1 + S12

1E1 =

1 = c11 + c12

1 - S12

1E2 = -

2 = c11 - c12

bonding

antibonding

Let overlap term go to zeroOverlap term is non zero

C CH

H H

HEthylene

system of ethylene

H2 C2H4

antibondingempty orbital

bondingfille with 2 e-

Now let us look at a more complicated system.

Three Orbitals!

H

H

H H

H

H

H

HH

H

Allyl anion

1 23 How do we calculate

the energies and coefficientsof the MO’s?

We are going to use the LCAOapproximation to make threeMO’s.

Use Symmetry!

The ends are the same so they must contibuteequally to any MO. So we can make linear combinationswhere atoms one and three contribute equally.1 = pz1 + pz3 2 = pz1 - pz3

3 = pz2 The second carbon is unique.

pza Ĥ pza = E = = 0Coulomb integral

pza Ĥ pzb = Eint = Interaction integral

Only if the two p orbitals are adjacentotherwise Interaction is 0.

pza pza = 1normalized

pza pzb = 0assume overlap = 0

Huckel Approximation

1 23 How do we calculate

the energies and coefficientsof the MO’s?

We are going to use the LCAOapproximation to make threeMO’s.

Use Symmetry!

The ends are the same so they must contibuteequally to any MO. So we can make linear combinationswhere atoms one and three contribute equally.1 = pz1 + pz3 2 = pz1 - pz3

3 = pz2 The second carbon is unique.

1 23

1 = pz1 + pz3

We need to normalizeour LCAOs 1

2 = 1

N2(pz1 + pz3 ) (pz1 + pz3 ) = 1

N2 (pz12

+ 2pz3 pz1 + pz32

) = 1 1 + 2 x 0 + 1

N2 x 2 = 1 So N = 12

1 23 How do we calculate

the energies and coefficientsof the MO’s?

We are going to use the LCAOapproximation to make threeMO’s.

Use Symmetry!

The ends are the same so they must contibuteequally to any MO. So we can make linear combinationswhere atoms one and three contribute equally.

1 = (pz1 + pz3 ) 2 = ( pz1 - pz3 )

3 = pz2 The second carbon is unique.

12

12

1 = (pz1 + pz3 )

2 = ( pz1 - pz3 )

3 = pz2

12

12

1 23

Same Symmetry

1 23

1 23

1 32

1 2 3

1 = (pz1 + pz3 ) 3 = pz2

12

1 3

1 23

1 23-

1 3

1 23

1 23+ Bonding MO

Antibonding MO

1 = (pz1 + pz3 )

3 = pz2

12

(pz1 + pz3 ) Ĥ pz2 = 12 21

2 = 2

1 23

3

+

Energy =

2

1 23

1

12

1 = (pz1 + pz3 )

-3 = -pz2

12

(pz1 + pz3 ) Ĥ (- pz2) = 12 - 21

2 = - 2

Energy =

2 -

1 23

3

-1 2

3

1

12

1 23

3

+

Energy =

1 23

1

12

1 23

2

Bonding

1 23

3

-

Energy =

1 23

1

12

2

2 -1 3

Antibonding

1 23

What about 2 ?

Thereis no overlapbetween ends so

E = 0

1 3

1 23

1 23

2 -

2

0

H

H

HH

H

H

H

H H

H

This is painful!

It makes the brain hurt.

So use a computer instead.

Simple Huckel MolecularOrbital Theory Calculator