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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 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).
22
Molecular Perturbations
i= r crr and Ei are solutions for a known system.We like to find solutions 'i= r c'rr 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 (Ei,i) and P=H’-H is an operator whose terms Pij=[<i|P|j>]* are weak. P is the operator defining the perturbation.
*data usually characterize atoms ppij=<i|P|j> which are few terms but no so small (a bond formation between i and j). Pij=<i|P|j> are derived from the ppij. Pij are significantly weaker than ppij due to delocalization of AOs within MO expressions. In general, results remain significant even though some ppij are important.
33
Molecular Perturbations
'I = ai I = r crr 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'rr it we be easier to solve the secular determinant directly.The advantage of using perturbations is to make it within approximations but simply.
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.
55
Molecular Perturbations
E1-E 0 0
= 0
0 E2-E 0
0 0 Ej-E
Secular determinant without perturbation: If known, the secular equation is solved (diagonal)!
66
Molecular Perturbations
E1+P11-E P12 P1j
= 0
P21 E2+P22-E P2j
Pj1 Pj2 Ej+Pjj-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.
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 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
(E’i+Pii-E) = 0 → E’j = Ej+ Pjj
88
Orbital associated with the First order Energy solution (Zero order)
Searching for ’j close to j
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 order ’j= j
99
E1+P11-E P12 P1j
= 0
P21 E2+P22-E P2j
Pj1 Pj2 Ej+Pjj-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 Ej, The secular determinant is
(E’i+Pii-E) - i ( (E’i+Pii-E)P*ijPji + = 0
[(E’i+Pii-E) (E’j+Pjj-E)]
1010
Second order Energy solution
(E’i+Pii-E) - i
( (E’i+Pii-E)P*ijPji + = 0
[(E’i+Pii-E) (E’j+Pjj-E)]
(E’i+Pii-E) –i
P*ijPji = 0
[(E’i+Pii-E) (E’j+Pjj-E)]
-(E’j+Pjj-E)= - i P*ijPji
(E’i+Pii-E)
E – Ej – Pjj = -i P*ijPji
(Ei-Ej)
1111
Second order Energy solution
E = Ej + Pjj +i P*ijPji
(Ej-Ei)
E’j = Ej + Pjj +i P*ijPji
(Ej-Ei)
For real terms E’j = Ej + Pjj +i Pij2/ E
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 energyIncreases the energy gap
Dimension of an energy:Square of an energy over an energy
1212
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 j, the ith line has small terms of the first order. ai = Pij/E aj
The orbital is ’j= j +
ij Pij/Ei-Ej) 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 ij
First order
not small terms if ij
First order
no dimension
In phase for the lowest combination
1313
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
A B
A B
B
A
Niveau liant
Niveau AntiliantAntibonding
Bonding
’j= j +
ij Pij/E i Ei’-Ei = Pij
2/ E
1414
2 OA interaction degenerate or close in energy
1 2
1 /2 -E = 0
2 /2 -E
(/2 -E) (-/2 -E) = 2 E2 - (2 /4) = 2
E2 = 2 + (2 /4)
E = ±√[2 + (2 /4)] general
• If , E+ - E =
• If <<, E+ = (/2) (1 + 4 2/ 2) 0.5
E+ = (/2) (1 + 2 2/ 2) = /2 (1 + 2 2/
E+ - E = 2/ 2nd order Perturbation
term
√ (2+2/4) The geometric mean
of and
1515
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/ E
Kenichi FukuiJapanNobel 1981
1616
3rd order perturbationCase 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 + P1b b +
P1b P
b2 2 E1-Eb
(E1-Eb) (E1-E2)
1
2
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
1717
3rd order perturbationanother view point:excitation
’1 = ’1 + P1b b +
P1b P
b2 2 E1-Eb
(E1-Eb) (E1-E2)
1
2
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
1818
3rd order perturbation’1 = 1 + P1b/(E1-Eb) b
At variance with1, ’1 is not orthogonal to 2
’’1 = ’1 + P1’2/(E1-E2) 2
Let express P1’2 by developing ’1 P1’2 = < ’1 IPI 2 > = < 1 IPI 2 > + P1b/(E1-Eb) < b IPI 2 >
P1’2 = 0 + P1b Pb2/(E1-Eb)
’’1 = ’1 + P1b Pb2/ [(E1-Eb)(E1-E2)] 2
’’1 = 1 + P1b/(E1-Eb) b + P1b Pb2/ [(E1-Eb)(E1-E2)] 2
1
2b
A Fragment :
B Fragment :
The third order terms account for the polarization of the fragment
1919
Eb below E1
-
+
-
-
12
b
1b
' '
2
+-
’1 = ’1 - b +
2 ’2 = ’1 - b + 2
Increases decreases Decreases increases
2020
+
-
-
12
b
1b
' '
2
+
+ +
Decreases increases Decreases increases
’1 = ’1 + b +
2 ’2 = ’1 - b+2
Eb intermediate between E1 and E2
2121
Eb above E1
+
+
+
2
''
b1
b2
1
-
+ -
Decreases increases Increases decreases
’1 = ’1 + b +
2 ’2 = ’1 - b-2
2222
ExamplesTo judge the method, we will search for known systems (easily calculated
without perturbation)
• Cyclobutadiene from butadiene
23230.3717
0.3717
0.6015
0.6015
0.6015
-.60150.6015
0.6015
-.6015
0.6015
-.3717
-.3717
-.3717
0.3717
-.3717
1.618
0.618
-.618
-1.618
Cyclobutadiene from Butadiene
C3
C4
C2
C1
C2
C2 C1
C4
0.3717
2424
MO Energies from perturbation
S S
A A
Only orbitals of the same symmetry mix together
What we know is <1IPI4> = 1: there is a bond between atoms 1 and 4 that did not exist for butadiene
C3
C4
C2
C1
C3
C2 C1
C4
A Mirror symmetry is preserved
2525
Conservation of Orbital Symmetry
H C Longuet-Higgins E W Abrahamson
Hugh Christopher Longuet-Higgins1923-2004
The Molecular orbitals are solution of the symmetry operators of the molecule.MOs from different symmetry groups do not mix.
2626
Orbitals ’1 from perturbation
2727
New lowest orbital
’1ABCD 1IPI 3>/A-B-CD
The + sign means in phase between 1 and 4
The new orbital looks like 1
but with modulation of amplitude indicated by red arrows
2828
Exact solutions First set
1/2
1/2 1/2
1/2
2929
Exact solutions Second set
1/2
1/2
1/√2 1/√2
3030
Solutions are very accurate even though <iIPIj> is large (introducing a bond as strong as others). < iIPIj> MOs are less large due to delocalization of AOs with MOs.
3131
Butadiene from 2 ethylenes
1/√ (A +B) 1/√ (C +D)
1/√ (A - B) 1/√ (C - D)
3232
Butadiene from 2 ethylenesfirst order terms
1/√ (A +B) 1/√ (C +D)
1/√ (A - B) 1/√ (C - D)
1/√ (A +B)IPI1/√ (c +D)> = 0.5
3333
Butadiene from 2 ethylenessecond order terms
1/√ (A +B) 1/√ (C +D)
1/√ (A - B) 1/√ (C - D)
1/√ (A +B)IPI1/√ (c +D) >]2 / (-(-))= 0.125
P has to be calculated on the unperturbed MO expressions (before taking into account 1st order perturbation)
3434
New lowest orbital
’1ABCD 1IPI 3>/A-B-CD
The + sign means in phase between 2 and 3
without normalization normalized
3535
The total energy stabilization comes only from second order terms
(interaction between occupied and vacant orbitals)
E = 4 * 0.125 = 0.5 This is the energy due to the delocalization.
3636
Butadiene from 2 ethylenesminimizing 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 butadieneAvoiding closing ring to cyclobutadiene
3737
Cyclobutadiene from 2 ethylenes
C3 C4
C2 C1
C3
C2 C1
C4
Only one mirror symmetry is preserved by perturbation
3838
cyclobutadiene from 2 ethylenesfirst 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
3939
Conservation of Orbital Symmetry
H C Longuet-Higgins E W Abrahamson
Hugh Christopher Longuet-Higgins1923-2004
The Molecular orbitals are solution of the symmetry operators of the molecule.MOs from different symmetry groups do not mix.
4040
Non-crossing ruleThe 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
En
erg
y
4141
Non-crossing ruleThe 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
En
erg
y o
f sta
tes
4242
Non-crossing rule
Atomic or covalent
ionicionic
Mixed ionic and covalent
ionic
atomic
4343
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
4444
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 andNormalization.
√
√
4545
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, symmetrizedWhat 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 andNormalization.
√
2√
2
1/√12: 1/3 +2 (1/2)2 + 2 c2 =11/2: 2 (1/2)2 + 2 c2 + 0 =1
1/√3: 1/3 +2 c2 + 0 =1
1/√12
2
√
2
2
1/√12
2
√
Develop <IHI>4 [(1/√12)(1/2)+(1/2)(1/√3)]= √3
√
√
4646
First order term.
√
2√
2
1/√12
2
√
2
2
1/√12
2
√
√
√
√
Benzene from pentadienyl + C radicals
4747
First order term.
S
Radical chain + C radical atomcomparing the chain with the ring: Aromaticity
0 for the ring2/√(N-1) for the chain
4/√(N-1) for the ring2/√(N-1) for the chain
A
4848
When the SOMO is symmetricThe ring is more stable than the chainThe polyene is AROMATICN-1 is evenN = 4n +2
S
Radical chain + C radical atomcomparing the chain with the ring: Aromaticity
A
Aromaticity accordingto Dewar
When the SOMO is antisymmetricThe ring is less stable than the chainThe polyene is ANTIAROMATICN-1 is oddN = 4n
4949
Second order terms.
1/√12
2
√
1/√12
2
√
√
√
√12)2/√3=0.1924
Benzene from pentadienyl + C radicals
5050
Second order terms.
√
2√
2
1/√12
2
√
2
2
1/√12
2
√
√
√
(2/√12)2/√3=0.1924
Benzene from pentadienyl + C radicals
√
√
5151
Benzene from butadiene+ethylene
0.3717
0.3717
0.6015
0.6015
0.6015
-.60150.6015
0.6015
-.6015
0.6015
-.3717
-.3717
-.3717
0.3717
-.3717
1.618
0.618
-.618
-1.618
C3
C4
C2
C1
+
C6
C5C3
C2
C1
C6
C5
C4
0.3717
5252
Benzene from butadiene+ethylene
-1.618
-0.618
0.618
1.618
-1.0
1.0 -0.447
-0.447
0.447
-0.447
0.447 -0.447
0.447
0.447
-2.065
-1.065
1.065
2.065
A
A
AS
S
S
Black arrow:√
Whire arrow:√
5353
Fulvene
-0.44
0.15 0.15
-0.44
0.66
-0.36
E = -1.86
-0.35
0.28 0.28
-0.35
-0.19
0.75
E = -0.25
0.0
0.5 0.5
0.0
-0.5
-0.5
E = 1.
0.43
0.39 0.39
0.43
0.52
0.25
E = 2.12
Here are 4 MOsFind 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?
5454
The two antisymmetric orbitals are from a butadiene (no contribution on the C that belong to the mirror plane)
5555
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 CH2=C(OMe)2
LUMO:
5656
Let call 1= a1 + b2 + c3 + c44 + c55 + c6 the lowest orbital1. Identify a, b and c to the numbers 0.232, 0.418 and 0.521.2. Express c4, c5 and c6 in terms of a, b and c 3. What relationship connects a, b and c?4. Give energy E1 of 1 as a function of a, b and c. Calculate E1.5. The coefficients of the other orbitals are obtained by the following
permutations: c1 c2 c3
3 c a b2 b c a1 a b c
Give c2 et c3 the sign of the coefficients (c1 will be conventionally chosen as positive).
We give E2=1.247 et E3=0.445 .
Hexatriene
5757
Let call 1= a1 + b2 + c3 + c44 + c55 + c6 the lowest orbital
1. a=0.232 b=0.418 c=0.5212. Express c4, c5 and c6 in terms of a, b and c: c6=c1 c5=c2 and c4=c3
3. What relation link a, b and c? a2+b2+c2 = 0.5
4. Give the energy E1 of 1 as a function of a, b and c. Calculate E1.
4ab+4bc+2c2 = E1 = 1.802 .5. The coefficients of the other orbitals are obtained by the following
permutations: c1 c2 c3
3 c a b
2 b c a
1 a b c
2 c2 and c3 are positive3 c2 is positive an c3 is negative.
E2=1.247 and E3=0.445 .
Hexatriene
5858
Ring C14 and pyrene
by perturbation from hexadienes
2'
3'
4'
5' 5
4
3
2
6'B
6
1
A
1'
y
x
pyrène
2 hexatrienes + atoms A and B
5959
The antisymmetric combinations remain unperturbed
2
-2.
- 1.802
- 1.247
- 0.445
1.802
1.247
0.445
en pointillé les correspondances entre orbitalesantisymétriques par rapport au plan contenant A et B
S
A
S
A
A+S
S
A
S
A
HEXATRIENES ANNULENE EN C14 A,B
2'
3'
4'
5' 5
4
3
2
6'B
6
1
A
1'
y
x
A
Hexatriene C14 ring A,B
6060
The Symmetric combinations interact
2
-2.
- 1.802
- 1.247
- 0.445
1.802
1.247
0.445
en pointillé les correspondances entre orbitalesantisymétriques par rapport au plan contenant A et B
S
A
S
A
A+S
S
A
S
A
HEXATRIENES ANNULENE EN C14 A,B
2'
3'
4'
5' 5
4
3
2
6'B
6
1
A
1'
y
x
S
2'
3'
4'
5' 5
4
3
2
6'B
6
1
A
1'
y
x
A
Orbitals are degenerate (except the lowest and highest levels); correlation jumps from one energy level to the next.
6161
Interaction with C=C2 stabilizing interactions between Frontier Orbitals
2'
3'
4'
5' 5
4
3
2
6'B
6
1
A
1'
y
x
S2'
3'
4'
5' 5
4
3
2
6'B
6
1
A
1'
y
x
A
SS
LUMO
HOMO LUMO
HOMO
pyrène
6262
C
CH2
CH2H2CTrimethylenemethane
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/√3
1/√6
1/√6
√2
√
-√
6363
•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 Ag6. 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 (Ag5 will always be assumed with 5-fold symmetry). (Ag5 +Ag); (Ag5
+ +Ag-) et (Ag5
- +Ag+).
Exercise
6464
• 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?
6565
cyclopentadiene
1.618
0.618
-.618
-1.618
6666
C10H8 bicyclic compoundswith a common CC bondNaphtalene 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
6767
Comparison with C10H10 : a ring closure
√2= 0.4253
√2= 0.2628
√(1/5-0.42532)=0.1383
√(1/5-0.26282)=0.3618
1/√5=0.4472
6868
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)
6969
HOMO and LUMO of azulene
HOMO
LUMOA S
High HOMO, Low LUMO: this is why it is blue!
7070
Polarization of the Frontier orbitalsmixing 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
7171
4n+2 e :Aromaticity C7+ C5
-
3 bonding orbitals to accommodate 6e
7272
The anion is stable (aromatic)The cation is not (antiaromatic-
Jahn-Teller situation)
Thorium+4
7373
Azulene is not an alternant hydrocarbon
C7 is positively charged and C5 is negatively charged.
There is a large dipole moment!
7474
H3
1/√3
1/√2 -1/√2
-
√(2/3)
-1/√6 -1/√6
7575
Benzene built from 3 double bonds
7676
First Order pertubation: Construction of symmetry orbitalsFrom and
One MO generates a set of 3 MOs
The interaction is ’= (1/√2)(1/√2) = 0.5
7777
’
-’
’’
-’’
’= -’’=
-1/2
1/2
-1/2
-1/2
1/√3
1/√6
1/√12
-1/√12
-1/√12
-1/√12
1/√6
1/√3
7878
Energies: 2nd order perturbationEach orbital interacts with two * orbitals
E’ = 2 P2/E
E’ = /2
P has to be calculated on the unperturbed MO expressions (before taking into account 1st order perturbation)
P2/E= ’2/(+1(-1))= /4
’= /
The interaction P is ’= (1/√2)(1/√2) = 0.5
7979
S
AS
A
8080
√12)(1/2)=1/√12
S
S
√3)(1/2)=1/√12 √3)(1/2)=1/√12
P/E =(3/√12)/[0.5-(-0.5)]= √3
Mixing of SS orbitalsAfter solving 3x3 determinant for degenerate MOs
1/√3
-1/√12
-1/√12
-1/2
1/2
8181
S
S
Mixing of SS orbitalsAfter solving 3x3 determinant for degenerate MOs
1/√3
-1/√12
+ √3 *
-1/2
1/2S
S
1→1/2
*- √3
-1→ -1/2Before normalization → After normalization
00
-1/√3→ -1/√12
2/√3→1/√3
Before normalization → After normalization
0
-1/2
1/2
-1/√121/√12
1/√3
1/√12
8282
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 2nd order.
P/E = √3 /1
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Exercice The Bow-tie molecule
83
C6H4 molecule is planar with three planes as mirror symmetry, so called
the molecule plan, the first mirror (containing C1 and C2) and the
second mirror (perpendicular to C1-C2) respectively. We call it the bow-
tie molecule hereafter.
C6
C5
C1 C2
C4
C3First Mirror
Second Mirror
8484
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
SS C1 C6
C1 -x+1 2
C6 1 -x+1
SA C1 C6
C1 -x-1 2
C6 1 -x+1
AS C6
C6 -x-1
AA C6
C6 -x-1
8585
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 -0.414)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 -1.732)7. What is the total () energy of the molecule?E= 2 (2.414 + 1.732-0.414)=7.4648. Show a Molecular diagram (the energy levels with values in units and label of symmetry).
86
Bow-tie
II-1. The energy levels for the hexatriene (the linear C6H8 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 6.988 . 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
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
Bow-tie
II-6. Calculate the second order shifts and show the resulting MO diagramP= (1/√3)(√2/√3) P2/E=2/27=0.074
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 = 1.414The 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.
SA1/√2 (C1-C2) 1/2
(C3+C4+C5+C6)
1/√2 (C1-C2) -x-1 √2
1/2 (C3+C4+C5+C6) √2 -x+1
Bow-tie
90
III-4. Why second order perburbation theory fails for the SA interaction?Because P=√2 is not small relative to E=1The application of second order theory for SA would lead to 2 and -2 instead of 1.732 and -1.732
Bow-tie
III-3. make the full diagram.
