1 Variational principle The complete set of exact eigenfunctions of H define an orthogonal complete basis set for the total space of wave functions. H.

1

Variational principle

The complete set of exact eigenfunctions of H define an orthogonal complete basis set for the total space of wave

functions.

H is Hermitic. Let be a and b two normalized wavefunctions (associated with two different values Ea and Eb ). We have therefore (due to hermiticity)

< a |H| b > = Ea< a | b > < a |H| b > = Eb < a | b >

Wherefrom (Ea - Eb) < a | b > = 0and since Ea Eb

< a | b > = 0

If so, it is possible to express any function as a linear combination of the exact eigenfunctions, .

Statement 1:

demonstration:

Consequence:

2

Variational principle

The energy associated with a function is always above that of the eigenfunction of lowest energy: E0.

is not a eigenfunction; it is associated with an energy <E> that is an average energy (mean value)

A mean value is always intermediate relative to extreme:Greater than the smallest!

<E> > E0

Statement 2:

3

Mean value

• If 1 and 2 are associated with the same eigenvalue o: O(a1 +b2)=o(a1 +b2)

• If not O(a1 +b2)=o1(a1 )+o2(b2)

we define <O> = (a2o1+b2o2)/(a2+b2)

Dirac notations

4

Variational principle

The energy associated with a function is always above that of the eigenfunction of lowest energy: E0.

and are two eigenfunctions < |H| = < | E and |H| > = E| >

wherefrom < |H| > = E From statement 1, is a linear combination of s

| >= c| >

let multiply the left by < |; this leads to only one term c= < | > and similarly c*= < | >

then | >= < | > | >

Statement 2:

5

Variational principle

and are two eigenfunctions associated to E and E

From statement 1, is a linear combination of s

| >= c| >

then

< | > = c < |> c

< | > = c 2

< |> = c 2

= 1

Statement 2, normalization:

6

< |H | > = c < |H | > c

< |H | > = E < | > c

< |H | > = E c c

< |H | > = E c 2 > Ec

2 = E

E > E

Variational principleStatement 2, Demonstration:

An non-exact solution has always a higher energy than the lowest exact solution

7

and are two eigenfunctions < |H| = < | E and |H| > = E| > wherefrom < |H| > = E

From statement 1, is a linear combination of s | >= c| >

Let multiplie the left by < |; this leads to only one term c= < | >

and similarly c*= < | > then < | > = c < |> c

< | > = < | > < |> < | > = < | > < | >

< | > = |< | >| 2 =1

Demonstration < |H | > = < | > < |H | > < | >

< |H | > = E < | > < | > < | >

< |H | > = E < | > < | >

< |H | > = E |< | >| 2 > E|< | >| 2 = E

< |H | > > E

Using Dirac notation

8

Variation principleGiven an approximate expression that depends on

parameters, we must determine the parameters to minimize the energy.

Within LCAO, an MO is a linear combination of AOs: We have then to chose the coefficients so to minimize the energy.

Note that the variational principle is not restricted to cases of linear combinations.

9

c02

c12

c22

c32

A mean value is always higher than the lowest value !

10

Slater exponent for He

where Z* is a parameter

<E> = - Z*2/2 u.a. <T> = + Z*2/2 u.a.

<V> = <(1,Z*)|-Z*/r|(1,Z*)>

using <(1,Z*)|1/r|(1,Z*)> = Z* u.a.

<V> = -Z*<(1,Z*)|1/r|(1,Z*)> = -Z*2u.a.

Next slide

11

<(1,Z*)|1/r|(1,Z*)> = Z* u.a.

demonstration

12

Slater exponent for He where Z* is a parameter

13

Slater exponent for He

where Z* is a parameter

Search for the energy minimum

14

LCAO

is a linear combination of N s

| >= c| >

an approximate function

Using parameters: c

Atomic orbitals

Worse description than

There are N coeffcients to determineLet minimize E with respect to each of themTherefore, there is a set of N equations, E/c = 0

This seems fine; however there is a problem; Which is the problem?

15

LagrangianThere is a constraint due to normalization: the

N coefficients are not independent.

Joseph Louis de Lagrange French1736 - 1831

We search the minimum of a function L of N+1 variablesWhich “makes no difference with H”

L = < | > - E [< | > - 1 ]

There is a set of linear expressions to derive

L/c = 0 Should be 0

16

Secular determinant

Solving | Hij-ESij| =0, we find the Eis.

This is a N degree equation of E to be solved.

From N AOs, we find N energies and N MOs.

We find simultaneously E and (eigenfunction and eigenvalues).

17

LCAO

Assuming no change in the AOs *, the MOs correspond to a unitary change of the basis set of AOs.

MOs are orthogonal and normalized.

Neglecting overlaps, rc2ir=1 for a MO, i

and ic2ir=1 for a AO, i ; the AOs are distributed

among the MOs

* This is not the case for self consistent methods.

Sum over r atoms

for a given iorbital

Sum over iorbitals for a given r atom

18

Schrödinger equation for LCAO

Erwin Rudolf Josef Alexander SchrödingerAustrian 1887 –1961

H = E

H icii = Eicii Let multiply on the left by j, integrate and develop

ici< |H | i > = Eici < j |i >

ici (Hij-E Sij ) = 0This is the set of linear equation solved for | Hij-E S ij |= 0 secular determinant

The complete set of MOs are solution (not only the lowest).Our first approach from the variational principle emphasized the solution of lowest energyEvery extreme (derivative=0) is a physical solution!

19

Hückel Theory

Erich Armand Arthur Joseph Hückel German 1896-1980Never awarded the Nobel prize!Also known for Debye-Hückel theory of electrolytic solutions

In the 1930's a theory was devised by Hückel to treat the electrons of conjugated systems such as aromatic hydrocarbon systems, benzene and naphthalene. Only electron MO's are included because these determine the general properties of these molecules and the electrons are ignored. This is referred to as sigma-pi separability. The extended Hückel Theory introduced by Lipscomb and Hoffmann (1962) applies to all the electrons.

20

Hückel Theory

Erich Armand Arthur Joseph Hückel German 1896-1980Never awarded the Nobel prize!Also known for Debye-Hückel theory of electrolytic solutions

We consider only 2pZ orbitalsTwo parameters:The atomic 2p energy level:

E(2pZ) = (~ -11.4 eV)The resonance integral for adjacent 2p orbitals

Hij(C=C) = (~ -3 eV)

E = + x x= ( – E)/

We chose defining as negative

Sij = ij (the overlap is neglected).

21

SLATER KOSTER

Slater Koster for metal atoms : 9 orbitals s, p et d

Several overlaps

ss, sp, sd,

pp, pp, pd, pd,

dd, dd, dd.

Phys. Rev B 94 (1954) 1498

22

EHT Theory

• Only valence orbitals are described• AOs are Slater orbitals

– Parameters are the Hii (atomic energy levels)– The Slater exponents

• The Overlap, Sij, are rigorously calculated from the geometry and the AOs

• The resonance integrals, Hij, are derived from the Sij. Wolfsberg-Helmoltz formula

Hij = 1.75 (Hii+Hjj)/2 Sij

23

For linear and cyclic systems (with n atoms), general solutions exist.

Linear:

                                                  Cyclic:                                    For linear polyenes the energy gap is given as:

Polyenes

Charles Alfred Coulson (1910-1974), English

24

Extension to heteroatomsparameter Coulombic

IntegralResonance

Integral

Sulfurwith 2 electrons

S = + 0,5

CS = 0,4

Sulfurwith 1 electron

S = + 0,2

CS = 0,6

Fluorine F = + 3

CF = 0,7

Chlorine Cl = + 2

CCl = 0,4

Bromine Br = + 1,5

CBr = 0,3

CH3 grouphyperconj

ugation

Me = + 2

CMe = 0,7

parameter CoulombicIntegral

Resonance

Integral

Carbon C = CC =

Oxygen 1 electron

O = +

CO =

Oxygen 2 electrons

O = + 2

CO = 0,8

Nitrogen 1 electron

N = + 0,5

CN =

Nitrogen 2 electrons

N = + 1,5

CN = 0,8

Sulfur 2 electrons

S = + 0,5

CS = 0,4

Sulfur 1 electron

S = + 0,2

CS = 0,6

25

| Hij-E Sij |= 0

1 2 3 4 5 6 …

1 -x 1 or 0 1 or 0 1 or 0 1 or 0 1 or 0

2 1 or 0 -x 1 or 0 1 or 0 1 or 0 1 or 0

3 1 or 0 1 or 0 -x 1 or 0 1 or 0 1 or 0 = 0

4 1 or 0 1 or 0 1 or 0 -x 1 or 0 1 or 0

5 1 or 0 1 or 0 1 or 0 1 or 0 -x 1 or 0

6 ::

1 or 0 1 or 0 1 or 0 1 or 0 1 or 0 -x

1 means that the atoms 2 and 4 are connected, The determinant is symmetric relative to the diagonal

26

| Hij-E Sij |= 0

1 2 3 4 5 6 …

1 -x 1 or 0 1 or 0 1 or 0 1 or 0 1 or 0

2 1 or 0 -x 1 or 0 1 or 0 1 or 0 1 or 0

3 1 or 0 1 or 0 -x 1 or 0 1 or 0 1 or 0 = 0

4 1 or 0 1 or 0 1 or 0 -x 1 or 0 1 or 0

5 1 or 0 1 or 0 1 or 0 1 or 0 -x 1 or 0

6 ::

1 or 0 1 or 0 1 or 0 1 or 0 1 or 0 -x

This is an N degree equation of x

27

| Ei-x |= 0solving is a diagonalization problem

1 2 3 4 5 6 …

1 E1-x 0 0 0 0 0

2 0 E2-x 0 0 0 0

3 0 0 E3-x 0 0 0 = 0

4 0 0 0 E4-x 0 0

5 0 0 0 0 E5-x 0

6 ::

0 0 0 0 0 E6-x

A set of first degree equations: E=Ei associated with i

28

The diagonalization is a unitary transformation

OMs are orthogonal thus rcricrj= ijIf i=j, the sum of the coefficient’s squares for a given atom is 1 (normalization)If i≠j, the sum of the products of coefficients is 0 (orthogonality)

The matrix of coefficients is unitary

wherefrom = icricsj= ijIf i=j, the sum of the coefficient’s squares for a given orbital is 1AOs are distributed among all the MOsIf i≠j, the sum of the products of coefficients is 0

If all the orbitals were filled, there would be no interaction: all the r-s bond indices should be zero.

29

C2H4 2 e in 2 orbitals as in H2

We have solved the problem using symmetry

and without solving the secular equation.

1 2

1 -x 1 = 0

2 1 -x

Inputs are what is in the secular determinant : and connectivity

Outputs are orbitals, energies, total energies, charges and bond indices

30

C2H4 without overlap

1 2

1 -x 1 = 0

2 1 -x

u Bonding g antibonding

Energy x=1 x=-1

coefficients

-x c1 + c2 = 0 C=1/√2

-c1 + c2 = 0

c1 = c2

c1 + c2 = 0

c1 = -c2

Charge on atoms:

1 – i i cri2

0 0

Bonding

lrs= i i cricsi

1 for u2 0 for ug

-1 for u2

N-1 linear equations; the last one is redundant.

31

C2H4 including overlap 4e repulsion

1 2

1 -E -ES = 0

2 -ES -E

u Bonding g antibonding

Energy / (1+S) / (1-S)

coefficients c1 = c2 = c

c = 1 /√(2+2S)

c1 = -c2 = c

c = 1 /√(2+2S)Charge on atoms:

1 – i i (cri2 –s

cri csi )

0 0

Bonding

OPrs= i i cricsiSrs

S/(1+S)

for u2

S/(2+2S)- S/(2-2S)

= -4S2/(4-4S-4S22) ~ - S2

for ug

E2 = (-ES)2 E = ± (-ES)

Mulliken

Overlap populations

32

2 OA interaction modifying

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)] •If , 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/

The geometric mean

of and

33

ButadieneC1

C2

C3

C4

The topology is C1 bond C2, C2 bond to C3 and C3 bond to C4

A linear model contains the information with more symmetry (the topology does not distinguish between cis and trans)

1 2 3 4

1 -x 1 0 0

2 1 -x 1 0 = 0

3 0 1 -x 1

4 0 0 1 -x

C1 C2 C3 C4

34

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.

35

S

A

0

0

Butadieneusing symmetry for the

topologyC1 C2 C3 C4

S or A

36

ButadieneSymmetric C1 C2 C3 C4

S

x2 - x - 1 = 0 x = (1 ± √5)/2 Golden numbers 1.618 and -0.618Coefficients - (1 ± √5)/2 c1 + c2 = 0 and normalization c1

2 + c22 = ½

c = 0.3717 and 0.6015

Mind that symmetry 1 is for all the atomsNot the reduced part

-0.6181.618

= 0

37

Golden ratio Golden ratio conjugate                               

38

2b

a= √(b2+(2b)2) =b(1+√5)

b

a= b(1+

√5)

Golden ratio

Golden ratio conjugate                               

39

Polyclete and Durer

40

x2 = 1+x

The medial right triangle of this "golden" pyramid (see diagram), with sides              is interesting in its own right, demonstrating via the Pythagorean theorem the relationship or

41

Fibonacci recursion,irrational numbers (incommensurable)

In the "zeroth" month, there is one pair of rabbitIn the first month, the first pair begets another pairIn the second month, both pairs of rabbits have another pair, and the first pair dies.In the third month, the second pair and the new two pairs have a total of three new pairs, and the older second pair.

Rabbit population, assuming that:

42

Spirale constructions,

in nature,in music..

Related ?The Doctrine of the Mean (中庸 ,

py Zhōngyōng) is one of the Four Books, part of the Confucian

canonical scriptures.

43

=1.618

=1/ =0.618

2

Pentagon, icosahedra

44

ButadieneSymmetric C1 C2 C3 C4

S

Recipe to build the same determinant

Express along the AOs and gather interactions with equivalent atomsMind that symmetry reduction makes the normalization incomplete.

45

ButadieneAntisymmetric C1 C2 C3 C4

A

x2 + x - 1 = 0 x = - (1 ± √5)/2 Golden numbers 0.618 and -1.618Coefficients (1 ± √5)/2 c1 + c2 = 0 and normalization c1

2 + c22 = ½

c = 0.3717 and 0.6015

Mind that symmetry 1 is for all the atomsNot the reduced part

0.618 -1.618

46

ButadieneAntisymmetric C1 C2 C3 C4

A

Recipe to build the same determinant

Express along the AOs and gather interactions with equivalent atomsMind that symmetry reduction makes the normalization incomplete.

47

ButadieneSmall 0.3717

Large 0.6015Small 0.3717

Large 0.6015

Large 0.6015

Small 0.3717

The number of nodes increases, the amplitude is large at the middle for the extreme; it is large on the edges for the Frontier orbitals.

480.3717

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

Butadiene

49

Bonding orders Ground state

Butadiene

0.894 0.8940.447

No initial statement that distinguishes 1-2 from 2-3

50

Bonding orders Ground state

Butadiene

1.35Ả

No initial statement that distinguishes 1-2 from 2-3Matches the Lewis formula and indicates the stabilization due to delocalization

1.35Ả1.46Ả

C=C – C=C

51

Bonding orders Excited state

Butadiene

0.7240.447

C – C = C – C

0.447

52

It is no use to iterate

The Hückel method is powerful since predictive.It is possible to make the parameterization

dependent on the differences in bond lengths choosing 12 = 34 larger than 23 or modifying according to the charge.

This does not make the calculation predictive.It is possible to make iterations ( technique) to

make the results “self consistent”; this is not recommended. This does not allow controlling the parameters.

53

Annulenes

1 2 3 N-1 N-1

1 -x 1 0 0 0 12 1 -x 1 0 0 0

3 0 1 -x 1 0 0 = 0

0 0 1 -x 1 0

N-1 0 0 0 1 -x 1

N 1 0 0 0 1 -x

cr-1 -E cr + cr+1 = 0Secular equation:

54

Delocalization

55

Annulenessolutions for CN

R (r) = r+s angle = 2s/N

An annulene is a periodic system;

Solutions for annulenes are those of any periodic system: crystals (Bloch sums)

The new coefficient for r is the old one multiplied by a constant

56

The system allows solutions Rj with angle (j=2j/N).

There is N rotations possible including identity (j=0 ou N) : the « simple » rotation and the « multiple» rotations j

Functions satisfying to rotations j=2j/N:

We build linear combinations of AOs,(crr) satisfying rotation j

(crr) = e-ijj (crr) with cr (j) = N-1/2 eijrj = e –i2jr . There are N solutions, with a quantum number: j..

i is the square root of -1; i2=-1

annulenes

57

r=00

r=5

r=4

r=1

r=2

r=3

r=40

r=3

r=2

r=5

r=0

r=1

N

N exp(i2 2/6)

N exp(i2 4/6)

N

N exp(i2 8/6)

N exp(i2 10/6)

C5 x exp(-ijj)=

N exp(i2 10/6) x exp(-i2 2/6) =

exp(i2 2/6)

Rotation by 120° = 2/6(j=2, orbital 2)

The 5th AO takes the place of the first

C5 x exp(i2 2/6)

N

N exp(i2 8/6)

N exp(i2 10/6)

N

N exp(i2 4/6)

58

They are the LCAO functions satisfying the rotations j

There are j solutions; the eigenvalues are e-ijs

annulenes

j= crr

59

Benzene

The complexe solutions are associated with rotations only; they are not eigenfunctions of the other symmetry operators and must be the degenerate

annulenes

+

+

Two real solutions and 4 complexesDegenerate by pairs

60

j = 3, AS

j = ± 2, AA j = ± 2, SS

j ± 1, SA j = ± 1, AS

j = 0, SS

annulenesReal solutionsSatisfying two mirror symmetry

61

j = 3, AS

j = ± 2, AA j = ± 2, SS

j ± 1, SA j = ± 1, AS

j = 0, SS

annulenesOrbitals from symmetry only

All the coefficients are equal c=1/√6

E=< 1/√61+.. .IHI1/√61+…>= 12*(< 1/√61IHI1/√62>+…) = 2

All the coefficients are equal c=1/2

E = 4*(< 1/22IHI1/23>+…) =

62

j = 3, AS

j = ± 2, AA j = ± 2, SS

j ± 1, SA j = ± 1, AS

j = 0, SS

annulenesOrbitals from symmetry only

The total density in E orbital must respect symmetry

Degenerate orbitals: energy must be

02+c2=2/6 → c= 1/√3

1/22+c2=2/6 → c= 1/√12

Normalization is satisfied 4*1/12+2*1/3=1

63

For any annulene of large size …the number of OMs remains in the gap

from 2 to -2

2

2

-2j=3

j=±2

j=±1

j=0

64

j = 3, AS

j = ± 2, AA j = ± 2, SS

j ± 1, SA j = ± 1, AS

j = 0, SS

annulenesChosing another set of planes; rotation by 60°

Combining orbitalsS’A’ = ½ [SA + √3 AS ]A’S’ = ½ [SA - √3 AS ]New symmetry appearsWhile old ones desappears

65

Hückel determinant for CyclobutadieneNotations: x = ↔ E =  + x units , origin 12 3 4

1 -x 1 0 12 1 -x 1 0 = 03 0 1 -x 14 1 0 1 -x

Two possibilities of using mirror symmetries

66

First set

1/2

1/2 1/2

1/2

67

Second set

1/2

1/2

1/√2 1/√2

68

The Jahn-Teller Theorem 1937 "any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy thereby removing the degeneracy"

There is a symmetry reduction (a geometry distortion) when the a set of HOMOs is not completely filled.

Edward Teller 1908-2003Hungarian-American

Hermann Arthur Jahn 1907-1979 English

69

4e-8e 0-6e 2e-8e3e-5e-9e 3e-9e

In an octahedral crystal field, the t2g orbitals occur at lower energy than the eg orbitals. The t2g are directed between bond axes while the eg point along bond axes.

The full octahedral symmetry takes place only when the t2g set is fully occupied.

70

C4H4 is unstable, it exists stabilized by asymmetric ligand (one eg orbital is stabilized).It dimerizes easily.

Cyclobutadiene: C4H4

Chain of Vn-benzenen+1: The HOMO gap vanishes for n=4; then there is a rotation of the successive benzene loosing the full D5h symmetry.

V V V

If the rings are eclipsed, there is no gap; the rotation of ring opens a gap.

71

B5 Boron compounds

72

Jahn-Teller effects are grouped into two categories. The first arises from incomplete shells of degenerate orbitals. It includes the first-order Jahn-Teller effect and the pseudo Jahn-Teller effect. The second arises from filled and empty molecular orbitals that are close in energy and is the second-order Jahn-Teller effect. The two categories have quite different physical bases. As a result, geometric distortions produced by the first are quite small and normally lead to dynamic effects only.

In favorable cases, the second-order Jahn-Teller effect produces very large distortions, including complete dissociation of a molecule. This can occur even when the relevant molecular orbitals are separated in energy by as much as 4 eV.

73

CuCl2 Cu2+ 9eNiCl2 Ni2+ 8e

The shielding effect this has on the electrons is used to explain why the Jahn-Teller effect is generally only important for odd number occupancy of the eg level. The effect of Jahn-Teller distortions is best documented for Cu(II) complexes (with 3 electrons in the eg level) where the result is that most complexes are found to have elongation along the z-axis.

The Jahn-Teller Theorem 1937

74

Apparent exceptions to the theorem are probably examples of what has been called the "dynamic Jahn-Teller effect". In these cases either the time frame of the measurement does not allow the distortion to be seen because of the molecule randomly undergoing movement or else the distortion is so small as to be negligible.

For one of the copper complexes, the bond lengths are apparently identical. If the X-ray structure of the sample is redone at varying temperatures it is sometimes possible to "freeze" a molecule into a static position showing the distortions.

75

Alternant conjugated hydrocarbons

Definition: Atoms of conjugated molecules could be divided into two sets of atoms (starred and unstarred atoms) so that a member of one set is formally bonded only to members of the other set. Many compounds are alternant; they are not when their structure contains an odd-member ring.

76

Alternant conjugated hydrocarbons

-E cr + s cs = 0

if j = r* cr r + s° cs s is an eigenfunction

‘j = r* cr r - s° cs s is another one with Ej'=-Ej

•Orbital energy levels are symmetric relative to a level.

•If there is an odd number of orbital, one of them is nonbonding

77

Alternant conjugated hydrocarbonsNon bonding orbital, SOMO

The nonbonding orbital has non-zero coefficient only at the starred atoms, and the sum of the coefficient of the starred atoms attached to a given unstarred atom must equal zero.

The secular equation -E cr + becomes s cs = 0 with E=0.

Applied on a starred atoms s cs = 0 it gives again that cs = 0

Applied on a unstarred atoms r cr = 0 it tells that the unstarred atoms belong to nodal planes.It is easy to find the coefficients from there.

78

Allyle radical

The SOMO is localized on the terminal atoms

Allyle anionThe charge is -1/2 on each terminal atom

Allyle cationThe charge is +1/2 on each terminal atom

a = 1/√

H2C CH2

a = -1/√

species):

79

Dewar method for estimating aromaticity

a

-a

Two bonds close the cycleOne bond makes the polyene

S

A

If the SOMO is symmetric, it is better to make 2 bonds:

If the SOMO is antisymmetric, it is better to make only one bond:

4n+2 electrons aromatic 4n electrons antiaromatic

SS

S

Singlecarbon

S or A

S 2n+1A 2n-1

Polyeneradical

80

Dewar method for estimating aromaticity

-3a

-b

-2a

3a

-a

a

-a

a

0

b- b

b -b

b

- b

5a

pyrène

There is an energy gainby forming a new cycle

Two bonds close the cycleOne bond makes the polyene

81

The net charge of alternant hydrocarbons is zero.The electron density on r is 1

| coeff2 of occupied MOs| = | coeff2 of unoccupied MOs | wherefrom ii cri

2=1

demonstration :

occ cr,occ2 + nb cr,nb

2 + vac cr,vacc2 =1

occ cr,occ2 + nb cr,nb

2 = 1 ii cr,i

2 = 1

82

The bond indices between atoms of the same set is zero.(first order perturbation between atoms of the same set is zero).

OMs are orthogonal thus rcricrj= ijIf i=j, the sum of the coefficient’s squares for a given atom is 1 (normalization)

If i≠j, the sum of the products of coefficients is 0 (orthogonality)The matrix of coefficients is unitary

wherefrom = icricsj= ijIf i=j, the sum of the coefficient’s squares for a given orbital is 1 If i≠j, the sum of the products of coefficients is 0

occ cr,occcs,occ + nb cr,nbcs,nb

+ vac cr,vacccs,vacc

=rs occ cr,occcs,occ

+ nb cr,nbcs,nb = 0

occ cr,occcs,occ + nb cr,nbcs,nb

= 0 ricricsj=

830.3717

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

Butadiene is an alternant

84

j = 3, AS

j = ± 2, AA j = ± 2, SS

j ± 1, SA j = ± 1, AS

j = 0, SS

annulenesBenzene is

alternant

E = 2

E = -2

E = 1

E = -1

85

Importance of the non bonding orbital

• for radical species, it represents the unpaired electron

•For anions or cations it monitors the charged= ii cr,i2

For anions d= occ cr,occ2 + cr,nb

2

d= (occ cr,occ2 + cr,nb

2 ) + cr,nb2

d= 1 + cr,nb2

For cations d= occ cr,occ2

d= (occ cr,occ2 + cr,nb

2 ) - cr,nb2

d= 1 - cr,nb2

86

Heteroatoms:CO

parameter CoulombicIntegral

Resonance

Integral

Carbon C = CC =

Oxygen 1 electron

O = +

CO =

Oxygen 2 electrons

O = + 2

CO = 0,8

1 electron for CO

C O

C -x 1 = 0

O 1 -x+1

x2-x-1=0x= (1±√5)/2

0

1

1.618

-0.618

0.8506 -0.5257

0.5257 0.8506

C O

Bonding, large amplitude on O

Antibonding, large amplitude on C

O more electronegative

87

allyle1 2 3

1 -x 1 0

2 1 -x 1

3 0 1 -x

x3-2x=0 Solutions:x=- √2x=0x= √2

Sym 1 2

1 -x 1

2 2 -x

Sym 1/√2(1+3) 2

1/√2(1+3) -x √ 2

2 √ 2 -x

AntiSym 1

1 -x

88

Allyle coefficients

x=0

-xc1+c2=0 c2=0

c1-xc2+c3=0 c3=-c1 this orbital is found antisymm

c2-xc3=0 c2=0 no new informationx=√2

-√2 c1+c2=0 c2= √2 c1

c1-√2 c2+c3=0 c3 = c1 this orbital is found symm

c2-xc3=0 1/√2 c3= 1/√2 c1 = c2

These expressions have be normalized

89

-√2

0

√2

-1/√2 1/√2

1/2 1/√2 1/2

1/2 -1/√2 1/2

radical cation anion

q11-2(1/2)2-1(1/√2)2 = 0 1-2(1/2)2

= 1/2

1-2(1/2)2-2(1/√2)2

= -1/2

q21-2(1/√2)2 = 0 1-2(1/√2)2

= 01-2(1/√2)2 = 0

Charges are controlled by the filling of the nonbonding MO.They appear on the terminal atoms.The sum of charges is the global charge

90

-√2

0

√2

-1/√2 1/√2

1/2 1/√2 1/2

1/2 -1/√2 1/2

radical cation anion

l12=l232(1/2)(1/√2) = 0.707 = 0.707 = 0.707

bond indices are independent from the filling of the nonbonding MO.

91

Exercises

Find, using symmetry, the orbitals for

• the trimethylenemethane H2C=C(CH2)2

• the pentadiene*

• the cyclopentadiene*

*Explain the origin of the solutions x=±1 for the pentadiene*Explain the origin of the golden number solutions for the cyclopentadiene

92

Trimethylene-methane

Symmetry orbitals : MOs

0 1/√2 [0 + 1/√3 (1+ 2+3)]

1/√3 (1+ 2+3) 1/√2 [0 - 1/√3 (1+ 2+3)]

1/√2 (2-3) 1/√2 (2-3)

√(2/3) 1+ 1√6 ( 2+3) √(2/3) 1+ 1√6 ( 2+3)

Energies 0 energies ± √3

}

C1 C0

C2

C3

93

Trimethylene-methaneC1 C0

C2

C3

1/√2 [0 - 1/√3 (1+ 2+3)]

1/√2 [0 + 1/√3 (1+ 2+3)]

√(2/3) 1+ 1√6 ( 2+3) 1/√2 (2-3)

94

Cyclopentadienyl radical and ions

Sym 1 2 3

1 -x +2 0

2 1 -x 1

3 0 1 -x+1

-x2(1-x)+2(1-x)+x=0 x3-x2-x+2=0 x=2 solution (x-2) (x2-x-1) x= (1±√5)/2

The antisymmetric solutions are those from butadiene

2

3

1

5

4

95

- =-1.618

=1/ =0.618

2

Pentagon, icosahedra

96

-1.618

0.618

2

Pentagon, icosahedra

This level is bonding, preferably filled.

97

The anion is stable (aromatic)The cation is not (antiaromatic-

Jahn-Teller situation)

Thorium+4

98

Exercise: naphtalene

2) Solve the SS secular determinant 3) Deduce from the SS energy levels those for AS 4) What must be the levels for SA and AA solutions ?5) Draw the non-bonding level of the pentadienyl radical (C5 chain)?

1) Write the Huckel determinant for each symmetry group of the naphatlene.

6) Explain from there what are the levels SS and AS at x=±1

99

naphtaleneSS 1 2 3

1 -x+1 +1 0

2 1 -x 1

3 0 2 -x+1

SS -x (1-x) 2-3(1-x)=0 (1-x)(-x2+x+3)=0 x=1 solution (-x2+x+3)=0 x= (1±√13)/2x = 2.30277 and -1.30277AS solutions have opposite energies (alternant) -1, x=-2.30277 and +1.30277The SA and AA solutions are the 4 MOs from butadiene ± 0.618 and ± 1.618The levels at ±1 are duplication of the non bonding MO from the pentadienyle in-phase and out of-phase.

100

Exercises• Calculate the energy for C3 and C5

– Linear or cyclic– Positively charge and negatively charged.

• Calculate the MO energies for the biphenyle.Repeat the calculation assuming that the central bond is only half of a C=C bond.