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?