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1 + S 12

1. E 1 =. b. 1 + S 12. 1. E 2 = -. b. 1 - S 12. Y 2 = c 1 F 1 - c 1 F 2. antibonding. Y 1 = c 1 F 1 + c 1 F 2. bonding. Let overlap term go to zero. Overlap term is non zero. Ethylene. p system of ethylene. C 2 H 4. H 2. p bonding fille with 2 e -. p * antibonding

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1 + S 12

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  1. 1 E1 = b 1+ S12 1 E2 = - b 1- S12 Y2 = c1F1 - c1F2 antibonding Y1 = c1F1 + c1F2 bonding

  2. Let overlap term go to zero Overlap term is non zero

  3. Ethylene p system of ethylene

  4. C2H4 H2

  5. p bonding fille with 2 e- p*antibonding empty orbital

  6. Now let us look at a more complicated system. Three Orbitals! Allyl anion

  7. 2 1 3 We are going to use the LCAO approximation to make three MO’s. How do we calculate the energies and coefficients of the MO’s? Use Symmetry! The ends are the same so they must contibute equally to any MO. So we can make linear combinations where atoms one and three contribute equally. F1 = pz1 + pz3 F2 = pz1 - pz3 F3 = pz2 The second carbon is unique.

  8. pzapza = 1 pzapzb = 0 normalized assume overlap = 0 pzaĤ pza = E = a = 0 Coulomb integral pzaĤ pzb = Eint = b Interaction integral Huckel Approximation Only if the two p orbitals are adjacent otherwise Interaction is 0.

  9. 2 1 3 We are going to use the LCAO approximation to make three MO’s. How do we calculate the energies and coefficients of the MO’s? Use Symmetry! The ends are the same so they must contibute equally to any MO. So we can make linear combinations where atoms one and three contribute equally. F1 = pz1 + pz3 F2 = pz1 - pz3 F3 = pz2 The second carbon is unique.

  10. 2 1 3 We need to normalize our LCAOs F12 = 1 F1 = pz1 + pz3 N2(pz1 + pz3 ) (pz1 + pz3 ) = 1 N2 (pz12+ 2pz3 pz1 + pz32) = 1 1 + 2 x 0 + 1 N2 x 2= 1 So N =

  11. 2 1 3 We are going to use the LCAO approximation to make three MO’s. How do we calculate the energies and coefficients of the MO’s? Use Symmetry! The ends are the same so they must contibute equally to any MO. So we can make linear combinations where atoms one and three contribute equally. F1 = (pz1 + pz3 ) F2 = (pz1 - pz3 ) F3 = pz2 The second carbon is unique.

  12. 2 1 3 2 1 3 2 1 2 3 1 3 F1 = (pz1 + pz3 ) F2 = (pz1 - pz3 ) Same Symmetry F3 = pz2 F1 F2 F3

  13. 2 + 1 3 2 1 3 2 1 - 3 2 1 3 F1 = (pz1 + pz3 ) F3 = pz2 Bonding MO F1 F3 Antibonding MO F1 F3

  14. 2 1 3 2 1 3 F1 = (pz1 + pz3 ) F3 = pz2 (pz1 + pz3 ) Ĥpz2 = = 2b b Energy = + b F1 F3

  15. 2 1 3 2 1 3 F1 = (pz1 + pz3 ) - F3 = -pz2 (pz1 + pz3 ) Ĥ (- pz2) = = - - 2b b Energy = - - b F1 F3

  16. Energy = 2 1 3 b Bonding 2 1 + 3 2 1 3 F1 F3

  17. Energy = 2 - 1 3 b Antibonding 2 1 - 3 2 1 3 F1 F3

  18. What about F2 ? There is no overlap between ends so E = 0 2 1 3

  19. - b 1 3 2 1 3 0 2 1 3 b

  20. This is painful! It makes the brain hurt. So use a computer instead. Simple Huckel Molecular Orbital Theory Calculator

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