Rogers Computational Chemistry Using the PC

174	COMPUTATIONAL CHEMISTRY USING THE PC

Exercise 6-1

Write the Hamiltonian for the helium atom, which has two electrons, one at a distance r1 and the other at a distance r2.

Solution 6-1

^	1 2		1 2	2		2		1
H ¼ 2r1			2r2		r1		r2	þ	r12

Notice the interelectronic repulsion term þ1=r12

These equations are mathematically identical to longer forms such as

^	h2	2		h2	2		Ze2		Ze2		1
H ¼	2m	r1		2m	r2		4pe0r1		4pe0r2	þ	4pe0r12

for helium.

Using atomic units, the Schroedinger equation for ground-state hydrogen is

21r2	1	1s ¼ E 1s	ð6-6Þ
	r

where r2 is now a kinetic energy operator in 3-space. We can write the Hamiltonian operator for the hydrogen atom in Cartesian 3-space, but the resulting Schroedinger equation is very difficult to solve. Instead, it is converted to spherical polar coordinates by routine but somewhat lengthy manipulations (Barrante, 1998). The Schroedinger equation in the new coordinate system is separated, and the three resulting equations are solved for R(r), ðyÞ, and ðfÞ. A similar procedure is followed to obtain the exact solution for the hydrogen molecule ion in confocal ellipsoidal coordinates.

By extension of Exercise 6-1, the Hamiltonian for a many-electron molecule has a sum of kinetic energy operators 12 r2, one for each electron. Also, each electron moves in the potential field of the nuclei and all other electrons, each contributing a potential energy V,

X X

^ ¼ 1r2 þ ð Þ

H 2 i Vi 6-7

where the terms Vi may be or þ according to whether the electron is attracted (to nuclei) or repelled (by other electrons). One can split up the attractive and repulsive terms

	¼ X½ 21 ri2		1
H^		þ Vi& þ	21 X rij	ð6-8Þ

where the Vi terms are now all negative and the repulsive sum is multiplied by 12 because we do not want to count repulsions twice, once as i-j repulsions and again as j-i repulsions. Equation (6-8) makes molecular problems look like nothing but a bunch of hydrogen atom problems plus a small interelectronic repulsion term. Unfortunately, this picture is deceptively simple, but it does lead to a useful approximation.

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2. The Independent Particle Approximation

By analogy to the atomic orbitals that Schroedinger gave us for the hydrogen atom, we assume that molecules will have orbitals too, that they will define certain electron probability distributions in space, and that they will have specific energies. We call them molecular orbitals. Indeed, the exact solution of the Schroedinger equation for the hydrogen molecule ion Hþ2 has exactly these characteristics, a lowenergy bonding orbital in which the electron probability density is relatively high between the nuclei and a high-energy antibonding orbital in which there is a node in electron probability density between the nuclei.

The Hamiltonian operator in Eq. (6-8) is called an n-electron Hamiltonian

^ ¼ ^ ð ; ; ; . . . ; Þ

H H 1 2 3 n

The reason the Schroedinger equation for molecules cannot be separated appears in the last term, 1 P 1 , involving a sum of repulsive energies between electrons. To

2rij

obtain this sum, or even one term of it, say for the ith electron, one must know exactly where the jth electron is. This is because repulsive force is dependent on distance. The position of the jth electron depends on the position of the ith electron, however, which is what we are trying to find. Knowing either of these positions exactly (as required by the problem) is a violation of the Heisenberg uncertainty principle for electrons having a kinetic energy within finite limits. The problem is insoluble.

In the independent particle approximation, the simplifying assumption is made that V0ðiÞ is an average potential due to a core that consists of the nuclei and all electrons other than electron i

H i i V i 6-9

^ ð Þ ¼ 1r2ð Þ þ 0ð Þ ð Þ

2

With this new approximation, the rij term does not appear [it is hidden in V0ðiÞ] and the Schroedinger equation becomes separable into n equations, one for each electron

^	ð6-10Þ
HðiÞcðiÞ ¼ EðiÞcðiÞ

^ ð Þ 0ð Þ

where H i includes the new potential energy term V i . This term is unknown (the rij problem hasn’t gone away); hence, no closed solution exists for Eq. (6-10). We are using the symbol c for a wave function for which we do not have an exact solution and c for an exact wave function. Throughout, there is an implicit assumption that c exists even though it may not be mathematically accessible.

The Basis Set

One way to obtain an approximate solution of Eq. (6-10) is to select any set of basis functions,

X

c ¼ aifi ð6-11Þ

176 COMPUTATIONAL CHEMISTRY USING THE PC

as an approximation to the true wave function. If the basis set of the fs is judiciously chosen c may be a good approximation to the true . As might be expected, a poorly chosen basis set gives a poor approximation to . After yet one more simplifying assumption, we will look at one way of choosing an

appropriate basis set and we will develop an iterative procedure to obtain the

P best c from any set aifi.

3. The p-Electron Separation Approximation

It has been known for more than a century that hydrocarbons containing double bonds are more reactive than their counterparts that do not contain double bonds. Alkenes are, in general, more reactive than alkanes. We call electrons in double bonds p electrons and those in the much less reactive C C or CH bonds s electrons. In Huckel theory, we assume that the chemistry of unsaturated hydrocarbons is so dominated by the chemistry of their double bonds that we may separate the Schroedinger equation yet again, into an equation for s electrons and one for p electrons. We assume that we can ignore the s electrons, as we did the nuclei, except for their contribution to the potential energy. We now have an equation of the same form as Eq. (6-8), but one in which the Hamiltonian for all electrons is replaced by the Hamiltonian for p electrons only

	n	1
	¼ X½ 21 r2ðiÞ þ VpðiÞ& þ
H^		21 X rij	ð6-12Þ

with somewhat different meanings for the symbols. Now, n is the number of p electrons of kinetic energy 12 r2ðiÞ, and the potential energy term VpðiÞ represents the potential energy of a single p electron in the average field of the framework of nuclei and all electrons except electron i. There is one p electron for each C atom participating in a system of C C double bonds. In Huckel theory, this is now the core potential.

In summary, we have made three assumptions 1) the Born–Oppenheimer approximation, 2) the independent particle assumption governing molecular orbitals, and 3) the assumption of p-s separation. The first two assumptions are characteristic of any molecular orbital theory, but the third is unique to the Huckel molecular orbital method.

The Huckel Method

In the late 1920s, it was shown that the chemical bond existing between two identical hydrogen atoms in H2 can be described mathematically by taking a linear combination of the 1s orbitals f1 and f2 of the two H atoms that are partners in the molecule (Heitler and London, 1927). When this is done, the combination

c ¼ a1f1 þ a2f2

ð6-13Þ

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is a new solution of the Schroedinger equation that has the characteristics of a chemical bond. Specifically, the energy calculated from Eq. (6-13) as a function of internuclear distance R goes through a minimum when R approximates the bond distance in H2. The Heitler–London method is known as the valence bond approximation (VB). The right side of Eq. (6-13) is called a linear combination of atomic orbitals (LCAO). The orbitals f1 and f2 are members of a small basis set. There is also a negative combination of f1 and f2 that produces an antibonding solution. Using the VB method, one arrives at a description of chemical bonding from a somewhat different logical premise from that of the molecular orbital (MO) method, but, in their more extended forms, the two methods approach each other, as they must, because they are attempts to describe the same thing. The LCAO method is only one of infinitely many ways that a molecular orbital can be approximated.

A few years later, Huckel (1931, 1932) showed that the LCAO approximation can be applied to the single electrons of the p atomic orbitals of the carbon atoms that are partners in a C C double bond. The p orbitals are considered to be independent of the s bonded framework except for the potential energy of charge interaction [Eq. (6-12)]. The linear combination is

X

c ¼ aipi ð6-14Þ

where the pi are the unhybridized p orbitals of the double-bonded carbon atoms (Fig. 6-2) over two or more conjugated carbon atoms. When this sum is taken for ethylene,

cþ ¼ a1p1 þ a2p2

ð6-15aÞ

and

c ¼ a1p1 a2p2

ð6-15bÞ

It is a property of linear, homogeneous differential equations, of which the Schroedinger equation is one, that a solution multiplied by a constant is a solution and a solution added to or subtracted from a solution is also a solution. If the solutions p1 and p2 in Eq. set (6-15) were exact molecular orbitals, cþ and c would also be exact. Orbitals p1 and p2 are not exact molecular orbitals; they are exact atomic orbitals; therefore, c is not exact for the ethylene molecule.

	H	π	H
	C		C	Figure 6-2 The p orbital of Ethylene. Both interactions ------ between pz
	H	π	H
				orbitals of carbon contribute to a single p orbital of ethylene.

178	COMPUTATIONAL CHEMISTRY USING THE PC

The Expectation Value of the Energy: The Variational Method

Premultiplying each side of the Schroedinger equation by c gives

c c ¼ c ^ c

E H

but E is a scalar; hence,

c c ¼ c ^ c

E H

for one electronic configuration in ethylene. For all electronic configurations, one must integrate over all space dt

ðð

E c c dt ¼		^
		c Hc dt
or	Ð cc2dt	ð6-16Þ
E ¼
	Hc dt

Ð

The term configuration is used in this context to designate a specific pair of locations for the two electrons in the p bond of ethylene. Even though we may not know exactly where the electrons are, we can know the probability density that they will be in a specific finite region of space by integrating over that region. From Lewis theory (see, e.g., Ebbing and Gammon, 1999) we are accustomed to thinking of an electron pair as constituting a chemical bond. The p bond in ethylene arises from such a pair of electrons, one each from the unhybridized pz orbitals of carbon, which are not engaged in s bonding.

It is a fundamental postulate of quantum mechanics that E in Eq. (6-16) is the expectation value of the energy for wave function c. If the values of are exact, E is exact. If the c values are approximate, as they are in this case, E is an upper bound on the true energy. Of two calculated E values, the higher one must be farther from the true value than the lower one, so it is discarded. Minimizing E, which is called the variational method, will be used to obtain the best value of c (the one that gives the lowest energy) from a given basis set. Criteria other than the energy may be selected, leading to different estimates of how closely c approximates its exact value. All properties of the system approach their true values as c approaches its exact value.

Exercise 6-2

In 1913 Bohr showed, by an argument that was essentially a combination of classical mechanics and quantum mechanics as it was known at that time, that the energy spectrum (ordered set of energy values) of hydrogen is given by

E ¼	me4	ð6-17Þ
	8e02n2h2

HUCKEL MOLECULAR ORBITAL THEORY I: EIGENVALUES

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where n ¼ 1 for the ground state and e0 ¼ 8:854 10 12 C2N 1m 2 is a physical constant called the vacuum permittivity. Substitute for the constants in the Bohr equation to obtain a ground state value for E. Give units.

Solution 6-2

The energy of an electron attracted to a proton is negative relative to infinite particle separation,

me4

9:109

10 31kg 1:602

10 19C

4

E

ð

Þ

¼ 8e02n2h2 ¼

8ð8:854 10 12C2N 1m 2Þ2ð6:626 10 34J sÞ2

¼ 2:179 10 18J

Units:

kg C4

¼

kg m N2m3

¼

N3m3

¼ N m ¼ J

C4N 2m 4J2s2

s2

J2

J2

where the charge is in coulombs (C). The vacuum permittivity e0, a constant from Coulomb’s law of force f resulting from charge q interaction f ¼ ð1=4pe0Þðq2=r2Þ (Young and Friedman, 2000), is used here essentially as a factor to convert the potential energy between two charges from coulombs2 per meter to joules. As indicated, the units of e0 are e0 : C2m 1=J ¼ C2N 1m 2. In atomic units, E ¼ 2:179 10 18J ¼ exactly 12 hartree.

Exercise 6-3

Schroedinger (1926) showed that the wave function or orbital for the hydrogen atom in its ground state can be written

¼ e ar

ð6-18Þ

where r is the radial distance between the proton and the electron and a contains several constants. Find the upper bound of the energy for the hydrogen atom by the variational method.

Solution 6-3

Choosing appropriate units for the charge, the Hamiltonian for radial motion can be written

^	h2 1 d						2 d		e2		ð6-19Þ
H ¼	2m		r2		dr	r		dr		r

where the form of the kinetic energy operator arises from the transformation of r2 from Cartesian coordinates to spherical polar coordinates (Barrante, 1998). This leads to an expression for the upper limit on E [Eq. (6-16)] from the wave function e ar

Ð

1

ar

2

E

¼

r dr

ar 2

2

Ð01ðe

Þ

4pr dr

0

e

2m 2

r

dr

r

e

4pr

dr

180	COMPUTATIONAL CHEMISTRY USING THE PC

where the factor 4pr2 dr accounts for the radial part of the Schroedinger equation. (The radius vector from the nucleus may be in any direction; hence, its point may be anywhere on the surface of a sphere of area 4pr2.) After factoring and canceling 4p,

Ð

1

ar

2 r

ar

E

¼

2m

r

dr

ar 2

r

ar

Ð01ðe

Þ

dr

0

e

2

r

dr

e

r

dr

The kinetic energy operator operating on e

gives

2m r2 dr r2 dr e ar ¼ 2m a2 ra e ar

h2

1 d d

h2

2

and ðe arÞ2¼ e 2ar, so

E

¼

Ð

2ar

2

dr

1

e

ar

h2

a

2

2a

e

ar

e2

e

ar

r

2

dr

0

2m

r

r

Ð0

Expanding the numerator we get

ð6-20Þ

ð6-21Þ

ð6-22Þ

E ¼

Ð

1

h2a2 2

2ar

þ 0Ð1 r2e 2ar dr

Ð

1 2

2ar

ð6-23Þ

0

2m

r

e

dr

0

m

r e

dr

0 e

r e

dr

Ð

This looks messy, but we really only need to evaluate two integrals, one with r and the other with r2. The integrals are of the known form

ð0

xne axdx ¼ anþ1

1

n!

so we get

4a3

¼ 2m

þ

m

ð Þ

¼

h2a2

1

h2a

1

2 1

2

2

2

þ m

e

h

a

h

a

a e2a

E

2m 4a3

4a2

4a2

6-24

1

or

h2a2

2

E ¼

e a

ð6-25Þ

2m

We can minimize the energy of the system with respect to the minimization parameter a by simply taking the first derivative and setting it equal to zero

	da	2m		e2a	¼	m e2	¼ 0
	d	h2a2				h2a
so that
a ¼		me2
		h2

ð6-26Þ

ð6-27Þ

HUCKEL MOLECULAR ORBITAL THEORY I: EIGENVALUES

181

(In more complicated cases, we shall have to verify that the extremum is a minimum.) Once we know that a ¼ me2=h2 we can substitute it back into the energy equation and find

E ¼ 2m e2a ¼

2m

h2

e2

h2

h2a2

h2

me2

2

me2

¼ 2

h2

h2

¼ 2

h2

ð6-28Þ

1

me4

me4

1

me4

or, for charge separation in a vacuum and E to be expressed in joules (which brings back 4pe0),

	¼ 2		ð4pe0Þ2			h2!		¼ 32p2e02		h2
E		1			me4				me4
Recognizing that h ¼ h=2p, we get
E ¼		me4								ð6-17Þ
		8e02h2

which is the energy expression we solved for in Exercise 6-2 with n ¼ 1 in the ground state. Going back to Eq. (6-27) m ¼ e ¼ h ¼ 1 in atomic units, so E ¼ 12 hartree.

The upper bound in Exercise 6-3 turns out to be exactly the energy of the hydrogen atom in its ground state. This should not come as a surprise, because we started with an exact ground-state orbital. In the general case we will not know but we will always be able to identify the better of two trial functions because it will give the lower energy. The simple hydrogen orbital e ar is not normalized. If it had been normalized, we would have had the form Ne ar where both N and a are collections of constants.

COMPUTER PROJECT 6-1 j Another Variational Treatment of the Hydrogen Atom

Part A. In Exercise 6-3 we found that the closed solutions for the integral Eq. (6-23)

E ¼

Ð

1

h2a2 2

2ar

þ 0Ð1 r2e 2ardr

Ð

1 2

2ar

ð6-23Þ

0

2m

r

e

dr

0

m

r e

dr

0 e

r e

dr

Ð

lead to the least upper bound for the energy of the hydrogen atom in the ground state: E ¼ 12 hartree. Instead of solving three integrals in the numerator and one in the denominator, carry out one numerical integration of the numerator and one integration of the denominator in the expression

E

¼ Ð

0

ð

01 r2e 2ardr

Þ

e

2ar

ð

6-29

Þ

2 a

r

2

þ ar r

dr

1

1 2

Ð

182	COMPUTATIONAL CHEMISTRY USING THE PC

at several values of a to find out which value of a gives the most negative value of E. Note that Eq. (6-29) is Eq. (6-23) with atomic units m ¼ e ¼ h ¼ 1. We shall expect E to be in atomic units of hartrees. In this determination, one can follow a procedure similar to the method of steepest descent used in iterative computer searches for an energy minimum. Pick two values of a and get the direction of descent by going from the more positive to the more negative of the two resulting energies. Now pick another a in the direction of descent and repeat this as many times as needed to find the minimum. Of course, the energy must be a well-behaved function of a for this to work (and it is). The size of the steps to be taken is determined by trial and error; if you overshoot the minimum, go back and take smaller steps.

To make an informed guess for your first value of a, you may wish to reread the section on the Bohr theory of the hydrogen atom and the Schroedinger wave functions for the hydrogen atom in a good physical or general chemistry book (see Bibliography).

A sample determination is

Mathcad

a :¼ 1:3

ð10

p :¼ :5 a2 r2 þ a r r expð 2 a rÞ dr

0 ð10

q :¼ r2expðt 2 a rÞ dr

0
p	E ¼ 0:455
E :¼ q

This is not Emin, of course; you must find the minimum energy by systematic variation of a. Alternatively, a QBASIC or TBASIC program can be written to integrate Eq. (6-28) by Simpson’s rule.

Complete Part A of this project by determining about 10 energies at various values of a over a range that is sufficient to prove that E is a well-behaved function of a with a minimum. Report the least upper bound of E and the value of a at which it is found.

Part B. Repeat the entire process of Part A using a Gaussian approximation to the wave function for the ground state of the hydrogen atom

c ¼ e g r2

ð6-30Þ

First decide what the integral equation corresponding to Eq. (6-29) is for the approximate wave function (6-30), then integrate it for various values of g. Report both g at the minimum energy and Emin for the Gaussian approximation function. This is a least upper bound to the energy of the system. Your report should include a

HUCKEL MOLECULAR ORBITAL THEORY I: EIGENVALUES

183

drawing of E as a function of g (SigmaPlot or equivalent) for enough values of g to make a clear picture of what the minimization function looks like and whether it is well behaved in the vicinity of Emin. Comment on the comparison between Emin for the ground state approximation function in Part A and Emin for the Gaussian approximation function in Part B.

Huckel Theory and the LCAO Approximation

Returning to Huckel theory for ethylene,								and substituting the first LCAO
[Eq. (6-15a)] for c, we have								ð		Þ
	¼ Ðð	þ a1p1 Þ		^	a2p2		2dt
E		a1p1	a2p2 Hða1p1				þ a2p2Þdt		6-31
		Ðð	þ			Þ

which is Eq. (6-16) with the linear combinations of atomic p orbitals substituted for the approximate molecular orbitals c. Expanding this equation yields four integrals in the numerator and four in the denominator. This takes a lot of space, so we use the notation

ðð

^	^	ð6-32aÞ
p1Hp1 dt ¼	p2Hp2 dt ¼ a

ðð

^	¼	^	ð6-32bÞ
p1Hp2 dt		p2Hp1 dt ¼ b
ð p1p1 dt ¼	ð p2p2 dt ¼ S11 ¼ S22		ð6-32cÞ

and

ð	ð
p1p2 dt ¼	p2p1 dt ¼ S12 ¼ S21	ð6-32dÞ

We have assumed that the order of the subscripts on the atomic orbitals p is immaterial in writing a, b, and S. In the general case, these assumptions are not self-evident, especially for b. The interested reader should consult a good quantum mechanics text (e.g., Hanna, 1981; McQuarrie, 1983; Atkins and Friedman, 1997) for their justification or critique.

The expression for the energy, after all assumptions and notational simplifications have been made, is

	E	a12a þ 2a1a2b þ a22a		6-33
		¼ a12S11 þ 2a1a2S12 þ a22S22	ð		Þ

If we could evaluate a, b, and S, which are called the coulomb, exchange, and overlap integrals respectively, we could compute E.