Molecular Orbitals

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Transcript Molecular Orbitals

Molecular Orbitals
Chapter 5
Molecular Orbital Theory
• 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.
Molecular Orbital Theory
Approximations to the MO Theory
• Born-Oppenheimer approximation
– (r,R)=el(r)(R)
• el(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
Molecular Orbital Theory
Approximations to the MO Theory
• 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
– el(e1,e2…en)=1(e1)2(e2)…,n(en) where i are
the molecular orbitals of the system.
• In this expression, electron-electron interactions are
neglected. More complicated expressions treat the
interactions between electrons for a many-electron system.
Molecular Orbital Theory
Approximations to the MO Theory
• Linear combinations of atomic orbitals, LCAO
– i   cij j where cij is the weight or amplitude of
the atomic orbital, j, in the molecular orbital, i.
– Usually the atomic orbitals are known and only the
coefficients need to be determined.
The coefficients can be positive
or negative.
Molecular Orbital Theory
Approximations to the MO Theory
• Linear combinations of atomic orbitals, LCAO
– 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
Three Conditions for Overlap/Combining of
Atomic Orbitals
• 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
• The distance between overlapping orbitals
must be short to be effective.
Construction of Molecular Orbitals
• The molecular orbital for diatomic hydrogen, H2.
– +=N[ca(1sa) + cb(1sb)] LCAO for bonding
– -=N[ca(1sa) - cb(1sb)] LCAO for antibonding
• ca and cb are adjustable coefficients to reflect the
contribution to bonding. Here, we assume that the
coefficients are the same for the bonding and antibonding.
Since the orbitals are idential, ca=cb, and these can be set
equal to 1.
• The normalization constant, N, is introduced to verify that
the integral,  *d  1 , is equal to 1.
– The probability density (location of the electron), is obtained by
squaring the wavefunction. The electron has to be located
somewhere in space.
Construction of Molecular Orbitals
• +=(ca(1sa) + cb(1sb))2 =
– ca22(1sa) is the probability of finding the
electron on the 1sa atomic orbital.
– 2cacb(1sa)(1sb) is the interaction term and
relates to the probability of finding the electron
between the atoms.
• The positive term indicates bonding between the
Construction of Molecular Orbitals
• +=(ca(1sa) - cb(1sb))2 =
– ca22(1sa) is the probability of finding the electron
on the 1sa atomic orbital.
– In this situation, -2cacb(1sa)(1sb) 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.
Finding the Normalization
• The electron must be somewhere in space.
N 2  caΨ (1sa )  cbΨ (1sb)  d  1
N 2  ca2 2 (1sa )  cb2 2 (1sb)  2ca cb (1sa ) (1sb) d  1
Note : The individual wavefuncti ons are already normalized .
ca  cb  1. Therefore , N 2 1  1   2 (1sa ) (1sb)d  1
Let the overlap int egral , 2 (1sa ) (1sb)d , equal S . N 2 
so N 
2(1  S )
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
Nonbonding Orbitals
• 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
• 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.
Formation of Molecular Orbitals
from p Atomic Orbitals
• Illustrate with the different p orbitals.
–  and  bonding molecular orbitals.
• The symmetry properties of the orbitals must match
(e.g. C2 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
• 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.
Formation of Molecular Orbitals
from d Atomic Orbitals
• The same rules apply for the combination of d
• Types of bonding with d orbitals.
–  bonding with dz2 orbitals
–  bonding with dyz and dxz orbitals
–  bonding (new) with d x 2  y 2 and dxy 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.
Homonuclear Diatomics
• 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.
Orbital Mixing
• The diagram shown in Fig. 5-5 is not entirely
correct homonuclear diatomics since other orbitals
of appropriate symmetry may interact.
i=ci,2sa(2sa) + ci,2sb(2sb) + ci,2pa(2pa) + ci,2pb(2pb)
How many molecular orbitals would be produced from this combination?
1=c1,2sa(2sa)+c2,2sb(2sb) + c3,2pa(2pa) +c4,2pb(2pb)
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.
Electron Configurations for
Homonuclear Diatomics
• The mixing effect decreases as a progression is made
across the periodic table.
– The order of g and u switch back at O2.
• The electron configurations for some homonuclear
– H2, C2, and O2
• The frontier orbitals (HOMO and LUMO)
– Important when considering bonding and reactivity.
• Paramagnetic versus diamagnetic.
Photoelectron Spectroscopy
• Method for determining orbital energies.
O2(g) + h(photon)O2+(g) + eIonization 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.
Molecular Orbitals of Polar Bonds
• 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 Molecular Orbitals for CO
• The symmetry of the molecule will be reduced to C2v for
easier discussion.
– The px and py orbitals have this symmetry.
• Combination of orbitals with the same symmetry.
– s and pz have A1 symmetry (character table).
– px and py possess B1 and B2 symmetry, respectively.
• Compare with Cv (px and py, together, behave like the E1 representation).
– Energies of the orbitals and MO orbitals were derived from
Spartan using semi-empirical calculations.
• Examine the MO diagram and PES.
The Molecular Orbitals for CO
• Molecular orbitals derived from Spartan (semi).
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
The Molecular Orbitals for CO
• 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.
The Molecular Orbitals for LiF
• 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.
Molecular Orbitals for Larger
• Determine the point group of the molecule.
– Linear Dh  D2h and Cv  C2v
• Assign x, y, and z axes.
– 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
– This determines the groups orbitals or SALCs.
Molecular Orbitals for Larger
• 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.
Linear Molecule,
• The symmetry is Dh but we shall use D2h.
– This retains the symmetry of the p orbitals.
– Examine the D2h 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
Linear Molecule,
• For a linear species, these representations do not need to
be reduced to IRs (but you can do it his way).
• Only the 1s atomic orbital is considered for hydrogen,
and it has an Ag type symmetry.
• Two SALCs have the correct symmetry to interact with
the the hydrogen atom.
– The SALC from the 2s atomic orbital is too low in energy.
The H 1s atomic orbital interacts most strongly with the SALC
from the 2pz 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.
Linear Molecule,
• 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.
Bonding in the CO2 Molecule
• The molecule reduces to D2h 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
• A large energy difference indicates that the interaction is
probably insignificant.
Bonding in the CO2 Molecule
• 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
A1g versus alg
H2O – A Nonlinear Molecule
• The point group is C2v.
• C2 axis is determined as the z axis and as the xz plane of
the molecule.
– Only the 1s 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).
H2O – A Nonlinear Molecule
• 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.
– Find the Irs and normalize the SALCs.
  ( R)  i ( R)
h R
# of IRs 
(# of operations in class )  (character of RR )  (character of IR )
order R
ai 
H2O – A Nonlinear Molecule
• Determine the central atomic orbitals that can combine
with the SALCs.
– The 2pz and 2s possess A1 symmetry and the px orbital
possesses B2 symmetry. What about the py central atomic
• Combine the central atomic orbitals with the SALCs
considering the differences in potential energy.
– How many MOs will form?
– How many a1orbitals will form? Why? Roughly determine
the contribution of the SALC and the center pz and 2s atomic
orbitals to each al MO (discuss this in some detail).
H2O – A Nonlinear Molecule
• 1=-0.88(O2s)+-0.11(O2pz)+-0.33(Ha)+-0.33(Hb)
or = c1(O2s)+c2((Ha)+(Hb)) (ignoring the O2pz contrib.)
• 2=0.77(O2px)+0.45(Ha)+-0.45(Hb)
Or = c3(O2s)+c4((Hb)-(Ha))
• 3=-0.33(O2s)+0.83(O2pz)+0.31(Ha)+0.31(Hb)
• 4=1.00(O2py)
• 5=-0.34(O2s)+-0.54(O2pz)+0.54(Ha)+0.54(Hb)
• 6=-0.64(O2px)+0.54(Ha)+-0.54(Hb)
The 2py atomic orbital is nonbonding (1b2).
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.
Ammonia – NH3
• Point group is C3v
• The C3 axis is determined as the z-axis.
• Find the RR (only consider the 1s atomic orbitals on the
• Determine the IR components of the RR.
– There are three SALCs, one with A1 symmetry and two
(considered together) with E symmetry. What does the E
representation indicate? The A1 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.
Ammonia – NH3
• Determine the symmetry representations of the atomic
orbitals on the central atom.
– s and pz have A1 symmetry and px and py (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.
2 (Ha)  (Hb)  (Hc)
 (Hb)  (Hc)
Ammonia – NH3
• MO diagram in Figure 5-31
– A1 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
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.
BF3 species
The Pi Bonding in C4H4
• 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.
Using the Projection Technique
Pj 
  ( R) j R
h R
• 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 C4H4.
This reveals how the atomic group orbitals combine to form
the SALCs of a given symmetry determined earlier.
Phosphorus Pentafluoride, PF5
• What is the point group?
• Find the z-axis and determine the y-axes for the fluorine
• 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?
Phosphorus Pentafluoride, PF5
• 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.
Molecular Shapes
• 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.
Hybrid Orbitals vs. Molecular Orbitals
• 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.
• Td example in the book and [PtCl4]2-