Crystal Field Stabilization Energy

Octahedral Symmetry (Oh)

If you put an electron into the t2g, like that for Ti3+,then you stabilize the barycenter of the d orbitals by0.4 Do.

Each additional electron you put in the t2gorbitals stabilizes the complex by an additional0.4.

However, as you put electrons in the eg orbitals,then you destabilize the crystal field.

This is because the upper orbitals have anti-bondingproperties so you destabilize the bonds.

How will we determine if a complexis high spin or low spin?

It all depends on the relative energyof the 10Dq and the pairing energy.

Remember that the 10Dq is set by the strength of the ligand.

Week 2-1

Weak ligands give high spin (hs) complexes,while strong ligands give low spin (ls) complexes.

Note that the pairing energy is the inherent repulsion ofelectrons and the electrons obey Hund’s rule.

The pairing energy is constant along a period.

However, the bigger the period the lower the pairing energy(n = 4 << n=3) so many heavy elements are low spin (Fe versus Os).

If the pairing energy is equal to the 10Dq then thehs/ls determination is dependent on temperature.

This complex is low spin at low temp,s = 1/2 but changes to high spin at room temperature

SN

SFe(III)( (

3-[ [

Week 2-2

Can we see this experimentally?

From the x-ray expt. We can see short bonds on z-axis when thereare no electrons in eg orbitals.

As increase temp, electron goes to egand bond lengths increase.

Tetraheral symmetry (Td).

With Td symmetry, the repulsion along the d-orbitals reverses fromOh.

Also note that the crystal field of Td is half thatof a cubic field because less repulsion.

The d-orbitals of the other coordinationenvironments can be reasoned by elongationand/or compression along particular axis.

For example, elongation along the z-axis of anoctahedral results in square planar geometry

(See Fig 11.12, next pg.).

Week 2-3

Fig 11.12

Factors that effect 10Dq

1) Electrostatic interaction:

a)Pulling the charge in closer,increases the 10 Dq.

b)Oxidation state of metal increases 10Dq, M+2 to M+3 is an approx. 50% increase in 10 Dq.

2) As you go down a column, 10Dq increases 50% with each element down.

This is because orbitals increase in size so they bump intonegative charged ligands and the perturbation is more.

3) Number and geometry of ligandsa) Td does not lie in any ligand axis so ligandperturbation on 10 Dqis less.(No direct interaction with orbitals)

b) As increase number of ligands, increaseelectrostatics and thus 10 Dq.

c) Generally, more ligands increase 10 Dq but thenature of the ligand can lead to Td (too much chargeaccumulation or steric hinderence)

Week 2-4

4) Nature of the ligands

a) Stronger ligands increase 10Dq, spectrochemical seriesindicates this, [TiCl6]3- versus [Ti(CO)6]3+ is good example.

b) CO and CN are strong ligands due to back bonding.

Fig 11.2

How does this concept of ligand strength affect us?

In biology, oxidative phosphorylation (ATP to ADP) createsenergy in mitochondria (proton gradient).

Oxygen is converted to water, 400 ml made each day in a human.

Cytochrome c oxidase is the terminal electron acceptorof the electrons to reduce oxygen to water.

However, cyanide is such a strong ligand it binds to cytochrome coxidase and stops last electron donation (i.e. death).

The spectrochemical series tests the crystal field theory very closely and the concept of a point charge incrystal field is not accurate (too simple).

Week 2-5

If the above rules hold for crystal field thenCl- should be a strong ligand but it is actually weak field.

What about the neutral ligands that have negative chargeon the bonding atoms (lone pairs)?

Ammonia versus water

Oxygen is more electro-negative so should have morecharge and thus be a stronger ligand.

This is not the case.

This will lead us to a new theory that includes orbitaloverlap, Molecular Orbital Theory, that we will discusslater.

Now lets compare Crystal Field to Valence Bond Theoryand see which explains chemical reactivity best.

Valence Bond Theory predicts:

[Co(H2O)6]3+ , d6, d2sp3-inner sphere-not reactive.ooooo o ooo ooooo 3d 4s 4p 4d

[Ni(NH3)6]2+, d8, sp3d2-outer sphere-reactive.ooooo o ooo ooooo 3d 4s 4p 4d

Pauling places the Ni electrons in the outer sphere because aspreviously discussed with CN- and Cl- , NH3 is a strong ligand. Bydoing this one would predict that the complex should be reactive. Itis not and thus his theory started to lose favor.

Week 2-6

An example of Crystal Field Theory working is the following.

E = -1.84 V [Co(H2O)6]2+ , d7, Oh, high spin

E = -0.1 V [Co(NH3)6]2+, d7, Oh, weakly low spin

E = 0.83 V[Co(CN)6]4-, d7, Oh, low spin

Crystal Field Theory (spectrochemical series) says that asthe ligand gets stronger, the lone electron of high spinpushes into the eg orbital (anti-bonding character) and soit reacts easily to get to d6.

Now lets ask the question:

How can one tell whether a complex will beOh or Td or h.s. or l.s.?

For Oh versus Td the key factor is size.

I- and Br- tend to form Td because they are large.

The larger central atoms can hold more ligands so they go Oh

Remember that there is d-orbital contraction

Because d-orbitals are like big ballons theyhave poor sheilding (not covering the nuclearcharge very well) so they contract.

Ionic radius decreases across the dand f series.

_ _ __ _

_ _ _

_ _

[Co(H2O)6]2+ [Co(NH3)6]2+

Week 2-7

However, f-orbitals are bigger then the d orbitals so they can handle even more than 6ligands.

[M(H2O)6]n+ versus [Ln(H2O)10-12]n+ due to increased size.

Also, low spin ions (just t2g orbitals) are smaller in radiusthan high spin ions (both t2g and eg orbitals) because theelectrons are more confined in fewer orbitals

CFSE is smaller for Td than Oh due to lessrepulsion between ligands and the metal.

d0, d5 h.s., d10 all have no CFSE dueto symmetrical distribution of the d-electrons.

When there is no CFSE then Td is the moststable to conserve space around metal center.

Mn2+ , d5 h.s. has no CFSE, unless you put a very strong ligand.

Fe3+ same as Mn2+ (both are d5) but Fe3+ has CFSE.

This is because the higher charge on metal makes forgreater CFSE and higher splitting of the d-orbitals.

Now lets put all this to practice andcalculate the CFSE for a particular metal ion.

Co(III), d6, l.s., diamagnetic

Free Co(III) pairing energy = -282.6 KJ/mol(see Table 11.4)

Week 2-8

However, for a particular complex thepairing energy is -197.8 KJ/mol

From spectroscopy we determine that 10Dq = 272 KJ/molspectroscopy

The equation to calculate CFSE is

CFSE = -24Dq - 3P = -658 + 593 KJ/mol= -64 KJ/mol

The –24Dq is from Table 11.3 and the 3P isfrom 3 electron pairs.

Now remember that for the complexto be l.s. the CFSE has to be negative.

If not then the pairing energy is too great and it goes h.s.

Now lets compare oxidation change: M2+ versus M4+

M4+ has a larger CFSE due togreater repulsion to ligands.

Therefore a metal can change spin state (h.s. tol.s.) as charge increases.

D-orbital relative energies are dependent of the Coord.Environment.

Start with Oh orbitals yet pull z-axis ligand away

Week 2-9

Jahn-Teller TheoremFor a non-linear molecule in an electronically degeneratestate, distortion must occur to lower symmetry, removedegeneracy and lower the energy of whole system.

Cu2+ d9 configuration in Oh geometryhas J-T distortion (see above).

Ti(H2O)63+

d1 configuration in Oh geometryhas J-T distortion (z-compression).

__ __ __ goes to what? Electron in the dxy orbital!

This is why the absorption spectra for Ti3+ has a shoulder (Fig. 11.8).

The electron jumps up to both eg orbitals thatare different in energy and thus 2 transitions

An example of the power of J-T distortion:

If you add ethylenediamine (en) to Fe3+, it forms [Fe(en)3]3+ immediately. However, if you add en to Cu2+

then you have trouble.

[Cu(H2O)6]2+ + en = [Cu(H2O)4(en)]2+ (Forms)

[Cu(H2O)4(en)]2+ + en = [Cu(H2O)2(en)2]2+ (Forms)

[Cu(H2O)2(en)2]2+ + en = [Cu[(en)3]2+

(does not form!!)Week 2-10

The [Cu[(en)3]2+ has the J-T effect which puts strain onthe en coordination so the last complex will not form.

Square Planar Coordination (See Fig. 11.12)

D8 complexes usually form sq. planar complexes (Ni2+, Pd2+, Pt2+)

[Ni(CN)4]2- is sq. planar while [NiCl4]2- is tetrahedral.

This is due to ligand strength.

Moving down a family the CFSEincreases so Pd2+ and Pt2+ are always sq. planar.

Note that due to larger atomic size there is less ligandrepulsion so the close packed ligands of sq. planar is fine.

There is also out of plane d(π) to p(π) interaction for Pd2+ and Pt2+.This increases stabilization of sq. planar geometry.

Note that [NiCl4]2- does not have thisπ-obital overlap because Ni2+ d-orbitals are smaller thus no stabilization.

Coinage metals

Cu(II), Ag(II) and Au(II), all are d9

However, note that 2 Au(II) = Au(I) and Au(III)

_ _ __ _ _

__

_ _ xz, yz

z2

xy

x2-y2

z2, x2-y2

xy, xz, yz

Pt Cl

Week 2-11

Why is there this instability?All have JT but as go down family, the 10Dq increases.

As the 10Dq for Au increases, the lone electron getspushed into a high energy anti-bonding orbital whichmakes it reactive.

Au ___Ag ___

Cu dx2-y2 ___So it will be either oxidized or reduced to form d8 or d10.

Molecular Orbital Theory (MO, pg. 413-433)

CFT is good for many chemistry problemsbut it falls apart in trying to explain strength of ligands.

For this reason, chemists developed theMolecular Orbital Theory (MO)

CFT predicts that charge is the main issue (such as Cl-,Br-, etc.) but neutral CO is the strongest ligand.

Therefore, CFT had to be modified to account for this.

To explain this better the orbitals must overlapbetween the metal and the ligand.

Orbital overlap between the ligand and the metal allowsthe electron cloud to expand and thus stabilize the bond.

Fig 11.17

Week 2-12