In the chapter on cyanine dyes, you learned several important concepts. You saw that a simple model, the particle in a box, could be used to understand the spectra of conjugated molecules. The model clearly showed the relationship of chain length and absorption wavelength and let you compute longest absorption wavelength by using the energy difference between the highest energy occupied molecular orbital, the HOMO, and the lowest unoccupied molecular orbital, the LUMO. You also learned that certain transitions were allowed and others not allowed and were able to predict these by using symmetry arguments and the transition moment integral.
In this chapter we will explore another model. This new model is called the Hückel model or Hückel Molecular Orbital (HMO) theory. Through HMO theory we will extend our understanding of molecular orbital methods. The HMO theory is an excellent vehicle for this extension as it employs some of the mathematical concepts and techniques that are used in current high-level ab initio molecular orbital calculations, which are considered in the next chapter.
We will use linear combination of atomic orbitals (the LCAO method) to form each molecular orbital. We will create and solve the secular equation for computing the energies of the molecular orbitals. Matrix methods, like those employed for vibrational normal mode analysis, will be used to solve the secular equation for large molecular systems. Charge densities for each atom in the molecule and bond orders for bonded pairs of atoms will be computed. Also, molecular orbitals will be used for computing chemically interesting properties such as the site of electrophilic or nucleophilic attack in a reaction.
12.1 Application to Conjugated Molecules
Conjugated double bonds are double bonds that alternate with single bonds along a molecular chain or in a molecular ring. Conjugated molecules are those that contain this pattern of alternating double and single bonds. These molecules have equivalent resonance structures and interesting chemical properties. Benzene is the prime example. Conjugated compounds were the first class of polyatomic molecules studied by molecular orbital theory because of their extended pi system of electrons. Pi electrons are those electrons found in pi orbitals. Pi orbitals are composed of pz atomic orbitals that are perpendicular to a plane of symmetry in the molecule. Pi orbitals change sign upon reflection in this plane, whereas sigma orbitals do not.
Several conjugated molecules are shown in Figure 12.1.1. Notice that they include both non-cyclic and cyclic molecules and heteroatoms such as nitrogen and oxygen. Hetero means different from carbon and hydrogen. One of the most important members of this group of molecules is benzene, which is given a special place in organic chemistry because of its stability.
Figure 12.1.1. Sample conjugated molecules. 
We can address a number of questions pertaining to benzene and other conjugated molecules using Hückel molecular orbital theory. What is the cause of benzene's stability? Can we quantify this stability? Since all the bonds in benzene have the same length, can we account for the fact that the bonds in benzene are characterized as intermediate between single and double bonds? Can we write molecular orbitals for benzene and systematically explain some of its chemistry using a quantitative molecular orbital model? Can we explain the chemical reactivity and spectroscopic properties of other diverse conjugated molecules with the same model?
Although the Hückel model is simple, is it sufficiently robust to predict trends and produce insights that are not possible through the very simple particle-in-a-box model that was used to describe the cyanine dyes. We will explore these questions in the following sections of this chapter and along the way introduce some of the mathematical tools needed in this molecular orbital approach and the molecular orbital methods used by scientists today. Just to keep things in perspective, remember that Hückel theory was developed at the time when computers were not available to chemists, and a simple mathematical approach was a very important development. Although HMO theory includes some drastic assumptions, it enabled the early calculations to be done with mechanical calculators or by hand.
12.2. Describing the Sigma and P Electrons Separately.
One of the important ideas used over and over again in Quantum Mechanics is the separation of variables. In the case of conjugated molecules, we have a chemical system where the separation of variables naturally falls out of our studies in organic chemistry. When one considers the pi and sigma electrons in a molecule like ethylene or 1,3-butadiene, it is easy to organize the electrons into sigma electrons and pi electrons. This separation follows from the molecular symmetry and the fact that hybrid atomic orbitals are used to describe bonds in molecules. The orbitals that form the sigma bonds are the sp2-hybrid orbitals on carbon and the s-orbital on hydrogen as shown in Figure 12.2.1. The 2pz orbital on carbon is perpendicular to the plane of these three hybrid orbitals and forms a pi bond through sideways overlap with a pz-orbital on a neighboring atom. This picture of bonding is very successful when carbon is bonded to three other atoms because it provides three orbitals that are highly directional.

Figure 12.2.1. Diagram of sp2 hybrid orbitals.
In quantum chemistry, this picture of hybridization and sigma/pi bonding is translated into a mathematical statement about the molecular orbitals in a molecule. The total wavefunction of a molecule is written as a product of a sigma part and a pi part.
12.2.1
where is the wavefunction describing the electrons
in sigma orbitals and
is the wavefunction describing the electrons
in pi orbitals. Such a product function would be an eigenfunction of the
molecular Hamiltonian if the sigma and pi electrons did not interact.
The wavefunction for the pi electrons then is described as a product of all the pi molecular orbitals.
12.2.2
Each ,
with j = 1…N, represents a molecular orbital,
i.e. a wavefunction for one electron moving in the electrostatic field of the
nuclei and the other electrons. Two
electrons with different spin are placed in each molecular orbital so that the
number of occupied molecular orbitals N is half the number of electrons, n,
i.e. N = n/2.
Each molecular orbital, ,
is written as a linear combination of atomic orbitals (LCAO).
with
, 12.2.3
where is the 2pz atomic orbital on atom r
of the conjugated pi system. The number
of molecular orbitals that one obtains by this procedure is equal to the number
of atomic orbitals. Consequently, the
indices j and r both run from 1 to N.
The cjr are the weighting coefficients for the atomic
orbitals in the molecular orbital. These
coefficients are not necessarily equal, or in other words, the orbital on each
carbon atom is not used to the same extent to form each molecular orbital. Different values for the coefficients give
rise to different net charges at different positions in a conjugated
molecule. This charge distribution is
very important when discussing spectroscopy and chemical reactivity.
The energy of the jth molecular orbital is given by a one-electron Schrödinger equation using an effective one electron Hamiltonian, heff, which expresses the interaction of an electron with the rest of the molecule.
12.2.4
is the energy eigenvalue of the jth
molecular orbital, corresponding to the eigenfunction
. The beauty of this method, as we will see
later, is that the exact form of heff is not needed. The total pi energy of the molecule is the
sum of the single electron energies.
12.2.5
where nj is the number of electrons in orbital j.
The expectation value expression for the energy for each molecular orbital is used to find εj and then Eπ.
. 12.2.6
The notation ,
which is called a bra-ket, just simplifies writing the expression for the
integral. Note that the complex
conjugate now is identified by the left-side position and the bra notation
and not by an explicit *.
After substituting Equation 12.2.3 into 12.2.6, we obtain for each molecular orbital
12.2.7
which can be rewritten as
12.2.8
where the index j for the molecular orbital has been dropped because this
equation applies to any of the molecular orbitals.
Exercise 12.2.1. Consider a molecular orbital made up of three atomic orbitals, e.g. the three carbon 2pz orbitals of the allyl radical, where the internuclear axes lie in the xy-plane. Write the LCAO for this MO. Derive the full expression, starting with Equation 12.2.6 and writing each term explicitly, for the energy expectation value for this LCAO in terms of heff. Compare your result with Equation 12.2.8 to verify that Equation 12.2.8 is the general representation of your result.
Exercise 12.2.2. Write a paragraph describing how the Variational Method could be used to find values for the coefficients cjr in the linear combination of atomic orbitals.
To simplify the notation we use the following definitions. The integrals in the denominator of Equation 12.2.8 represent the overlap between two atomic orbitals used in the linear combination. The overlap integral is written as Srs. The integrals in the numerator of Equation 12.2.8 are called either resonance integrals or coulomb integrals depending on the atomic orbitals on either side of the operator heff as described below.
is the overlap integral.
because we use normalized atomic
orbitals. For atomic orbitals r and s on
different atoms,
has some value between 1 and 0: the further
apart the two atoms, the smaller the value of
.
is the Coulomb
Integral. It is the kinetic and potential
energy of an electron in, or described by, an atomic orbital,
r,
experiencing the electrostatic interactions with all the other electrons and
all the positive nuclei.
is the Resonance
Integral or Bond Integral. This integral
gives the energy of an electron in the region of space where the functions
r
and
s
overlap. This energy sometimes is
referred to as the energy of the overlap charge. If r and s are on adjacent bonded atoms, this
integral has a finite value. If the
atoms are not adjacent, the value is smaller, and assumed to be zero in the
Hückel model.
In terms of this notation, Equation 12.2.8 can be written as
. 12.2.9
We now must find the coefficients, the c's. One must have a criterion for finding the coefficients. The criterion used is the Variational Principle. Since the energy depends linearly on the coefficients in Equation 12.2.9, the method we use to find the best set of coefficients is called the Linear Variational Method. This method is similar to the nonlinear variational method described in Chapter 10 for optimizing the parameters in a wavefunction.
12.3. The Linear Variational Method.
The Variational Principle tells us that the energy we compute for any trial wavefunction used in equation 12.2.6 will be greater than the exact energy for that Hamiltonian. In more mathematical terms, any energy we compute is an upper bound to the true energy for the molecular system. We write this statement as
12.3.1
If the trial wavefunction we use has adjustable parameters, then we can vary the parameters in a systematic way to find their optimum values. The optimum values give the lowest energy for the trial function. Initial guess values of the parameters will always give energies that are greater than the best energy that one could obtain from a given function unless you are extremely lucky and happen to guess the optimum values. This problem is equivalent to finding the minimum in a potential energy function or any other mathematical function. To accomplish this minimization, rather than making trial guesses, it is more efficient to use the calculus approach of taking the derivative of the energy function with respect to each parameter and setting each derivative equal to zero. If you have two parameters, you then will have two simultaneous equations to solve. If you have N parameters you will have N simultaneous equations.
The task is to minimize the energy with respect to all the
coefficients by solving the N simultaneous equations produced by
differentiating Equation 12.2.9 with respect to each coefficient.
for t
= 1, 2, 3, … N 12.3.2
Actually we also should differentiate Equation 12.2.9 with respect to the ,
but this second set of N equations is just the complex conjugate of the first
and produces no new information or constants.
To carry out this task, rewrite Equation 12.2.9 to obtain Equation 12.3.3 and then take the derivative of Equation 12.3.3 with respect to each of the coefficients.
12.3.3
Actually we don't want to do this differentiation N times, so consider the
general case where the coefficient is ct. Here t represents any number between 1 and N.
This differentiation is relatively easy, and the result, which is
shown by Equation 12.3.4, is relatively simple because some terms in Equation
12.3.3 don't involve ct and others depend linearly on ct. The derivative of the terms that don't
involve ct is zero (e.g. ). The
derivative of terms that contain ct is just the constant factor that
multiples the ct, (e.g.
).
Consequently, only terms in Equation 12.3.3 that contain ct
contribute to the result, and whenever a term contains ct, that term
appears in Equation 12.3.4 without the ct because we are
differentiating with respect to ct.
The result after differentiating is
. 12.3.4
If we take the complex conjugate of both sides, we obtain
. 12.3.5
Since ε = ε*, ,
and
,
this equation can be reversed and written as
12.3.6
or upon rearranging as
. 12.3.7
There are N simultaneous equations that look like this general one; N is the number of coefficients in the LCAO. Each equation is obtained by differentiating
Equation 12.3.3 with respect to one of the coefficients.
Exercise 12.3.1. Explain why the energy ε = ε*, show that (write out the integral expressions and take
the complex conjugate of Srt), and show that
(write out the integral expressions, take the
complex conjugate of Hrt, and use the Hermitian property of quantum
mechanical operators).
Exercise12.3.2: Rewrite your solution to Exercise 12.2.1 3 for the 3-carbon pi system found in the allyl radical in the form of Equation 12.3.3 and then derive the set of three simultaneous equations for the coefficients. Compare your result with Equation 12.3.7 to verify that Equation 12.3.7 is a general representation of your result.
This method is called the linear variational method because the variable parameters affect the energy linearly unlike the shielding parameter in the wavefunction that was discussed in Chapter 10. The shielding parameter appears in the exponential part of the wavefunction and the effect on the energy is nonlinear. A nonlinear variational calculation is more laborious than a linear variational calculation.
Equations 12.3.6 and 12.3.7 represent a set of homogeneous linear equations. As we discussed for the case of normal mode analysis in Chapter 6, a number of methods can be used for solving these equations to obtain values for the energies, ε's, and the coefficients, the cr’s.
Matrix methods are the most convenient and powerful. First we write more explicitly the set of simultaneous equations that is represented by Equation 12.3.6. The first equation has t=1, the second t=2, etc. N represents the index of the last atomic orbital in the linear combination.
12.3.8
This set of equations can be represented in matrix notation.
12.3.9
Here we have square matrix H and S multiplying a column vector C' and a scalar ε. Rearranging
produces:
|
|
12.3.10 |
Exercise 12.3.3. For the three atomic orbitals you used in Exercises 12.2.1 and 13.3.2, write the Hamiltonian matrix H, the overlap matrix S, and the vector C'. Show by matrix multiplication according to Equation 12.3.9 that you produce the same Equations that you obtained in Exercise 12.3.2.
The problem is to solve these simultaneous equations, or the matrix equation, and find the orbital energies, which are the ε's, and the atomic orbital coefficients, the c's, that define the molecular orbitals.
Exercise 12.3.4. Identify two methods for solving simultaneous equations and list the steps in each.
12. 4. The Hückel Method
In the Hückel method only the atoms that are part of the conjugated bond system and their pz-π orbitals are used in the LCAO. To greatly simplify the calculation, some approximations are introduced for the Coulomb, resonance, and overlap integrals. While we could evaluate the overlap integrals in Equation 12.3.7, there is no way we can evaluate the other integrals involving the effective Hamiltonian operator since we cannot write an explicit expression for this operator. It is only a conceptual tool. The HMO method treats these integrals as parameters that are evaluated empirically by fitting the theory to experimental results.
The Coulomb integrals like H11, H22, etc represent the energy of an electron on a particular atom in the molecule. These integrals are all taken to be same because the carbon atoms in a pi system are similar, though not identical, and are represented by the symbol α.
The resonance integrals like H12, H13, H23, etc are taken to be zero unless the atoms are neighbors, i.e. bonded together by a sigma bond. For these cases, it is assumed that the resonance integrals are all equal and are represented by the symbol β.
Since the atomic orbitals are normalized, the Sii
=1. Although the overlap between
functions on neighboring atoms is around 0.3, all the Sij with are taken to be zero.
With these approximations, Equations 12.3.8 can be rewritten in matrix form as
12.4.1
where β’s
only appear for neighboring atoms.
Exercise 12.4.1: Write the equivalent of Equation 12.4.1 for the case of the allyl radical that you have been describing in previous exercises e.g. Exercise 12.3.3.
We now need to say a bit more about the relationship of the matrix
in Equation 12.4.1 to the structure of the molecule that it is describing. For any molecule, you could label the carbon atoms with integer numbers: 1,2,3,4, etc. These numbers then correspond to the rows and columns in the H matrix. The diagonal elements (H11, H22, H33, etc.) refer to particular carbon atoms (1,2,3, etc., respectively). The off-diagonal elements (H12, H13, H23, etc.) refer to pairs of atoms, 1 and 2, atoms 1 and 3, atoms 2 and3, etc., respectively. If any two atoms are bonded neighbors, then the corresponding matrix element is β, otherwise it is 0.
It also is common in HMO theory, to divide both sides of Equation 12.4.1 by β and make the following substitution.
. 12.4.2
Equation 12.4.1 then becomes
12.4.3
Rearranging produces
12.4.4
This matrix equation now is in the form of an eigenvalue problem, where x is the eigenvalue and the column vector of coefficients, C', is the eigenvector. The square matrix of ones and zeros is called the topology matrix, T, because it specifies the topology or connectedness of the molecule. The elements of the topology matrix are 1 if the corresponding atoms are bonded or connected and 0 otherwise.
Putting Equation 12.4.4 into matrix notation produces
TC′ = C′
x 12.4.5
where T is the topology matrix and C′
is the eigenvector of coefficients for the x eigenvalue. More generally we can write, similar to what
was done in Chapter 6 for the vibrational normal mode analysis,
TC=CX 12.4.6
where C is a square matrix where
columns are the eigenvectors, one of which was C′,
and X is a diagonal matrix with the eigenvalues
x as the diagonal elements. The columns
in matrix C contain the coefficients
of the atomic orbitals used in the LCAO approximation.
If we left multiply each side of Equation 12.4.6 by the inverse of C we obtain
C-1TC=X. 12.4.7
The problem then is to find the transformation matrix C that diagonalizes T. Computer routines generally are available for
this purpose.
At this point it becomes informative and possible to consider a real molecule. As an example take the butadiene molecule. We need only write down the topology matrix, T, and ask a computer to produce the eigenvalues and eigenvectors. The eigenvalues are values for x, which we then use in Equation 12.4.2 in order to solve for ε in terms of α and β.
Numbering the 4 carbon atoms in butadiene 1 through 4 from left to right, produces the following T matrix, and Mathcad gives us the eigenvalues and eigenvectors. The resulting energies and molecular orbitals are given in Table 12.4.1 in order of increasing energy. The energies are expressed in terms of α and β. All the information about bonding is in the β parameter. A numerical value for β is not needed to answer many questions. When a numerical value is needed, it is obtained from spectroscopic or thermodynamic data.
12.4.8

Table 12.4.1. Hückel Energies and Wavefunctions for Butadiene
|
Energy |
Molecular Orbital |
|
ε1 = α + 1.618 β |
ψ1 = 0.372 |
|
ε2 = α + 0.618 β |
ψ2 = 0.602 |
|
ε3 = α - 0.618 β |
ψ3 = 0.602 |
|
ε4 = α - 1.618 β |
ψ4 = - 0.372 |
Exercise 12.4.2. Identify the number of nodes in each of the butadiene orbitals along a line above the plane of the molecule. What is the correlation between the number of these nodes and the energy of the orbital? Which mo is lower in energy, ε1 or ε2? Why?
Exercise 12.4.3. Write the topological matrices for the allyl radical, benzene, and azulene.
Exercise 12.4.4. Use matrix diagonalization to determine the Hückel energies and wavefunctions for the allyl radical, benzene, and azulene. Construct energy level diagrams, and identify the bonding, antibonding, and non-bonding orbitals for each of these molecules.
Exercise 12.4.5. Based on the values of the atomic orbital coefficients, sketch graphs of the amplitudes of the molecular orbitals along a line above the plane of the allyl radical. Draw a diagram of the allyl radical and for each of the molecular orbitals show the location of the nodes by lines and the sign of the wavefunction by + and - signs.
Exercise 12.4.6. Draw a hexagon to represent benzene and for each of the six pi molecular orbitals show the location of the nodes by lines and the sign of the wavefunction by + and - signs.
12.5 Molecular Properties and Chemical Reactivity.
One of the important things molecular orbital models and chemical computations do is provide the chemist with insights into the reactivity of molecules. At the most basic level we get insight by just knowing the charge on the atom. If an atom in a molecule is slightly negative then electrophiles are attracted to it. Similarly nucleophiles are attracted to positive centers in a molecule.
The charge at an atom in a molecule is computed by summing up the contribution of all of the molecular orbitals to the electron density at that atom. The contribution of a MO to the electron density at an atom is given by the square of the coefficient of atomic orbital used for the atom in the LCAO for that MO. The total electron density is then the sum of all contributions from all occupied molecular orbitals. This sum is written as
12.5.1
where qr is the total charge at atom r and crj is the
coefficient from molecular orbital j for the atomic orbital of atom r. The sum is taken over all occupied molecular
orbitals and nj is the number of electrons in an occupied molecular
orbital. The coefficient squared is called the atomic population (of atomic orbital r from one electron in
molecular orbital j), and sum qr is called the net atomic population (of atomic orbital r from all the electrons
in occupied molecular orbitals).
Exercise 12.5.1. Using the data in Table 12.4.1, computer the total pi electron charge at each atom of butadiene. What do you observe and what would be the consequences for chemical reactivity?
Exercise 12.5.2. Using the results of Exercise 12.4.4, compute the total pi electron charge at each atom in the allyl radical, in benzene, and in azulene. What does the result tell you about preferred sites for nucleophilic and electrophilic attack in these molecules?
Azulene and benzene are different in that the former is a nonalternant hydrocarbon and latter is an alternant hydrocarbon. You can see this difference by placing a star next to every other (or alternating) carbon in benzene and in azulene. In azulene you can not do this labeling, you must eventually put two stars together. All alternant hydrocarbons have equal charge distributions like benzene. All nonalternant hydrocarbons have unequal charge distributions like azulene. All alternate hydrocarbons have equal charge distributions like benzene.
Exercise 12.5.3. Write brief statements characterizing the pi charge distribution in an alternate hydrocarbon and in a nonalternant hydrocarbon.
Another property that we can compute from the LCAO coefficients is the bond order. This property gives the strength of the pi bonding between pairs of adjacent atoms. The pi bond order, prs, between two atoms r and s is computed by summing the product of the coefficients between pairs of atoms.
12.5.2
where crj is the coefficient of the atomic orbital for atom r in
molecular orbital j.
Exercise 12.5.4. Using the data in Table 12.4.1 and the results of Exercise 12.4.4, compute the pi bond orders for the allyl radical, benzene, and azulene. Do all bonds in each of these molecules have the same bond order? Which bonds have greater double bond character and which have lesser double bond character? Discuss the results with respect to the nature of the molecular orbitals. What is the consequence of this with respect to chemical reactivity?
Predicting the stability of molecules is often a key in a successful synthesis. An important step in this direction was taken by Hückel when, by using his theory, he showed that monocylclic conjugated polyenes have the greatest stability and degree of aromaticity when they obey the “4n+2” rule. In this rule, n is an integer and 4n+2 gives the number of pi electrons in the molecule. Benzene follows this rule for n=1. This rule is so powerful and elegant in its simplicity that it successfully predicted the stability of the cyclopentadienyl anion.
The stability of a molecule is determined to a first approximation by the number of bonding orbitals that are doubly occupied. Benzene has six electrons in three bonding pi molecular orbitals. Cyclobutadiene on the other hand has 2 electrons in one bonding orbital and 2 electrons in two degenerate nonbonding molecular orbitals. Cyclobutadiene is very unstable and has only transient existence at extremely low temperatures.
Exercise 12.5.5. Determine the resonance stabilization energy for benzene by comparing the total energy of the 6 pi electrons in benzene to three times the energy for the two pi electrons in ethylene. What conclusion can you draw from this calculation?
The HMO method has been surprisingly successful. It led to many insights about chemical reactivity. Many calculations were possible to do by hand, and with the arrival of computers large and larger hydrocarbons were studied. Various corrections were made to the method to accommodate heterocyclic atoms in the conjugated systems. Other corrections were made to account for the interaction of pi electrons with the sigma framework. Correlations were made between the reactivity of sites in a molecule and computed parameters based on the coefficients of the molecular orbitals. The rapid growth in computing power quickly made more extensive and accurate calculations possible. It became possible to include integrals in the calculations and diagonalize larger matrices. The Extended Hückel Method is a result of these developments.
12.6. The Extended Hückel Method
The Extended Hückel Molecular Orbital Method (EH) grew out of the need to consider all valence electrons in a molecular orbital calculation. By considering all valence electrons, chemists could determine molecular structure, compute energy barriers for rotation about bonds, and even determine energies and structures of transition states for reactions. The computed energies could be used to choose between proposed transitions states to clarify reaction mechanisms.
In the EH method one writes the wavefunction as a product of a valence wavefunction and a core wavefunction.
. 12.6.1
The total valence electron wavefunction is described as a product of the one-electron wavefunctions.
12.6.2
where n is the number of electrons and j identifies the molecular orbital as
was done for the HMO method. Each
molecular orbital is again is given as an LCAO
12.6.3
where now the r
are the valance atomic orbitals chosen to include the 2s, 2px, 2py,
and 2pz of the carbons and heteroatoms in the molecule and the 1s
orbitals of the hydrogens. The set of
orbitals defined here is called a basis
set. Since this basis set contains
only the atomic-like orbitals for the valence shell of the atoms in a molecule,
it is called a minimal basis set.
In the EH method we use an effective one electron Hamiltonian, heff, and then we proceed with determining the energy of a molecular orbital as we did above for the simple HMO method. Using the same notation as in Section 12.2 we have again
12.6.4
where and
.
Minimization of the energy with respect to each of the
coefficients again yields a set of simultaneous equations just like Equation
12.3.7.
12.6.5
As before, these equations can be written in matrix form
. 12.6.6
Equation 12.6.6 accounts for one molecular orbital. It has energy ε, and it is defined by the elements in the C' column vector, which are the coefficients that multiply the atomic orbital basis functions in the linear combination of atomic orbitals.
As before (see Equation 12.4.6 and the vibrational analysis problem in chapter 6), we can write one matrix equation for all the molecular orbitals.
HC=SCE 12.6.7
where H is a square matrix
containing the Hrs, the one electron energy integrals, and C is the matrix of coefficients for the
atomic orbitals. Each column in C is the C' that defines one molecular orbital in terms of the basis
functions. In extended Hückel theory,
the overlap is not neglected, and S
is the matrix of overlap integrals. E is the diagonal matrix of orbital
energies. All of these are square
matrices with a size that equals the number of atomic orbitals used in the LCAO
for the molecule under consideration.
As you should recognize from the discussion of Hückel Molecular
orbital theory and the normal coordinate analysis in Chapter 6, Equation 12.6.7
represents an eigenvalue problem. For
any extended Hückel calculation
we need to set up these matrices and then find the eigenvalues and
eigenvectors. The eigenvalues are the
orbital energies, and the eigenvectors are the atomic orbital coefficients that
define the molecular orbital in terms of the basis functions.
Exercise 12.6.1. What is the size of the H matrix for HF? Write out the matrix elements in the H matrix using symbols for the wavefunctions appropriate to the HF molecule. Consider this matrix and determine if it is symmetric by examining pairs of off-diagonal elements. In a symmetric matrix, pairs of elements located by reflection across the diagonal are equal, i.e. Hrc = Hcr where r and c represent the row and column, respectively. Why are such pairs of elements equal? Write out the S matrix in terms of symbols, showing the diagonal and the upper right portion of the matrix. This matrix also is symmetric, so if you compute the diagonal and the upper half of it, you know the values for the elements in the lower half. Why are pairs of S matrix elements across the diagonal equal?
12.7. Extended Hückel Calculation Details.
The elements of the H matrix are assigned using experimental data. This approach makes the extended Hückel method a semi-empirical molecular orbital method. The basic structure of the method is based on the principles of physics and mathematics while the values of certain integrals are assigned by using educated guessing and experimental data. The Hrr are chosen as valence state ionization potentials with a minus sign to indicate binding. The values used by R. Hoffmann when he developed the extended Hückel technique were those of H.A. Skinner and H.O. Pritchard (Trans. Faraday Soc. 49 (1953), 1254). These values for C and H are listed in Table 12.7.1. The values for the heteroatoms (N, O, and F) are taken from Pople and Beveridge (reference). The Hrs values are computed from the ionization potentials according to
12.7.1
The rationale for this expression is that the energy should be proportional to
the energy of the atomic orbitals, and should be greater when the overlap of
the atomic orbitals is greater. The
contribution of these effects to the energy is scaled by the parameter K. Hoffmann assigned the value of K after a
study of the effect of this parameter on the energies of the occupied orbitals
of ethane. The conclusion was that a
good value for K is K=1.75.
Exercise 12.7.1. Fill in numerical values for the diagonal elements of the Extended Hückel Hamiltonian matrix for HF using the ionization potentials given in Table 12.7.1.
The overlap matrix also must be determined. The matrix elements are computed using the
definition where
r
and
s
are the atomic orbitals. Slater-type
orbitals (STO’s) are used for the atomic orbitals rather than hydrogenic
orbitals because integrals involving STO's can be computed more quickly on
computers. Slater type orbitals have the
form
12.7.2
where zeta, ,
is a parameter describing the screened nuclear charge. In the extended Hückel calculations done by
Hoffmann, the Slater orbital parameter ζ was 1.0 for the H1s and 1.652 for
the C2s and C2p orbitals.
Exercise 12.7.2. Describe the difference between Slater-type orbitals and hydrogenic orbitals.
Overlap integrals involve two orbitals on two different atoms or
centers. Such integrals are called
two-center integrals. In such integrals there are two variables to consider,
corresponding to the distances from each of the atomic centers, rA
and rB. Such integrals can be
represented as
12.7.3
but elliptical coordinates must be used for the actual integration. Fortunately the software that does extended
Hückel calculations contains the programming code to do overlap integrals. The interested reader will find sufficient
detail on the evaluation of overlap integrals and the creation of the
programmable mathematical form for any pair of Slater orbitals in Appendix B4
(pp. 199 - 200) of the book Approximate
Molecular Orbital Theory by Pople and Beveridge. The values of the overlap integrals for HF
are given in Table. 12.7.2.
Exercise 12.7.3. Using the information in Table 12.7.2, identify which axis (x, y, or z) has been defined as the internuclear axis. Fill in the missing values in Table 12.7.1; which requires no calculation, only insight.
Table 12.7.2 Overlap Integrals for HF
|
|
F 2s |
F 2px |
F 2py |
F 2pz |
H 1s |
|
F 2s |
|
|
|
|
0.47428 |
|
F 2px |
|
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0 |
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F 2py |
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0.38434 |
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F 2pz |
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0 |
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H 1s |
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Exercise 12.7.4. Using the information in Tables 12.7.1 and 12.7.2 write the full Hückel H matrix and the S matrix that appears in Equation 12.6.7 for HF.
Our goal is to find the coefficients in the linear combinations of atomic orbitals and the energies of the molecular orbitals. For these results, we need to transform Equation 12.6.7, HC=SCE, into a form that allows us to use matrix diagonalization techniques. We are hampered here by the fact that the overlap matrix is not diagonal because the orbitals are not orthogonal. Mathematical methods do exist that can be used to transform a set of functions into an orthogonal set. Essentially these methods apply a transformation of the coordinates from the local coordinate system describing the molecule into one where the atomic orbitals in the LCAO are all orthogonal. Such a transformation can be accomplished through matrix algebra, and computer algorithms for this procedure are part of all molecular orbital programs. The following paragraph describes how this transformation can be accomplished.
If the matrix M has an inverse M-1 then MM-1 = 1, and we can place this product in a matrix equation without changing the equation. When this is done for Equation 12.6.7, we obtain
H MM-1C=SMM-1CE 12.7.4
Next multiply on the left by M-1and
determine M so the product M-1SM
is the identity matrix, i.e. a matrix that has
1's on the diagonal and 0's off the diagonal is the
case for an orthogonal basis set.
M-1HMM-1C=M-1SMM-1CE
12.7.5
which then can be written as
H"C"=C"E 12.7.6
where C"=M-1C. The identity matrix is not included because
multiplying by the identity matrix is just like multiplying by the number
1. It doesn’t change anything. The H"
matrix can be diagonalized by multiplying on the left by the inverse of C" to find the energies of the
molecular orbitals in the resulting diagonal matrix E.
E = C"-1H"C"
The matrix C" obtained in
the diagonalization step is finally back transformed to the original coordinate
system with the M matrix, C=M C" since C"=M-1C.
Fortunately this process is automated in some computer software. For example, in Mathcad, the command genvals(H,S) returns a list of the eigenvalues for Equation 12.6.7. These eigenvalues are the diagonal elements of E. The command genvecs(H,S) returns a matrix of the normalized eigenvectors corresponding to the eigenvalues. The i-th eigenvalue in the list goes with the i-th column in the eigenvector matrix. This problem, where S is not the identity matrix, is called a general eigenvalue problem, and gen in the Mathcad commands refers to general.
Exercise 12.7.5. Using your solution to Exercise 12.7.4, find the orbital energies and wavefunctions for HF given by an extended Hückel calculation. Construct an orbital energy level diagram, including both the atomic and molecular orbitals, and indicate the atomic orbital composition of each energy level. Draw lines from the atomic orbital levels to the molecular orbital levels to show which atomic orbitals contribute to which molecular orbitals. What insight does your calculation provide regarding the ionic or covalent nature of the chemical bond in HF?
12.8. Mulliken Populations
Mulliken populations (R.S. Mulliken, J. Chem. Phys. 23, 1833,
1841, 23389, 2343 (1955)) can be used to characterize the electronic charge
distribution in a molecule and the bonding, antibonding, or nonbonding nature
of the molecular orbitals for particular pairs of atoms. To develop the idea of these populations,
consider a real, normalized molecular orbital composed from two normalized
atomic orbitals.
12.8.1
The charge distribution is described as a probability density by the square of
this wavefunction.
12.8.2
Integrating over all the electronic coordinates and using the fact that the
molecular orbital and atomic orbitals are normalized produces
12.8.3
where Sjk is the overlap integral involving the two atomic orbitals.
Mulliken's interpretation of this result is that one electron in
molecular orbital ψι contributes to the electronic charge in atomic orbital
j,
to the electronic charge in atomic orbital
k,
and 2cijcikSjk to the electronic charge in the
overlap region between the two atomic orbitals.
He therefore call
and
,
the atomic-orbital populations, and
2cijcikSjk, the overlap population. The
overlap population is >0 for a bonding molecular orbital, <0 for an
antibonding molecular orbital, and 0 for a nonbonding molecular orbital.
It is convenient to tabulate these populations in matrix form for
each molecular orbital. Such a matrix is
called the Mulliken population matrix. If there are two electrons in the molecular
orbital, then these populations are doubled.
Each column and each row in a population matrix is corresponds to an
atomic orbital, and the diagonal elements give the atomic-orbital populations,
and the off-diagonal elements give the overlap populations. For our example, Equation 12.8.1, the
population matrix is
. 12.8.4
Since there is one population matrix for each molecular orbital,
it generally is difficult to deal with all the information in the population
matrices. Forming the net population matrix decreases the
amount of data. The net population matrix is the sum of all
the population matrices for the occupied orbitals.
12.8.5
The net population matrix gives the atomic-orbital populations and overlap
populations resulting from all the electrons in all the molecular
orbitals. The diagonal elements give the
total charge in each atomic orbital, and the off-diagonal elements give the
total overlap population, which characterizes the total contribution of the two
atomic orbitals to the bond between the two atoms.
The gross population matrix
condenses the data in a different way.
The net population matrix combines the contributions from all the
occupied molecular orbitals. The gross
population matrix combines the overlap populations with the atomic orbital
populations for each molecular orbital.
The columns of the gross population matrix correspond to the molecular
orbitals, and the rows correspond to the atomic orbitals. A matrix element specifies the amount of charge,
including the overlap contribution, that a particular molecular orbital
contributes to a particular atomic orbital.
Values for the matrix elements are obtained by dividing each overlap
population in half and adding each half to the atomic-orbital populations of
the participating atomic orbitals. The
matrix elements provide the gross charge that a molecular orbital contributes
to the atomic orbital. Gross means
that overlap contributions are included.
The gross population matrix therefor also is called the charge matrix for the molecular orbitals. An element of the gross population matrix (in
the jth row and ith column) is given by
12.8.6
where Pi is the population matrix
for the ith molecular orbital,
Pijj is the atomic-orbital
population and the Pijk
is the overlap population for atomic orbitals j and k in the ith
molecular orbital.
Further condensation of the data can be obtained by considering atomic and overlap populations by atoms rather than by atomic orbitals. The resulting matrix is called the reduced-population matrix. The reduced population is obtained from the net population matrix by adding the atomic orbital populations and the overlap populations of all the atomic orbitals of the same atom. The rows and columns of the reduced population matrix correspond to the atoms.
Atomic-orbital charges are obtained by adding the elements in the rows of the gross population matrix for the occupied molecular orbitals. Atomic charges are obtained from the atomic orbital charges by adding the atomic-orbital charges on the same atom. Finally, the net charge on an atom is obtained by subtracting the atomic charge from the nuclear charge adjusted for complete shielding by the 1s electrons.
Exercise 12.8.1. Using your results from Exercise 12.7.5 for HF, determine the Mulliken population matrix for each molecular orbital, the net population matrix, the charge matrix for the molecular orbitals, the reduced population matrix, the atomic orbital charges, the atomic charges, the net charge on each atom, and the dipole moment. Note: The bond length for HF is 91.7 pm and the experimental value for the dipole moment is 6.37 x 10-30 C m.
Activity 12.1
This Mathcad document (Activity 12.1 Formaldehyde.mcd) summarizes an Extended Hückel calculation on formaldehyde and provides the data you need to complete the following analysis. The calculation was done using the HyperChem software package. The geometry for formaldehyde was provided by a HyperChem SCF 6-31G** calculation. Such ab initio calculations are discussed in the next chapter. A minimum in the calculated energy for formaldehyde was found at a CH bond length of 109.3 pm, a CO bond length of 118.4 pm, and a HCH bond angle of 115.7 degrees.
(1) Use the Coordinate Table to determine the orientation of the molecule with respect to the Cartesian coordinate system. Draw a diagram of the molecule relative to the axes, and label and number the atoms according to the numbering used in the table.
(2) Use the Overlap Matrix to identify the signs of the overlap integrals for the carbon and oxygen px and py orbitals with the two hydrogen 1s orbitals. Add the carbon and oxygen px and py orbitals to your diagram. Explain in terms of your diagram why these four overlap integrals have the signs that are given in the Overlap Matrix.
(3) In the Overlap Matrix, each number in the first row is either 1, 0, a positive number between 0 and 1, a negative number between 0 and 1, or almost 0. Explain for each of these matrix elements in the first row specifically why such values are obtained.
(4) From the Hückel Hamiltonian matrix identify the values that are used for the atomic-orbital valence ionization energies for C, O, and H in this calculation.
(5) Explain why all the elements in the column for PzO1 in the Hückel Hamiltonian matrix are 0 except for two, and show that the correct values for those two indeed are -14.80000 and -5.27682 eV.
(6) Use the Eigenvector Matrix to draw sketches of the molecular orbitals numbered 3, 4, and 5. Base your sketches on the coefficients in the Eigenvector Matrix. Identify which of these orbitals are CH bonding and which are CO bonding. Note that the molecular orbitals are numbered 0 through 9 not 1 through 10.
(7) Calculate the atomic-orbital populations and the overlap populations for the molecular orbitals numbered 3 and 5 from data in the Eigenvector Matrix and the Overlap Matrix. What do the values for these populations tell you about the electronic charge on each atom, in the overlap region, and the bonding, nonbonding, or antibonding characteristics of these orbitals? What happens to the bond lengths if an electron is removed from molecular orbital number 3? What happens to the bond lengths if an electron is removed from molecular orbital number 5?
(8) Use the eigenvectors to identify each of the 10 molecular orbitals as π or σ.
(9) Draw an energy-level diagram of the atomic and molecular orbitals. Draw dotted lines connecting the molecular-orbital levels to the atomic-orbital levels from which the molecular orbitals are derived. Are all the π molecular orbitals higher in energy than the σ molecular orbitals? Is the highest energy molecular orbital σ or π? Explain why the molecular orbitals have the energy ordering that they do.
(10) Write the ground-state electronic configuration for formaldehyde identifying the molecular orbitals by their symmetry.
(11) How many degenerate molecular orbitals are there?
(12) Determine the partial charge at each atom due to electrons in the highest occupied molecular orbital.
(13) For molecular orbital number 3, verify that the entries in the Gross Population Matrix for the Molecular Orbitals tabulate the gross charge (including the overlap contributions) that a particular molecular orbital contributes to an atomic orbital.
(14) What is the physical significance of the elements in the Reduced Population Matrix. Use the Reduced Population Matrix to verify that the sum of all the atomic orbital populations and overlap populations is equal to the number of valence electrons.
(15) Verify that that the first (1.718) and last (0.977) values in the list of atomic orbital charges are correct given the values in the Gross Population Matrix.
(16) What is the physical meaning of the net atomic charge? What is the net charge at each atom calculated for formaldehyde? Discuss the chemical reactivity of formaldehyde in terms of this charge distribution.
(17) Calculate the dipole moment for formaldehyde predicted by this calculation in units of debye (D), which are the traditional units for dipole moments. Which end of the molecule is negative? One debye is equal to 10-18 esu cm where esu is a non-SI unit for electric charge. The charge of a proton is 4.803x10-10 esu. What is the conversion factor between D and C m (coulomb meters, which is the SI unit for a dipole moment)?
(18) How many molecular orbitals are produced in an Extended Hückel calculation on acetone?
Problems
Problem 12.1. Use the results in Table 12.5.1 for butadiene and symmetry arguments to determine the electric dipole allowed transitions between electronic energy levels in the molecule.
Problem 12.2. Compute the resonance energy for stilbene. Compare your results to the resonance energy for benzene. What is the significance of this with respect to the presence of the double bond in the center of the molecule? Include the structure of stilbene.
Problem 12.3. The compound shown below has never been synthesized. Use HMO theory to explain why.
Problem 12.4. Use a computational chemistry software package of your choice and draw the structure of ethane. Compute the Extended Hückel energy for ethane as you rotate one methyl group in increments of 10°. Plot the results. How large is the potential energy barrier to rotation about the CC single bond in ethane. Repeat the calculations for n-butane rotating about the central CC bond. Interpret your data in terms of potential energy barriers and identify the most stable conformations.
Problem 12.5. Draw cyclohexane in both the chair and boat conformation. Use an Extended Hückel calculation to determine the most stable conformation. How does this result compare to what you learned in organic chemistry? Calculate the energy of the planar geometry, which should approximate the transition state. What temperatures correspond to the energy differences between the stable conformations and the transition state? In the liquid state, do you expect cyclohexane to convert freely between the two forms at some temperature?
Problem 12.6. Benzo-[a]-pyrene has been extensively studied because of the connection between this compound and cancer caused by soot and smoke. A proposed intermediate step in the metabolic conversion of this compound to a highly mutagenic metabolite is through epoxidation. Use the Hückel method to predict potential sites of epoxidation of benzo-[a]-pyrene and the likely hood of reaction at each potential site.
Problem 12.7. Consider the formic acid molecule and the internal H-bond shown here as a suitable model for the transition state for proton transfer from one oxygen to the second oxygen in the gas phase. Compute the energy of this transition state and compare it to the energy for the ground state structure of formic acid. Above what temperature should such proton transfer occur freely?

Problem 12.8. Carry out an extended Hückel calculation on the Series A cyanine dyes listed in Activity 5.1. Compare the experimental transition energies with your results and determine which model (particle-in-a-box or extended Hückel) best accounts for the electronic energy level structure of these dyes.
Problem 12.9. Write the topological matrices for the following molecules: ethylene, cyclobutadiene, and the cyclopentadienyl anion. Diagonalize the topological matrices to determine the Hückel energies and wavefunctions, construct an energy level diagram, sketch pictures of the molecular orbitals, and identify the bonding, antibonding, and non-bonding orbitals for each molecule. Discuss the stability of these moleulules using resonance energy as a criterion in comparison to benzene.
Self-study and Review Questions
· How are values in the Hückel Hamiltonian matrix determined, and how are the eigenvalues and eigenvectors derived from this matrix?
· How are the atomic orbital populations and the overlap populations calculated for a molecular orbital, and what is their physical meaning?
· How are values in the Mulliken population matrices obtained?
· How are values in the Net Population Matrix obtained?
· How are values in the Charge Matrix (aka Gross Population Matrix) obtained?
· How are values in the Reduced Population Matrix obtained?
· How are Atomic Orbital Charges obtained?
· How are the Atomic Charges and the Net Charges determined?
· How does the overlap population characterize a molecular orbital as bonding, antibonding or nonbonding?
· How can you determine the amount of charge that a particular molecular orbital contributes to a bond between two atoms?