If you put an electron into the t 2g, like that for Ti 3+, then you stabilize the barycenter of the d orbitals by 0.4 D o.

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1 Crystal Field Stabilization Energy Week 2-1 Octahedral Symmetry (O h ) If you put an electron into the t 2g, like that for Ti 3+, then you stabilize the barycenter of the d orbitals by 0.4 D o. Each additional electron you put in the t 2g orbitals stabilizes the complex by an additional 0.4. However, as you put electrons in the e g orbitals, then you destabilize the crystal field. This is because the upper orbitals have anti-bonding properties so you destabilize the bonds. How will we determine if a complex is high spin or low spin? It all depends on the relative energy of the 10D q and the pairing energy. Remember that the 10D q is set by the strength of the ligand.

2 Week 2-2 Weak ligands give high spin (hs) complexes, while strong ligands give low spin (ls) complexes. Note that the pairing energy is the inherent repulsion of electrons 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 10D q then the hs/ls determination is dependent on temperature. [ Fe(III)( - S S N ( 3 [ This complex is low spin at low temp, s = 1/2 but changes to high spin at room temperature

3 Can we see this experimentally? Week 2-3 From the x-ray expt. We can see short bonds on z-axis when there are no electrons in e g orbitals. Tetraheral symmetry (T d ). As increase temp, electron goes to e g and bond lengths increase. With T d symmetry, the repulsion along the d-orbitals reverses from O h. Also note that the crystal field of T d is half that of a cubic field because less repulsion. The d-orbitals of the other coordination environments can be reasoned by elongation and/or compression along particular axis. For example, elongation along the z-axis of an octahedral results in square planar geometry (See Fig 11.12, next pg.).

4 Week 2-4 Fig Factors that effect 10D q 1) Electrostatic interaction: a)pulling the charge in closer, increases the 10 D q. b)oxidation state of metal increases 10D q, M +2 to M +3 is an approx. 50% increase in 10 D q. 2) As you go down a column, 10D q increases 50% with each element down. This is because orbitals increase in size so they bump into negative charged ligands and the perturbation is more. 3) Number and geometry of ligands a) T d does not lie in any ligand axis so ligand perturbation on 10 D q is less. (No direct interaction with orbitals) b) As increase number of ligands, increase electrostatics and thus 10 Dq. c) Generally, more ligands increase 10 D q but the nature of the ligand can lead to T d (too much charge accumulation or steric hinderence)

5 4) Nature of the ligands Week 2-5 a) Stronger ligands increase 10D q, spectrochemical series indicates this, [TiCl 6 ] 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) creates energy in mitochondria (proton gradient). Oxygen is converted to water, 400 ml made each day in a human. Cytochrome c oxidase is the terminal electron acceptor of the electrons to reduce oxygen to water. However, cyanide is such a strong ligand it binds to cytochrome c oxidase 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 in crystal field is not accurate (too simple).

6 If the above rules hold for crystal field then Cl - should be a strong ligand but it is actually weak field. Week 2-6 What about the neutral ligands that have negative charge on the bonding atoms (lone pairs)? Ammonia versus water Oxygen is more electro-negative so should have more charge and thus be a stronger ligand. This is not the case. This will lead us to a new theory that includes orbital overlap, Molecular Orbital Theory, that we will discuss later. Now lets compare Crystal Field to Valence Bond Theory and see which explains chemical reactivity best. Valence Bond Theory predicts: [Co(H 2 O) 6 ] 3+, d 6, d 2 sp 3 -inner sphere-not reactive. ooooo o ooo ooooo 3d 4s 4p 4d [Ni(NH 3 ) 6 ] 2+, d 8, sp 3 d 2 -outer sphere-reactive. ooooo o ooo ooooo 3d 4s 4p 4d Pauling places the Ni electrons in the outer sphere because as previously discussed with CN - and Cl -, NH 3 is a strong ligand. By doing this one would predict that the complex should be reactive. It is not and thus his theory started to lose favor.

7 An example of Crystal Field Theory working is the following. E = V [Co(H 2 O) 6 ] 2+, d 7, O h, high spin Week 2-7 E = -0.1 V [Co(NH 3 ) 6 ] 2+, d 7, O h, weakly low spin E = 0.83 V[Co(CN) 6 ] 4-, d 7, O h, low spin Crystal Field Theory (spectrochemical series) says that as the ligand gets stronger, the lone electron of high spin pushes into the e g orbital (anti-bonding character) and so it reacts easily to get to d 6. Now lets ask the question: How can one tell whether a complex will be O h or T d or h.s. or l.s.? For O h versus T d the key factor is size. _ I - and Br - tend to form T d because they are large. _ [Co(H 2 O) 6 ] 2+ [Co(NH 3 ) 6 ] 2+ The larger central atoms can hold more ligands so they go O h Remember that there is d-orbital contraction Because d-orbitals are like big ballons they have poor sheilding (not covering the nuclear charge very well) so they contract. Ionic radius decreases across the d and f series.

8 Week 2-8 However, f-orbitals are bigger then the d orbitals so they can handle even more than 6 ligands. [M(H 2 O) 6 ] n+ versus [Ln(H 2 O) ] n+ due to increased size. Also, low spin ions (just t 2g orbitals) are smaller in radius than high spin ions (both t 2g and e g orbitals) because the electrons are more confined in fewer orbitals CFSE is smaller for Td than Oh due to less repulsion between ligands and the metal. d 0, d 5 h.s., d 10 all have no CFSE due to symmetrical distribution of the d- electrons. When there is no CFSE then T d is the most stable to conserve space around metal center. Mn 2+, d 5 h.s. has no CFSE, unless you put a very strong ligand. Fe 3+ same as Mn 2+ (both are d 5 ) but Fe 3+ has CFSE. This is because the higher charge on metal makes for greater CFSE and higher splitting of the d-orbitals. Now lets put all this to practice and calculate the CFSE for a particular metal ion. Co(III), d 6, l.s., diamagnetic Free Co(III) pairing energy = KJ/mol (see Table 11.4)

9 However, for a particular complex the pairing energy is KJ/mol Week 2-9 From spectroscopy we determine that 10Dq = 272 KJ/mol spectroscopy The equation to calculate CFSE is CFSE = -24D q - 3P = KJ/mol= -64 KJ/mol Now remember that for the complex to be l.s. the CFSE has to be negative. The 24D q is from Table 11.3 and the 3P is from 3 electron pairs. If not then the pairing energy is too great and it goes h.s. Now lets compare oxidation change: M 2+ versus M 4+ M 4+ has a larger CFSE due to greater repulsion to ligands. Therefore a metal can change spin state (h.s. to l.s.) as charge increases. D-orbital relative energies are dependent of the Coord.Environment. Start with O h orbitals yet pull z-axis ligand away

10 Jahn-Teller Theorem For a non-linear molecule in an electronically degenerate state, distortion must occur to lower symmetry, remove degeneracy and lower the energy of whole system. Ti(H 2 O) 6 3+ Cu 2+ d 9 configuration in O h geometry has J-T distortion (see above). d 1 configuration in O h geometry has J-T distortion (z-compression). goes to what? Electron in the d xy orbital! This is why the absorption spectra for Ti 3+ has a shoulder (Fig. 11.8). The electron jumps up to both e g orbitals that are different in energy and thus 2 transitions An example of the power of J-T distortion: If you add ethylenediamine (en) to Fe 3+, it forms [Fe(en) 3 ] 3+ immediately. However, if you add en to Cu 2+ then you have trouble. [Cu(H 2 O) 6 ] 2+ + en = [Cu(H 2 O) 4 (en)] 2+ (Forms) [Cu(H 2 O) 4 (en)] 2+ + en = [Cu(H 2 O) 2 (en) 2 ] 2+ (Forms) Week 2-10 [Cu(H 2 O) 2 (en) 2 ] 2+ + en = [Cu[(en) 3 ] 2+ (does not form!!)

11 Week 2-11 The [Cu[(en) 3 ] 2+ has the J-T effect which puts strain on the en coordination so the last complex will not form. Square Planar Coordination (See Fig ) D 8 complexes usually form sq. planar complexes (Ni 2+, Pd 2+, Pt 2+ ) [Ni(CN) 4 ] 2- is sq. planar while [NiCl 4 ] 2- is tetrahedral. This is due to ligand strength. _ Moving down a family the CFSE increases so Pd 2+ and Pt 2+ are always sq. planar. xy, xz, yz z 2, x 2 -y 2 _ x 2 -y 2 xy z 2 xz, yz Note that due to larger atomic size there is less ligand repulsion so the close packed ligands of sq. planar is fine. There is also out of plane d(π) to p(π) interaction for Pd 2+ and Pt 2+. This increases stabilization of sq. planar geometry. Pt Coinage metals Cl Note that [NiCl 4 ] 2- does not have this π-obital overlap because Ni2+ d- orbitals are smaller thus no stabilization. Cu(II), Ag(II) and Au(II), all are d 9 However, note that 2 Au(II) = Au(I) and Au(III)

12 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 gets pushed into a high energy anti-bonding orbital which makes it reactive. Au Ag Cu d x2-y2 So it will be either oxidized or reduced to form d 8 or d 10. Molecular Orbital Theory (MO, pg ) CFT is good for many chemistry problems but it falls apart in trying to explain strength of ligands. For this reason, chemists developed the Molecular 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 overlap between the metal and the ligand. Orbital overlap between the ligand and the metal allows the electron cloud to expand and thus stabilize the bond. Fig Week 2-12