Physical Chemistry (II)

Lecture 14

CHEM 3172-80

Lecturer: Hanning Chen, Ph.D.03/20/2017

Semi-empirical Hückel Theory

the Major Deficiency of Molecular Orbital Theory

H2O molecule:

O

H2H1

atomic orbitalscO

cH1 cH2

ϕ = cO fO + cH1 fH1 + cH2 fH2molecular orbital:

fO , fH1 , fH2

HΨ = EΨ cO ,cH1 ,cH2{ }

Construction of Molecular Orbitals:

ϕn = cnij fijj=1

N

∑i=1

M

∑ atomic orbitals

A molecular orbital is a linear combination of atomic orbitals LCAO

M : the number of atomsN : the number of orbitals in each atom

hundreds or even thousands of primitive functions

construction and diagonalization of a million by million matrix

Complexity of Some Popular MO Thoeries

Correlation Theory Complexity

Configuration Interaction Single (CIS) N5

Configuration Interaction Single-Double (CISD) N6

Coupled Cluster Single-Double (CCSD) N7

Coupled Cluster Single-Double-Triple-Quadruple (CCSDTQ) N9

Second-order Moller-Plesset Perturbation (MP2) N5

Fourth-order Moller-Plesset Perturbation (MP4) N6

Hartree-Fock (HF) N4

Is there any way to avoid the expensive computational cost without losing the most important chemical and physical information ?

N : number of atoms

BenzeneBenzene: C6H6

C

C

C

C

C

CH

H

H

H

H

H

C : 1s2 2s2 2p2 H : 1s1

C

H

C C

3 σ bonds1 unpaired electron

6 π electrons

in total6 × 4 + 6 ×1= 30 valence electrons

24 σ electrons

σ πσ electrons: inner sphere

π electrons: outer sphere

chemically inert

chemically active

Is it possible to investigate the

chemically active electrons only?

12 σ bonds3 π bonds

Electron ApproximationAn atom is decomposed into a core part and a valence part:

π

Core: nucleus + electronsσValence: electronsπ

valence electron Hamiltonian:

Hπ = H jcore i( )

j=1

ncore

∑ + 1rijj=1

j<i

∑i=1

nπ

∑i=1

nπ

∑ valence-valence electronic repulsion

valence-core Hamiltonian:H j

core(i) : interaction between the i th valence electron and the j th core

ijelectron-electron repulsion ( - )π σelectron-nucleus attraction ( - N)π

Hückel Molecular Oribital Method

Hπ = H jcore i( )

j=1

ncore

∑ + 1rijj=1

j<i

∑i=1

nπ

∑i=1

nπ

∑valence electron Hamiltonian:

= 0The first assumption:

no valence-valence

repulsion !

Hπ = H jcore i( )

j=1

ncore

∑i=1

nπ

∑The second assumption:

Ψ = ϕii=1

nπ

∏ =ϕ1ϕ2...ϕnπ

Ψ : many-electron wavefunctionϕi : molecular orbitals

The anti-symmetric requirement of the many-electron wavefunction is NO LONGER satisfied !

no electron exchange interaction no electron correlation

P r1,r2( ) = P r1( )P r2( )first electron at r1second electron at r2

simultaneously product of individual probabilities

only valence-electron interactions

Linear Combination of Atomic Orbitals

H jcore i( )

j=1

ncore

∑ ϕi = Eiϕi

Solution for Hückel Hamiltonian:

C

C

C

C

C

CH

H

H

H

H

H

ϕi = cij f jj=1

nπ

∑ f j : atomic orbitalsf1

f2

f3f4

f5

f61

2

345

6

Benzene: C6H6

ϕi = ci1 f1 + ci2 f2 + ci3 f3+ci4 f4 + ci5 f5 + ci6 f6

following the procedure in Hartree-Fock theory:

HC = ESCH : Hamiltonian matrixS : Overlap matrixC : Coefficient vectorE : molecular orbital energy

the molecular orbitals, entirely dependent on

ϕi

valence− core interaction

Matrix Representation of Hückel HamiltonianHamiltonian matrix:

Hij = drfi*(r)Hπ

core∫ f j r( ) =α for i = j

β for bonded atom i and j0 for unbonded atom i and j

⎧

⎨⎪

⎩⎪

energy of non-interacting orbitalscoupling between bonded atomsno coupling between non-bonded

atomsThe third assumption:

The heterogeneity of atom types and chemical environments are ignored !

Overlap matrix:Sij = drfi

*(r)∫ f j r( ) = 1 for i = j0 for i ≠ j

⎧⎨⎪

⎩⎪The Fourth assumption:

The atomic orbitals are orthonormal !

Simplified Schrödinger equation: HC = EC

identity matrix

core-mediated electron interactions

Butadiene as an Example

Butadiene:CH2 CH CH CH2

1 2 3 4

HC = EC

α β 0 0β α β 00 β α β0 0 β α

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= E

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

Secular equation:

det

α − E β 0 0β α − E β 00 β α − E β0 0 β α − E

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 0

continuant determinant

α : energy of an isolated valence electron

β : coupling energy between chemically bonded valence electrons mediated by cores

4 π electrons

α − E β 0 0β α − E β 00 β α − E β0 0 β α − E

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 0nontrivial

solutions

General Solution of Continuant Determinantcontinuant determinant:

a b 0 0c a b 00 c a b0 0 c a

0

0

a b 0 0c a b 00 c a b0 0 c a

= a − 2 bc cos jπn +1

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥j=1

n

∏

det

α − E β 0 0β α − E β 00 β α − E β0 0 β α − E

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= α − E( ) + 2β cos( jπ4 +1

)⎡⎣⎢

⎤⎦⎥j=1

4

∏ = 0

α − E + 2β cos jπ5

⎛⎝⎜

⎞⎠⎟ = 0, j = 1,2,3,4

E1 =α +1.618βE2 =α + 0.618βE3 =α − 0.618βE4 =α −1.618β

⎧

⎨

⎪⎪

⎩

⎪⎪

decomposition

only the diagonal terms and their adjacent elements are nonzero !

β < 0

Energy Levels of ButadieneE

nerg

y

α +1.618β

α + 0.618β

α − 0.618β

α −1.618β

E1

E2

E3

E4

ΔE1→2 = −β

ΔE3→4 = −β

ΔE2→3 = −1.236β

bonding orbitals

anti-bonding orbitals

Etotal = 2 × α +1.618β( ) + 2 × α + 0.618β( ) = 4α + 4.472β

Delocalization EnergyButadiene:

CH2 CH CH CH2

artificial wall to block electron delocalizationtruncated block of ethene

secular equation

detα − E ββ α − E

⎛

⎝⎜

⎞

⎠⎟ = 0 α − E( )2 − β 2 = 0

E1 =α + β

E2 =α − β

Eethene = 2α + 2β

delocalization energy of butadiene

ΔEdelocalization = Ebutadiene − 2Eethene = 4α + 4.472β − 2 2α + 2β( ) = 0.472βButadiene is stabilized by electron delocalization !

1 2 3 4ϕi = c1 f1 + c2 f2 ϕi = c3 f3 + c4 f4

β < 0

Solutions of Molecular Orbitals

α β 0 0β α β 00 β α β0 0 β α

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= E

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

E1 =α +1.618βE2 =α + 0.618βE3 =α − 0.618βE4 =α −1.618β

⎧

⎨

⎪⎪

⎩

⎪⎪

c12 + c2

2 + c32 + c4

2 = 1normalization condition:

orthonormal atomic orbitals

ϕ1 = 0.372 f1 + 0.602 f2 + 0.602 f3 + 0.372 f4ϕ2 = 0.602 f1 + 0.372 f2 + 0.372 f3 + 0.602 f4ϕ3 = 0.602 f1 − 0.372 f2 − 0.372 f3 + 0.602 f4ϕ4 = 0.372 f1 − 0.602 f2 + 0.602 f3 − 0.372 f4

⎧

⎨

⎪⎪

⎩

⎪⎪

two central atoms

two terminal atoms

two central atoms

Visualization of Molecular Orbitals

ϕ1

ϕ2

ϕ3

ϕ4

α +1.618β

α + 0.618β

α − 0.618β

α −1.618β# of nodal planes

1

2

3

4

Higher-energy orbitals have more nodal planes !horizontal molecular axis

CyclobutadieneCyclobutadiene:

C

C

C

C

H

H

H

H

12 3

4

Hückel Hamiltonian matrix:

α β 0 ββ α β 00 β α ββ 0 β α

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= E

c1c2c3c4

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

det

α − E β 0 ββ α − E β 00 β α − E ββ 0 β α − E

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 0

Secular equation: Circulant determinant:

a1 a2 a3 ... anan a1 a2 ... an−1an−1 an a1 ... an−2... ... ... ... ...a2 a3 a4 ... a1

shift of array elements by one position between two

adjacent rows

Expansion of Circulant Determinant

a1 a2 a3 ... anan a1 a2 ... an−1an−1 an a1 ... an−2... ... ... ... ...a2 a3 a4 ... a1

= a1 +ω ka2 +ω k2a3 + ...+ω k

n−1an( )k=1

n

∏ω k = e

2πkni

= cos 2π kn

⎛⎝⎜

⎞⎠⎟ + isin

2π kn

⎛⎝⎜

⎞⎠⎟

k = 1,2,...,n

phase factor:

det

α − E β 0 ββ α − E β 00 β α − E ββ 0 β α − E

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= α − E( ) + e2πk4iβ + 0 + e

(4−1)2πk4iβ

⎛⎝⎜

⎞⎠⎟k=1

4

∏α − E + 2β cos π k

2⎛⎝⎜

⎞⎠⎟ = 0

k = 1,2,3,4

E1 =α + 2βEnergy levels:

E2 = E3 =α E4 =α − 2β

Energy Levels of CyclobutadieneE

nerg

y

α + 2β

α

α − 2β Cyclobutadiene is very unstable !

Ecyclobutadine = 2α + 2 α + 2β( ) = 4α + 4β

Eethene = 2α + 2β

ΔEdelocalization = Ecyclo − 2Eethene = 0

Benzene

C

C

C

C

C

C

H

H

H

H

f1

f2

f3f4

f5

f61

2

345

6

Benzene: C6H6 Hückel Hamiltonian matrix:

α β 0 0 0 ββ α β 0 0 00 β α β 0 00 0 β α β 00 0 0 β α ββ 0 0 0 β α

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

c1c2c3c4c5c6

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

= E

c1c2c3c4c5c6

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

6X6 circulant determinant:

α − E( ) + e2πk6iβ + 0 + e

(5−1)2πk6iβ

⎛⎝⎜

⎞⎠⎟j=1

6

∏ = 0

E1 =α + 2βE2 = E3 =α + βE4 = E5 =α − βE6 =α − 2β

⎧

⎨

⎪⎪

⎩

⎪⎪

H

H

degeneratedegenerate

Energy Levels of Benzene

E1 =α + 2β

E2 = E3 =α + β

E4 = E5 =α − β

E6 =α − 2β

ΔEdelocalization = 2 α + 2β( ) + 4 α + β( )− 3(2α + 2β ) = 2β

Benzene is very stable !

three fragments of ethene

Hückel [4n+2] RuleFor a cyclic ring molecule with 4n+2 electrons,π

it is stabilized by electron delocalization

For a cyclic ring molecule with 4n electrons,π

aromaticity

non-aromaticityanti-aromaticityit is NOT stabilized by electron delocalization

Homework 14

Reading assignment: Chapters 10.6

Homework assignment: Problems 10.12

Homework assignments must be turned in by 5:00 PM, March 21st, Tuesday

to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall