Any group of atoms:
•Molecules,
•Molecular ions,
•Fragments,
•Supermolecules
1

Molecular orbital theory is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.
2

Molecular orbital is a function that describes the wave-like behavior of a single electron in a molecule. A polyelectronic wave-function is expressed in terms (as a product or a determinant) of MOs. The use of the term "orbital" was first used in English by Robert S. Mulliken in 1925 as the
English translation of Schrödinger's use of the German word,
'Eigenfunktion'.
Robert Sanderson
Mulliken
1996-1986
Nobel 1966
3

Generalizing the
A linear combination of atomic orbitals or LCAO
Sir John Lennard-Jones
1894-1954
Linus Carl Pauling
1901-1994 Nobel 1962
It was introduced in 1929 by Lennard-Jones with the description of bonding in the diatomic molecules of the first main row of the periodic table, but had been used earlier by Pauling for H
2
+ .
4

Any calculation starts by inputs and provides outputs
Input
Definition of the system
Total number of electrons
Choice of state (ground state or excited state)
Nature of atoms involved (Z)
Atomic Orbitals
Geometry to define Potential
Output
Energy,
Molecular Orbitals
Electronic density
Spin, Magnetism
Search for minimization of energy
The calculated geometry is the minimum found by optimization
5

Structure and Reactivity
How the energy of a system respond to a structural variation?
Changing distance or angle.
6

Symmetry
To describe a molecule is describing
•Its geometry and its symmetry
•Its energy
MOs are eigenfunctions of the symmetry operator and of H
For H
2
, we have use symmetries without solving the Schrodinger equation,
For a larger system, symmetry helps simplifying or analyzing.
One common symmetry is the plane for planar molecules; this causes s and p separation.
To be considered when some parameter varies (structure modification, structure optimization, reactivity) symmetry has to be preserved during the variation
7

Usually, the atomic orbitals that participate to the MOs are valence orbitals.
These are the valence orbitals of the neutral atom. For a cation in its highest oxidation state, this might be the unoccupied shell whereas the size of the cation is given by the outermost occupied shell
See the case of Li +
8

+
Size of the cation Li + according to Slater?
The last occupied shell is the
1s
Z=(3-0.31) <r> = 1.5/2.69 a
0
= 0.558 a
0
=
0.295 Ả
This is small (The Pauling ionic radius is also small, 0.60 Ả)
The valence orbital
2s is the one involved in forming bonds
Z=(3-2*0.85) <r> = 5/1.3 a
0
= 3.846 a
0
=
2.034
Ả
Atomic radius (empirical) : 1.45 Ả
Atomic radius (calculated) : 1.67 Ả
Covalent radius (2008 values) : 1.28 Ả
Covalent radius (empirical) : 1.34 Ả van der Waals radius: 1.82 Ả
The covalent radius for the 2s orbital is such that the orbitals overlap; it is less than the <r> value
9

• Determine the symmetry elements of the molecule
• Make the list of the functions involved (valence atomic orbitals)
• Classify them according to symmetry (build symmetry orbitals if necessary by mixing in a combination the set of orbitals related by symmetry)
• Combine orbitals of the same symmetry (whose overlap is significant and whose energy levels differ by less than 10 eV).
10

Valence orbitals of H
2
O
2sO
2pxO
2pyO
2pzO
1s
1
1s
2 yOz
Molecular plane
S
A
S
S
S
S xOz symmetry plane
S
S
A
S
-
-
C2 z axis
S
S
-
A
A
-
Symmetry relative to the molecular plane, sigma-pi separation sigma pi sigma sigma sigma sigma
11

Valence orbitals of H
2
O
2sO
2pxO
2pyO
2pzO
1s
1
1s
2 yOz
Molecular plane
S
A
S
S
S
S xOz symmetry plane
S
S
A
S
-
-
C2 z axis
S
S
-
A
A
-
Symmetry relative to xOz, necessitates building symmetry orbitals
Build them ! What are their energy levels?
12

Valence orbitals of H
2
O
2sO
2pxO
2pyO
2pzO
1
2
(1s
1
+1s
2
)
1
2
(1s
1
-1s
2
) yOz
Molecular plane
S
S
S
A
S
S xOz symmetry plane
S
S
S
S
A
A
Redundancy: Three groups; A2 is not present
C2 z axis
A
S
A
A
S
S
A1
B1
B2
A1
A1
B2
13

C2v
A1
A2
B1
B2
E
1
1
1
1 yOz molecular plane
1
-1
1
-1 xOz symmetry plane
1
-1
-1
1
C2 axis
-1
-1
1
1
14

1
1
2
(1s
1
+1s
2
)
( 1s
1
-1s
2
)
2
The H-H distance is
2*sin(105/2)*1.09 Ả = 1.73 Ả
This is long relative to 0.74 Ả !
The energy level of the symmetry orbital is the atomic level for H, -13.6 eV
15

1s
2s
H
13.6
Li
5.4
He
24.25
Be
10
B
15.2
C
21.4
N
26
O
32.3
F
40
2p
Roald Hoffmann
Nobel 1981
3s
3p
3.5
Na
5.1
3.
6
Mg
9
4.5
8.5
Al
12.3
6.5
11.4
Si
17.3
9.2
13.4
P
18.6
14
14.8
S
20
13.3
18.1
Cl
30
15
William Nunn Lipscomb, Jr.
American 1919-
Nobel 1976
16

2sO
2pxO
2pyO
1
2
2pzO
1
(1s
1
+1s
2
)
2
(1s
1
-1s
2
)
1 Valence orbital of H
2
O
Symmetry B1
A1
B1
B2
A1
A1
B2
The atomic orbital
2pxO is a lone in its group,
It is a molecular orbital
It has p symmetry.
E
2p
(O)= -14.8 eV
17

2sO
2pxO
2pyO
2pzO
1
2
(1s
1
+1s
2
)
1
2
(1s
1
-1s
2
)
2 Valence orbitals of H
2
O
Symmetry B2
A1
B1
B2
A1
A1
B2
E
2p
(O)= -14.8 eV and
E
1S
(H) = -13.6 eV
-13.6eV
s
A
-s
B s *
A s
A
2py
0
-14.8 eV
18

2sO
2pxO
2pyO
2pzO
1
2
(1s
1
+1s
2
)
1
2
(1s
1
-1s
2
)
A1
B1
B2
A1
A1
B2
3 Valence orbitals of H
2
O
Symmetry A1
E
2s
(O)= -32.3 eV
E
2p
(O)= -14.8 eV
And E
1S
(H) = -13.6 eV
E
2s is low in energy an d does not mix s *
S
-13.6eV
s
A
+s
B
2pz
0
-14.8eV
s
S
19

6 AO → 6
MOs
3 A1 → 3 A1
2B2 → 2B2
1 B1 → 1 B1
What is the ordering of the levels?
20

y
° y z z
Overlap involved in B2
Overlap involved in A1
At 45 ° same overlap angle HOH/2 = 105/2 = 52 °5 > 45°
Splitting in B2 > Splitting in A1
21

2B2
6 AO → 6 MOs s
*
A
3 A1 → 3 A1
2B2 → 2B2
1 B1 → 1 B1
O H
H
2p x s
*
S
In the Lewis formula, two electron pairs, two bonds
The electron pairs are 2s and 2p
Spectroscopy shows different electron levels
The bonds are the result of two contributing MO
A1 and B2
Orbital numbering: 1A1 for the core?
2 s
0 s
S s
A
4A1
3A1
1B2
2A1
22

Hybridization sp 3 agreement with VSEPR and Total density
MOs sp 2 closer to MOs and spectroscopy: two different pairs 23

The Walsh diagram
E(eV)
2px
O s
S s
S s
A s
A
2s
O

90°
-13.6
-14.8

°
-33.3
24

25

Typical Bond Angles in AH
2
Molecules molecule electronic configuration H-A-H bond angle
BeH
2
(1a
1
) 2 (1b
2
) 2 180 °
BH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 1 127 °
CH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 1 (b
1
) 1 134 °
CH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 2 102 °
NH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 2 (b
1
) 1 103 °
OH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 2 (b
1
) 2 104 °
MgH
2
(1a
1
) 2 (1b
2
) 2 180 °
AlH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 1 119 °
SiH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 1 (b
1
) 1 118 °
SiH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 2 93 °
PH
2
(1a
1
) 2 (1b
2
) 2 (2a
1
) 2 (b
1
) 1 92 °
SH
2
1a
1
= 1 s g ;
1b
2
= 1 s u
(1a
1
) 2 (1b
2
) 2 (2a
1
) 2 (b
1
) 2 92 °
26

F. Walsh diagram for H
2
S
27

This is called an orbital correlation diagram :
28

C
3 Valence orbitals of CH
2
Symmetry A1
E
2s
(C)= -21.4 eV
E
2p
(C)= -11.4 eV
And E
1S
E
2sC
(H) = -13.6 eV
– E
1sH
< 10 eV
2sC
2pxC
2pyC
2pzC
1
2
(1s
1
+1s
2
)
1
2
(1s
1
-1s
2
)
A1
B1
B2
A1
A1
B2
Hybrid orbitals
Not atomic eigenfunctions
A
But part of the MOs eigenfunctions for the molecule
2s-2p
Z
2A1 is bonding 3A1 is non-bonding 4A1 is antibonding
2s+2p
Z
29

C
2sC
2pxC
2pyC
2pzC
1
2
(1s
1
+1s
2
)
1
2
(1s
1
-1s
2
)
A1
B1
B2
A1
A1
B2
3 Valence orbitals of CH
2
Symmetry A1
E
2s
(C)= -21.4 eV
E
2p
(C)= -11.4 eV
And E
1S
E
2sC
(H) = -13.6 eV
– E
1sH
< 10 eV
A
-13.6eV
s
A
+s
B
Among the 3 A1 valence MO, the lowest one is bonding and the middle one non-bonding; this differs from H
2
O!
s *
S s
S
2pz
-11.4eV
C
30

Justification:
• s and p
Z both are of same symmetry and appear in the same linear combinations (MOs); hybridization allows anticipating.
• hybridization aims combining AOs, each hybrid maximizing the interaction with a partner (here the symmetry orbital on Hs) and minimizing those with others. The idea is that we can thus separate the interactions with different partners. If correct, this is useful and allows transferability (replacing a substituent by another one involving the same hybrid).
• Mathematically a set of hybrids is equivalent to a set of canonic orbitals
Failure:
• It is not possible to rigorously separate the interactions with different partners. Hybrids interact with each other.
• Existing symmetries reappear if interactions between hybrids is included.
31

The MO orbital description conflicts with sp 3 hybridization
32

33

<t i
It j
>= δ ij
<t i
IHIt j
>= (E
2s
-E
2p
)/4 t
3 t
4 t
1 t
2
3
2s 2p x
2p y
2p
1/2 1/2 1/2 1/2 z
1/2 1/2
1/2
1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2 1/2
E ti
= <t i
IHIt i
> = (E
2s
+3E
2p
)/4
34

x b b b b x b b b b x b b b b x
3
= 0
With b
=(E
2s
-E
2p
)/4
E
2p
(E
2s
+3E
2p
)/4
1(E
2s
-E
2p
)/4
3 (E
2s
-E
2p
)/4
E
2s
35

¼ of 2s
3/4 of 2p
3
3 t
3 t
4 t
1 t
2
2s 2p x
2p y
2p z
√ 
/2 1/2 0 0
1/2 √(2/3)
0 1/√12
1/2 -1/ √6 1/ √2 -1/√12
1/2 1/ √6 1/ √2 -1/√12
36

• Pictorial
• Anticipating AO combinations within MOs
• Global density
• Transferable: Analysis through substituents
• Writing VB structures
• Not describing symmetry
• Not good for spectroscopy
• Lack of orthogonality between hybrid orbitals belonging to the same center (tails of localized orbitals)
37

3
2 t
3 t
4 t
1 t
2
2s 2p x
1/2 1/ √2
2p
0
1/2 -1/ √2 0 y
1/2 0
2p
1/2 z
1/2
1/ √2 -1/2
1/2 0 1/ √2 -1/2
38

39

40

Mixing 2s and 2p: requires degeneracy to maintain eigenfunctions of AOs.
Otherwise, the hybrid orbital is an average value for the atom, not an exact solution.
This makes sense when ligands impose directionality: guess of the mixing occurring in
41
OMs.

q
Expression along C
2 t = a s + √(1-a 2 ) [p z t = a’s + √(1-a’ 2 ) [p z
(z is the main axis) cos q ± p cos q ’ ± p x y sin sin q q
]
’]
2a 2 +2a’ 2 =1
√(1-a 2 ) sin q  / √2
√(1-a’ 2 ) sin q ’  / √2
:
q ’
: q 5.5/ (<9 ° )  5.75
° → a = 0.459
a’ = √(1/2-a 2 ) = 0.548
sin q ’  (/ √2)/(1/ √(1-a’ 2 )=0.8453 q ’  57 °   4.7/ > 9 ° 
42

q
:
q ’
: a 2 +a’ 2 = ½
√(1-a 2 ) sin q  / √2
√(1-a’ 2 ) sin q ’  / √2
(1-a 2 ) sin q
2
 /
2

(1-a ’ 2 ) sin q ’ 2
1-
/
2sin q
2
 a 2
1-
/
2sin q ’ 2
 a ’ 2
1-
/
2sin q
2 + 1-
/
2sin q ’ 2
 a 2
+ a ’ 2
2 =
/
2sin q
2 +
/
2sin q ’ 2
+ ½
3 =
/ sin q
2 +
/ sin q ’ 2
When q decreases q ’ increases!
43

q
= 105 °5 water a is smaller than ½ a’ is larger than ½
The weight of s is larger in the lone pairs.
The s level is lower in energy than the p level. It has to participate to the stabilization of lone pairs.
L , CH
3
+
90 ° pure p sp 3 ligand
109 °28’
¼ s ¾ p
CH
3
, L +
>109 ° 28’
>1/4 s
44

Si
Si
Si Si
Si
Si
Si
Si
Si
+
Si
Si
Si
45

1/ √2 -1/√2
1/ √3
b
H
3
b
The orbitals of H
3
(equilateral triangle in the xy plane)
Are symmetry orbitals of NH
3 and CH
3 matching p orbitals
2p x 2p y
2p z
46

1/ √2 -1/√2
1/ √3
b
H
3
The orbitals of H
3
(equilateral triangle in the xy plane) are symmetry orbitals of NH
CH
3 matching p orbitals
3 and
b
2p x 2p y
2p z
47

1/ √2 -1/√2
H
3
Pseudo symmetry : for H
3
OM are degenerate.
the
2/3 = 1/ √2 + c 2
They remain degenerate interacting with the 2p orbitals (E)
√(2/3)
b
2p x
Normalization
1 = (2/ √3) 2 +2 c 2
2p y
48

H
3
1/ √2 -1/ √2 -1/ √6 -1/ √6
b
√(2/3)
b
1/ √3
The orbitals of H
3
(equilateral triangle in the xy plane) are symmetry orbitals of NH
CH
3 matching p orbitals
3 and
2p x 2p y
2p z
49

NH
3 s system
Lone pair
Bonding
50

1/ √2 -1/ √2
-1/ √6 -1/ √6
√(2/3)
NH
3 p system
2p x
2p y
51

52

Free rotation of a methyl group adjacent to a p system
Hyperconjugation
CH
3 always has one orbital conjugating with the p system
53
Mirror symmetry pseudosymmetry

X Z
X Z x p y s x s y p
54

2
2
+
D
+
+ or
D
55

+
D
D x
+
A
CH
2
With a substituent, the 2 orbitals are not equivalent.
That of highest energy level interacts more strongly with the CH
2 group
56

57

58

59

BF
3
60

Octet rule, eighteen electron rule
N in NH
3
4 atomic orbitals
4 bonding orbitals or non bonding accommodating 8 electrons
3 antibonding orbitals
C in CH
4
4 atomic orbitals
4 bonding orbitals accommodating 8 electrons
4 antibonding orbitals
W in W(CO)
6
9 atomic orbitals
9 bonding orbitals or non bonding accommodating 8 electrons
6 antibonding orbitals
61

Ligand field for octahedral environment
W in W(CO)
6
9 atomic orbitals
9 bonding orbitals or non bonding accommodating 8 electrons
6 antibonding orbitals
The s interaction raises the d
X
2 levels
The p with p
*
CO levels” and d
X
2 -d
Y
2 stabilizes the “non bonding
62

Ligand field for octahedral environment
W in W(CO)
6
9 atomic orbitals
9 bonding orbitals or non bonding accommodating 8 electrons
6 antibonding orbitals
The s interaction raises the d
X
2 and d
X
2 -d
The p with p
*
Y
2 levels
CO stabilizes the “non bonding levels”
63

Ligand field for octahedral environment
Ti in Ti(H
2
O)
2
(HO)
4
Or Ti in TiO
2
Ti in 9 atomic orbitals
9 bonding orbitals or non bonding accommodating 8 electrons
6 antibonding orbitals
The s interaction raises the d
X
2 and d
X
2 -d
Y
2 levels
The p interaction with p
O
(oxygen pairs) is a stabilizing interaction.
The same occurs with
PH
3 ligands
Rutile TiO
2
These 6 electrons count for the eighteen rule

Ligand field for various environments
65

3
2
2
2
Z
D
D
Ni
D
D
Z
Ni nickel (Z= 28)
NiD
2 where D is a donor substituent (group PH
3
). Ni will be only represented by the set of d (x 2 -y 2 , z 2 , xy, xz et yz) orbitals taking E(3d) = -
12 eV. D is modeled by an s orbital with 2 electrons: E(D)= - 15 eV.
What is the best orientation for the C=C bond?
1) Draw an energy diagram for NiD
2
2) Why the ethylene molecule is represented above by its p
* orbital?
3) Draw an energy diagram for the interaction and tell which orientation is the best.
66

Ni(PH
3
)
2
...CH
2
=CH
2 x
2
-y
2 z
2
Z
D
D
Ni xy, xz, yz
Interaction with dx 2 -y 2
Largest interaction
D
D
Z
Ni
Interaction with d(x+y),z weaker
67

z y
Formaldehyde
 s
A s
CH
, A
1
 s
S
2p x
 s
S
 s
A s
CH
, A
1 s
CO
, A
1 p
CO
, B
2
2p y
, B
1 p
*
CO
, B
2 s
CO
, A
1 s
CH
, B
1
2p
s
S s
CH
, A
1
2s
O
, A
1
CH
2
CH
2s
O
2
H
CH
2
O O
C=O O
2
68

B s orbital
With p overlap
69