If an exercise asks for the hybridisation state of a coordinate complex (ion in solution) consisting of a central cobalt atom surrounded by $\ce{NH_3}$ 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 $\ce{[Co(NH_3)_6]^{3+}}$ 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.


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 π effects.

orbital scheme octahedral complex
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.

  • $\begingroup$ You'd perhaps be surprised to know that in Indian curriculum we are supposed to use hybridization even in CFT. Like in hexaaquacobalt(III), after splitting into t2g and eg, cobalt's six electrons fill the t2g and eg is empty. We are then supposed to use the two empty d orbitals of eg, plus 4s and 4p to give it a d2sp3 hybridization, which are then filled with electrons from aqua ligands. Sounds pretty horrible to you, I expect ;) $\endgroup$ – TRC yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.