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Hückel Molecular Orbital (HMO) Theory

A simple quantum mechanical concept that gives important insight

into the properties of large molecules

Why HMO theory

• The first MO theory that could be applied to large molecules.

• A theory that can be implemented without the aid of a computer.

• Has been the starting point for very successful theoretical interpretation of electronic (UV-vis) spectra of conjugated and aromatic hydrocarbons.

• Has been the starting point for more advanced theories.

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Important conjugated and aromatic compounds

BenzeneCarotene

Chlorophyl

Photosynthetic apparatus

N

N N

N

Mg

H3C

CH3

CH3H3C

CH3

OO

O

CH3

H3C H3C

3

CO2CH3

Molecular electronics

Image of graphene taken with TEAM 0.5NANO LETTERS, 2008, 8, 3582-3586

OCH3

H3CO

S

S

CH3

CH3

FFF

F

FFOCH3

H3CO

Molecular memory

Graphene

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The basis of the Hückel approach 1

• The geometries and special properties of conjugated and aromatic compounds may be interpreted in terms of the sp2-hybridization.

• The unhybridized 2p AOs perpendicular to the sp2 hybrideorbitals are available for the formation of the π-electron structure.

• The σ-electrons are disregarded, except for their role in establishing the molecular geometry.

Ethene σ-skeleton and unhybredized 2p-orbitals

2p orbital

σ-skeleton

S. M. Lindsay Introduction to Nanoscience pp 279-284B.H. Bransden and C.J. Joachain Physics of Atoms and Molecules 2nd Edn. pp 533-537

The basis of the Hückel approach 2

• There is no bonding interaction between σ- and π-orbitals in planar conjugated molecules.• The σ-bonds between the sp2-hybridized atoms and their neighbours

(the σ-orbitals) lie in the molecular plane. The σ-orbitals are symmetrical with respect to reflection in the plane of the molecule.

• The unhybridized 2p AOs are perpendicular to the molecular skeleton made up by the sp2 hybride atoms. Any MO (π-orbital) formed from them will therefore be antisymmetric with respect to reflection in the plane of the molecule.

• The two types of electrons move in different regions of space: the σ-electrons occupy the space in the plane of the conjugated molecule, the π-electrons are located above and below the that plane.

• (The energies of the σ-electrons are on the whole lower than those of the π-electrons. The σ-electrons are involved in strong bonds and the π-electrons in weaker ones.)

Benzene

C C C CCCH HH

HH

Hσ-skeleton

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= ==1

Hückel theory

Where rI is the number of electrons in molecular orbital I.

CCC CC CC CC C C C

x y1 2

i j

a b n

2pz (φ) orbital

The total π-electron energy Eπ is:

Hückel theory

CCC CC CC CC C C C

x y1 2

i j

a b n

2pz (φ) orbital

= 1 1 + 2 2 + 3 3 + ⋯ + =

The problem: find the set of π-electron MOs (ΨI) formed as linear combinations of the 2pz(φ) orbitals at the atoms in the conjugated system.

The variation theorem gives that the best wave function is the one that gives the lowest energy EI.

S. M. Lindsay Introduction to Nanoscience pp 279-284B.H. Bransden and C.J. Joachain Physics of Atoms and Molecules 2nd Edn. pp 130-135

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Expectation value of the MO energy εIThe expectation value of the energy εI is:

= ℋ| = ∑ ℋ ∑∑ ∑ = =1 ℋ =1 =1 =1

= ∗=1 ℋ=1 ∗=1 =1 ≡ ∗=1 =1 ∗=1 =1

Rearrange and drop the index I since the equation applies to any ofthe π molecular orbitals

= ∗=1 =1 ∗=1 =1

Where = ℋ is the resonans or interaction integral and = is the overlap integral.

Variational approach

= 0

∗ = ∗

Rearrange and differentiate

∑ ∗ ∑ = ∑ ∗ ∑

To find the minimum energy, differentiate both sides of the expressionfor εI with respect to the coefficients ca, where a runs from 1 to n.

∗=1 =1 + ∑ ∗ ∑ = ∑ ∗ ∑

S. M. Lindsay Introduction to Nanoscience pp 63 and 268-270

= 0, ∗ = 0 , thus

The derivative will be non-zero only for those terms where j = a.

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∗ = ∗

∗ ∗ = ∗

=

( − )=1 = 0

Take the complex conjugate of each side.

But = ∗, = ∗ = ∗

Rearrange.

There are n simultaneous equations, one for each coefficient – a setof homogeneous linear equations – the secular equations.

Secular equationsThere are n simultaneous equations, one for each coefficient – a setof homogeneous linear equations – the secular equations.

( − ) ′ = 0

1( 11 − 11)1( 21 − 21)⋮1( 1 − 1)+++

2( 12 − 12)2( 22 − 22)⋮2( 2 − 2)+++

⋯⋯⋱⋯+++

( 1 − 1 )( 2 − 2 )⋮( − )= 0= 0= 0

Rewritten in matrix notation( 11 − 11) ( 12 − 12)( 21 − 21) ( 22 − 22) ⋯ ( 1 − 1 )⋯ ( 2 − 2 )⋮ ⋮( 1 − 1) ( 2 − 2) ⋱ ⋮⋯ ( − )12⋮ = 0

or

where C’ is a column vector.

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Hückel’s assumptions

• The overlap integral Sij is a measure of the extent of overlap between the 2pz AOs on carbon atoms i and j.• i = j: Sji = 1, because the overlap of an AO with itself is unity for

a normalized wave function.• i ≠ j: Sij = 0. A drastic assumption*

• The resonance integral Hij represents the energy of a single electron moving in the average field of all the nuclei, all the σ-electrons and all other π-electrons.• i = j: Hii = α. The integral represents the energy of an electron in

a “free” carbon 2pz AOs in a molecule.• Carbon i adjacent to j = i ± 1: Hi(i ± 1) = β.**• Hij negligable (= 0) if the two carbons i and j are separated by

two or more bonds.*** The overlap integral for two 2pz orbitals, one each on two adjacent carbon atoms in a conjugated compound, is approximately 0.25. It falls of rapidly with increasing distance.** The molecular structure (connectivity) has to be known on beforehand.

Apply Hückel’s approximationsThe secular equations.

( − ) ( − ) ⋯ /0⋯ /0⋮ ⋮/0 /0 ⋱ ⋮⋯ ( − )12⋮ = 0

( − ) ( − )( − ) ( − ) ⋯ ( /0 − 0)⋯ ( /0 − 0)⋮ ⋮( /0 − 0) ( /0 − 0) ⋱ ⋮⋯ ( − 1)12⋮ = 0

Simplifies to:

( 11 − 11) ( 12 − 12)( 21 − 21) ( 22 − 22) ⋯ ( 1 − 1 )⋯ ( 2 − 2 )⋮ ⋮( 1 − 1) ( 2 − 2) ⋱ ⋮⋯ ( − )12⋮ = 0

Equals:

Divide by and define = − ( − ).

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Matrix algebra− 11 − ⋯ 1/0⋯ 1/0⋮ ⋮1/0 1/0 ⋱ ⋮⋯ −

12⋮ = 0

0 11 0 ⋯ 1/0⋯ 1/0⋮ ⋮1/0 1/0 ⋱ ⋮⋯ 0 − 00 ⋯ 0⋯ 0⋮ ⋮0 0 ⋱ ⋮⋯12⋮ = 0

0 11 0 ⋯ 1/0⋯ 1/0⋮ ⋮1/0 1/0 ⋱ ⋮⋯ 0 − 1 00 1 ⋯ 0⋯ 0⋮ ⋮0 0 ⋱ ⋮⋯ 112⋮ = 0

Extrapolate to solve for all orbital coefficients cIj and orbital energies εI [all x(n)] for all n π-orbitals simultaneously.

Topology matrix0 11 0 ⋯ 1/0⋯ 1/0⋮ ⋮1/0 1/0 ⋱ ⋮⋯ 0 − 1 00 1 ⋯ 0⋯ 0⋮ ⋮0 0 ⋱ ⋮⋯ 112⋮ = 0

Extrapolate to solve for all orbital coefficients cIj and orbital energies εI [all x(n)] for all n π-orbitals simultaniously.0 11 0 ⋯ 1/0⋯ 1/0⋮ ⋮1/0 1/0 ⋱ ⋮⋯ 0 − 1 00 1 ⋯ 0⋯ 0⋮ ⋮0 0 ⋱ ⋮⋯ 1

(1)0⋮00(2)⋮0

⋯⋯⋱⋯00⋮( )

(1)1(1)2⋮(1)(2)1(2)2⋮(2)

⋯⋯⋱⋯( )1( )2⋮( ) = 0

Topology matrixT

I Eigenvalue matrix

X

Eigenvector matrix

C

( − ) = ⇒ = − =

orSolved according to

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Butadiene as an example

CH

CC

HC

H H

HH

1 2

3 4

The topology matrix:

C no. 1 2 3 4

1 S11 = 1H11 = α

S12 = 0H12 = β

S13 = 0H13 = 0

S14 = 0H14 = 0

2 S21 = 0H21 = β

S22 = 1H22 = α

S23 = 0H23 = β

S24 = 0H24 = 0

3 S31 = 0H31 = 0

S32 = 0H32 = β

S33 = 1H33 = α

S34 = 0H34 = β

4 S41 = 0H41 = 0

S42 = 0H42 = 0

S43 = 0H43 = β

S44 = 1H44 = α

= 0 11 0 0 01 00 10 0 0 11 0

The π-orbital energies

ε3 = α – 0.61803β

ε2 = α + 0.61803β

ε1 = α + 1.61803β

ε4 = α – 1.61803β

Bonding orbitals

Antibonding orbitals

Next, determine the shape of the π-orbitals

= − (α − ) = ± 0.61803 and ± 1.61803

= ± 0.61803 and ± 1.61803

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Orbital coefficientsI

Coefficient CIi for carbon number

1 2 3 4 Energy

ΨI

4 0.3718 –0.6015 0.6015 –0.3718 α – 1.61803β3 0.6015 –0.3718 –0.3718 0.6015 α – 0.61803β2 0.6015 0.3718 –0.3718 –0.6015 α + 0.61803β1 0.3718 0.6015 0.6015 0.3718 α + 1.61803β

• β is a negative quantity representing the (π) bonding energybetween two adjacent carbon atoms.

• The order of the energy levels follows the number of nodes.• The term α is common to all energies. It is the energy of anon-bonded 2pz electron IN the molecule.

• MOs of energy of lower energy than α are bonding and those ofenergy greater than α are antibonding .

Orbital shapesE

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Add in the electrons

E

• Each carbon contributes with one electron. The stabilization due tothe π-electrons is: 4α – 2(α + 1.61803β) – 2(α + 0.61803β)= –4.47212β

Total energy is equal to the sum of orbital energies times the occupation

Add in electrons according to:1. Aufbau principle2. Pauli principle3. Hund’s rule

= ==1

S. M. Lindsay Introduction to Nanoscience pp 35-36 and 58.

General solutionsLinear polyenes

Cyclic polyenes

CCC CC CC CC C C C

x y1 2

i j

a b n= + 2 ( + 1) , = 2+ 1 1/2 ( + 1)

= + 2 2( − 1) , = 1 1/2 − 2( −1)( −1)

Nb (i)2 = −1I = 1, 2, 3,… n (MO number)

j = atom number in linear π-system.I = 1, 2, 3,… n (MO number).n = total number of atoms in π-system

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= 0, ±1, ±2, … , 2 n even: = 0, ±1, ±2, … , ( − 1)2 n odd:

General solution – Cyclic polyenes

= + 2 2 , = 1 1/2 − 2( −1)

Nb (i)2 = −1

αβ

α − β

α − 2β

α + 2β

α + β

0 ππ/3 2π/3− π − 2π/3 −π/3

π 4π/3 5π/3 Phaseangle

Energy

Orbital shapesE

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Cyclic polyenes

Benzene

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AromaticityFor a molecule to experience the stabilization of aromaticity it must fulfill the following requirements:• It must contain a cyclic sequence of atoms.• All the atoms of the cyclic sequence must lie in

the same plane, or nearly so.• Each atom in the cyclic sequence must

“contribute with a p-orbital” aligned perpendicular to the plane of the cyclic structure.

• The cyclic, conjugated system must contain (4n + 2) electrons.

Parameterized method

• The parameter β is composite parameter representing the electron-nucleus attraction, the interelectronic repulsion and the kinetic energy.

The atomic 2p energy level forcarbon:E(2pZ) = α (~ -11.4 eV)

The resonance integral for adjacent2p orbitalsHij(C=C) = β (~ -3 eV)