Crystal Field Theory

how to work out the dⁿ configuration of a metal ion in a complex

Subtract the oxidation state of the metal from the group number

What is crystal field theory

A purely ionic description of bonding in TM complexes.

It predicts the splitting of d-orbitals in complexes of known geometry.

It can be used to predict spectroscopic and magnetic properties in TM complexes.

What assumptions does crystal field theory make

The metal centre and ligands are point charges

Bonding arises solely via electrostatic interactions between these charges

What results in the splitting of energy of the d-orbitals

We consider the ligands as negative point charges - hence they repel the electrons in the metal d-orbitals.

What are the e_g and t_2g orbitals

e_g - 2 destabilised orbitals that are higher in energy. e stands for equal and g is a symmetry label (gerade)

t_2g - 3 stabilised lower energy orbitals

What do the orbital energy levels look like

The barycentre is the energy of the orbitals if we consider the ligands to be a sphere of negative charge that equally destabilises each d-orbital.

What is crystal field stabilisation energy

A measure of how much more stable a complex is thanks to the splitting of the d-orbitals in a particular arrangement.

what is ∆_oct

The Crystal Field Splitting Parameter for an octahedral complex. It is the energy gap between the stabilised and destabilised orbitals.

The average energy of the orbitals is the barycentre energy

Overall diagram of crystal field theory

How do you calculate CFSE_oct (stabilisation energy)

What determines the colour of a complex

The energy gap between the split d-orbital energies (wavelength of absorbed light)

What can happen when we add a 4th d-electron

The electron can sometimes go in the higher energy orbital rather than pairing up if the energy gap is small enough that the repulsion pairing energy exceeds it.

This gives rise to high and low spin configurations

High vs low spin

High spin complexes form if ∆_oct < P (pairing energy)

So ∆_oct and P are the factors affecting whether a complex is high spin or low spin.

P is dependent on the metal and its oxidation state and ∆_oct is additionally affected by the ligands.

How do you calculate CFSE for d⁴+ compounds

Same as normal but you need to add the pairing energy for each pair of electrons.

For d⁶+ don’t forget the pairing energy present in the spherical ligand field at the start as CFSE relies on the difference between the energies

What is different about d^8-10

They have no high or low spin states as the configuration ends up the same either way as electrons will be paired in the lower energy orbitals first every time.

d¹⁰ actually has no CFSE as the stabilisations and destabilisations cancel out as the orbital energies must cancel out

How does ∆ change with oxidation state of the metal

Increases with increasing oxidation state of the metal as the ligands are pulled closer so there is a larger repulsion between electrons in d-orbitals and ligands, increasing the d-orbital splitting.

How does going down the group affect ∆_oct

It increases as d-orbitals get bigger so interact more with the ligands so more destabilising.

Also, the pairing energy decreases as there is more room to pair electrons.

Hence, all of the second and third row octahedral complexes are low spin

What determines ligand field strength

Ordered by periodic table based on electronegativity. More electronegative ligands hang onto their electrons more strongly and negative ligands are better sigma donors as they want to donate away the extra electrons.

What is the determinant of ligand field strength

How good a pi acceptor they are. This generally follows sigma donation but not perfectly

What is the problem with this determinant

Crystal Field theory cannot explain this as we assume point charges hence there are no covalent interactions

F^- is a pi donor due to the presence of its lone pairs while CN^- is a pi acceptor due to the presence of a pi bond in the ligand resulting in an empty pi* orbital being present which can accept a lone pair from the metal centre.

Why is I^- a weaker field ligand than F^-

Because its lone pairs are in larger orbitals it can donate them more easily hence it is a stronger pi donor

Why is P typically a stronger field ligand

It has low-lying d-orbitals hence it is a pi acceptor

How do things look for tetrahedral ligand fields

How does the energy diagram look for tetrahedral ligand fields

note that the symmetry labels are gone and e and t₂ have swapped

How do you calculate CFSE_tet

Differences between octahedral and tetrahedral ligand fields

What is a consequence of ∆_tet values being smaller than ∆_oct

Tetrahedral complexes are always high spin as pairing an electron is not worth the energy penalty compared to promoting an electron

How can we easily work out the square planar orbital energy diagram

Remove the axial ligands from an octahedral complex

When are complexes tetrahedral or square planar

Tend to be tetrahedral as it is better for sterics

But square planar is electronically very stable for d⁸ complexes. The complex will be square planar if ∆ is large:

• 3d⁸ with strong field ligands (strong field gives large ∆)

• 4/5d⁸ with any ligand (larger d-orbitals give larger ∆)

what happens in d⁹, high spin d⁴ and low spin d⁷ complexes

Degeneracy leads to being able to place the paired electron in one of two orbitals of equal energy

Jahn-Teller effect

How does the complex distort