Question

a. Would increasing the oxidation number, say Fe(II) to Fe(III), result in any change in crystal field splitting for an octahedral complex? Explain your choice.

b. In a tetrahedral geometry all oxidation states are high spin (HS). What does this reveal about crystal field splitting Δtetrahedral versus Δoctahedral? Explain (n.b. There is only one known Co complex that does not follow this generalization and no one has been able to explain why this one complex does not follow the rule.)

Answer 1

a.

Yes. In fact, by increasing the oxidation number of the metal complex, there is always a change in the crystal field splitting in octahedral complex. The change is almost always positive too, in other words, the crystal field splitting increases.

By going from Fe(II) to Fe(III), the effective nuclear charge on the metal increases. Hence, the metal ions pull the ligand electron more. Because of this increasd attraction, the ligand electrons feel more repulsion from the ligand d orbitals, specifically the orbitals that have their lobes along the ligand axes.

The orbitals (they comprise the set)are the ones that have their lobes along the axes of ligands. Hence, these orbitals are increased in energy. This increase in energy of the set results in the increase in the crystal field splitting.

For example,

b.

In tetrahedral geometry, we find that all the complexes are high spin. The splitting in tetrahedra; geometry results in a lower doubly degenerate e level and higher triply degenerate t2 level.

Lets assume we have a d3 complex. After filling the two electron in the lower e level, the 3rd electron either pairs off with an existing e electron or goes to the upper t2 level.

The energy consideration for both is that it need to cross the crystal splitting to go to t2 level (which results in a high spin complex) or spend the pairing energy to pair off in e level.

But, since we already know that almost all the time the electron chooses to go to the t2 level by crossing the crystal field splitting , then must be less than the pairing energy or it must be small.

This actually reveals that the tetrahedral splitting energy is small and it is lower than the pairing energy.