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11 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.

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Presentation on theme: "11 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."— Presentation transcript:

1 11 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).

2 22 Molecular Perturbations  i =  r c r  r and E i are solutions for a known system. We like to find solutions  ' i =  r c' r  r and E' i for an unknown system that is resembling (whose Hamiltonian, H', does not differ by much from H). If so, solutions (E' i,  ' i ) are close to (E i,  i ) and P=H’-H is an operator whose terms P ij =[ ]* are weak. P is the operator defining the perturbation. p p p p * data usually characterize atoms p ij = which are few terms but no so small (a bond formation between  i and  j ). P ij = are derived from the p ij. P ij are significantly weaker than p ij due to delocalization of AOs within MO expressions. In general, results remain significant even though some p ij are important.

3 33 Molecular Perturbations  ' I =  a i  I =  r c r  r unknown MOs,  ' i, are linear combinations of known MOs,  i, that are linear combinations of AOs,  i ; they are hence linear combinations of AOs,  i. If we were to find exact solutions  ' i =  r c' r  r it we be easier to solve the secular determinant directly. The advantage of using perturbations is to make it within approximations but simply.

4 44 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.

5 55 Molecular Perturbations E 1 -E00 = 0 0E 2 -E0 00E j -E Secular determinant without perturbation: If known, the secular equation is solved (diagonal)!

6 66 Molecular Perturbations E 1 +P 11 -EP 12 P 1j = 0 P 21 E 2 +P 22 -EP 2j P j1 P j2 E j +P jj -E Secular determinant with perturbation: We search to develop the determinant, retaining only the largest terms. One term is larger than any other: the diagonal.

7 77 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 E j, every term of the diagonal is large except one: E’ i - E j = E i - E j which is not small (assuming no degeneracy). Only one term is small; that for j: E’ i - E j = P jj 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  (E’ i +P ii -E) = 0 → E’ j = E j + P jj

8 88 Orbital associated with the First order Energy solution (Zero order) Searching for  ’ j close to  j Secular equation: P j1 a 1 + P j2 a 1 + …+ (Ej+Pjj-E) a j + …+ P jn a n = 0 All the terms are small except one. All the a i are zero except a j. a j =1 The orbital is unperturbed to this order  ’ j =  j

9 99 E 1 +P 11 -EP 12 P 1j = 0 P 21 E 2 +P 22 -EP 2j P j1 P j2 E j +P jj -E Secular determinant without perturbation: 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 E j, The secular determinant is  (E’ i +P ii -E) -  i (  (E’ i +P ii -E)P* ij P ji +   = 0 [(E’ i +P ii -E) (E’ j +P jj -E)]

10 10 Second order Energy solution  (E’ i +P ii -E) -  i (  (E’ i +P ii -E)P* ij P ji +   = 0 [(E’ i +P ii -E) (E’ j +P jj -E)]  (E’ i +P ii -E)  –  i P* ij P ji  = 0 [(E’ i +P ii -E) (E’ j +P jj -E)] -(E’ j +P jj -E)= -  i P* ij P ji  (E’ i +P ii -E) E – E j – P jj = -  i P* ij P ji (E i -E j )

11 11 Second order Energy solution E = E j + P jj +  i P* ij P ji (E j -E i ) E’ j = E j + P jj +  i P* ij P ji (E j -E i ) For real terms E’ j = E j + P jj +  i P ij 2 /  E Second order term (2 orbital mixing) If E j < E i stabilization of the orbital of lower energy If E j >E i destabilization of the orbital of higher energy Increases the energy gap Dimension of an energy: Square of an energy over an energy

12 12 Orbital associated with the second order Energy solution (first order) Secular equation: P i1 a 1 + P i2 a 2 + …+ (Ei+Pii-E) a i + P ij a j + …+ P jn a n = 0 Searching a solution close to  j, the i th line has small terms of the first order. a i = P ij /  E a j The orbital is  ’ j =  j +  i  j P ij /  E i -E j )  i Such expression is not orbital is not normalized and has to be multiplied by 1/√N Product of 2 small terms Second order not small terms if i  j First order not small terms if i  j First order no dimension In phase for the lowest combination

13 13 Combination of 2 AOs of same symmetry 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  E small 10 eV rule Frontier orbitals Antibonding Bonding  ’ j =  j +  i  j P ij /  E  i E i ’-E i = P ij 2 /  E

14 14 2 OA interaction degenerate or close in energy 11 22 11  /2 -E  = 0 22  /2 -E (  /2 -E ) (-  /2 -E ) =  2  E 2 - (  2 /4 ) =  2 E 2 =  2 + (  2 /4 ) E = ±√[  2 + (  2 /4 )] general If , E + - E =  If  << , E + = (  /2 ) (  2 /  2 ) 0.5 E + = (  /2 ) (  2 /  2 ) =  /2 (1 + 2  2 /  E + - E =  2 /   2 nd order Perturbation term   √ (  2 +  2 /4) The geometric mean of  and  

15 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. E i ’-E i = P ij 2 /  E Kenichi Fukui Japan Nobel 1981

16 16 3 rd 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:  ’ 1 =  ’ 1 + P 1b  b + P 1b P b2  2 E 1 -E b (E 1 -E b ) (E 1 -E 2 ) 11 22 bb A Fragment : The orbital 1 and 2 belong to the same fragment and are orthogonal inside A S. Imagaki, H. Fujimoto K. Fukui J. Amer. Chem. Soc 98 (1976) 4054 L. Libit R. Hoffmann J. Amer. Chem. Soc 96 (1974) 1370 B Fragment : A single orbital  1 and  2 both interact with  b ; they mix, the phase relation being imposed by their relation with B. This causes polarization

17 17 3 rd order perturbation another view point:excitation  ’ 1 =  ’ 1 + P 1b  b + P 1b P b2  2 E 1 -E b (E 1 -E b ) (E 1 -E 2 ) 11 22 bb A Fragment : The orbital 1 and 2 belong to the same fragment and are orthogonal inside A S. Imagaki, H. Fujimoto K. Fukui J. Amer. Chem. Soc 98 (1976) 4054 L. Libit R. Hoffmann J. Amer. Chem. Soc 96 (1974) 1370 B Fragment : A single orbital  1 and  2 mix.  1 is partially depopulated and  2 becomes partially populated. This corresponds to mix with an excited state: IC

18 18 3 rd order perturbation  ’ 1 =  1 + P 1b /(E 1 -E b )  b At variance with  1,  ’ 1 is not orthogonal to  2  ’’ 1 =  ’ 1 + P 1’2 /(E 1 -E 2 )  2 Let express P 1’2 by developing  ’ 1 P 1’2 = = + P 1b /(E 1 -E b ) P 1’2 = 0 + P 1b P b2 /(E 1 -E b )  ’’ 1 =  ’ 1 + P 1b P b2 / [(E 1 -E b )(E 1 -E 2 )]  2  ’’ 1 =  1 + P 1b /(E 1 -E b )  b + P 1b P b2 / [(E 1 -E b )(E 1 -E 2 )]  2 11 22 bb A Fragment : B Fragment : The third order terms account for the polarization of the fragment

19 19 E b below E 1  ’ 1 =  ’ 1 -  b +  2  ’ 2 =  ’ 1 -  b +  2 Increases decreasesDecreases increases

20 20 Decreases increases  ’ 1 =  ’ 1 +  b +  2  ’ 2 =  ’ 1 -  b +  2 E b intermediate between E 1 and E 2

21 21 E b above E 1 Decreases increasesIncreases decreases  ’ 1 =  ’ 1 +  b +  2  ’ 2 =  ’ 1 -  b -  2

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

23 Cyclobutadiene from Butadiene

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

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

26 26 Orbitals  ’ 1 from perturbation

27 27 New lowest orbital  ’ 1  A  B  C  D  1  IPI  3 >/      A  -  B -  C  D   The + sign means in phase between 1 and 4 The new orbital looks like  1 but with modulation of amplitude indicated by red arrows

28 28 Exact solutions First set 1/2   

29 29 Exact solutions Second set 1/2 1/√2   

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

31 31 Butadiene from 2 ethylenes 1/√  (  A +  B )1/√  (  C +  D ) 1/√  (  A -  B ) 1/√  (  C -  D )

32 32 Butadiene from 2 ethylenes first order terms 1/√  (  A +  B )1/√  (  C +  D ) 1/√  (  A -  B ) 1/√  (  C -  D )      1/√  (  A +  B )IPI1/√  (  c +  D )> = 0.5 

33 33 Butadiene from 2 ethylenes second order terms 1/√  (  A +  B )1/√  (  C +  D ) 1/√  (  A -  B ) 1/√  (  C -  D )      1/√  (  A +  B )IPI1/√  (  c +  D ) >] 2 / (  -(-  ))=  P has to be calculated on the unperturbed MO expressions (before taking into account 1st order perturbation)

34 34 New lowest orbital  ’ 1  A  B  C  D  1  IPI  3 >/      A  -  B -  C  D   The + sign means in phase between 2 and 3  without normalization  normalized

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

36 36 Butadiene from 2 ethylenes minimizing the repulsion 1/√  (  A +  B )1/√  (  C +  D )   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        

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

38 38 cyclobutadiene from 2 ethylenes first order terms 1/√  (  A +  B )1/√  (  C +  D ) 1/√  (  A -  B ) 1/√  (  C -  D )      1/√  (  A +  B )IPI1/√  (  c +  D )> =  No terms from second order

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

40 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. Reaction coordinate or any transformation Energy

41 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. distance Energy of states

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

43 43 Benzene from pentadienyl + C radicals 1.What are the coefficients of the SOMO? 2.What are the coefficients of the bonding MO antisymmetric relative to the mirror? 3.What is the corresponding Energy level? 4.What are the coefficients of the bonding MO symmetric relative to the mirror? 5.Deduce from them the MO Energy level? pentadienyl radical without calculations

44 44 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? pentadienyl radical using only symmetry and Normalization.     √   √ 

45 45 What are the coefficients of the SOMO? Use alternant property What are the coefficients of the bonding MO antisymmetric relative to the mirror? a double 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? pentadienyl radical using only symmetry and Normalization.  √   2  √    2  1/√12: 1/3 +2 (1/2) c 2 =1 1/2: 2 (1/2) c =1 1/√3: 1/3 +2 c =1 1/√12  2  √   2   2 1/√12  2  √  Develop 4 [(1/√12)(1/2)+(1/2)(1/√3)]= √3  √   √ 

46 46 First order term.  √   2  √    2  1/√12  2  √   2   2 1/√12  2  √   √   √    √    Benzene from pentadienyl + C radicals

47 47 First order term.  S Radical chain + C radical atom comparing the chain with the ring: Aromaticity  0 for the ring 2/√(N-1) for the chain 4/√(N-1) for the ring 2/√(N-1) for the chain A 

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

49 49 Second order terms. 1/√12  2  √  1/√12  2  √   √   √    √12) 2 /√3=     Benzene from pentadienyl + C radicals

50 50 Second order terms.  √   2  √    2  1/√12  2  √   2   2 1/√12  2  √   √   √  (2/√12) 2 /√3=     Benzene from pentadienyl + C radicals   √   √      

51 51 Benzene from butadiene+ethylene

52 52 Benzene from butadiene+ethylene Black arrow:  √    Whire arrow:  √   

53 53 Fulvene Here are 4 MOs Find the missing ones (coefficients and Energy) Justify the energy 1  for the orbital at the bottom left Is fulvene more likely a donor or an acceptor?

54 54 The two antisymmetric orbitals are from a butadiene (no contribution on the C that belong to the mirror plane)  

55 55 A non bonding orbital made of two  CC  Is fulvene more likely a donor or an acceptor? The LUMO level  is low in energy Fulvene is an acceptor (interacting with an electron rich dienophile such as  CH 2 =C(OMe) 2 LUMO: 

56 56 Let call  1 = a  1 + b  2 + c  3 + c 4  4 + c 5  5 + c   6 the lowest orbital 1.Identify a, b and c to the numbers 0.232, and Express c 4, c 5 and c 6 in terms of a, b and c 3.What relationship connects a, b and c? 4. Give energy E 1 of  1 as a function of a, b and c. Calculate E 1. 5.The coefficients of the other orbitals are obtained by the following permutations: c 1 c 2 c 3  3 cab  2 bca  1 abc Give c 2 et c 3 the sign of the coefficients (c 1 will be conventionally chosen as positive). We give E 2 =1.247  et E 3 =0.445 . Hexatriene

57 57 Let call  1 = a  1 + b  2 + c  3 + c 4  4 + c 5  5 + c   6 the lowest orbital 1. a=0.232 b=0.418 c= Express c 4, c 5 and c 6 in terms of a, b and c: c 6 =c 1 c 5 =c 2 and c 4 =c 3 3.What relation link a, b and c? a 2 +b 2 +c 2 = Give the energy E 1 of  1 as a function of a, b and c. Calculate E 1. 4ab+4bc+2c 2 = E 1 = . 5.The coefficients of the other orbitals are obtained by the following permutations: c 1 c 2 c 3  3 ca b  2 bc a  1 ab c  2 c 2 and c 3 are positive  3 c 2 is positive an c 3 is negative. E 2 =1.247  and E 3 =0.445 . Hexatriene

58 58 Ring C 14 and pyrene by perturbation from hexadienes 2 hexatrienes + atoms A and B

59 59 The antisymmetric combinations remain unperturbed A Hexatriene C 14 ring A,B

60 60 The Symmetric combinations interact S A Orbitals are degenerate (except the lowest and highest levels); correlation jumps from one energy level to the next.

61 61 Interaction with C=C 2 stabilizing interactions between Frontier Orbitals S A S S LUMO HOMO LUMO HOMO

62 62 Trimethylenemethane From symmetry, 4 degenerate orbitals; 2 of them remain non bonding, The two other mix (3(1/√  )  ). This is a diradical  √2 √(2/3) -1/√6 1/√3 1/√6  √2 √ √  -√  

63 63 What are Mo’s coefficients energy level of the p orbital of lowest energy level for the cyclopentadienyl radical? Two MOs from cyclopentadienyl are MOs from butadiene? Which ones? Give all the energy levels for the cyclopentadienyl radical without any calculation. Considering Ag 6. Each Ag will be described by a single valence eorbital, 5s, occupied by a single electron. Among the 3 below which one is the most stable when fragments are located far from each other (Ag 5 will always be assumed with 5-fold symmetry). (Ag 5 +Ag); (Ag 5 + +Ag - ) et (Ag 5 - +Ag + ). Exercise

64 64 Using PMO theory based on Frontier orbitals, tell which among the 3 structures which is the most stable. (every Ag-Ag distance will be supposed equal). Exercise 6A 6B 6C Show (using properties of alternant systems) that this structure has an energy close to that of an hexagonal structure. Knowing that this structure is more stable guess another structure encore plus structure even more stable?

65 65 cyclopentadiene

66 66 C 10 H 8 bicyclic compounds with a common CC bond Naphtalene vs. Azulene Are these structures stable? Which one has the lowest energy? Why azulene is named azulene? (22-8)/2=7 insaturations: 2 rings 5 double bonds

67 67 Comparison with C 10 H 10 : a ring closure  √2=  √2= √(1/ )= √(1/ )= /√5=0.4472

68 68 Naphtalene vs. Azulene Naphtalene formation is better than azulene formation Between 1 and 6 large lobes Between 1 and 5 small (zero considering the rule for alternants: 1 and 6 are both starred atoms)

69 69 HOMO and LUMO of azulene HOMO LUMO A S High HOMO, Low LUMO: this is why it is blue!

70 70 Polarization of the Frontier orbitals mixing with the orbital with the closest energy level HOMO mixes out of phase with blue LUMO in phase with blue LUMO On C7 HOMO On C5

71 71 4n+2 e :Aromaticity C 7 + C bonding orbitals to accommodate 6e

72 72 The anion is stable (aromatic) The cation is not (antiaromatic- Jahn-Teller situation) Thorium +4

73 73 Azulene is not an alternant hydrocarbon C7 is positively charged and C5 is negatively charged. There is a large dipole moment!

74 74 H3H3  1/√3 1/√2-1/√2 -- √(2/3) -1/√6 

75 75 Benzene built from 3 double bonds

76 76              First Order pertubation: Construction of symmetry orbitals From and One MO generates a set of 3 MOs The interaction is  ’= (1/√2)(1/√2) = 0.5 

77 77               ’ -’-’  ’’ -  ’’  ’= -  ’’=        -1/2 1/2 -1/2 1/√3 1/√6 1/√12 -1/√12  1/√6 1/√3

78 78 Energies: 2 nd order perturbation Each  orbital interacts with two  * orbitals  E’ = 2 P 2 /  E  E’ =  /2 P has to be calculated on the unperturbed MO expressions (before taking into account 1st order perturbation) P 2 /  E  =  ’ 2 /(+1(-1))=  /4  ’=  /  The interaction P is  ’= (1/√2)(1/√2) = 0.5 

79 79                   S A S A  

80 80  √12)(1/2)=1/√12     S S  √3)(1/2)=1/√12 P/  E =(3/√12)/[0.5-(-0.5)]= √3 Mixing of SS orbitals After solving 3x3 determinant for degenerate MOs 1/√3 -1/√12 -1/2 1/2

81 81 S S Mixing of SS orbitals After solving 3x3 determinant for degenerate MOs 1/√3 -1/√12  + √3  * -1/2 1/2 S S 1→1/2  *  - √3  -1→ -1/2 Before normalization → After normalization 00 -1/√3→ -1/√12 2/√3→1/√3 Before normalization → After normalization 0 -1/2 1/2 -1/√12 1/√12 1/√3 1/√12

82 82 The perturbation is large We are building half of the bonds which is not a small perturbation. Formula are accurate because there is no other terms beyond 2 nd order. P/  E = √3 /1

83 83 Exercice The Bow-tie molecule 83 C 6 H 4 molecule is planar with three planes as mirror symmetry, so called the molecule plan, the first mirror (containing C 1 and C 2 ) and the second mirror (perpendicular to C 1 -C 2 ) respectively. We call it the bow- tie molecule hereafter. First Mirror Second Mirror

84 84 1.Explain what partition of the molecular orbitals the molecule plane performs. (Give the labels of the molecular orbitals that this mirror symmetry performs) the s and p separation: s orbitals are symmetric and p orbitals are antisymmetric. 2. Write down the secular determinant for each symmetry group of the  orbitals by considering the two other mirror symmetries ( “ first ” and “ second “ mirror as indicated on the figure above). 3. Two orbitals are pure symmetry orbitals. What are their energies? x = -1 (unit  ) 4. Make a drawing of these two molecular orbitals and give their atomic coefficients. Coefficients are 0.5. Bow-tie SSC1C1 C6C6 C1C1 -x+12 C6C6 1 SAC1C1 C6C6 C1C1 -x-12 C6C6 1-x+1 ASC6C6 C6C6 -x-1 AAC6C6 C6C6 -x-1

85 85 Bow-tie 5. Solve the secular determinant involving orbital symmetric relative to the two mirrors and give the energy levels. SS (1-x) 2 =2 x=1±√2 (2.414 and ) 6. Solve the secular determinant involving orbital antisymmetric relative to the two mirrors and give the energy levels. SA (-1-x) (1-x) =2 x=±√3 (1.732 and ) 7. What is the total (  ) energy of the molecule? E= 2 ( )= Show a Molecular diagram (the energy levels with values in  units and label of symmetry).

86 86 Bow-tie II-1. The energy levels for the hexatriene (the linear C 6 H 8 chain) are E1=1.802 , E2=1.247 , E3=0.445  E4=1.802 , E5=1.247 b and E6=0.445 . Compare the stability of the bow- tie molecule with that of the hexatriene molecule. Is the Bow-tie molecule aromatic? It is aromatic: Its energy is larger an that of the chain . However 6e does not correspond to an optimal count since the HOMO is antibonding. II-2.What would be the ideal charge for the Bow-tie molecule? Positive (formally 2+ would allow removing the two antibonding electrons but would induce electrostatic repulsion). II-3. Remind what are the energy levels and the coefficients of the orbitals of C3H3 (a single equilateral triangle)

87 87 Bow-tie II-4. What would be the ideal electron count for such a ring? What would then be the charge for the ring? The ideal number of electrons is 2 (filling the bonding level and not the antibonding one); this implies a total charge +1. II-5. We are now building the Bow-tie molecule from two three member rings. Make a schematic drawing for the energies at the first order considering perturbations.

88 88 Bow-tie II-6. Calculate the second order shifts and show the resulting MO diagram P= (1/√3)(√2/√3) P 2 /  E=2/27=0.074

89 89 III-1. We now search finding the MOs using symmetry orbitals? What are the energy levels of the symmetry orbitals? Describe them? These are double bonds whose energy levels are ±1 (the horizontal one and the combinations of the two vertical ones) III-2. Use perturbation theory to calculate the MOs from symmetry orbitals of the SS symmetry. The interaction P is 4x(1/2)x(1/√2) = √2 = The shifts are then ±1.414 for SS (first order term) The resulting levels are 1±1.414 for SS (first order term) III-3. Write the secular determinant for the SA orbitals and make the full diagram. SA 1/√2 (C 1 -C 2 )1/2 (C 3 +C 4 +C 5 +C 6 ) 1/√2 (C 1 -C 2 )-x-1√2 1/2 (C 3 +C 4 +C 5 +C 6 )√2-x+1 Bow-tie

90 90 III-4. Why second order perburbation theory fails for the SA interaction? Because P=√2 is not small relative to  E=1 The application of second order theory for SA would lead to 2 and -2 instead of and Bow-tie III-3. make the full diagram.


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