How do you determine the hybridisation state of a coordinate complex?
Again, I feel a bit like a broken record. You should not use hybridisation to describe transition metal complexes.

Instead, you should discuss transition metal complexes using molecular orbital theory; a simplified version of it such as the crystal field theory may also suffice for simple complexes. Below, I have attached the orbital scheme of an octahedral coordination complex of a random transition metal not including &#x3c0; effects.


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 $\ce{M\bond{-&gt;}L}$ 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.