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Triatomics and Beyond

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 )

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Triatomics and Beyond

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  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. BeH2 Be H • BeH2is 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 pMOs? • Which will not combine remaining σ or pnonbonding MOs? 2pz 2pxy 2s 1s 1s Along bond axis

  3. BeH2 MO Diagram Along bond axis H * 2 Be 2pz 2pxy Lewis Structure? Electron Configuration? BO? HOMO? LUMO? Lewis Acid?

  4. CO2 Lewis Structure? Shape Family? Valence atomic orbitals on C and O: 2s and 3 x 2p Consider s and p MO’s formed separately. 6 s and 6 p MO’s will be formed (12 possile for each) O * 2 C 2pz 2pxy Order of energies: 2pz 2pxy 2s (O) + 2s(C) small 2s (O) + 2pz(C) smallest 2pz(O) + 2s(C) large 2pz(O) + 2pz(C) largest Along bond axis Along bond axis

  5. Valence MO Diagram for CO2 2s (O) + 2s(C) small 1s,2s, 3s*, 2s 2s (O) + 2pz(C) smallest 3s, 2s, 3s, 4s* 2pz(O) + 2s(C) large 4s, 5s*, 4s, 3s 2pz(O) + 2pz(C) largest 6s*, 5s, 4s,5s, 2px(O) + 2px(C) largest 1p , 2p, 3p*, 2p 2py(O) + 2py(C) largest 1p , 2p, 3p*, 2p 6s* 5s* 4s Free atom 3s* Free atom 2s 1s

  6. BH3 B H * 3 Lewis structure? 2pxy Shape Family? 2pz Along Bonding Plane What orbital combinations are possible now?

  7. BH3 MO Diagram H * 3 B 2pxy 2pz Along Bonding Plane CH4 - The third dimension…

  8. Frontier MO Theory BH3 H- 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? Free atom Free atom BH3 + H- —> BH4- 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”.

  9. Electron delocalization (Resonance) In resonance structures, the only electrons that move are: Delocalizedelectrons 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*’s 2p* 1p 5s C: 2*(2s + 3*(2p)) => 8 AO’s 4s 3s H: 4*(1s ) => 4 AO’s => 12 AO’s => 12 MO’s 2s 1s

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

  12. Butadiene Ethylene LUMO 1.2 2 0.2 -7.3eV LUMO 12.2 ev 9.7ev HOMO -9.5 -11 HOMO -12 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.

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

  14. The -MOs of Benzene How many pi-electrons? Nodes…(Cuts?) Aromatic Stabilization (1,3,5-hexatriene) 3 2 1 Benzene can’t be considered to have “three double bonds and three single bonds”. It has three pbonds 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). 0

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