Methodology:
1) Choose a geometry for the molecule (Born Oppenheimer approx.)
2) Choose a Basis Set of N atomic orbitals   n
3) Obtain values of the Hamiltonian integral
H
and overlap integrals mn
S mn
4) Set up the N x N Secular Determinant det H mn
  mo
S mn

0
and solve for N values of
 mo
, i.e.
 a mo
5) For each value of
 mo
LCAO-MO a
 
solve the Secular Equations for
 
 a
  n n a
 n a c which gives the n mo c
This effectively solves the molecular orbital Schrödinger equation for N mo.s of the molecule.
Example:
Li H
Geometry R = 3.0 a
0
Basis Set: Li [1s] 2s

To calculate a c a
, c b
in
and H 1s

  mo b c a
 a
 c b
 b
and
 mo
Li 2s :
 a

1



2


2
5 / 3

1 /

2 r exp
  
2 r
  
1


1
5 /

1 /

2 exp
  
1 r

1 .
10 ,

2

0 .
18 ,

1
 from Hartee-Fock calculations on the Li atom
2 .
69 ,

2

0 .
64 and
H 1s :
 b
  
1 / 2 exp
  from these (using molecular Hartree-Fock Hamiltonian) calculate:
H
S ab aa


0

0 .
102 ,
.
463 (
H
S bb aa



0 .
267 ,
S bb

1 )
H ab
 
0 .
223 can now substitute in the Secular Equations:

 m c m

H mn n c a


H a aa n c a


H b ab
  mo
S mn
  mo
S aa
  mo
S ab
 
0
  c b

H ba
  c b

H bb for all
  mo
S ba n
 
  mo
S bb
 
0
0
(NOTE: 2 equations, 3 unknowns!!) c c a a

 

0
0
.
102
.
102


   c b
0 .
463





0 .
c b
223



0 .
463

0 .
267
  
 

0
0 which give the Secular Determinant:
 
 
0 .
102
0 .
223

   
0 .
463


 
0 .
223


0 .
267
0 .
463

  


0 which multiplies out to give:
0 .
7856

2

0 .
1625

  
0 .
302 or

0 .
00225
0 .
095

0
Substitute the -0.302 value into the Secular Equations. Both equations show: c a

0 .
416 c b need another equation.
Normalization:
Then:
 mo
 c a c a
2

 a c b 
2
 c b
2 c a  b c b 
S ab

1
0 .
333
 a the bonding orbital.
Summarize results as a molecular orbital diagram:

0 .
801
 b

0.0
E/au
-0.1
H aa
ψ = 1.078φ a
- 0.792φ b
H bb
-0.3 ψ = 0.333φ a
+ 0.801φ b
The use of values for Hartree-Fock for
H aa
, H bb
, H ab
, S ab is particular to this example.
The problem of how to ‘Obtain values of the Hamiltonian integral
H
and overlap mn integrals
S
’ is considered next. mn
Approximations for:
H
S mn mn





 m m
(
( i ) i )

H n
( i mo
)
( i d

)
 n
( i ) d

when m = n:
H aa


 a
( i ) H mo
( i )
 a
( i ) d
 and is an energy ~ the energy of the a.o
 in the molecule ~ the energy of
 a a
in the atom.
Hamiltonia
Overlap n integrals integrals

Empirical approximation to
H
is the ‘binding energy’ of an electron in aa

= -Valence State Ionization Energy (VSIE) of
 a
The average I.P. for states of the atom produced when
 a
is ionized. a
H (1s
1
)

H
+
I.P. = 13.6 eV (=VSIE)
H
(H 1s) = -13.6 eV aa
Boron is a more complicated case
B (1s
2
2s
2
2p
1
) (
2
P)

B
+
(1s
2
2s
2
) I.P. = 8.5 eV but for the 2s two states of the atom arise:
B (1s
2
2s
2
2p
1
) (
2
P)

B
+
(1s
2
2s
1
2p
1
)
(
1
P) I.P. = 15.8 eV , (
3
P) I.P. = 15.0 eV
VSIE = {(15.8 x 3) + (15.0 x 9)}/12 = 15.2 eV
H
(B 2s) = -15.2 eV aa
H
(B 2p) = -8.5 eV aa
Atom H aa
(s)
N
O
F
Na
Mg
Al
Si
P
S
Cl
H
Li
Be
B
C
-13.6
-5.4
-10.0
-15.2
-21.4
-26.0
-32.3
-40.0
-5.1
-9.0
-12.3
-17.3
-18.6
-20.0
-30.0
H
(p) aa
-3.5
-6.0
-8.5
-11.4
-13.4
-14.8
-18.1
-3.0
-4.5
-6.5
-9.2
-14.0
-13.3
-15.0
H
(d) aa
-6.0
-7.0
-8.0
-9.0

(s,p)
1.300
0.650
0.975
1.300
1.625
1.925
2.275
2/425
0.733
0.950
1.167
1.383
1.600
1.817
2.033

(d)
1.383
1.400
1.500
2.033