# How do you determine the hybridisation state of a coordinate complex?

If an exercise asks for the hybridisation state of a coordinate complex (ion in solution) consisting of a central cobalt atom surrounded by $\ce{NH3}$ ligands, and neither coordination number or complex ion is given, how do you decide which configuration will occur?

The answer is that in a solution this will form a $\ce{[Co(NH3)6]3+}$ complex ion. I know that this means the hybridisation is $\mathrm{d^2sp^3}$ but can't figure out how to get there since cobalt is found in different oxidation states.

Figure 1: molecular orbital scheme of an octahedral $\ce{[ML6]}$ complex. Originally taken from Professor Klüfers’ coordination chemistry course’s web scriptum (in German).
Forcing hybridisation onto this system is very bad and very wrong; look at the huge energy differences between the 3d, 4s and 4p orbitals. Hybridisation would also lead us to assume six equal bonds, which is not the case: the six $\ce{M\bond{->}L}$ bonds transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ which leads to three different energies. And finally, hybridisation is not able to explain crystal field split, and why $\mathrm{e_g}$ — the orbitals that should take part in hybridisation and thus form bonds — are raised in energy (they are $\mathrm{e_g^*}$ in MO terms).
In most cases, coordination complexes would default to octahedral or derived. The main catch case is low-spin $\mathrm{d^8}$ in which case square-planar should be assumed. This is relatively independent of the type and number of ligands and most exceptions are due to sterics. Thus, octahedral complex is a safe first guess. Note that the only difference between two complexes of the same metal and ligands but with different metal oxidation states is typically whether a Jahn-Teller distortion is relevant and how strong it is.