) complexes, so even there it is not a clear-cut rule.