Lecture 27
Molecular orbital theory III
Applications of MO theory



Previously, we learned the bonding in H2+.
We also learned how to obtain the energies
and expansion coefficients of LCAO MO’s,
which amounts to solving a matrix eigenvalue
equation.
We will apply these to homonuclear
diatomic molecules, heteronuclear
diatomic molecules, and conjugated πelectron molecules.
MO theory for H2+ (review)
j±k
E± = E1s +
4pe 0 R 1± S
e2
j - k φ– = N–(A–B)
E1s +
4pe 0 R 1- S anti-bonding
e2
E1s +
j + k φ+ = N+(A+B)
4pe 0 R 1+ S bonding
e2
-
E1s
φ– is more anti-bonding
than φ+ is bonding
R
MO theory for H2+ and H2

MO diagram for H2+ and H2 (analogous to
aufbau principle for atomic configurations)
H2+
H2
Reflecting: anti-bonding orbital is more anti-bonding than
bonding orbital is bonding
Matrix eigenvalue eqn. (review)
Y = cA A+ cB B
L = E-l
{
¶ E- l
(
)
*
Y
ò Y dt -1
¶L ¶L ¶L
=
=
=0
¶cA ¶cB ¶l
( ò Y Y dt -1)} = ¶{ E - l ( ò Y Y dt -1)} = 0
*
*
¶cA
¶cB
æa
b
A
æ
æ
æ b aB
ææ c æ
æ c æ
æ
æ
ææ A æ = E æ 1 S ææ A æ
æ
æ S 1 ææ
ææ
æ cB æ
æ
æ cB æ
æ
(a A - E)(a B - E) - (b - ES)(b - ES) = 0
MO theory for H2
 A  B  
(a - E)(a - E) - ( b - ES)( b - ES) = 0
(a - E) = ±( b - ES)
a + b = (1+ S)E or a - b = (1- S)E
a +b
a -b
E=
1+ S
or E =
1- S
α is the 1s orbital energy.
β is negative.
anti-bonding orbital is more anti-bonding
than bonding orbital is bonding.
MO theory for H2
æ c æ
æ a b ææ c æ
æ
æ
A
1
S
A
æ = Eæ
æ
æ
æ
ææ
æ
æ
æ
æ
æ
æ
b
a
c
S
1
c
æ
æ
æ
ææ
B
B
æ
æ
æ
æ
E=
æ c æ æ
A
cA = cB ® æ
æ= æ
æ
æ cB æ
æ æ
æ
a +b
1+ S
1
2
1
2
æ
æ
æ
æ
or E =
a -b
1- S
æ c æ æ 1
2
A
cA = -cB ® æ
æ= æ
1
æ
æ
æ
c
æ B æ æ
2
æ
æ
æ
æ
MO theory for He2 and He2+

He2 has no covalent bond (but has an
extremely weak dispersion or van der Waals
attractive interaction). He2+ is expected to be
bound.
He2
He2+
σ and π bonds

A π bond is weaker than σ bond because of
a less orbital overlap in π.
σ bond
π bond
MO theory for Ne2, F2 and O2
Hund’s rule
O2 is magnetic
Ne2
F2
O2
MO theory for N2, C2, and B2
Hund’s rule
B2 is magnetic
N2
C2
B2
2
Z
Eµ- 2
n
Polar bond in HF


The bond in hydrogen
fluoride is covalent but
also ionic (Hδ+Fδ–).
H 1s and F 2p form the
bond, but they have
uneven weights in LCAO
MO’s .
Y = CH1sy H1s + CF2 py F2 p
2
2
2
2
2
Y = CH1s y H1s + CF2 p y F2 p +…
Hδ+Fδ–
Polar bond in HF

Calculate the LCAO
MO’s and energies of
the σ orbitals in the
HF molecule, taking
β = –1.0 eV and the
following ionization
energies (α’s): H1s
13.6 eV, F2p 18.6 eV.
Assume S = 0.
Matrix eigenvalue eqn. (review)

With S = 0,
æa
b
A
æ
æ
æ b aB
ææ c æ
æ c æ
A
A
ææ
æ = Eæ
æ
æ
æ
ææ
æ cB æ
æ
æ cB æ
æ
( E - a )( E - a ) - b
A
a
(
E=
(a
=
+aB) ±
A
B
(a
2
=0
2
a
+
4
b
)
A
B
2
2
A
+ a B ) ± (a A - a B )
2
2
æ 2b æ
1+ æ
æa A - a B æ
æ
Polar bond in HF

Ionization energies give us the depth of
AO’s, which correspond to −αH1s and −αF2p.
(
E=
) (
a H1s + a F2 p ±
)
2
a H1s - a F2 p + 4b 2
2
= -18.8 eV and -13.4 eV
æ c
H1s
æ
æ
æ cF2 p
æ æ
æ
0.19
æ= æ
æ
æ
æ æ 0.98 æ
æ c
H1s
æ
æ
æ cF2 p
æ æ
æ
0.98
æ= æ
æ
æ
æ æ -0.19 æ
Hückel approximation




We consider LCAO MO’s constructed from
just the π orbitals of planar sp2 hybridized
hydrocarbons (σ orbitals not considered)
We analyze the effect of π electron
conjugation.
Each pz orbital has the same a = ò pz Ĥpz dt .
Only the nearest neighbor pz orbitals have
nonzero b = ò pz Ĥpz¢ dt .
Centered on the nearest neighbor carbon atoms
Ethylene (isolated π bond)
Coulomb integral of
2pz
Resonance integral
(negative)
 c1 
     c1 

   E  
     c2 
 c2 
α
α
β
Ethylene (isolated π bond)
(
= a + b ® ( c ,c ) = (
Ep * = a - b ® ( c1 ,c2 ) =
Ep
1
2
1
2
1
2
,,
E  2      2  2
1
2
1
2
)
)
Butadiene
2
β
2
1
1
1
4
3
3
4
β
β
æ
æ
æ
æ
æ
æ
æ
2
3
4
a b 0 0 æ
æ
b a b 0 æ
0 b a b æ
æ
0 0 b a æ
æ
Butadiene
Two conjugated π bonds
E  2   1.62   2   0.62   4  4.48
extra 0.48β
Two isolated π bonds
stabilization =
E = 2 2a + 2b = 4a + 4b
π delocalization
(
)
Cyclobutadiene
β
1
1
β
β
3
4
4
2
2
3
1
β
æ
æ
æ
æ
æ
æ
æ
2
3
4
a b 0 b æ
æ
b a b 0 æ
0 b a b æ
æ
b 0 b a æ
æ
Cyclobutadiene
E  2   2   2  4  4
No delocalization energy; no aromaticity
β2
β2
4
3
β1
1
1
2
3
a
b1
0
2
1
2
b1
a
b2
3
β1
0
b2
a
4
Cyclobutadiene
b2
0
b1
æ
æ
æ
æ
æ
æ
æ
4
b2 æ
æ
0 æ
æ
b1 æ
a æ
æ
b1 < b < b2 < 0
Cyclobutadiene
E = 2 (a + b1 + b2 ) + 2 (a + b1 - b2 ) = 4a + 4b1
Spontaneous distortion from square to rectangle?
Homework challenge #8



Is cyclobutadiene square or rectangular? Is it
planar or buckled? Is its ground state singlet
or triplet?
Find experimental and computational
research literature on these questions and
report.
Perform MO calculations yourself (use the
NWCHEM software freely distributed by
Pacific Northwest National Laboratory).
Summary



We have applied numerical techniques of
MO theory to homonuclear diatomic
molecules, heteronuclear diatomic
molecules, and conjugated π electron
systems.
These applications have explained molecular
electronic configurations, polar bonds, added
stability due to π electron delocalization in
butadiene, and the lack thereof in
cyclobutadiene.
Acknowledgment: Mathematica (Wolfram Research) &
NWCHEM (Pacific Northwest National Laboratory)