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Advanced Physical ChemistryChem5350

MOLECULAR STRUCTURE

Professor Angelo R. Rossi

http://homepages.uconn.edu/rossiDepartment of Chemistry, Room CHMT215

The Univerity of Connecticut

Fall Semester 2013

[email protected]

Last Updated: October 20, 2013 at 8:18pm

Molecular Electronic Structure

Fall Semester 2013 2

The Born-Oppenheimer ApproximationBorn and Oppenheimer pointed out that since nuclei are thousands oftimes more massive than electrons, they move much more slowly andcan be treated as being stationary in considering the motions ofelectrons in molecules.

Thus, a simpler electronic Schrodinger equation can be solved for eachfixed internuclear distance R.

The molecular vibrations can then be obtained by solving a Schrodingerequation for the motion of the nuclei in the average potential fieldproduced by the electrons.


The Hydrogen Molecule Ion H2+


The hydrogen molecule ion H2+ is the simplest molecule and will be

discussed in some detail as an introduction to the treatment of morecomplicated molecules.

Hψ(rrr1,RRRA,RRRB) = Eψ(rrr1,RRRA,RRRB)

rrr1 is the vector locating the electron, and RRRA and RRRB are the vectorslocating the nuclei in a coordinate system.




Coordinates for the hydrogen molecule ion H2+. The protons are

labeled A and B




The Hamiltonian for the hydrogen molecule ion is given by

H = − ~22M

(∇2A +∇2

B)− ~22me

∇2e − e2

4πǫ0r1A− e2

4πǫ0r1B− e2

4πǫ0R

where M is the mass of each nucleus, me is the mass of the electron,r1A is distance between the electron and nucleus A, and r1B is distancebetween the electron and nucleus B,The Born-Oppenheimer approximations is introduced to solve theSchrodinger equation for the H2

+ ion.

ψ(rrr1A, rrr1B,RRRA,RRRB) = ψe(rrrA, rrrB,RRR)× ψn(RRRA,RRRB)

where ψn(RRRA,RRRB) is the wave function for nuclear motion.




The Schrodinger equation for the electronic motion is then

Hψe = Ee(R)ψe

with the Hamiltonian for electronic motion given by

He = − ~2

2me

∇2e −

e2

4πǫ0r1A− e2

4πǫ0r1B− e2

4πǫ0R

containing the electronic kinetic energy, the electrostatic attractions ofthe electrons to each nucleus, and the nuclear electrostatic repulsion.The nuclear repulsion term is a constant since the internuclear distanceR is fixed, and the electronic Schrodinger equation can be solved for allR, giving the electronic energy Ee(R) as a function of R.




The Born-Oppenheimer approximation shows that the nuclei move inthe potential energy Ee(R) determined by the electronic motion, so theSchrodinger equation for nuclear motion becomes

[− ~2

2M(∇2

A +∇2B) + Ee(R)

]ψn(RRRA,RRRB) = Enψn(RRRA,RRRB)

where En is the energy of nuclear motion which contains contributionsfor translational, rotational, and vibrational motion.

The wave function for nuclear motion ψn(RRRA,RRRB) is the product of thewave functions for translational, rotational, and vibrational motions.


The Variational Method


The variational method provides an upper bound to the ground-stateenergy of a system.

The ground-state wave function ψ0 and energy E0 satisfy theSchrodinger equation

Hψ0 = E0ψ0

Multiply the above equation from the left by ψ∗0 and integrate over all

space to obtain

E0 =

∫ψ∗0Hψ0dτ∫ψ∗0ψ0dτ

The denominator is not set equal to unity to allow for the possibility thatψ0 is not normalized.




A theorem says that if we substitute another function φ for ψ0 andcalculate the corresponding energy according to

Eφ =

∫φ∗Hφdτ∫φ∗φdτ

then Eφ will be greater than the ground-state energy E0

This is the variational principle where Eφ ≥ E0, and the equality holdsonly if φ = ψ0, the exact wave function.




The variational principle states that an upper bound to E0 can becalculated by using any trial function we wish.

A trial function φ can be chosen such that it depends upon somearbitrary parameters, α, β, γ, · · · , called variational parameters.

The energy will also depend upon these variational parameters

Eφ(α, β, γ, · · · ) ≥ E0

Now, Eφ can be minimized with respect to each of the variationalparameters to determine the best possible ground-state energy fromthe trial wave function.


The Hydrogen Molecule Ion H2+: Molecular Orbital Theory


The approach used in solving the Schrodinger equation for thehydrogen molecule ion is to use a trial wave function made up ofhydrogenlike atomic orbitals This approach is called molecular orbitaltheory.

Molecular properties calculated in this way are are approximate, but thetreatment can be improved to any desired accuracy.

As trial wave functions for ψe(rrrA, rrrB,RRR) for the one-electron hydrogenmolecule ion, the following linear combination is used

ψ± = c11sA ± c21sB

where 1sA and 1sB are atomic hydrogen orbitals on protons A and B,and c1 and c2 are constants.




This type of function is referred to as a linear combination of atomicorbitals or an LCAO molecular orbital. Since the two nuclei areidentical, c1 = c2 = c.

For nuclei that are close, the 1s orbitals overlaps shown below:

The above diagram shows the magnitudes of the 1s wave functions on atoms A and B when they are 3a0apart but not interacting. The abscissa is the distance from atom A along the internuclear axis in units of a0.




In order to normalize the molecular orbital with the plus sign, the following integralmust have the value of unity: ∫

ψ∗+ψ+dτ = 1

Substituting the wave function yields

c2∫

(1s∗A + 1s∗B)(1sA + 1sB)dτ = 1

c2∫

1s∗A1sAdτ + c2∫

1s∗A1sBdτ + c2∫

1s∗B1sAdτ + c2∫

1s∗B1sBdτ = 1

• The first and last integrals are equal to 1 because the 1s orbitals are normalized.• The second and third integrals are equal because the wave function is real and are

called overlap integrals.




The normalization can be written in terms of the overlap integral

c2(2 + 2

∫1sA1sBdτ

)= c2 (2 + 2S) = 1

so thatc =

1

[2(1 + S)]12

The normalized wave function with a plus sign is given by

ψg =1

[2(1 + S)]12

(1sA + 1sB)

The normalized wave function with a negative sign is given by

ψu =1

[2(1− S)]12

(1sA − 1sB)


The Hydrogen Molecule Ion H2+: Even and Odd Parity


When a molecule has a center of symmetry, the wave function may ormay not change sign when it is inverted through the center of symmetry.

• If ψ(x, y, z) = ψ(−x,−y,−z), the wave function is said to have evenparity and is designated with a subscript g for gerade.

• If ψ(x, y, z) = −ψ(−x,−y,−z), the wave function is said to have oddparity and is designated with a subscript u for ungerade.


The Hydrogen Molecule Ion H2+: Even and Odd Parity


The parity of an orbital is even (g) ifthe wave function is unchanged un-der inversion through the center ofsymmetry.

The parity of an orbital is odd (u) ifthe wave function changes sign un-der inversion through the center ofsymmetry.

Homonuclear diatomic moleculeshave a center of inversion whileheteronuclear diatomic molecules donot.




The overlap integral S can be evaluated analytically as a function ofinternuclear distance R, and the result is given below

S = e−R

(1 +R +

R2

3

)

When R = 0, the two 1s orbitals overlap completely, and S = 1. As R isincreased to infinity, the overlap integral decreases asymptotically tozero.




(a) Magnitude of the normalized molecular orbital ψg when protons Aand B are separated by 3a0. (b) Magnitude of the normalized molecularorbital ψu. (c) Magnitude of ψ2

g . (d) Magnitude of ψ2u.




Using the molecular orbitals (ψg or ψu), the upper bound to the energyof H2

+ can be calculated using the variation principle:

E =

∫ψ∗Hψdτ∫ψ∗ψdτ

=

∫[c11sA + c21sB]H[c11sA + c21sB]dτ∫

[c11sA + c21sB]2dτ

E =c21HAA + 2c1c2HAB + c22HBB

c21SAA + 2c1c2SAB + c22SBB

=c21HAA + 2c1c2HAB + c22HBB

c21 + c1c2S + c22

HAA =∫1sAH1sAdτ =

∫1sBH1sBdτ = HBB

HAB =∫1sAH1sBdτ =

∫1sBH1sAdτ

SAA =∫1sA1sAdτ =

∫1sB1sBdτ = SBB = 1

SAB =∫1sA1sBdτ =

∫1sB1sAdτ = S




The integral HAA is called the Coulomb integral because thedifference between HAA and the energy of a single hydrogen atom isjust that of the Coulomb interaction of nucleus B with an electroncentered on nucleus A.

HAA = E1s +e2

4πǫ0R− J, where J =

∫1sA

(e2

4πǫ0rB

)1sAdτ

J is called the electron repulsion integral

J = e−2R

(1 +

1

R

)




The integral HAB is referred to as the resonance integral.

HAB = SAB

(E1s +

e2

4πǫ0R

)−K, where K =

∫1sB

(e2

4πǫ0rB

)1sAdτ

K is called the exchange integral.

K =S

R− e−R(1 +R)




To find the minimum energy. the derivatives of E with respect to c1 andc2 are set to zero.

To do this, the equation for the energy of H2+ is set rewritten in the form:

E(c21 + 2c1c2S + c22) = c1HAA + 2c1c2HAB + c22HBB

Differentiating this equation with respect to c1 equals

E(2c1 + 2c2S) +∂E

∂c1(c21 + 2c1c2S + c22) = 2c1HAA + 2c2HAB

and differentiating this equation with respect to c2 equals

E(2c1S + 2c2) +∂E

∂c2(c21 + 2c1c2S + c22) = 2c1HAB + 2c2HBB




Since ∂E∂c1

and ∂E∂c2

are equal to zero for the minimum energy, the aboveequations can be written as

c1(HAA − E) + c2(HAB − SE) = 0c1(HAB − SE) + c2(HBB − E) = 0

There is a nontrivial solution to these equations only if the determinantof the coefficients is equal to zero

∣∣∣∣HAA − E HAB − SEHAB − SE HBB − E

∣∣∣∣ = 0

This yields two solutions

Eg = E1s +J+K1+S

Eu = E1s +J−K1−S




The energy ∆Eg of the H2+ ion in the bonding orbital ψg relative to the

completely dissociated species H+ and H is given by

∆g = Eg − E1s =J +K

1 + S

The energy ∆Eu of the H2+ ion in the antibonding orbital ψu relative to

the completely dissociated species H+ and H is given by

∆u = Eu − E1s =J −K

1− S


The Hydrogen Molecule Ion H2+: Theory vs Experiment


Calculated ExperimentalEquilibrium Separation (Re) 2.50 a0 = 1.32 A 1.06 ABinding Energy (kJ mol−1) 170 258

The simple molecular orbital wave function which was chosen does notquantitatively explain the bonding in the hydrogen molecule ion, but thiswave function can be improved by adding more terms.

The next step is to add terms for the 2s and 2p orbitals on the twonuclei.

ψ = c(1sA + 1sB) + c′(2sA + 2sB) + c′′(2pzA + 2zB)

In the limit of adding more terms, the results obtained from the solutionof the Schrodinger equation are in good agreement with experiment.


The Hydrogen Molecule Ion H2+: Potential Energy Curves


Calculated and experimental molec-ular potential energy curves for H2

+

showing the variation of energy asthe bond length is changed.


The Hydrogen Molecule Ion H2+: Orbital Interaction Diagram


It is useful to represent the energies of molecular orbitals by use of anorbital interaction diagram

• Since HAA and HBB are negative, and S is positive, Eg is more negative than Eu

and is the energy of the more stable molecular orbital.• Since ψg is symmetrical around the internuclear axis, it is referred to as a σ, since

it is even, it is referred to as gerade, and since it is made up of two 1s orbitals, it isdesignated as σg1s.

• The less stable orbital is represented by the σ∗u1s.


Molecular Orbital Description of the H2 Molecule


Using the Born-Oppenheimer approximation, the electronic Hamiltonianfor the hydrogen molecule may written as

H = − ~2

2me

(∇21 +∇2

2) +e2

4πǫ0

(− 1

rA1

− 1

rA2

− 1

rB1

− 1

rB2

+1

r12

)+

e2

4πǫ0R

where the coordinates are defined in the figure below




When the electronic Hamiltonian for the hydrogen molecule is used in

Hψe = Ee(R)ψe

An exact solution cannot be obtained because of the 1r12

term.

This is the reason that the LCAO-MO method is used to obtain anapproximate solution.

According to the LCAO-MO approach, molecular hydrogen is formed byputting two electrons with opposite spin in the σg1s orbital.




Each electron can be assigned to an orbital, and the electronic wavefunction for the molecule is the product of two wave functions for the twoelectrons:

ψMO = ψi(1)ψj(2)

where i and j designate the different orbitals, and 1 and 2 designate thetwo electrons.

According to the Pauli principle, two electrons with opposite spin can beassigned to a given spatial orbital.

As a first approximation, we will assume that in the ground state of thehydrogen molecule, the two electrons are placed in a σg orbital.




The electronic configuration of H2 will be described as (1σg)2.

The wave function for electron 1 in the 1σg molecular orbital isrepresented by

1σg(1) =1

[2(1 + S)]12

[1sA(1) + 1sB(1)]

The wave function satisfying the antisymmetric requirement is given bya Slater determinant, and so the approximate wave function for theground state of the hydrogen molecule is given by the following Slaterdeterminant:

ψMO[(1σg)2] =

1√2

∣∣∣∣1σg(1)α(1) 1σg(1)β(1)1σg(2)α(2) 1σg(2)β(2)

∣∣∣∣


The H2 Molecule: Theory vs Experiment


The approximate energy of the hydrogen molecule is obtained bycalculating the expectation of the Hamiltonian using this wave function

E =

∫ψ∗MO[(1σg)

2]HelψMO[(1σg)2]dτ

Calculated ExperimentalEquilibrium Separation (Re) 0.84 A 0.741 ADissociation Energy (kJ mol−1) 255 458

Although simple MO theory does account for a large proportion of thebinding energy of the H2 molecule, it has to be extended to yieldaccurate results.


Deficiency in the Approximate MO Wave function


One can see the deficiency in the approximate wave function bymultiplying out the spatial part:

1sA(1)1sA(2)︸︷︷︸ionic

+1sA(1)1sB(2) + 1sB(1)1sA(2)︸︷︷︸covalent

+1sB(1)1sB(2)︸︷︷︸ionic

• The first and last term correspond to forms of the hydrogenmolecule with ionic bonding: H−

AH+B and H+

AH−B.

• Thus, at R = ∞, the molecular orbital wave function describes astate which is 50 % H+ + H− and 50 % H ·+H· which is clearly notcorrect.

• This problem can be reduced by introducing variable coefficients c1and c2 in

ψ = c1(R)ψcovalent + c2(R)ψionic


Deficiency in the Approximate MO Wave function


Schematic E(R) curves for the LCAO-MO (redcurve) and exact (blue curve) for H2.

Note that the LCAO-MO curve dissociates to amixture of neutral and ionic species.

The MO curve shows that the model does notcorrectly describe H2 in the limit of large R.

The horizontal dashed line is the zero of en-ergy corresponding to two H atoms at infiniteseparation.

The improved results are given in the table be-low:

Equilibrium Separation (Re) 0.749 A 0.741 ADissociation Energy (kJ mol−1) 386 458

The inclusion of one variational parameterleads to significant improvement.


Molecular Orbitals for Many-Electron Atoms


In MO theory, an electron is associated with a wave function that isdelocalized over the entire molecule.

The MOs (σi) are constructed from linear combinations of atomicorbitals (AOs) leading to the term LCAO-MO Model.

The many-electron molecule must satisfy the Pauli exclusion principleleading to the expression of a wave function of an n-electron moleculeas a Slater determinant

ψ(1, 2, 3, . . . , n) =1√n!

∣∣∣∣∣∣∣∣

σ1(1)α(1) σ1(1)β(1) . . . σm(1)β(1)σ1(2)α(2) σ1(2)β(2) . . . σm(2)β(2). . . . . . . . . . . .σ1(n)α(n) σ1(2)β(2) . . . σm(n)β(n)

∣∣∣∣∣∣∣∣

where m = n2

if n is even, and m = n+12

if n is odd.


Molecular Orbitals for Many-Electron Atoms


Each of the entries in the Slater determinant is an MO that is expressedas a linear combination of AOs denoted by φi:

σi(1) =∑

j

cijφj(1)

The sum extends over AOs on all the atoms in a molecule, and thenumber in parentheses refers to the electron under consideration.

For example, in H2, both electrons are in the σ1 molecular orbital

σ1(1) = (φ1,1sA + φ1,1sB)


How Are Observables for Molecules Calculated?


Observables for a diatomic molecule such as the dipole moment, thebond dissociation energy, the bond length, and the ionization energy areultimately all related to the MOs and to the expansion coefficients cij.

Therefore, we need a method to determine the cij in a given MO.

• Assume that the atomic orbitals have been chosen.• The variational method is used to determine the optimal values for

cij.


Computing Molecular Orbital Energies


Although the optimization of the cij is best carried out by a computer, we can carry outa simple calculation by having one atomic orbital on each atom:

σ1 = c11φ1 + c12φ2

where φ1 and φ2 are AOs on the different atoms.

To determine the optimal values for these coefficients, the MO energy is minimizedwith respect to the values of the AO coefficients.

The expectation value of the energy is given by

< ǫ > =∫σ∗1 Hσ1dτ∫σ∗1σ1dτ

=∫[c11φ1+c12φ2]H[c11φ1+c12φ2]dτ∫(c11φ1+c12φ2)∗(c11φ1+c12φ2)dτ

=(c11)

2∫φ∗1Hφ1dτ+(c21)

2∫φ∗2Hφ2dτ+2c11c21

∫φ∗1Hφ2dτ

(c11)2∫φ∗1φ1dτ+(c12)2

∫φ∗2φ2dτ+2c11c12

∫φ∗1φ2dτ

< ǫ > = (c11)2H11+(c12)

2H22+2c11c21H12

(c11)2+(c12)2+2c11c12S12


Solving for Molecular Orbital Energies and Coefficients


To find the minimum energy, the derivatives of ǫ with respect to c11 and c21 are set tozero.

Differentiating this equation with respect to c11 and c12 equals

(2c11 + 2c12S12)ǫ+∂ǫ

∂c11((c11)

2 + (c12)2 + 2c11c12S12) = 2c11H11 + 2c12H12

(2c12 + 2c11S12)ǫ+∂ǫ

∂c12((c11)

2 + (c12)2 + 2c11c12S21) = 2c12H22 + 2c11H12

Since ∂ǫ∂c11

and ∂ǫ∂c21

are equal to zero for the minimum energy, the above equationscan be written as

c11(H11 − ǫ) + c12(H12 − ǫS12) = 0c11(H12 − ǫS12) + c12(H22 − ǫ) = 0

There is a nontrivial solution to these equations only if the determinant of thecoefficients is equal to zero

∣∣∣∣H11 − ǫ H12 − ǫS12

H12 − ǫS12 H22 − ǫ

∣∣∣∣ = 0


MO ǫi and cij for Homonuclear Diatomic Molecules


For a homonumclear diatomic molecule such as H2,Be2, · · · ,F2, H11 = H22

leading to the following two solutions

ǫ1 = H11+H12

1+S12

ǫ2 = H11−H12

1−S12

ǫ1 and ǫ2 correspond to the bonding and antibonding MO energies, respectively.

The ǫi can be obtained from the following formulas:

• Assume H11 and H22 can be approximated by the ionization energy of the neutralatom.

• H12 can be calculated using the approximation: H12 = −1.75S12

√H11H22

The coefficients cij can then be obtained from the following equations:

c11(H11 − ǫ) + c12(H12 − ǫS12) = 0c11(H12 − ǫS12) + c12(H22 − ǫ) = 0


The H2+ and H2 Molecules


For both the H2+ and H2 molecules, H11 =

H22 = -13.6 eV, the ionization energy of H.

A reasonable value is assumed for the over-lap: S = 0.30. Using these values yieldsǫ1 = -16.0 eV and ǫ2 = -9.2 eV.

• The energy of the bonding orbital is low-ered relative to H11 by 2.4 eV.

• The energy of the antibonding orbital israised relative to H11 by 4.4 eV.

The MO energy results for H2 can be con-veniently summarized in a molecular orbitalenergy diagram.


Heteronuclear Diatomic Molecule: HF


In this case H11 6= H22.

Let φ1 be a hydrogen 1s orbital withH11 = -13.6 eV, and let φ2 be a fluo-rine 2pz orbital with H22 = -18.6 eV.

The two MOs have the form

σ1 = C11φH1s + c12φF2pz

σ2 = C21φH1s + c22φF2pz

Given an overlap S = 0.30, H12 =-8.35 eV.

Solving the equation for the energiesgives ǫ1 = -19.6 eV (the bondingMO), and ǫ2 = -10.3 eV (the anti-bonding MO).


The HF Molecule: Obtaining Coefficients for the MOs


For ǫ1 = -19.6 eVc11(H11 − ǫ1) + c12(H12 − ǫ1S12) = 0

Substitution of the appropriate values gives

c11 (-13.6 + 19.6) + c12 (-8.35 + 0.30 x 19.6) = 0

which leads toc11c12

= 0.41

Using this result in the normalization equation

c211 + c212 + 2c11c12S12 = 1

yields c11 = 0.34, c12 = 0.84 and σ1 = 0.34φH1s + 0.84φF2pz .


Obtaining MO Coefficients for the HF Molecule


For ǫ2 = -10.3 eVc21(H11 − ǫ1) + c22(H12 − ǫ1S12) = 0

Substitution of the appropriate values gives

c21 (-13.6 + 10.3) + c22 (-8.35 + 0.30 x 10.3) = 0

which leads toc21c22

= -1.58

Using this result in the normalization equation

c221 + c222 + 2c21c22S12 = 1

yields c21 = 0.99, c22 = -0.63 and σ2 = 0.99φH1s − 0.64φF2pz .


Obtaining MO Coefficients for the HF Molecule


The relative magnitude of the coefficients of the AOs gives information about thecharge distribution in the molecule.

Consider an electron in the bonding MO of HF given by σ1 = 0.34φH1s + 0.84φF2pz.

The probability density of this orbital can be calculated as∫

|ψ2|dτ =

∫σ∗1σ1dτ = c211 + c212 + 2c11c12S12 = 1

This gives

• c211 + c11c12S12 = 0.21 is the probability of finding the e− around the H atom.• c212 + c11c12S12 = 0.79 is the probability of finding the e− around the F atom.• 2c11c12S12 = 0.17 is the probability that is shared between the H and F atoms.


The Molecular Orbital Energy Diagram


By convention, the AOs for the atoms are shown onthe left and right side of the diagram.

The MOs are generated by combining the AOs areshown in the middle.

The molecular energy diagram allows us to deter-mine the energies of the MOs relative to the AOs.

The bonding MO is lower in energy than the lowerof the two AOs, and the antibonding MO is higher inenergy than the higher of the two AOs.

Empirical rules can be used to obtain a reasonablesize basis set to describe the electronic structure offirst and second row diatomic molecules at a quan-titative level.

For H and He, only 1s orbitals are used; for Li → Ne,

the 2s, 2px, 2py , and 2pz atomic orbitals included


Effect of Symmetry on Overlap


Only orbitals of the same symmetry will combine with one another.

Nonzero overlap between two AOs occurs

• if both AOs are cylindrically symmetric with respect to the molecularaxis (σ MOs) or

• if both AOs have a common nodal plane that coincides with themolecular axis (π orbitals).


Overlap as a Function of R


The adjacent figure indicates how over-lap varies with distance for different or-bitals.

As discussed earlier, a nonzero overlapof atomic orbitals is necessary for chemi-cal bond formation.

The molecular bond lies on the z-axis.

The overlap of two s or two px and py or-bitals is maximized when R = 0, but theoverlap of an s orbital with a pz orbital hasit maximum value at a larger internucleardistance.


Symmetry Operations for Diatomic Molecules


All MOs of diatomic molecules can be divided into two groups with regard to rotationabout the molecular axis.

• If this rotation leaves the MO unchanged, the MO has σ symmetry. Combining sorbitals always gives rise to σ MOs for diatomic molecules.

• If the molecule has one nodal plane containing the molecular axis, the MO has πsymmetry. Combining πx or πy AOs always gives rise to π MOs.

The second operation of inversion through the center of the molecule is valid onlyfor homonuclear diatomic molecules.

• If this operation leaves the molecule unchanged, the MO has g symmetry:σ(x, y, z) → σ(−x,−y,−z).

• If σ(x, y, z) → −σ(−x,−y,−z), the MO has u symmetry.


Molecular Orbitals for Homonuclear Diatomic Molecules


Contour plots of several bonding and antibonding orbitals. The yellowarrows show the transformation σ(x, y, z) → σ(−x,−y,−z).If the amplitude changes sign under this transformation, it has usymmetry. If it is unchanged, it has g symmetry.


Symmetry Labels for Homonuclear Diatomic Molecules


Two different notations are commonly used to describe MOs inhomonuclear diatomic molecules.

1. The MOs are classified according to increasing energy.For example, a 2σg orbital has the same symmetry, but a higherenergy than the 1σg orbital.

2. The MOs are classified by the AOs from which they weregenerated.Thus, the σg(2s) MO has a higher energy than the σg(1s) MO.

A superscript ∗ is used to designate antibonding MOs for bothclassifications.


σ and π Orbitals for Diatomic Molecules


Two types of MOs can be generated from2p AOs.

1. If the axis of the 2p orbital lies onthe intermolecular axis (by conven-tion the z-axis), a σ MO is generated.This specific MO is called 3σg orσg(2pz).

2. Combining 2px or 2py orbitals oneach atom gives a π MO because theMO has a nodal plane containing themolecular axis.These MOs are degenerate inenergy and are called 1πu orπu(2px), πu(2py) MOs.


Mixing Atomic Orbitals to Form Molecular Orbitals


In principle, it is possible to take a linear combination of all the basisfunctions of the same symmetry (either σ or π) when constructing MOs.

However, an additional criterion based on orbital energies, reduces thenumber of AOs which mix into a given MO.

• For example, the mixing between 1s and 2s AOs for second rowdiatomics can be neglected at our level of discussion.

• However, for these same molecules, the 2s and 2pz AOs both have σsymmetry and will mix if their energies are not significantly different.

Since the energy difference between 2s and 2pz orbitals increases ingoing from Li → F, the s-p mixing decreases for second-row diatomicsin the order Li2, B2, . . . , O2, F2.


Labeling Molecular Orbitals of Diatomic Molecules


The MOs that used to describe chemical bonding in first and secondrow homonuclear diatomic molecules is shown in the table below.

The AO that is the major contributor to the MO is shown in the lastcolumn, and the minor contribution is shown in parenthesis.


Energy Ordering of Molecular Orbitals


It is useful to think of MO formation as a two step process:

1. create separate MOs from the 2s and 2p AOs.

2. combine the MOs of the same symmetry to create the final MOsthat include s− p mixing.

For the sequence H2 → N2, the orbital energies increase in thesequence

1σg < 1σ∗u < 2σg < 2σ∗

u < 1πu < 3σg < 1π∗g < 3σ∗

u

Moving across the periodic table to O2 and F2, the order of the 1πu andthe 3σg changes.

1σg < 1σ∗u < 2σg < 2σ∗

u < 3σg < 1πu < 1π∗g < 3σ∗

u


Orbital Interaction Diagrams for H2+ and H2


The interaction of 1s orbitals on eachatom gives rise to a bonding and anti-bonding MO as shown schematically inthe adjacent figure.

Each MO can hold two electrons of oppo-site spin.

The configurations of H2 and He2 are(1σg)

2 and (1σg)2 (1σ∗

u)2, respectively.

For H2 both electrons are in the 1σg or-bital which is lower than the 1s AOs.

In the MO model for He2, there are twoelectrons each in the 1σg and 1σ∗

u or-bitals. Because the 1σ∗

u is greater thanzero, He2 is not a stable molecule in thismodel.


Schematic MO Energy Diagram for the Valence Electrons in F2


F2 has s and p valence electrons, but thes − p mixing can be neglected becausethe 2s AO lies 21.6 eV below the 2p AO.

The configuration for F2 is

(1σg)2(1σu)

2(2σg)2(2σ∗

u)2(3σg)

2(1πux)2(1πuy)

2(1π∗gx)

2(1π∗gy)

2

For this molecule, the 2σ MO is essen-tially described by a single 2s AO oneach atom while the 3σ MO is describedquite well by a single 2pz AO on eachatom. The degenerate p and π are shownslightly offset in energy.

The dominant contributions in energy areshown in solid lines, and minor contribu-tions of s− p mixing are neglected.


Schematic MO Energy Diagram for the Valence Electrons in N2


N2 also has s and p valence electrons, but unlikeF2, the s−p mixing is not negligible because the 2sAO lies below the 2p AO by only 12.4 eV.Therefore, the 2σ and 3σ MOs have significant con-tributions from both the 2s and 2pz AOs.The configuration for N2 is

(1σg)2(1σu)

2(2σg)2(2σ∗

u)2(1πux)

2(1πuy)2(3σg)

2

Note that the mixing has changed the shape of the2σ and 3σ MOs of N2.The 2σ∗

u has become less antibonding, and the 3σg

MO has become less bonding for N2 in comparisonwith F2.The degenerate p and π are shown slightly offset inenergy.The dominant contributions in energy are shown insolid lines, and minor contributions of s − p mixingare neglected.From the overlaps of the MOs, the triple bond in N2

arises from the occupation of the 3σg , 1πux, and1πuy MOs.


MO Energy Levels for Second-Row Homonuclear Diatomics


The MOs are filled in sequence of increasing en-ergy, and the number of unpaired electrons for adiatomic molecule can be predicted.The energy of the MOs tends to decrease with in-creasing atomic number which is a result of largereffective charge and smaller atomic size lower-ing the corresponding AO energies.The 3σg orbital energy falls more rapidly than the1πu orbital because the s − p energy splitting in-creases in going from Li2 to F2.The mixing which occurs between the 2pz orbitalscomposing the 3σg orbital and the 2s orbitals of the2σg decreases.Because the 2px and 2py AOs don’t mix with the 2sAO, the 1πu remains constant across this series ofatoms resulting in a switch in the energy ordering ofthe 3σg and 1πu MOs for O2 and F2.Both B2 and O2 are predicted to have two unpairedelectrons and, thus, to have a net magnetic moment(paramagnetic molecule) which is in good agree-ment with experiment.


Bond Order, Bond Energy, and Bond Length


The concept of bond order can be defined as

bond order = total bonding electrons−total antibonding electrons2

The bond energy is expected to be very small for a bond order of zero.

The trend in bond energies and force constants shows the same trend as the bondorders.

For a given atomic radius the bond length is expected to vary inversely with bondorder.

• The trend is approximately followed for the series Be2 → N2.• The trend is not followed for H2 → Li2 because the atomic orbital changes from 1s

to 2s and because the He2 molecule is not really bonded.


Bond Order, Bond Energy, and Bond Length


By looking at the trend in bond orders,MO theory can provide an understand-ing of the trends in the binding en-ergies, bond lengths, and vibrationalforce constants for diatomic molecules.

As the number of electrons increases,the bond energy has a pronounced max-imum for N2 and a smaller maximum forH2.

The bond length also increases in the se-ries H2 → F2, but exhibits a more compli-cated trend for smaller molecules.

The vibrational force constant k showsthe same trend as the bond energyabove.


Heteronuclear Diatomic Molecules


New issues arise when chemical bond formation in heteronucleardiatomic molecules is considered.

1. When two interacting AOs of different energies interact, thecoefficients of the AOs in the MOs have different magnitudes.

• The coefficient of the lower energy AO is greater than that ofthe higher AO in the bonding MO.

• The opposite is true for the antibonding MO.

2. The g and u symmetries do not apply and the MOs are numbereddifferently:

Homonuclear 1σg 1σ∗u 2σg 2σ∗

u 1πu 3σg 1π∗g 3σ∗

u . . .Heteronuclear 1σ 2σ 3σ 4σ 1π 5σ 2π 6σ . . .


The Molecular Orbital Energy Diagram for HF


Schematic energy diagram showingthe AO and MO orbital energylevels for the valence electrons inHF.

The 3σ, 4σ, and 1π MOs for HF canbe described as follows:

The 2s electrons are almost completelylocalized on the F atom.

The 1π electrons are completely local-ized on the F atom because the 2px and2py orbitals have a zero overlap with the1s orbitals of H.

The admixture of s and p in the 4σ and5σ∗ MOs changes the electron distribu-tion somewhat when compared with ahomonuclear diatomic molecule.


Huckel MO Theory for Conjugated Molecules


Consider planar, conjugated hydrocarbons:

• acyclic polyene: CnHn+2

• monocyclic polyene: CnHn

• polycyclic polyenes: CnHn−2, CnHn−4, . . .

Electronic Configurations of Carbon and Hydrogen6C: 1s22s22p2 - the C atom contributes four valence electrons.1H: 1s1 - the H atom contributes one valence electron.

Octet RuleAtoms other than H tend to be surrounded by 8 electrons either throughbonding or sharing.


Huckel MO Theory for Conjugated Molecules


Lewis Structure of C2H4

C2H4 - total valence electrons = 4× 1 + 2× 4 = 12

We will be mainly concerned with the pz orbitals, since they are thepredominant influence on reactivity.The underlying σ electrons provide an effective potential for the πelectrons.


Huckel MO Theory for Planar Conjugated Molecules


We are primarily interested in obtaining the electronic energies and wave functions forplanar unsaturated molecules like benzene.

HΨ = EΨ

In planar unsaturated molecules, the π MOs will not mix with the σ MOs because of

differences in their symmetry

Ψ = ψσ × ψπ

H = Hσ + Hπ + Hσ π

We assume that we know ψσ and can then treat the σ system as merely providing partof the effective potential in which the π electrons move.

Heffπ (σ)ψπ = Eπψπ where Heff

π (σ) = Hπ + Eσ


Huckel MO Theory for Ethylene


In general, for n carbon atoms, there are n basis functions or n× pzorbitals:

ψj =n∑

r=1

crjφr

where ψj is MO j.Thus,

Hπψj MO = ǫjψj MO

As an example, the ethylene wave functions (ψπ, ψπ∗) are taken aslinear combinations of pz orbitals on each carbon center (LCAO-MO).

ψ1(ψπ) = c11φ2pz1 + c21φ2pz2 ψ2(ψπ∗) = c21φ2pz1 + c22φ2pz2

where φ2pz1 and φ2pz2 are pz orbitals on C1 and C2, respectively.


Variational Principle for Huckel Model Applied to Ethylene


ǫj =

∫ψjH

effπ ψjdτ∫

ψjψjdτ

ǫj =(c11φ1 + c21φ2)H

effπ (c11φ1 + c21φ2)dτ∫

(c11φ1 + c21φ2)(c11φ1 + c21φ2)dτ

The following definitions are given

H11 =∫φ1H

effπ φ1dτ H12 =

∫φ1H

effπ φ2dτ

S11 =∫φ1φ1dτ S12 =

∫φ1φ2dτ

∂ǫj∂c11

= 0 : c11(H11 − ǫjS11) + c21(H12 − ǫjS12) = 0∂ǫj∂c21

= 0 : c21(H11 − ǫjS12) + c22(H22 − ǫjS22) = 0(H11 − ǫjS11 H12 − ǫjS12

H21 − ǫjS12 H22 − ǫjS22

)(c11c21

)= 0


The Huckel Approximation


Hrr =∫φrHφrdτ = α Coulomb Integral

Hrs =∫φrHφsdτ =

{β, r bonded to s0, otherwise Resonance Integral

Srs =∫φrφsdτ =

{1, if r = s0, if r 6= s

Overlap Integral

According to the Huckel approximation, the previous set of linearequations for ethylene become:

(α− ǫj ββ α− ǫj

)(c11c21

)= 0


The Huckel Approximation for Ethylene


The secular determinant for ethylene becomes∣∣∣∣α− ǫ ββ α− ǫ

∣∣∣∣ = 0

Solution of this determinant leads to the following results:

• The energies of the two MOs are ǫ = α± β.

• c11 = c21 = 1/√2 for ǫ = α + β

• c11 = −c21 = 1/√2 for ǫ = α− β

The parameter α plays the role of reference energy because it is theenergy of a 2pz AO in the absence of delocalization.


The Huckel Approximation for Ethylene


Energy levels and molecular orbitals for the π electrons in ethylene.


The Huckel Model for the Allyl System


Secular Determinant for Allyl

∣∣∣∣∣∣

α− ǫ β 0β α− ǫ β0 β α− ǫ

∣∣∣∣∣∣= 0

Now divide both sides of the equation by β∣∣∣∣∣∣

α−ǫβ

1 0

1 α−ǫβ

1

0 1 α−ǫβ

∣∣∣∣∣∣= 0




Define−x =

ǫ− α

β∣∣∣∣∣∣

−x 1 01 −x 10 1 −x

∣∣∣∣∣∣= 0

The solution of the above determinant leads to the following values for x

x3 − 2x = 0 x = ±√2, 0




Eigenvalues

ǫ1 = α−√2β

ǫ2 = α

ǫ3 = α +√2β




Eigenvectors

The following is obtained from the secular equations:

− c11x3 + c21 = 0

c11 − c21x3 + c31 = 0

c21x3 − c31x3 = 0

The normalization condition:

c211 + c221 + c231 = 1




Eigenvectors

c11 = c31 =1

2; c21 =

2√2


Generalization of Huckel Model to A Planar Moleculewith n Basis Functions


H11 − ǫjS11 H12 − ǫjS12 . . . H1n − ǫjS1n

H21 − ǫjS12 H22 − ǫjS22 . . . H2n − ǫjS2n...

...Hn1 − ǫjSn2 Hn2 − ǫjSn2 . . . Hnn − ǫjSnn

c11c21...cn1

= 0

HCj = ǫjSCj

H =

H11 . . . H1n

......

Hn1 . . . Hnn

S =

S11 . . . S1n

......

Sn1 . . . Snn

E =

ǫ1 . . . 0...

. . ....

0 . . . ǫn

HC = ESC




Hrr =∫φrHφrdτ = α Coulomb Integral

Hrs =∫φrHφsdτ =

{β, r bonded to s0, otherwise Resonance Integral

Srs =∫φrφsdτ =

{1, if r = s0, if r 6= s

Overlap Integral

The overlap matrix is now a unit matrix:

S =

(1 . . . 0...

. . ....

0 . . . 1

)

HC = EC




Now start with the eigensystem for the Schrodinger equation

HC = EC

in the Huckel approximation

The exists a unitary transformation U such that

U† = U−1, UU−1 = 1

ThenUHU−1︸︷︷︸

D

UC︸︷︷︸C′

= E UC︸︷︷︸C′

DC′ = EC′

D =

λ1 . . . 0...

......

0 . . . λn

= E =

ǫ′1 . . . 0...

. . ....

0 . . . ǫ′n

λn = ǫ′n = −α−ǫn

β


The Huckel Model for the 1,3-Butadiene


H Matrix for 1,3-Butadiene

H =

0 1 0 01 0 1 00 1 0 10 0 1 0

Perform diagonalization using mathematical software.



Fall Semester 2013 81 Last Updated: October 20, 2013 at 8:18pm



The total π energy for a planar conjugated hydrocarbon is given as

Eπ =

occupied∑

j=1

νjǫj

where νj is the number of electrons (0, 1, 2) in MO j.

For Butadiene

Eπ(Butadiene) = 2(α + 1.61804β) + 2(α + 0.61804β) = 4α + 4.7216β

α and β are negative quantities.


Charge and Bond Order Matrix


ChargeThe π electron density on atom r is given as

qr = prr =

occupied∑

j=1

νjc2rj

n∑

r=1

qr = N (Total Number of π Electrons)


Bond OrderThe bond order between two atoms r and s is given as

prs =

occupied∑

j=1

νjcrj csj



Aromaticity of Benzene: Valence Bond Theory


In terms of valence bond theory the benzene molecule is described as a resonancehybrid of two canonical Lewis structures containing three isolated C=C double bonds.

Both of these canonical resonance structures are shown below in the figure below:

❜❜

✧✧❜❜

✧✧

❜❜

✧✧

H

H✧✧

H❜❜

H

H ✧✧

H❜❜ ❜❜

✧✧❜❜

✧✧

❜❜

✧✧ ❜❜

✧✧❜❜

✧✧ ❜❜

✧✧

ΨI ΨII

For Valence Bond theory, delocaalization energy points to an ambiguity because thetotal wave function is constructed from localized wave functions:

ΨTot = AΨI +BΨII


Aromaticity of Benzene: Molecular Orbital Theory


The molecular orbital description of benzene provides a delocalized picture in whichthe electron density is smeared out in a π cloud above and below the ring.

This can be emphasized by placing a circle inside the benzene ring

❜❜

✧✧❜❜

✧✧

✖✕✗✔

ΨTotal

The molecular orbitals are considered to delocalized over the entire molecule, and thisextra stability of planar aromatic conjugated hydrocarbons which is called theresonance delocalization energy can be calculated using Huckel theory.

From experiment it is known that benzene does not have three isolated C-C doublebonds but consists instead of a delocalized π electron system which produces anextra stability called resonance stabilization energy or aromaticity.


Delocalized Molecular Orbitals of Benzene and π-Energies



Delocalization Energy


In general, the heat of hydrogenation of an isolated double bond (essentially anethylene double bond) in butene is about 30.3 kcal/mol.

Adding the heat of hydrogenation of localized double bonds appears to be violatedwhen the double bonds are conjugated.

• Thus, the heat of hydrogenation for trans-1,3-butadiene is 57.1 kcal/mol comparedwith the 60.6 kcal/mol for a pair of ethylene double bonds.

• Since the hydrogenation in each case is butane, the energy difference of 3.5kcal/mol must be due to the greater stability of conjugated double bonds.

There is a theoretical parallel to the additivity of localized π double bonds versusdelocalized π bonds.

The Huckel Molecular Orbital (HMO) energy for butadiene is 4 α + 4.472 β.

For a pair of isolated double bonds, the Huckel energy of ethylene is doubled toobtain 4 α + 4 β.

The HMO method indicates that the double bonds of the butadiene molecule arestabilized by 0.472 β called the delocalization energy.


Huckel (4n + 2 ) Rule


Monocyclic planar systems of trigonally hybridized atoms that contain(4n + 2) p-electrons (where n is a non-negative integer) will exhibitaromatic character.

The rule is generally limited to n = 0 - 5.

This rule is derived from the Huckel MO calculation on planarmonocyclic conjugated hydrocarbons (CH)m where m is an integerequal to or greater than 3 according to which (4n + 2) p-electrons arecontained in a closed-shell system.

Systems containing 4n p-electrons such as cyclobutadiene and thecyclopentadienyl cation are antiaromatic and are not considered asstable.


Huckel (4n + 2 ) Rule


Examples of systems that obey the Huckel rule include:


Delocalization Energies of Several Cyclic Polyenes


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