Chapter 5

• Molecular orbital theory uses the methods of group theory to describe bonding. Symmetry and relative energies largely determine how they interact to form molecular orbitals.
• If the total energy of the electrons in the molecular orbitals is less than in the atomic orbitals, the molecule is predicted to form and be stable.

• Born-Oppenheimer approximation
– 
( r , R )=
•  el nuclei.
 el
( r )

( R )
( r ) describes the electrons and

( R ) describes the
– ( R ) describes the nuclear coordinates and ( r ) describes the electron coordinates.
– Nuclei are much heavier and move slowly relative to the electrons. The motion of the electrons can be separated out.
•  el
( r ) will be uniquely considered

• Orbital Approximation
– It is not possible to find an exact solution for the electronic wavefunction in a many-electron system.
It can be expressed as a product of one-electron wavefunctions.
–  el
(e
1
,e
2
…e n
)=

1
(e
1
)

2
(e
2
)…,  n
(e the molecular orbitals of the system.
n
) where
 i are
• In this expression, electron-electron interactions are neglected. More complicated expressions treat the interactions between electrons for a many-electron system.

• Linear combinations of atomic orbitals, LCAO approximation.
–
 i
  j c ij
 j where the atomic orbital,
 j c ij is the weight or amplitude of
, in the molecular orbital,
 i
.
– Usually the atomic orbitals are known and only the coefficients need to be determined.

1
= c
11

2
= c
21


1
1
+ c
+ c
12
22 or negative.

2

2
The coefficients can be positive

• Linear combinations of atomic orbitals, LCAO approximation.
– Minimal basis set (how many AOs are used).
• Orbitals describing core electrons are ignored
(coefficients are extremely small).
• The sum includes all valence electron orbitals.
– H, O, and Na

• Symmetry of the orbitals must be such that regions with the same sign of
 overlap.
– Overlap of s atomic orbital with p atomic orbitals
• Energies of the overlapping orbitals must be similar.
• The distance between overlapping orbitals must be short to be effective.

• The molecular orbital for diatomic hydrogen, H
– 
+
= N [ c a


(1
(1 s s a
) + c b


(1
(1 s s b
)] LCAO for bonding
– 
-
=
• c a
N [ c a and c b a
) c b b
)] LCAO for antibonding are adjustable coefficients to reflect the contribution to bonding. Here, we assume that the
2
.
coefficients are the same for the bonding and antibonding.
Since the orbitals are idential, c a equal to 1.
= c b
, and these can be set
• The normalization constant,
N , is introduced to verify that

*

– The probability density (location of the electron), is obtained by squaring the wavefunction. The electron has to be located somewhere in space.

• 
+
=( c a

(1 sa ) + c b c a
2

– c a
2
2 (1 sa )+ c b
2

2 (1

(1 sb sb
)+2
)) c
2 a c
= b

(1 sa )

(1

2 (1 sa ) is the probability of finding the sb ) electron on the 1sa atomic orbital.
– 2 c a c b

(1 sa )

(1 sb ) is the interaction term and relates to the probability of finding the electron between the atoms.
• The positive term indicates bonding between the atoms.

•  c a
+
2
=( c a

(1 sa

2 (1 sa )+ c
) b
2 c b

(1 sb )) 2 =

2 (1 sb )-2 c
– c a
2 on the 1sa atomic orbital.
a c b

(1 sa )

(1 sb )

2 (1 sa ) is the probability of finding the electron
– In this situation, -2 c a c b

(1 sa )

(1 sb ) is the interaction term. The electrons in this orbital are excluded from the region between the atoms.
• The negative term indicates antibonding between the atoms. The surface where the electron is excluded is called a nodal surface. Make sure that you understand
Figure 5-1.

• The electron must be somewhere in space.
N
2

 c a
2

N
2
  c a
Ψ
( 1 sa )
 c b
Ψ
( 1 sb )

2 d
 
1
2
( 1 sa )
 c b
2

2
( 1 sb )

2 c a c b

( 1 sa )

( 1 sb )
 d
 
1
Note : c a
 c b
The

1 .
individual
Therefore , wavefuncti
N
2
1


1
  ons
2
 are already
( 1 sa )

( 1 sb ) normalized d



1
.
Let the overlap int egral , 2

( 1 sa )

( 1 sb ) d

, equal S .
N
2 
1
2 ( 1

S ) so N

1
2 ( 1

S )
If the overlap int egral was equal to zero ( i .
e .
orthogonal wavefuncti ons ), the normalizat ion cons tan t would equal
1
2
.

• Nonbonding orbitals occur when there is not a corresponding orbital of the correct symmetry.
– There is no net overlap between the wavefunctions or the wavefunctions are orthogonal.
– The energy of the nonbonding orbital is essentially equal to the atomic orbital.
– Illustrate the combining of an s orbital with the three p orbitals.
• Nonbonding orbitals may also result because of the energy differences of combining orbitals. It may also occur if the separation between atoms is too great.

p
• Illustrate with the different p orbitals.
–  and
 bonding molecular orbitals.
• The symmetry properties of the orbitals must match
(e.g. C
2 rotation) for the wavefunctions to combine.
– The character tables reveal the symmetry of particular orbitals indicating if they will or will not combine.
– If there is no symmetry match for an orbital, the orbital is nonbonding.
• When overlapping regions have the same sign, there will be an increased probability in the overlap region.
If opposite signs exist, the combination produces decreased electron probability in the overlap region.

d
• The same rules apply for the combination of d orbitals.
• Types of bonding with d orbitals.
–
–
–
 bonding with d z
 bonding with d yz
2 orbitals
 and d xz d x
2  y orbitals bonding (new) with and
2 d xy orbitals.
• This type of bonding is invoked to account for quadruple bonds.
Discuss nodal surfaces and their locations. These can be illustrated with orbital viewer software.

• Figure 5-5 illustrates a simplified molecular orbital diagram for homonuclear diatomics.
– Aufbau, Hund’s, and Pauli-exclusion
– Bond order = ½(# of bonding electrons - # of antibonding electrons.
– g and u stand for gerade and ungerade which relate to the inversion symmetry operation. Gerade, g , indicates that the orbital is symmetric with respect to the inversion operation.
• Illustrate this with a few orbitals.

• The diagram shown in Fig. 5-5 is not entirely correct homonuclear diatomics since other orbitals of appropriate symmetry may interact.
 i
= c i, 2sa

(2 sa ) + c i, 2sb

(2 sb ) + c i, 2pa

(2 pa ) + c i, 2pb

(2 pb )
How many molecular orbitals would be produced from this combination?

1
= c
1,2sa

(2 sa )+ c
2,2sb

(2 sb ) + c
3,2pa

(2 pa ) + c
4,2pb

(2 pb )
This is an example of one of the σ-type orbitals.
– The result of this mixing is that the lowest energy orbitals move lower and the higher energy orbitals move higher in energy. The mixing inverts the order of the
 g and
 u bonding orbitals (Examine Fig. 5-6).
When overlapping the AOs the valence shell orbitals are commonly the only orbitals considered. This is called the minimal basis set. Illustrate for N.

• The mixing effect decreases as a progression is made across the periodic table.
– The order of  g and
 u switch back at O
2
.
• The electron configurations for some homonuclear diatomics.
– H
2
, C
2
, and O
2
• The frontier orbitals (HOMO and LUMO)
– Important when considering bonding and reactivity.
• Paramagnetic versus diamagnetic.

• Method for determining orbital energies.
O
2
(g) + h

(photon)

O
2
+ (g) + e -
Ionization energy = h

-KE (of expelled electron)
Figures 5-10 and 5-11. The ionization energies of the low-energy orbitals are indicated.
The fine structure is a result of the interaction between the electron energy and the vibration energy. The peaks with pronounced vibration structure are involved strongly in bonding.

• A greater nuclear charge shifts the atomic energy levels lower in energy. The atomic orbitals have different energies and a given MO receives unequal contributions from the atomic orbitals.
– Orbital energies are given in Table 5-1 and Fig. 5-13.
– The contribution of an atomic orbital to a molecular orbital is directly related to the energy. Generally, the
MO has the most character of the AO closest to it in energy (i.e. it will have the greatest coefficient value).

• The symmetry of the molecule will be reduced to C
2 v easier discussion.
– The p x and p y orbitals have this symmetry.
• Combination of orbitals with the same symmetry.
for
– s and p z have A
1 symmetry (character table).
– p x and p y possess
• Compare with C
B
 v
1 and
( p x
B
2 symmetry, respectively.
and p y
, together, behave like the E
1 representation).
– Energies of the orbitals and MO orbitals were derived from
Spartan using semi-empirical calculations.
• Examine the MO diagram and PES.

• Molecular orbitals derived from Spartan (semi).


-40.0,2

=0.82

-20.7,2

*
=-0.48
(O
2s
)+-0.21

(O

(O
2s
)+-0.66
2 pz
)+0.41

(C

(O
2 pz
)+0.56
2 s
)+0.34

(C
2 s
)

(C
2 pz
)

-16.2,1

=-0.85


-16.2,1

=0.85

-13.0,3

=-0.46
1.0,1

*
=0.51

(O

(O

(O

(O
2 px
)+-0.51
2 py
)+0.51

(C
2 pz
)+-0.66
2 px
)+-0.85

(C
2 py
)

(C

(C
2 px
)
2 s
)+0.59
2 px
)

(C
2 pz
)
Discuss and understand the various contributions to the MOs.
Examine the MOs. The character of the MO is determined by the relative amounts of contribution from the combining orbitals.
Atomic orbital contributions with small coefficients are ignored
(<0.13).

• The MOs of greatest interest are the frontier orbitals.
Identify these in the diagram.
– HOMO contributes electrons in reactions and LUMO accepts electrons (future discussion).
• For the HOMO, the greater electron density is on the carbon (larger lobe).
– Actual bonding in most compounds is M-C-O. In fact, a higher overall electron density is on the carbon (show with
Spartan). Why?
• Examine the fine structure in the PES. Why is there the fine structure originating from the 1
 orbitals.

• Examine the diagram for LiF.
– Small interactions due to poor energy overlap.
– In a real compound, each Li is surrounded by six F ions. A modification to this simple picture is needed (developed later).
• Utilize Spartan to construct correct MO energy diagrams for HF and HI.

• Determine the point group of the molecule.
– Linear D
 h

D
2 h and C
 v
• Assign x , y , and z axes.

C
2 v
– The principal rotation axis is chosen as the z -axis.
– In non-linear molecules, the y axes of the outer atoms point toward the central atom.
• Find the characters of the representations for the combination of atomic orbitals on the outer atoms.
• Reduce each representation to its irreducible representations.
– This determines the groups orbitals or SALCs.

• Identify the appearance and character of the
SALCs by using the projection technique.
– For nonlinear molecules
• Find the orbitals on the central atom with the same symmetries as the SALCs.
• Combine the atomic orbitals of the central atom with the SALCs of the same symmetry to produce the MOs.
– Energy similarities are also considered to determine amount of contribution in a particular MO.

-
• The symmetry is D
 h but we shall use D
– This retains the symmetry of the p orbitals.
2 h
.
– Examine the
D
2 h character table and identify the IRs that the same symmetry as the AOs.
• The z -axis is down the internuclear axis.
• Fig. 5-16 illustrates the group orbitals (i.e. SALCs) that form on the fluorine atoms.
– Make sure that you can identify the symmetry or IR.
– There does not have to be direct boding between the outer orbitals.

-
• For a linear species, these representations do not need to be reduced to IRs (but you can do it his way).
• Only the 1 s atomic orbital is considered for hydrogen, and it has an A g type symmetry.
• Two SALCs have the correct symmetry to interact with the the hydrogen atom.
– The SALC from the 2 s atomic orbital is too low in energy.
The H 1s atomic orbital interacts most strongly with the SALC from the 2 p z orbitals on fluorine.
• (-13.6 eV and -18.7 eV).
– Five of the SALCs do not interact with the central atom.
These are essentially nonbonding.

-
• The MO picture illustrates a 3-center, 2-electron bond(s). This is different than the Lewis approach which utilizes a localized description of bonding between 2 atoms.
• Involving a large number of atoms coordinated to a central atom usually decreases the bonding orbitals even further.

2
• The molecule reduces to D
2 h analysis.
symmetry for easier
• Construct the group orbitals as before and determine their symmetry (Fig. 5-19).
• Determine the symmetry of the C atomic orbitals and group interactions to determine MOs.
– Label the interactions according to symmetry taking into account the energy differences. Discuss these interactions.
• A large energy difference indicates that the interaction is probably insignificant.

2
• Construct the MO diagram from the combinations determined previously.
– Notice the multitude of 3-center, 2-electron orbitals.
– All the bonding orbitals are occupied as well as two nonbonding MOs.
– Identify the 
-type and

-type bonds. Where is the electron density for the nonbonding electrons?
Note: Use capital when describing symmetry and lower case letter when describing the actual orbitals (see diagram).
A
1 g versus a l g

2
• The point group is C
2 v
.
•
C
2 axis is determined as the the molecule.
z axis and as the xz plane of
– Only the 1 s orbital will be considered on the hydrogens so it is not necessary to assign axes to hydrogen.
• Find the representation for the group orbitals.
– The book utilizes transformation matrices to find the reducible representation for the SALCS. We will use a slightly different approach throughout the semester which is simpler (especially for larger molecules).

2
• The book largely finds the IRs of  by inspection. We will reduce the RR into its component IRs by a systematic approach.
– Transformation matrix exist which will reduce the matrix to one consisting of blocks along the diagonal. Each of these matrices belongs to an IR.
a i
– Find the Irs and normalize the SALCs.

1 h

R

( R )
 i
( R )
# of IRs

1 order


(#
R of operations in class )

( character of RR )

( character of IR )


2
• Determine the central atomic orbitals that can combine with the SALCs.
– The 2 p z and 2 s possess A
1 possesses B orbital?
2 symmetry and the p x symmetry. What about the p y orbital central atomic
• Combine the central atomic orbitals with the SALCs considering the differences in potential energy.
– How many MOs will form?
– How many a
1 orbitals will form? Why? Roughly determine the contribution of the SALC and the center p z orbitals to each a l and 2
MO (discuss this in some detail).
s atomic

2
•
•
 or = c
1

(O

1
2
=-0.88
=0.77
Or = c


3

(O
• 
3
=-0.33
• 
4
=1.00
• 
5
=-0.34
• 
6
=-0.64

2 s
)+ c



(O
(O
(O
(O
(O
(O
2 s
2
(
2 px
2 s
2 py
2 s
2 px
)
)+-0.11
(O
2 pz
)+-0.33

(H a )+

(H b )) (ignoring the O
)+0.45
)+0.83
)+-0.54
)+0.54





(H
(O
(O
(H a
2 s
)+ c
4
(

(H b )-

(H a )) a
)+-0.45
2 pz
2 pz
)+0.31
)+0.54
)+-0.54





(H
(H
(H
(H b
(H a b
) a a
)
)+-0.33
2 pz
 contrib.)
)+0.31
)+0.54


(H
(H
(H b b
)
) b )
The 2 p y atomic orbital is nonbonding (1 b
2
).
Examine Table 5-3 and Fig. 5-29. All four MOs are different. What are the major differences when compared to the Lewis structure?
Can also view the MOs with Spartan.

3
• Point group is C
3 v
• The
C
3 axis is determined as the z-axis.
• Find the RR (only consider the 1 s atomic orbitals on the hydrogens).
• Determine the IR components of the RR.
– There are three SALCs, one with A
1 symmetry and two
(considered together) with E symmetry. What does the E representation indicate? The A
1
SALC is easy to visualize.
What about the two SALCs with E symmetry?
• Sum of the squares of the coefficients for each AO must equal 1.
• Symmetry of the central atom orbitals matches the symmetry of the
SALCs. There must be one nodal surface in each E SALC.

3
• Determine the symmetry representations of the atomic orbitals on the central atom.
– s and p z have symmetry.
A
1 symmetry and p x and p y
(as a pair) possess E
• Combine the central atomic orbitals with the SALCs of appropriate symmetry to form MOs.
– Spartan helps in understanding the SALCs and MOs.

1 e
=0.37

1 e
=0.63

(2

(2 px
)+-0.63
px
)+0.37

(2

(2 py
)+0.58
py
)+0.49

(H a )+-0.28

(H b )+-0.27

(H c )

(H b )+-0.48

(H c )
SALCs
1
6

2

( H a )
 
( H b )
 
( H c )
 1
2
 
( H b )
 
( H c )


3
• MO diagram in Figure 5-31
– A
1 symmetry orbitals
• A bonding, nonbonding, and antibonding MO (roughly)
– E symmetry orbitals
• These are doubly degenerate orbitals which means that there is a pair at low energies and a pair at high energies (the same energy).
Lone pair chemistry – there is a lone pair of electrons largely located on the nitrogen atom. This can act as a
Lewis base. The LUMO/HOMO chemistry will be discussed in detail later.
BF
3 species

4
4
• What is the point group?
• Construction of the SALCs.
– Find the RR and IRs.
– The appearance of the SALCs may not be obvious when there is more than a two group atoms. Obviously, Spartan can be used to determine the wavefunctions and appearance of the orbitals. There is another way!!! This involves the use of projection operators.


P j
 l j 
R

( R ) j

R h
• The most important and frequent use for projection operators is to determine the proper way to combine atomic wave functions on individual atoms into Mos that correspond to the molecular symmetry.
– A particular atomic orbital or wavefunction will be projected by the symmetry operations. Let’s perform this for C
4
H
4
.
This reveals how the atomic group orbitals combine to form the SALCs of a given symmetry determined earlier.

5
• What is the point group?
• Find the z -axis and determine the y -axes for the fluorine ligands.
• Determine the RR representing the group orbitals from the fluorine ligands.
– The axial and equatorial have to be considered separately since they are not interconverted by symmetry.
• Determine the IRs components contained in 
.
– How many SALCs will there be?
Website: http://www.mpip-mainz.mpg.de/~gelessus/group.html

5
• Use the projection technique to determine the appearance of the group orbitals.
• Determine the symmetry types of the central atom atomic orbitals.
• Combine the SALCs and the atomic orbitals on phosphorus to make the MOs.
• Draw the interaction diagram considering the potential energy differences.

• Determining the actual shapes of molecules using the MO approach usually involves the use of molecular modeling software (Spartan).
• The overall energy at different bond distances and angles is calculated until the minimum is found. Any energy that is calculated will be equal to or greater than the true energy.

• The hybrid orbitals point from a central atom toward surrounding atoms or lone pairs.
– Therefore, the symmetry properties of a set of hybrid orbitals will be identical to the properties of a set of vectors with origins at the nucleus of the central atom and pointing toward the surrounding atoms and lone pairs.
•
T d example in the book and [PtCl
4
] 2-