1. Triatomics and Beyond 1) Complex, so we deal with simple symmetrical molecules 2) Same principles apply to orbital combinations as with Diatomics: i)Compatible symmetry ii)Compatible energy (within 1 Rydberg, 1 Ry) 3) The number of valence AO’s must equal the number of Mo’s 4) MO’s must conform to the symmetry of the molecule. 5) Orbitals of the same energy and the same number of nodes mix.

2. BeH 2 BeH 2 is the simplest triatomic molecule. Linear the gas phase. The relative energies for the AO’s of Be and H are: 1s (Be) = -9.38 Ry 1s (H) = -0.99 Ry 2s (Be) = -0.61 Ry 2p (Be) = +0.14 Ry Which atomic orbitals will combine to make σ MOs? Which will combine to make  MOs? Which will not combine remaining σ or  nonbonding MOs? 2p z 2p xy BeH 2s 1s Along bond axis

3. BeH 2 MO Diagram Lewis Structure? Electron Configuration? BO? HOMO? LUMO? Lewis Acid? 2p xy 2p z Be H * 2 Along bond axis

4. CO 2 Lewis Structure? Shape Family? Valence atomic orbitals on C and O: 2s and 3 x 2p Consider  and  MO’s formed separately. 6  and 6  MO’s will be formed (12 possile for each) Order of energies: C O * 2 2p z 2p xy 2p z Along bond axis 2s (O) + 2s(C) small 2s (O) + 2pz(C) smallest 2pz(O) + 2s(C) large 2pz(O) + 2pz(C) largest

5. Valence MO Diagram for CO 2 Free atom 2s (O) + 2s(C) small 1  2  3  2  2s (O) + 2pz(C) smallest 3  2  3  4  2pz(O) + 2s(C) large 4  5  4  3  2pz(O) + 2pz(C) largest 6  5  4  5  2px(O) + 2px(C) largest   2py(O) + 2py(C) largest       

6. BH 3 What orbital combinations are possible now? 2p z 2p xy BH * 3 Lewis structure? Shape Family? Along Bonding Plane

7. BH 3 MO Diagram CH 4 - The third dimension… B H * 3 2p z 2p xy Along Bonding Plane

8. Frontier MO Theory BH 3 H-H- BH 3 + H - —> BH 4 - Reactions take place during collisions. Bonds are formed and/or broken. That must mean that there is some kind of orbital interaction. Which orbitals are most likely interact in forming the new bond? In general, reactions take place via the interaction of the HOMO of one component with the LUMO of the other because these are the closest in energy. These orbitals are known as the “frontier orbitals”. Free atom

9. Electron delocalization (Resonance) In resonance structures, the only electrons that move are: Delocalized electrons are always found in  orbitals. As  orbitals are usually found at higher energy than the  orbitals, the HOMO and LUMO of molecules with multiple bonds are usually  orbitals. As a result of this, we often look only at the  orbitals and construct  MO diagrams.

10. Ethylene         ’s C: 2*(2s + 3*(2p)) => 8 AO’s H: 4*(1s ) => 4 AO’s => 12 AO’s => 12 MO’s

11.  -MO diagram of Ethylene Nodes… Pi-bond order… Sigma bond order When Ethylene reacts… Ethyne? Ozone Nodes… Sigma & Pi-bond order… Total bond order Lewis BO Formal Charge: Total bond order =  bond order +  bond order

12. -7.3eV -11 1.2 -9.5 0.2 2 -12 Ethylene Butadiene Nodes… The importance of the HOMO/LUMO gap. Note: this is not two isolated double bonds but a single  -system spread out over four carbons. HOMO LUMO HOMO LUMO 12.2 ev 9.7ev

13. Benzene The polygon method for determining  -MOs of monocyclic unsaturated molecules: Works for any monocyclic molecule with contiguous atomic p orbitals.

14. Benzene can’t be considered to have “three double bonds and three single bonds”. It has three  bonds with bond order _____. Accordingly, all six C-C bonds in benzene are 140 pm (whereas pure C-C bonds are 154 pm and pure C=C bonds are 134 pm). The  -MOs of Benzene How many pi-electrons? Nodes…(Cuts?) Aromatic Stabilization (1,3,5-hexatriene) 0 1 2 3