11
$\begingroup$

Octahedral complexes have greater splitting in the d orbitals. Is it because octahedral complexes have more atoms and thus more interactions?

$\endgroup$
8
$\begingroup$

The answer is, as @orthocresol mentioned, due to two parts. Initially, if we are assuming the ligands are point charges (crystal field theory) then in octahedral complexes, the ligands are more closely associated with the d orbitals, as shown in this diagram:

But, in tetrahedral complexes, the ligands are not so closely associated with the d orbitals, as shown in this diagram: (Try to associate all orbitals seen in the first diagram with the point charges in this diagram)

By associating the d orbitals seen in the first diagram, with the tetrahedral point charges in the second diagram, you can see how close the point charges are to the d orbitals in the octahedral case compared to tetrahedral.

Also as otho mentioned, octahedral has 6 ligands interacting with 5 d orbitals, whereas tetrahedral has only 4 ligands interacting with 5 d orbitals. Octahedral has the more negative charge due to more electrons, therefore more repulsion.

$\endgroup$
3
$\begingroup$

The crystal field theory given in Benzene’s answer is a nice simple model, but we can get a deeper, maybe more logical explanation if we check out molecular orbital theory. For a general octahedric complex, the MO scheme looks like depicted in figure 1 (only σ-donors, π effects not included because I was too lazy to draw another image).

octahedral
Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

We realise that the $\mathrm{e_g^*}$ orbitals are raised in energy because they are antibonding to the metal-ligand bonds. $\mathrm{t_{2g}}$ orbitals are not affected. Including π basic effects on the ligands, they would be raised slightly because they again would be antibonding. These interactions always stem from 6 ligands. Also, $\mathrm{e_g^*}$ are actually directly pointing at the ligands as Benzene nicely showed.

A typical tetrahedral complex, this time including π interactions to a π basic ligand is shown in figure 2.

tetrahedral
Figure 2: Tetrahedral $\ce{[ML4]}$ complex including ligand to metal π donating orbitals. Image copied from this answer.

Here, the ligand orbitals that interact in a σ manner transform as $\mathrm{a_1 + t_2}$ so the stronger destabilisation is present on $\mathrm{t_2^*}$. $\mathrm{e^*}$ is also destabilised due to being antibonding but lesser so. It is a little bit hard to see in Benzene’s image, but all of the orbitals are now pointing away from the ligands they are interacting with. Thus, the interaction must be weaker than for the octahedral case above. Finally, we also need to consider that we are dealing with the mixing of less orbitals here: there are only four ligands this time further reducing the energy difference.

$\endgroup$

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.