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Molecular Perturbations

Molecular Perturbations. Within the frame of LCAO , we search to express an unknown system from an known system. The unknown system are in general real systems: molecule or supermolecule.

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Molecular Perturbations

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  1. Molecular Perturbations Within the frame of LCAO,we search to express an unknown system from an known system. The unknown system are in general real systems: molecule or supermolecule. Known systems may be real systems, simple molecules, fragments of real systems or partners within a reaction (fragments of a supermolecule: reactants). 1

  2. Molecular Perturbations i= Sr crfrand Ei are solutions for a known system. We like to find solutions Y'i= Sr c'rfrand E'i for an unknown system that is resembling (whose Hamiltonian, H', does not differ by much from H). If so, solutions (E'i,y'i) are close to (Ei,yi) and P=H’-H is an operator whose terms Pij=[<yi|P|yj>]* are weak. P is the operator defining the perturbation. *data usually characterize atoms pij=<fi|P|fj> which are few terms but no so small (a bond formation between fi and fj). Pij=<yi|P|yj> are derived from the pij. Pij are significantly weaker than pij due to delocalization of AOs within MO expressions. In general, results remain significant even though some pij are important. 2

  3. Molecular Perturbations Y'I = S aiYI = Sr crfrunknown MOs, Y'i, are linear combinations of known MOs, Yi, that are linear combinations of AOs, fi; they are hence linear combinations of AOs, Yi. If we were to find exact solutions Y'i= Sr c'rfrit we be easier to solve the secular determinant directly. The advantage of using perturbations is to make it within approximations but simply. 3

  4. Molecular Perturbations “Simply” means not rigorously that is somehow negative. However ° understanding (as opposed to modeling) is always a work of simplification. ° perturbation is a direct comparison. We search for the difference between 2 large numbers (total energies) with a large error bar each. The difference calculating them independently and making the difference may be large. A direct estimation of the difference even using simple theory may be more precise. 4

  5. Molecular Perturbations Secular determinant without perturbation: If known, the secular equation is solved (diagonal)! 5

  6. Molecular Perturbations Secular determinant with perturbation: We search to develop the determinant, retaining only the largest terms. One term is larger than any other: the diagonal. 6

  7. First order Energy solution Since perturbations are small, all the Pij are small and the largest terms are those from the diagonal: Searching for E’j close to Ej, every term of the diagonal is large except one: E’i - Ej= Ei - Ej which is not small (assuming no degeneracy). Only one term is small; that for j: E’i - Ej= Pjj is a “first order” small term. Expressing the determinant, the product of the terms of the diagonal is contains the smallest number of small terms and is the greatest. The secular determinant is P (E’i+Pii-E) = 0 → E’j =Ej+ Pjj 7

  8. Orbital associated with the First order Energy solution (Zero order) Searching fory’jclose toyj Secular equation: Pj1 a1 + Pj2 a1 + …+ (Ej+Pjj-E) aj + …+ Pjn an = 0 All the terms are small except one. All the ai are zero except aj. aj=1 The orbital is unperturbed to this ordery’j=yj 8

  9. Second order Energy solution • Developing the determinant is: • the diagonal • all the terms obtained by a single permutation of two terms in the diagonal • + negligible terms. • Searching for E’j close to Ej, The secular determinant is P(E’i+Pii-E)- Si (P (E’i+Pii-E)P*ijPji + e= 0 [(E’i+Pii-E) (E’j+Pjj-E)] Secular determinant without perturbation: 9

  10. Second order Energy solution P(E’i+Pii-E)- Si (P (E’i+Pii-E)P*ijPji + e= 0 [(E’i+Pii-E) (E’j+Pjj-E)] P(E’i+Pii-E)[1–Si P*ijPji ] = 0 [(E’i+Pii-E) (E’j+Pjj-E)] -(E’j+Pjj-E)= - Si P*ijPji ] (E’i+Pii-E) E – Ej – Pjj = -Si P*ijPji (Ei-Ej) 10

  11. Second order Energy solution E = Ej + Pjj +Si P*ijPji (Ej-Ei) E’j = Ej + Pjj +Si P*ijPji (Ej-Ei) For real terms E’j = Ej + Pjj +Si Pij2/ DE Dimension of an energy: Square of an energy over an energy Second order term (2 orbital mixing) If Ej< Ei stabilization of the orbital of lower energy If Ej>Ei destabilization of the orbital of higher energy Increases the energy gap 11

  12. Orbital associated with the second order Energy solution (first order) Secular equation: Pi1 a1 + Pi2 a2 + …+ (Ei+Pii-E) ai+ Pij aj + …+ Pjn an = 0 Searching a solution close to Yj, the ith line has small terms of the first order. ai = Pij/DE aj The orbital isy’j= yj +S ij Pij/(Ei-Ej)yi Such expression is not orbital is not normalized and has to be multiplied by 1/√N not small terms if ij First order not small terms if ij First order Product of 2 small terms Second order no dimension In phase for the lowest combination 12

  13. Combination of 2 AOs of same symmetry Ei’-Ei = Pij2/ DE y’j= yj +S ij Pij/DEyi • Two orbital mixing: a bonding combination and an antibonding one. • The bonding orbital • is the in-phase combination • is issued from the orbital of lowest energy • (larger coefficient of mixing) • The antibonding orbital • is issued from to the orbital of highest energy • The interaction widens the gap; • It is more important when S is large and DE small • 10 eV rule • Frontier orbitals Antibonding Bonding 13

  14. 2 OA interaction degenerate or close in energy • (D/2 -E) (-D/2 -E) = b2 E2- (D2/4) = b2 • E2 = b2 + (D2/4) • E = ±√[b2 + (D2/4)]general • If D = 0, E+ - E = b • If b <<D, • E+ = (D/2) (1+4b2/ D2) 0.5 • E+ = (D/2) (1+2b2/ D2) = D/2 (1 +2b2/ D) • E+ - E = b2/ D «2nd order Perturbation term -D/2 D/2 b √ (b2+D2/4) The geometric mean of b and D/2 14

  15. Frontier Orbitals • due to perturbation formula: • SOMO, HOMO and LUMO • The least stable but the most mobile electrons • The largest amplitude on the weakly bound atoms that are reactive sites. Ei’-Ei = Pij2/DE Kenichi Fukui Japan Nobel 1981 15

  16. 3rd order perturbation Case of two fragments interacting. The perturbation is the interaction between fragments (without perturbation within each fragment) Permutation between 3 lines of the determinant: Y’1 = Y’1 + P1bYb +P1b Pb2Y2 E1-Eb (E1-Eb) (E1-E2) Y2 Y1 Yb B Fragment : A single orbital A Fragment : The orbital 1 and 2 belong to the same fragment and are orthogonal inside A Y1 and Y2 both interact with Yb; they mix, the phase relation being imposed by their relation with B. This causes polarization 16 S. Imagaki, H. Fujimoto K. Fukui J. Amer. Chem. Soc 98 (1976) 4054 L. Libit R. Hoffmann J. Amer. Chem. Soc 96 (1974) 1370

  17. 3rd order perturbationanother view point:excitation Y’1 = Y’1 + P1bYb +P1b Pb2Y2 E1-Eb (E1-Eb) (E1-E2) Y2 Y1 Yb B Fragment : A single orbital A Fragment : The orbital 1 and 2 belong to the same fragment and are orthogonal inside A Y1 and Y2 mix. Y1 is partially depopulated and Y2 becomes partially populated. This corresponds to mix with an excited state: IC 17 S. Imagaki, H. Fujimoto K. Fukui J. Amer. Chem. Soc 98 (1976) 4054 L. Libit R. Hoffmann J. Amer. Chem. Soc 96 (1974) 1370

  18. 3rd order perturbation Y’1 = Y1 + P1b/(E1-Eb) Yb At variance with Y1, Y’1 is not orthogonal to Y2 Y’’1 = Y’1 + P1’2/(E1-E2) Y2 Let express P1’2 by developing Y’1 P1’2 = < Y’1IPI Y2> = < Y1IPI Y2>+ P1b/(E1-Eb) < YbIPI Y2 > P1’2 = 0 + P1b Pb2/(E1-Eb) Y’’1 = Y’1 + P1b Pb2/ [(E1-Eb)(E1-E2)]Y2 Y’’1 = Y1 + P1b/(E1-Eb) Yb + P1b Pb2/ [(E1-Eb)(E1-E2)]Y2 Y2 Yb Y1 B Fragment : A Fragment : The third order terms account for the polarization of the fragment 18

  19. Eb below E1 Increasesdecreases Decreasesincreases Y’1 = Y’1 -Yb+Y2 Y’2 = Y’1 -Yb+Y2 19

  20. Eb intermediate between E1 and E2 Decreasesincreases Decreasesincreases Y’1 = Y’1 +Yb+Y2 Y’2 = Y’1 -Yb+Y2 20

  21. Eb above E1 Decreasesincreases Increasesdecreases Y’1 = Y’1 +Yb+Y2 Y’2 = Y’1 -Yb-Y2 21

  22. ExamplesTo judge the method, we will search for known systems (easily calculated without perturbation) • Cyclobutadiene from butadiene 22

  23. Cyclobutadiene from Butadiene -.3717 -.6015 -1.618 0.3717 0.6015 0.6015 -.3717 -.618 0.6015 -.3717 -.6015 0.3717 0.618 -.3717 0.6015 0.3717 0.6015 1.618 0.6015 0.3717 23

  24. MO Energies from perturbation What we know is <f1IPIf4> = 1: there is a bond between atoms 1 and 4 that did not exist for butadiene A Mirror symmetry is preserved S S A A 24 Only orbitals of the same symmetry mix together

  25. Conservation of Orbital Symmetry H C Longuet-Higgins E W Abrahamson The Molecular orbitals are solution of the symmetry operators of the molecule. MOs from different symmetry groups do not mix. Hugh ChristopherLonguet-Higgins 1923-2004 25

  26. Orbitals Y’1 from perturbation 26

  27. New lowest orbital Y’1= (fA + fB+ fC+ fD ) + < Y1IPI Y3>/(E1-E3) (fA-fB-fC+ fD ) + 0.2 The + sign means in phase between 1 and 4 The new orbital looks like Y1 but with modulation of amplitude indicated by red arrows 27

  28. Exact solutions First set 1/2 -2 b 1/2 1/2 0 1/2 2 b 28

  29. Exact solutions Second set 1/2 - 2 b 1/√2 1/√2 0 1/2 2 b 29

  30. Solutions are very accurate even though <fiIPIfj> is large (introducing a bond as strong as others). < YiIPIYj>MOs are less large due to delocalization of AOs with MOs. 30

  31. Butadiene from 2 ethylenes 1/√ (fA - fB) 1/√ (fC - fD) 1/√ (fA +fB) 1/√ (fC +fD) 31

  32. Butadiene from 2 ethylenesfirst order terms -1.5 b 1/√ (fA - fB) 1/√ (fC - fD) -0.5 b 0.5 b 1/√ (fA +fB) 1/√ (fC +fD) 1.5 b < 1/√ (fA +fB)IPI1/√ (fc +fD)> = 0.5 b 32

  33. Butadiene from 2 ethylenessecond order terms -1.5 b - 0.125 b = -1.625 b 1/√ (fA - fB) 1/√ (fC - fD) -0.5 b- 0.125 b = -0.625 b 0.5 b+ 0.125 b= 0.625 b 1/√ (fA +fB) 1/√ (fC +fD) 1.5 b + 0.125 b = 1.625 b [< 1/√ (fA +fB)IPI1/√ (fc +fD) >]2/ (b-(-b))= 0.125 b P has to be calculated on the unperturbed MO expressions (before taking into account 1st order perturbation) 33

  34. New lowest orbital Y’1= (fA + fB+ fC+ fD ) + < Y1IPI Y3>/(E1-E3) (fA-fB-fC+ fD ) + 0.5/2.0 The + sign means in phase between 2 and 3 1.25 0.75 0.75 1.25 without normalization 0.606 0.364 0364 0.606 normalized 34

  35. The total energy stabilization comes only from second order terms(interaction between occupied and vacant orbitals) E = 4 * 0.125 = 0.5 b This is the energy due to the delocalization. 35

  36. Butadiene from 2 ethylenesminimizing the repulsion 0.5 b 1/√ (fA +fB) 1/√ (fC +fD) 1.5 b It is better to reduce the overlap: not making bond between atoms 1 and 4 ! Making trans rather than cis butadiene Avoiding closing ring to cyclobutadiene 1 1 2 2 3 3 4 4 36

  37. Cyclobutadiene from 2 ethylenes Only one mirror symmetry is preserved by perturbation 37

  38. cyclobutadiene from 2 ethylenesfirst order terms -2 b 1/√ (fA - fB) 1/√ (fC - fD) -0.0 b 0.0 b 1/√ (fA +fB) 1/√ (fC +fD) 2 b • 1/√ (fA +fB)IPI1/√ (fc +fD)> = b No terms from second order 38

  39. Conservation of Orbital Symmetry H C Longuet-Higgins E W Abrahamson The Molecular orbitals are solution of the symmetry operators of the molecule. MOs from different symmetry groups do not mix. Hugh ChristopherLonguet-Higgins 1923-2004 39

  40. Non-crossing rule The potential energy curves of two MOs do not cross unless they have different symmetry. If the MOs are of the same symmetry, they interact. The interaction increases when the energy levels are close. This opens a gap and prevents crossing. Energy Reaction coordinate or any transformation 40

  41. Non-crossing rule The potential energy curves of two electronic states of a diatomic molecule do not cross unless the states have different symmetry. If 2 states are of the same symmetry, they interact. The interaction increases when the energy levels are close. This opens a gap and prevents crossing. Energy of states distance 41

  42. Non-crossing rule ionic ionic Atomic or covalent atomic Mixed ionic and covalent ionic 42

  43. Benzene from pentadienyl + C radicals pentadienyl radical without calculations • What are the coefficients of the SOMO? • What are the coefficients of the bonding MO antisymmetric relative to the mirror? • What is the corresponding Energy level? • What are the coefficients of the bonding MO symmetric relative to the mirror? • Deduce from them the MO Energy level? 43

  44. pentadienyl radical using only symmetry and Normalization. What are the coefficients of the SOMO? What are the coefficients of the bonding MO antisymmetric relative to the mirror? What is the corresponding Energy level? What are the coefficients of the bonding MO symmetric relative to the mirror? Deduce from them the MO Energy level? E = -√3 b E = -b E = 0 E = b E = √3 b 44

  45. pentadienyl radical using only symmetry and Normalization. What are the coefficients of the SOMO? Use alternant property What are the coefficients of the bonding MO antisymmetric relative to the mirror? adouble bond, symmetrized What is the corresponding Energy level? What are the coefficients of the bonding MO symmetric relative to the mirror? Deduce from them the MO Energy level? E = -√3 b 1/√12 -1/2 -1/2 1/2 1/√3 E = -b 1/√ 1/√12: 1/3 +2 (1/2)2 + 2 c2 =1 1/2: 2 (1/2)2 + 2 c2 + 0 =1 1/√3: 1/3 +2 c2 + 0 =1 E = 0 1/2 -1/√ 1/2 1/√12 Develop <YIHIY> 4 [(1/√12)(1/2)+(1/2)(1/√3)]= √3 E = b 1/2 E = √3 b 1/√3 45

  46. Benzene from pentadienyl + C radicals First order term. E = -√3 b -1.155 b 1/√12 -1/2 -1/2 1/2 1/√3 E = -b 1/√ 2/√ = 1.155 b E = 0 E = 0 1/2 -1/√ 1/2 1/√12 1.155 b E = b 1/2 E = √3 b 1/√3 46

  47. Radical chain + C radical atomcomparing the chain with the ring: Aromaticity First order term. S A E = 0 E = 0 E = 0 E = 0 0 for the ring 2/√(N-1) for the chain 4/√(N-1) for the ring 2/√(N-1) for the chain 47

  48. Radical chain + C radical atomcomparing the chain with the ring: Aromaticity Aromaticity accordingto Dewar S A When the SOMO is antisymmetric The ring is less stable than the chain The polyene is ANTIAROMATIC N-1 is odd N = 4n When the SOMO is symmetric The ring is more stable than the chain The polyene is AROMATIC N-1 is even N = 4n +2 48

  49. Benzene from pentadienyl + C radicals Second order terms. (2/√12)2/√3=0.1924 -1.925 b E = -√3 b 1/√12 -1/2 -.962 b 1/√3 E = 0 0.962 b 1/√12 1/2 1.925 b E = √3 b 1/√3 49

  50. Benzene from pentadienyl + C radicals Second order terms. (2/√12)2/√3=0.1924 -1.925 b 0.1924 E = -√3 b -1.155 b E = -√3 b 1/√12 0.1924 -1/2 -b -1/2 1/2 -0.962 b 1/√3 E = -b 1/√ E = 0 0.962 b 1/2 -1/√ b 1/2 0.1924 E = √3 b 1/√12 1.155 b E = b 1/2 0.1924 E = √3 b 1.925 b 1/√3 50

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