1 + s 12 1 e 1 = 1 = c 1 1 + c 1 2 1 - s 12 1 e 2 = - 2 = c 1 1 - c 1 2 bonding antibonding
DESCRIPTION
Ethylene system of ethyleneTRANSCRIPT
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