CHE 110B: Handout 110a Review

8/18/2019 CHE 110B: Handout 110a Review

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CHE 110B 110A Review

1 � Dirac (Bra-ket) Notation

Wavefunctions:ψ →

|ψi

, ψ∗

→ hψ| , and |ψ|2 = ψ∗ψ

→ hψ| ψ

i.

Expectation values:

hÂi =Z ψ∗(~ r)Âψ(~ r)d~ r → hψ| Â |ψi .

The Schrödinger Equation in Dirac notation:

Ĥ |ψi = E |ψiĤ = T̂+ V̂ = − ~

2

2m∇2 + V̂

∇2 = ∂ 2

∂ x2 +

∂ 2

∂ y2 +

∂ 2

∂ d2

2 � Postulates of Quantum Mechanics (McQ Ch 3)

1. The state of a quantum system is completely specified by its wavefunction |ψi.2. To every observable quantity in classical mechanics, there exists a Hermitian quantum mechanical

operator (e.g., kinetic energy in CM is p2/2m, in QM is p̂2/2m = − ~2m∇2).3. Observable quantities in quantum mechanics are eigenvalues of their corresponding operators: Â |φni =

An |φni.4. The set of eigenfunctions of an operator are orthogonal: hφn| φmi = δ nm. The Kronecker delta δ nm is

1 if n = m, and zero otherwise. As a secondary point, the set of all eigenfunctions of an operator form

a complete set. So if |φni (n = 0, 1, 2 . . .) are the set of eigenfunctions of an operator ˆA, then anyother state in the same space can be expressed as a linear combination of these functions:

|ψi =Xn

cn |φni ,

where cn are complex numbers.5. If two operators do not commute, then their corresponding observables cannot be measured simul-

taneously to arbitrary precision. The most famous example is position and momentum: [x̂, p̂x] =x̂p̂x − p̂xx̂ = i~ , and as a result ∆ px∆x ≥ ~ /2.

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Wavefunctions (a) and probability densities (b) for a particle in a one-dimensional box. Note that the number of nodes isn− 1.

The one-dimensional particle in a box is sometimes used to estimate the energy of a π electron in a conjugatedpolyene.

3.2 | Particle in a 3D box

It is straightforward to show (see McQ 3-9) that a particle in a 3D box is simply an extension of the 1Dproblem. The wavefunction ψ(x,y,z) can be written as a product of independent one-dimensional functionsφ(x)φ(y)φ(z), and the energy can consequently be shown to be E = E x + E y + E z. This is actuallyan extremely important point– anytime a wavefunction can be expressed as a product of 1D functions, theindividual components can be treated separately, and the total energy computed from a sum of the individualenergies.

The particle in a 3D box, therefore, has (normalized) wavefunctions:

ψ(x,y,z) =

s 8

LxLyLzsin

nxπx

Lxsin

nyπy

Lysin

nzπz

Lznx, ny, nz = 1, 2, . . . (3)

and energies

E nx,ny,nz = h2

8m

n2xL2x

+n2yL2y

+ n2zL2z

! nx, ny, nz = 1, 2, . . . (4)

In the special case of a particle in a cube, Lx = Ly = Lz, and we find that many of the states have the sameenergy (i.e., they are degenerate). The energies are

E nx,ny,nz = h2

8mL2

n2x + n2y + n

2z

(5)

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so they only depend on the sum of the squares of the x, y, and z quantum numbers. If we denote thewavefunctions in Dirac notation as |nx, ny, nzi, the |1, 1, 2i, |1, 2, 1i, and |2, 1, 1i states all have energy(6h2)/(8mL2). These degeneracies arise naturally as a result of the symmetry of the system, and we willexplore this concept in more detail when we talk about group theory.

Energy levels and degeneracies of a particle in a cube.

The particle in a 3D box can be used to model the confinement energy of electrons as a function of nanoparticlesize.

3.3 | Harmonic oscillator

The harmonic oscillator is the model of a particle in a quadratic potential or, equivalently, two particleswhose relative displacement sits in a quadratic potential. From the standpoint of chemistry, the harmonicoscillator is the simplest model for the vibration of a diatomic molecule (and even vibrations of the normalmodes of a polyatomic molecule). If the potential between the two bodies is truly quadratic, then theSchrödinger equation can be solved exactly. However, any potential that has a smooth minimum can beapproximated as a harmonic potential near that minimum. For any potential V (x) where x = l − l0 (i.e.,the displacement relative to the separation at minimum potential l0), a Taylor expansion gives

V (x) = V (0) +

dV

dx

0

x + 1

2!

d2V

dx2

0

x2 + 1

3!

d3V

dx3

0

x3 + . . . (6)

The first term can be arbitrarily set to 0, and the second term is 0 because at x = 0, the derivative of thepotential is zero (it’s a minimum). Therefore, as long as x is small, we an truncate the series at the quadratic

term and define the force constant k =d2V dx2

0

:

V (x) ≈ 12

kx2.

For the quantum harmonic oscillator involving two bodies with reduced mass µ = m−11 + m−12 and force

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constant k with displacement x, the Hamiltonian is

Ĥ = − ~ 2

2µ

d2

dx2 +

1

2kx2,

and the Schrödinger equation tells us that the wavefunctions ψ(x) must be the solutions of

d2ψ

dx2 +

2µ

~ 2

E − 1

2kx2

ψ = 0.

It turns out that the solutions have the general form

ψv(x) = N vH v(α1/2x)e−αx

2/2, v = 0, 1, 2, . . . (7)

where

α =

r kµ

~ 2 , N v = (2

vv!)−1/2

α

π1/4

and H v(α1/2x) are the Hermite polynomials, which are a series of polynomials of order v . It is conventionalto use v as the quantum number for a harmonic oscillator.

If we define the angular frequency of the oscillator ω =p

k/µ or, alternatively, its frequency ν =p

k/µ/(2π),the energies are given by

E v = ~ ω

v +

1

2

= hν

v +

1

2

, v = 0, 1, 2 . . . (8)

A unique feature of the harmonic oscillator is that its energy levels are evenly spaced, and that the lowestenergy level is greater than the minimum of the potential. This means that even at absolute zero, theharmonic oscillator still has nonzero kinetic and potential energy, which is called zero-point energy.

Wavefunctions (a) and probability densities (b) of the harmonic oscillator, along with energies.

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3.4 | Rigid rotor

Model of a rigid rotor.

The rigid rotor is the model system for both a rotating diatomic molecule and for the angular distributionof the electron in a hydrogen atom. This model starts with two bodies with reduced mass µ and a fixedseparation r. Since there is no preferred orientation of the body in free space, the Hamiltonian only has akinetic energy term:

Ĥ = T̂ = − ~ 2

2µ∇2.

Because we’re dealing with rotation, it’s more natural to work in spherical coordinates, and after sometedious calculus it can be shown that:

∇2 = 1

r2∂

∂ r

r2 ∂

∂ rθ,φ

+ 1

r2 sin θ

∂

∂θ

sin θ ∂

∂θr,φ

+ 1

r2 sin2 θ ∂ 2∂φ2

r,θ

In the rigid rotor, r is constant, so we can eliminate the first term to get

∇2 = 1r2 sin θ

∂

∂θ

sin θ

∂

∂θ

r,φ

+ 1

r2 sin2 θ

∂ 2

∂φ2

r,θ

(9)

Recall from physics that angular momentum L = I ω, where I is the moment of inertia and ω is the angularfrequency. Relative to the center of mass, I = m1r

21 + m2r

22, and consequently in the center-of-mass frame

I = µr2. The kinetic energy of a rotating body is T = L2/(2I ) (just like the linear quantity T = p2/(2m)).If we now substitute equation (9) into the Hamiltonian, we see

Ĥ = −~ 2

2I

" 1

r2 sin θ

∂

∂θ

sin θ

∂

∂θ

r,φ

+ 1

r2 sin2 θ

∂ 2

∂φ2

r,θ

#, (10)

and thus we find the quantum mechanical form of the square of the angular momentum operator:

L̂2 = −~ 2"

1

r2 sin θ

∂

∂θ

sin θ

∂

∂θ

r,φ

+ 1

r2 sin2 θ

∂ 2

∂φ2

r,θ

#. (11)

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The eigenfunctions of the Hamiltonian (and L̂2) are the spherical harmonics Y ml (θ,φ), where l = 0, 1, 2, . . .and m = 0, ±1, ±2, . . . , ±l. These spherical harmonics have the form:

Y m

l (θ,φ) = s (2l + 1)(l − |m|)!

4π(l + |m|)! P

|m|

l (cos θ)eimφ, (12)

where P |m|l (cos θ) are the associated Legendre functions.

The energies of the rigid rotor depend only on the l quantum number

E l = ~ 2

2I l(l + 1), l = 0, 1, 2, . . . , (13)

and each energy level has a degeneracy of 2l + 1 arising from the possible values of the m quantum number.While the quantum numbers l and m (often ml) are used when describing the angular portion of the hydrogenatom wavefunction, typically the quantum numbers become J and mJ when discussing molecular rotation.They are equivalent.

Energy level diagram of a rigid rotor. The gaps between successive levels are evenly spaced in energy.

As mentioned before, L̂2 commutes with Ĥ, so the square of the total angular momentum has a definite value(l is a good quantum number). Looking at its components L̂y, L̂y, and L̂z, we find that only [L̂

2, L̂z] = 0,

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while [L̂i, L̂j ] 6= 0 if i 6= j . This means that we can know the total angular momentum l and its projectionm onto one three axes (under the coordinate system we’ve chosen, that axis is z, but diff erent coordinatechoices are possible).

Vector representation of total angular momentum l and its projection ml onto the z axis. (a) is a two-dimensional illustration,while (b) illustrates the uncertainty in the projection of the angular momentum onto the x and y axes.

| Hydrogen atom

For the hydrogen atom, we let the nucleus be defined as the center of our coordinate system, and theHamiltonian becomes:

Ĥ = − ~ 2

2me∇2 − e

2

4π0r,

where r is the electron distance, ∇2 is in spherical coordinates as before, e is the charge of the electron, andthe Coulombic term is negative to indicate that the proton-electron potential is attractive (i.e., the protonand electron have charges that are opposite in sign). The factor 4π0 comes from using SI units; the factordisappears if using atomic units. The solution to this equation is tedious, but it turns out that the methodof separation of variables can be used to solve this system in terms of radial and angular components:

ψ(r,θ,φ) = R(r)Y (θ,φ).

The angular solutions are the same as those for the rigid rotor: the spherical harmonics Y ml (θ,φ), so the

hydrogen atom wavefunctions have quantum numbers l and ml. You are already familiar with these fromgeneral chemistry, l = 0 corresponds to an s orbital, l = 1 is a p orbital, and so on.

It is found that the energy of the hydrogen atom depends only on a single quantum number n, even thoughfull specification of the wavefunction also requires the quantum numbers l and ml:

E n = − e2

8π0a0n2, n = 1, 2, . . . , (14)

where a0 is the Bohr radius. The allowed quantum numbers are n = 1, 2, 3, . . ., 0 ≤ l ≤ n, and 0 ≤ |ml| ≤ l.

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4 � Approximation methods

There are very few chemical systems for which the Schrödinger equation can be solved exactly, so mostof the time approximations must be made. The most common approximation is the Born-Oppenheimerapproximation, which says that the motion of the electrons is fast compared to the motion of the nuclei. Underthe Born-Oppenheimer approximation, nuclear coordinates are held fixed and the electron system is solvedexactly (one-electron systems) or approximately (multi-electron systems), and then the nuclear coordinatesare systematically varied and the electronic system recalculated. In this manner, a potential energy surface(PES) is created, giving the potential energy of the system as a function of the nuclear coordinates, and thePES can be used to solve the vibrational and rotational parts of the molecular wavefunction. The Born-Oppenheimer approximation is specific for molecules; the approximation methods that follow are general forquantum mechanics.

4.1 | Variational principle

The variational principle states that for a given operator, the expectation value of that operator with anynormalized function will be greater than or equal to the lowest eigenvalue of that operator. For a molecularHamiltonian, that means that the energy calculated with an arbitrary wavefunction will provide an upperbound for the ground state energy of the molecule. In equation form, the variational principle states that if |ψi is a trial function, then

hψ| Ĥ |ψihψ| ψi ≥ E 0, (15)

where E 0 is the lowest eigenvalue of Ĥ. The trial function may have adjustable parameters which can bevaried to minimize the calculated energy and therefore improve the estimate; this is usually done usingspecialized software packages.

A special type of linear variational calculation is called the Rayleigh–Ritz method. In this method, the trial

function is simply a linear combination of orthogonal basis functions

|ψi =Xn

cn |φni .

The solution to this problem involves calculating the cn coefficients that minimize the variational energy.After some calculus, it can be shown that this is essentially an eigenvalue problem, and that the values of the coefficients can be obtained by solving the secular determinant

hφ0| Ĥ |φ0i − E hφ0| φ0i hφ0| Ĥ |φ1i − E hφ0| φ1i · · · hφ0| Ĥ |φni − E hφ0| φnihφ1| Ĥ |φ0i − E hφ1| φ0i hφ1| Ĥ |φ1i − E hφ1| φ1i · · · hφ1| Ĥ |φni − E hφ1| φni

......

. . . ...

hφn| Ĥ |φ0i − E hφn| φ0i hφ1| Ĥ |φ1i − E hφn| φ1i · · · hφn| Ĥ |φni − E hφn| φni

= 0 (16)

Note that if the basis functions |φni are normalized, then hφn| φmi = δ nm, and the equation simplifies to

hφ0| Ĥ |φ0i − E hφ0| Ĥ |φ1i · · · hφ0| Ĥ |φnihφ1| Ĥ |φ0i hφ1| Ĥ |φ1i − E · · · hφ1| Ĥ |φni

......

. . . ...

hφn| Ĥ |φ0i hφ1| Ĥ |φ1i · · · hφn| Ĥ |φni − E

= 0 (17)

This equation will result in n values for E , and for each value E k, there will be a set of n coefficients ck thatgive rise to that energy. If you take this argument further, you’ll see that this is the general matrix formalism

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of quantum mechanics: wavefunctions |ψi can be represented by vectors of coefficients cn in the space of orthogonal unit vectors |φni, and operators become matrices whose elements are the integrals hφn| Â |φmi.

|ψi =

c0

c1...cn

, hψ| = c∗0 c∗1 · · · c∗n , and Ĥ =

hφ0| Ĥ |φ0i hφ0| Ĥ |φ1i · · · hφ0| Ĥ |φnihφ1| Ĥ |φ0i hφ1| Ĥ |φ1i · · · hφ1| Ĥ |φni... ... . . . ...hφn| Ĥ |φ0i hφ1| Ĥ |φ1i · · · hφn| Ĥ |φni

Solving for the cn coefficients that minimize the energy is equivalent to diagonalizing Ĥ in this vector space;the eigenvalues are the energies and the eigenvectors are the coefficients on each basis function that make upthe corresponding wavefunction. One can, for instance, estimate the ground state energy and wavefunction of a harmonic oscillator by using a trial function which is a linear combination of particle in a box wavefunctions.

4.2 | Perturbation theory

To use perturbation theory eff ectively, the problem must be expressed in terms of a separate problem thatcan be solved exactly together with a small correction term. That is, if Ĥ is the problem, then we must beable to express it as

Ĥ = Ĥ0 + V̂

where the eigenvalues and eigenfunctions of Ĥ0 are known, and V̂ is relatively small. If we do this, then theeigenvalues n and eigenfunctions |Ψni can be expressed as

n = (0)n +

(1)n +

(2)n + . . . , |Ψni =

ψ(0)n E+ ψ(1)n E+ ψ(2)n E+ . . .where the superscripts indicate the “order” of the approximation. You can think of this as similar to a typeof expansion like a Taylor series.

The zeroth order terms in each of these equations are the eigenvalues E n and eigenfunctions |φni

of Ĥ0,which are known exactly. It can be shown that the first order terms are

(1)n = hφn| V̂ |φni (18)and ψ(1)n E = X

k6=n

−hφk| V̂|φni

E k − E n |φki . (19)

In this way, we can use the solutions to exactly-solvable problems and correct them for small perturbations.This is a second reason we spent so much time deriving the solutions to the model systems above; not onlydo the systems give us a mental picture of what’s going on, they also provide a starting point for performingmore accurate calculations while still maintaining the same qualitative physical picture. Perhaps the mostcommon example of the use of perturbation theory is to calculate energies of anharmonic oscillators. Recall

our earlier Taylor expansion of an arbitrary potential near the minimum, eq (6). If we truncate the series atthe fourth term instead of the third, we see that the Hamiltonian becomes

Ĥ = − ~ 2

2µ∇2 + 1

2kx2 +

1

6γ x3,

where γ is the third derivative of the potential at the minimum. This problem can be treated with pertur-bation theory if we let

Ĥ0 = − ~ 2

2µ∇2 + 1

2kx2, and V̂ =

1

6γ x3.

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� Multielectron atoms

◦ Hydrogen atom wavefunctions are used as a basis for multielectron atoms.◦ Electrons in s orbitals “shield” the nuclear charge from p and higher orbitals, causing the s orbitals to

lie lower in energy.◦ Electrons have intrinsic spin angular momentum ±~ /2. There is no classical analog for this.◦ Pauli exclusion principle– wavefunctions must be antisymmetric with respect to permutations of

fermions (particles with half-integer spin, such as electrons). Implication is that only two electronsmay share the same wavefunction, and then only if they have opposite spin.

◦ Slater determinants of hydrogen atom wavefunctions can be used to construct antisymmetric totalwavefunctions: these are called linear combinations of atomic orbitals.

◦ Term symbols (2S +1L2J +1) describe the coupling of electron angular momenta– distinguish betweenotherwise identical electron configurations.

◦ Hund’s rules (McQ 8-10) predict the ground-state term symbol (i.e., ground-state electron configura-tion).

� Bonding in diatomic molecules

◦ While not 100% accurate, molecular orbital theory provides intuitive and qualitative picture of bonding.◦ When atomic orbitals overlap, they can do so in-phase (bonding orbitals) or out-of-phase (antibonding

orbitals).◦ Net stabilization occurs when atomic orbitals overlap in-phase from an “exchange integral” term–

purely quantum-mechanical eff ect.◦ Many atomic orbitals can be combined to produce bonding, antibonding, and non-bonding orbitals

(LCAO–MO); these can be sorted by energy and filled in like atomic orbitals.◦ Antibonding orbitals have nodal planes perpendicular to the internuclear axis.◦ Bond order is estimated as the number of electrons in bonding orbitals minus number of electrons in

antibonding orbitals, all divided by two. A bond order of 0 means that the bond is not predicted tobe stable.

◦ σ orbitals are cylindrically symmetric about the internuclear axis, π orbitals have one nodal planealong the internuclear axis, δ orbitals have two nodal planes along the internuclear axis, etc...

◦ Molecular term symbols, like atomic term symbols, describe electron configurations. L is replaced by aGreek letter Λ in a molecular term symbol, and the 2J +1 is gone because the total angular momentumnow depends on the rotational state of the molecule in addition to the electron angular momentum.

◦ For homonuclear diatomics, the parity subscripts g and u tell whether the wavefunction is centrosym-metric (g is, u is not).

◦ For Σ states, a superscript +/− indicates the symmetry with respect to a plane containing the inter-nuclear axis. This only applies when π or higher MOs are involved. For each set of degenerate π MOs,one is symmetric and the other is not, so the net symmetry can be calculated by looking at the numberof singly-occupied π orbitals. Similar logic applies for the δ and higher orbitals.



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CHE 110B Lecture 1 Notes

way we’ll do this is with these ABCD cards, which will let you give me real-time feedback on how you’relearning. Hand out ABCD sheets, explain how to use them, and ask multiple-choice questions at the end of this document.

4 � Matrix operations and Dirac notation

4.1 | Matrix operations

A matrix is simply a rectangular grid of numbers or operators, such as

a b

c d

.

A special type of matrix has only one dimension: such a matrix is called a vector (specifically, a row vectoror column vector).

Row vector:

a b

Column vector:

cd

In general, matrices can be added, subtracted, and multiplied. Matrix addition is done element-by-element:

a b

c d

+

w x

y z

=

a + w b + xc + y d + z

.

Subtraction is exactly the same; it is also done element-by element:

a b

c d

−

w x

y z

=

a − w b − xc − y d − z

.

There are two diff

erent rules for matrix multiplication: the first is multiplication by a scalar, which goeselement-by-element:

2

a b

c d

=

2a 2b2c 2d

.

Note that this follows directly from the addition rule:

2

a b

c d

=

a b

c d

+

a b

c d

=

a + a b + bc + c d + d

=

2a 2b2c 2d

.

However, multiplying one matrix by another is a little more tricky:

a b

c d

w x

y z

=

aw + by ax + bzcw + dy cx + dz

.

The matrix multiplication rule is a little easier to understand if we break it down: we’ll look at vectors first.The rule for vector multiplication is:

a bw

y

= aw + by.

That is, the product of a row vector with a column vector is just a number, and it’s given by the “dotproduct:” the sum of the products of each element in sequence. However, the number of columns in the row

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vector (more generally, the matrix on the left) must be equal to the number of rows in the column vector(more generally, the matrix on the right)! It would be invalid to multiply:

a b cw

y

If we reverse the operation, we get something rather diff erent:

a

b

w y

=

aw ay

bw by

.

However, because this time the number of columns for the matrix on the left is always equal to the numberof columns on the right, the lengths do not need to be equal anymore:

ab

c

w y =

aw aybw by

cw cy

.

Now that we know how to multiply vectors, we can break down the general matrix multiplication. Theelement in the product matrix at row n column m is just a vector product of row n from the matrix on theleft with column m from the matrix on the right:

1,1:

a b

w

y

=

aw + by

1,2:

a b

x

z

=

ax + bz

2,1:

c d

w

y

=

cw + dy

2,2:

c d

x

z

=

cx + dz

While it is possible using these rules to multiply matrices with diff erent dimensionalities (so long as thenumber of columns in the left matrix is equal to the number of rows of the right matrix), the matrices wewill deal with will almost always be square.

As a final note on matrix multiplication, you should note that it is not, in general, commutative. That is,for two matrices A and B , AB 6= BA. Compare the reverse of our earlier multiplication:

a b

c d

w x

y z

=

aw + by ax + bzcw + dy cx + dz

w x

y z

a b

c d

=

wa + xc wb + xdya + zc yb + zd

There are two other matrix operations we’ll use in this class are the trace and the determinant of a matrix.The trace is easy: it is simply the sum of the “diagonal” matrix elements in a square matrix. An element is“diagonal” if the row number is equal to the column number, so

Tr

a b ci j k

x y z

= a + j + z.

The determinant of a matrix is denoted with vertical bars on the sides, and for a 2 × 2 matrix, it is definedas a bc d

= ad − bc.

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That is, it’s the product of the two diagonal elements minus the product of the two antidiagonal elements.For larger matrices, it is easy to calculate the determinant by a method called cofactor expansion. It’s amethod that is difficult to write out in words, but easy to see visually:

a b c

i j k

x y z

= a j ky z

− b i kx z

+ c i jx y

.

What we’ve done is choose a row or column in the matrix. In this case, I chose row 1. We then go elementby element in that row, and we form a cofactor by multiplying the element by a determinant formed byremoving that row and column from the matrix. The total determinant is then formed by adding “odd”cofactors and subtracting “even” cofactors, where “evenness” and “oddness” is determined from the productof the row and column number that were eliminated. In the example above, we see that we took a, whichis in row 1, column 1, and multiplied it by the determinant formed by removing row 1 and column 1 fromthe original matrix. Since 1 × 1 is odd, that term contributes positively to the sum. For the next term , wetook b, at row 1, column 2, and multiplied it by the determinant formed by removing row 1 and column 2from the original matrix. Since 1 × 2 is even, that term contributes negatively to the sum. We can chooseany row or column to do the cofactor expansion, so long as we’re careful to follow these rules.

If we have an even larger determinant, it’s still no problem. Each time we apply the cofactor method, we geta sum of determinants that are all one dimension smaller than we started. For each of those determinants, weapply the cofactor method and reduce the dimensionality by one, and so on until we have 2 × 2 determinants.Since the number of terms goes up pretty rapidly, it’s a good idea to make smart choices about which rowsand columns to use in the cofactor expansion. If you have to calculate by hand, you should always chooseto expand along the row or column with the greatest number of zeroes. We will be using determinants tosolve linear systems of equations, as we’ll see in the next class and on the problem set.

4.2 | Dirac notation

We’ll illustrate Dirac notation first by example, and then we’ll lay out more details. First, recall theSchrödinger equation:

H ψ = E ψ.

We’ll do some manipulations with it and see how the notation feels. You know this can be written out in3D as

− h̄2

2m

∂ 2

∂ x2 +

∂ 2

∂ y2 +

∂ 2

∂ z2

+ V (x,y,z)

ψ(x, y, z) = E ψ(x, y, z)

where V (x, y, z) is the potential energy and the first term with the partial derivatives is the kinetic energy.If we let the kinetic energy operator be T , then we could write:

(T + V )ψ = E ψ

We know that if we want the average energy of the system, we can multiply both sides by ψ∗ and integrateover all space to get:

Z ψ∗(T + V )ψ dτ =

Z ψ∗T ψ dτ +

Z ψ∗V ψ dτ =

Z ψ∗E ψ dτ = E

Z ψ∗ψ dτ

The dτ is a shorthand notation meaning that the integration covers all space (it includes all variables as wellif the integral is multidimensional). Of course, this is exactly the same as writing

Z ψ∗H ψ dτ = E

Z ψ∗ψ dτ .

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If, as is usually the case, ψ is normalized we knowR ψ∗ψ dτ = 1, and that integral can be dropped. This

notation seems easy enough, why would we need a diff erent notation?

Well, many times in this class, and in quantum mechanics in general, we don’t know ψ. However, we can

guess that ψ is a linear combination of other functions, and perhaps these other functions are easier to workwith for one of many reasons (the first of which is that we know what they are!). Let’s say we think ψ is alinear combination of two functions φ1 and φ2.

ψ = c1φ1 + c2φ2,

where c1 and c2 are weighting coefficients that might be complex numbers. The normalization condition isthen:

1 =

Z ψ∗ψ dτ = c∗

1c1

Z φ∗

1φ1 dτ + c

∗

1c2

Z φ∗

1φ2 dτ + c

∗

2c1

Z φ∗

2φ1 dτ + c

∗

2c2

Z φ∗

2φ2 dτ .

If we choose φ1 and φ2 so that they are orthogonal and normalized, thenR φ∗kφn dτ = δ kn (δ kn = 1 if k = n

and 0 if k 6= n), and we get:1 = c∗1c1 + c

∗

2c2.

The whole Schrödinger equation is then

c1H φ1 + c2H φ2 = E (c1φ1 + c2φ2).

If we multiply both sides by ψ∗ and integrate, we get:

c∗1c1

Z φ1H φ1 dτ + c

∗

1c2

Z φ1H φ2 dτ + c

∗

2c1

Z φ2H φ1 dτ + c

∗

2c2

Z φ2H φ2 dτ = E (c

∗

1c1 + c∗

2c2).

Now this is starting to get cumbersome, especially because those integral signs start getting hard to visuallyparse.

In Dirac notation, we represent a wavefunction ψ as |ψi; this is called a “ket.” The complex conjugate of a wavefunction, ψ∗ is denoted hψ|, called a “bra.” When a bra and a ket come together, it represents an“overlap integral”: hψ| ψi =

R ψ∗ψ dτ . Anytime there is a bra and a ket, that represents integration, even

if there is something in between. Operators act on kets. The Schr̈odinger equation in Dirac notation justlooks like:

Ĥ |ψi = E |ψi

where Ĥ is the Hamiltonian operator, defined the same way as above. We can perform the first set of operations that we did again: multiplying both sides by ψ∗ and integrating is the same as left-multiplyingby hψ|:

hψ| Ĥ |ψi = E hψ| ψi .

When we let |ψi = c1 |φ1i + c2 |φ2i, we then write:

c1 Ĥ |φ1i + c2 Ĥ |φ2i = E (c1 |φ1i + c2 |φ2i),

which we can left-multiply by hψ| and simplify (hφk| φni = δ kn) to get:

c∗1c1 hφ1| Ĥ |φ1i + c∗

1c2 hφ1| Ĥ |φ2i + c∗

2c1 hφ2| Ĥ |φ1i + c∗

2c2 hφ2| Ĥ |φ2i = E (c∗

1c1 + c∗

2c2).

The notation is a little more crisp, and lets you focus more on the physical picture than the mathematicaldetail.

The most important part of Dirac notation is that it corresponds directly to matrix and vector operationswhen the wavefunction is a linear combination, as above. In that case, a wavefunction in an orthonormalbasis |ψi = c1 |φ1i + c2 |φ2i is represented by a column vector:

|ψi =

c1c2

,

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and its corresponding “bra” hψ| is a row vector:

hψ| =

c∗1

c∗2

.

The overlap integral hψ| ψi is evaluated by the matrix multiplication:

hψ| ψi =

c∗1

c∗2

c1c2

= c∗1c1 + c

∗

2c2,

just like we saw before. An operator like Ĥ becomes a matrix whose elements are hφk| Ĥ |φni. In matrixform, the Schrödinger equation looks like:

hφ1| Ĥ |φ1i hφ1| Ĥ |φ2i

hφ2| Ĥ |φ1i hφ2| Ĥ |φ2i

c1c2

= E

c1c2

.

Now, left-multiplying by hψ| gives

c∗

1 c∗

2hφ1| Ĥ |φ1i hφ1| Ĥ |φ2i

hφ2| Ĥ |φ1i hφ2| Ĥ |φ2ic1

c2

= E

c∗

1 c∗

2c1

c2

.

You can easily verify that this is exactly the same equation as before.

What we’ve gained from all of this is extensibility: if we choose to let our state be a linear combination of 3 functions, we just increase the dimensionality of our matrices and vectors by 1. Thus, we have a conciserepresentation of the Hamiltonian in matrix form even as we greatly increase the complexity of the system.For a typical polyatomic molecule like methane, our wavefunction will be a linear combination of at least 9hydrogenic orbitals, so Dirac notation and its matrix representation saves a lot of time.

5 � Reading Assignment

McQ 9-1 – 9-5. Pre-lecture questions will be posted to Smartsite on Sunday afternoon!

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Which of the following best describes the Born-Oppenheimer approximation?

A) For a wavefunction ψ and an operator Â that corresponds to an observable, the average

value of the observable hAi is R ψ∗Âψ dτ .B) For a wavefunction ψ describing a particle, the probability that the particle lies in theregion dr is

R ψ∗ψ dr.

C) When solving the Schrödinger equation for an atom or molecule with more than oneelectron, let the nuclei remain fixed because the electrons are much lighter.

D) The wavefunction for a molecule can be approximated as a linear combination of atomicorbitals.

The energy of a 2s orbital is lower than the energy of a 2 p orbital.

A) True

B) False

I have a quantum system in a state described by: ψ = c1φ1 + c2φ2 + c3φ3. Now, I wishto make a single measurement Â on the system. If each φn is an eigenfunction of Â witheigenvalue n and is orthonormal with respect to the others, what will be the result of themeasurement?

A)R φ∗Âφ dτ .

B) One of 1, 2, or 3 with respective probabilities |c1|2, |c2|

2, and |c3|2.

C) |c1|2

1 + |c2|2

2 + |c3|2

3

D) 1 + 2 + 3


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CHE 110B Molecular Symmetry

1 ‖ Overview

In this section of the course, we will discuss how to make use of the symmetry of molecules to simplifyquantum-mechanical calculations. In essence, symmetry can give a quick answer to the question: “Is thisintegral zero or nonzero?” Many times, quite a large number of integrals turn out to be zero, so screeningthose out not only provides for faster computer calculations, but it also helps our qualitative understandingof which quantum-mechanical interactions are important in a molecule and which are not. And as we willdiscuss further in the next section, symmetry allows us to determine which spectroscopic transitions areallowed and which are forbidden for a molecule.

The formal mathematics of symmetry arise from the properties of mathematical structures called groups,and so we must learn about group theory in order to make use of symmetry. Group theory is a type of abstract algebra, and many, many theorems can be proved about the relationships that come out of objectsin groups. We do not have time to derive and prove all of the relationships we will make use of. They are allvery simple and straightforward, though clever, and are actually very beautiful, mathematically speaking.Most groups we’ll be interested in for molecular applications contain just a handful of elements (usually lessthan 20), and the calculations involving these elements boil down to basic arithmetic on integers. However,there are two challenges people face when trying to learn and understand group theory:

1. In chemistry courses, we use what are called spatial point groups, which are groups of operations inthree-dimensional space (rotations, reflections, inversions). Typically, people who do not have greatspatial intuition struggle with “seeing” the symmetry operations, and this creates a problem withunderstanding the basic properties of groups. We will attempt to get around this by introducing theprinciples of group theory with permutation groups, which are much easier to visualize and manipulate.

2. Mathematically, the notation gets very complex, because mathematicians like to make their equationsvery general. There will be a ton of summations over various indices, and terms with multiple subscriptsand superscripts. It is very important not to get overwhelmed by the notation. Almost every equationturns out to be extremely simple to use.

By the end of this section, you should know the following:

◦ The four defining properties of a mathematical group◦ What are the elements, classes, and order of a group◦ How to derive a Cayley table for the elements of a group◦ What is a representation◦ What is a character table◦ The great orthogonality theorem◦ Given a partial character table, how to complete it◦ How to determine the dimensionality of a representation◦ How to calculate a group multiplication table◦ The vanishing integral rule◦ The diff erence between a reducible representation and an irreducible representation◦ How to decompose an arbitrary representation into its irreducible components◦ What is a symmetry element◦ What are the implications for the solutions to the Schrödinger equation when the Hamiltonian is

invariant under certain symmetry operations?◦ What is a point group◦ How to determine the point group of a molecule◦ How to compute symmetry-adapted linear combinations (SALCs)

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CHE 110B Molecular Orbital Theory

1 � Overview

The overall goal for this section is to understand the quantum mechanical nature of chemical bonds andthe general structure of molecules in terms of molecular orbital theory. We’ve already studied the basicpostulates of quantum mechanics and solved several model systems. We’ve also looked at multi-electronatoms, and have an understanding of atomic term symbols. Now we are going to look at what happenswhen we have multiple nuclei and multiple electrons. These types of problems cannot be solved exactly,but variational methods and perturbation theory methods can be used to treat the system. We will use anapproximate method called MO theory because it provides a useful qualitative description of bonding andmakes many accurate predictions about energetics and structure, despite not being exact.

By the end of this section, you should know the following:

◦ What the Born-Oppenheimer approximation means in the context of the molecular Hamiltonian◦ The justification for making the Born-Oppenheimer approximation◦ What a potential energy surface is and how it is calculated◦ How the potential energy surface is related to molecular structure and vibration◦ What makes a chemical bond stable◦ What bonding and antibonding orbitals are◦ How to calculate the bond order◦ How to calculate the energy of a simple set of molecular orbitals, both from the variational principle

and using matrix formalism◦ How the number of molecular orbitals relates to the set of basis functions chosen◦ What approximations are made for MO theory in multielectron molecules◦ What the wavefunctions and energies are for LCAO-MOs◦ How to calculate appropriately antisymmetrized wavefunctions from a populated MO diagram◦ What is the Hartree-Fock limit◦ How are integrals evaluated on computers, and what tradeoff s are made in regard to basis set size and

quadrature◦

What basis sets are used in quantum chemical calculations◦ What is hybridization, and how to calculate hybrid orbitals◦ Limitations of hybridization theory◦ What are the approximations made in Hückel theory◦ How to calculate Hückel energies and orbitals, and what their general properties are

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2 � Origins of MO Theory - Diatomic molecules

Coordinate system used for diatomic (H2

) calculation.

The Hamiltonian of the diatomic molecule H2 in atomic units is:

Ĥ = −12

∇21 + ∇22 +

∇2A1823

+ ∇2B1823

− 1

r1A− 1

r1B− 1

r2A− 1

r2B+

1

r12+

1

R. (1)

Term breakdown (in order):

−1

2∇21 + ∇22 + ∇

2

A

1823 + ∇2B

1823

Kinetic energies of electrons and protons (proton mass is ∼1823me).1r1A

, 1r1B , 1r2A

, 1r2B Electron-proton attraction terms for each pair.1r12

Electron-electron repulsion.1R Proton-proton repulsion.

We make the Born-Oppenheimer approximation to neglect the proton kinetic energy terms and hold Rconstant. Now we have the electron terms to deal with, and that seems hard. So we’ll start by simplifyingeven further, and removing one of the electrons:

Ĥ = −12∇2 − 1

rA− 1

rB+

1

R. (2)

We want to find wavefunctions ψ(r1, r2; R) that are eigenfunctions of the operator. It is possible to do soexactly (under the BO approximation, at least), but we can learn something more fundamental by using thevariational method. The solutions to the hydrogen atom problem are known, so as a start, we’ll take a linearcombination of hydrogen atom 1s orbitals as trial functions:

|ψi = c1 |1sAi + c2 |1sBi . (3)

From symmetry arguments, we can guess that c1 = c2. We’ll have to normalize these at some point, but fornow let our trial functions be

|ψ±i = |1sAi ± |1sBi . (4)

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What is the variational energy of |ψ+i?

E + = hψ+| Ĥ |ψ+i

hψ+| ψ+i . (5)

We can expand the denominator:

hψ+| ψ+i = h1sA| 1sAi + h1sB| 1sBi + h1sA| 1sBi + h1sB | 1sAi . (6)

The individual H-atom wavefunctions are normalized, so the first and second terms are 1. The other twoterms represent the overlap integrals of hydrogen 1s wavefunctions located at diff erent positions separatedby R. We can visualize this: think of multiplying them and integrating; it is a function that has a peakin between the atoms. Those terms are denoted S (R) (the R indicates that the exact value of the integraldepends on R). Technically, one is S and the other S ∗, but 1s orbitals are real, so they are equal. Thedenominator evaluates to 2(1 + S (R)). Mathematically evaluating the integral shows that in these units,0 < S (R) ≤ 1. It is equal to 1 if R = 0, and decreases with increasing R.

Now let’s look at the numerator of (5) and expand:

hψ+| Ĥ |ψ+i = hψ+| ĤA0 + V̂A |1sAi + hψ+| ĤB0 + V̂B |1sBi . (7)

where the Ĥ0 terms are hydrogen atom Hamiltonians for atoms A and B, and the V terms contain thepotential energy from the electron interacting with the other nucleus. We know that Ĥ0 |1si = E 1s |1sialready, so the terms involving Ĥ0 become E 1s hψ+| 1sAi + E 1s hψ+| 1sBi, which is just E 1s hψ+| ψ+i =2E 1s(1 + S (R)). Now let’s look at this again:

hψ+| Ĥ |ψ+i = 2E 1s(1 + S (R))+ h1sA| V̂A |1sAi + h1sB | V̂B |1sBi+ h1sB | V̂A |1sAi + h1sA| V̂B |1sBi . (8)

The terms on the second line represent the average Coulombic interaction between a hydrogen atom anda proton, so call these J AA and J BB . On the third line, the terms are called exchange integrals K AB andK BA, and have no classical analog. They represent a quantum interaction of the 1s wavefunction on oneatom with the electron as if the positions of the nuclei were swapped (hence, “exchange integral”). Becauseboth nuclei are the same and the H atom wavefunctions are real, J AA = J BB and K AB = K BA. In the endthe numerator of (5) is

hψ+| Ĥ |ψ+i = 2E 1s(1 + S (R)) + 2J + 2K, (9)and the variational energy is:

E + = hψ+| Ĥ |ψ+i

hψ+| ψ+i = E 1s + J + K

1 + S (R). (10)

If we repeat the same procedure, we find that the other wavefunction |ψ−i has energy

E − = hψ−| Ĥ |ψ−i

hψ−| ψ−i = E 1s + J − K 1 − S (R) . (11)

Defining ∆E ± = E ± − E 1s, we compare the energy of the molecular orbitals with the energy of a protonand hydrogen atom separated infinitely. We see that |ψ+i actually has lower energy than E 1s over a certainregion and is a “bound” potential, while |ψ−i has greater energy and is not bound. We call |ψ+i a bondingorbital and |ψ−i an antibonding orbital. Breaking down the contributions from J and K to E +, we see thatit is the exchange term K that gives rise to the stability of a bond: it’s purely quantum-mechanical!

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Energies of |ψ+i and |ψ−i compared to E 1s (left), and contributions to ∆E + from J and K (right).

This is an example of a potential energy curve that results from making the Born-Oppenheimer approxi-mation. It assumes that the nuclei eff ectively experience a static potential from the electrons that variesparametrically with the nuclear position. If we had more nuclei, then the dimensionality of the curve in-creases, and we have a potential energy surface. These are generated by calculating the energy of the

electronic wavefunctions at many nuclear positions. The potential energy surface is typically used as aneff ective nuclear potential for calculating the nuclear geometry (the equilibrium geometry is the minimumof the PES) and the vibrational energy (related to the derivative of the PES near a minimum).

All that’s left to do is normalize our wavefunctions. We already calculated hψ+| ψ+i = 2(1 + S (R)) andhψ−| ψ−i = 2(1 − S (R)), so the normalized wavefunctions, which we’ll call |ψbi and |ψai for bonding andantibonding, respectively, are:

|ψbi =s

1

hψ+| ψ+i |ψ+i =s

1

2(1 + S (R)) (|1sAi + |1sBi) (12)

|ψai =s

1

hψ−| ψ−i |ψ−i =s

1

2(1 − S (R)) (|1sAi − |1sBi) (13)

These look like the following:

Wavefunctions (top) and probability densities (bottom) for |ψbi (left) and |ψai (right).

It is easier in general to express all of this in matrix formalism, especially as we increase the number of atomsin the system. For this example, with |ψni = c1n |1sAi + c2n1sB , we have (remember that operators act “to

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the right”):

Ĥ =

h1sA| Ĥ |1sAi h1sA| Ĥ |1sBih1sB | Ĥ |1sAi h1sB| Ĥ |1sBi

=

h1sA| ĤA0 + V̂A |1sAi h1sA| ĤB0 + V̂B |1sBih1sB| ĤA0 + V̂A |1sAi h1sB| ĤB0 + V̂B |1sBi

=

E 1s + J E 1sS (R) + K E 1sS (R) + K E 1s + J

(14)

The goal is to solve the equation H 11 H 12H 21 H 22

c11 c12c21 c22

=

h1sA| 1sAi h1sA| 1sBih1sA| 1sBi h1sB | 1sBi

c11 c12c21 c22

E 1 0

0 E 2

, (15)

where cij are the coefficients in the jth eigenvalue/eigenvector. Writing this out gives a system of linearequations, and linear algebra tells us that a nontrivial solution exists only if the eigenvalues satisfy theequation:

E 1s + J − h1sA| 1sAi λ E 1sS (R) + K − h1sA| 1sBi λE 1sS (R) + K − h1sB| 1sAi λ E 1s + J − h1sB| 1sBi λ

=

E 1s + J − λ E 1sS (R) + K − S (R)λE 1sS (R) + K − S (R)λ E 1s + J − λ = 0 (16)

where λ is a dummy variable that represents all eigenvalues. Note that if the basis functions we were usingwere orthogonal, the off -diagonal elements of the determinant would not contain λ: the overlap matrixbecomes the identity matrix.

Solving for λ requires some tedious algebra (that you will do on a problem set), but ultimately gives the twoeigenvalues

λ± = E 1s + J ± K

1 ± S . (17)

Plugging in these values for λ back into (15) and solving, we find that c11 = c21 and c12 =

−c22. Normalize

(choose coefficients that are real and positive, though it’s arbitrary):

1 = hψ1| ψ1i = c211(h1sA| 1sAi + h1sB | 1sBi + h1sA| 1sBi + h1sB| 1sAi)1 = c211(2 + 2S )

c11 =

s 1

2(1 + S ), and (18)

1 = hψ2| ψ2i = c212(h1sA| 1sAi + h1sB | 1sBi − h1sA| 1sBi − h1sB| 1sAi)1 = c212(2 − 2S )

c12 =

s 1

2(1 − S ) (19)

To summarize, the matrix form gives us exactly the same answer as before, but it is easier to extend thematrix form to larger systems:

|ψ1i =s

1

2(1 + S (R))(|1sAi + |1sBi), E 1 = E 1s + J + K

1 + S (R) (20)

|ψ2i =s

1

2(1 − S (R)) (|1sAi − |1sBi), E 2 = E 1s + J − K 1 − S (R) (21)

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In the previous section, we chose to take a simple linear combination of H-atom 1s orbitals. But we didn’thave to: could choose, for instance:

|ψi = c1 |1sAi + c2 |2sAi + c3 |1sBi + c4 |2sBi (22)

Then Ĥ has the general form:

Ĥ |1sAi |2sAi |1sBi |2sBi|1sAi Atom A Atom B|2sAi E & J ES(R) & K|1sAi Atom A Atom B|2sAi ES(R) & K E & J

and solving the eigenvalue equation would give 4 eigenvalues and 4 eigenfunctions, because our linear com-bination involves 4 functions. We can make this as large as we want, and we will get more and more energylevels and wavefunctions. This is one of the tenets of general MO theory: the number of orbitals thatcome out of a MO calculation is equal to the number of orbitals put in. Still working with H+2 , let’s look

qualitatively at what happens when we include the |2 pi H-atom wavefunctions for each atom as well (10orbitals total). We find that the linear combinations that come out look like (in order of increasing energy,and taking only the most dominant terms):

|ψ1i ∝ |1sAi + |1sBi (23)|ψ2i ∝ |1sAi − |1sBi (24)|ψ3i ∝ |2sAi + |2sBi (25)|ψ4i ∝ |2sAi − |2sBi (26)|ψ5i ∝ |2 pz,Ai − |2 pz,Bi (27)|ψ6i ∝ |2 px,Ai + |2 px,Bi (28)|ψ7i ∝ |2 py,Ai + |2 py,Bi (29)|ψ8

i ∝ |2 px,A

i −|2 px,B

i (30)

|ψ9i ∝ |2 py,Ai − |2 py,Bi (31)|ψ10i ∝ |2 pz,Ai + |2 pz,Bi (32)

There are some new types of orbitals in there, and we need some nomenclature:

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Pictures for |ψ5i through |ψ10i. Note the nodal planes.

We call molecular orbitals that are cylindrically symmetric about internuclear axis σ orbitals, and ones thathave one nodal plane containing the internuclear axis π orbitals. We further denote them as bonding (nospecial notation) and antibonding (superscript ∗) to identify them. Using this nomenclature, we can relabel|ψ1i through |ψ10i: |σ1si, |σ1s∗i, |σ2si |σ2s∗i, |σ2 pzi, |π2 pxi, |π2 pyi, |π2 p∗xi,

π2 p∗y, and |σ2 p∗zi. The πorbitals are degenerate, as are the π∗ orbitals.

Another way of visualizing this is with a MO diagram:

MO diagram for H+2

(no electrons added).

The energy of the system is determined by which orbital the electron occupies.

3 � General MO theory - Polyatomic molecules

Now we’ve done all this work on H+2 and found out how to calculate the energies of many diff erent energy

levels that electron can occupy, at least approximately under the variational principle. The solutions we’vecome up with are not exact, but they do qualitatively explain the basic structure of the molecular orbitals thatH+2 has. But what about H2? If we look back at the Hamiltonian, we see that if we pretend that the electronsdon’t interact with one another (that is, ignoring the r12 term), then we can write the Hamiltonian underthe Born-Oppenheimer approximation as a sum of two single-electron Hamiltonians at a fixed internuclear

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separation (R is a constant):

Ĥ = Ĥ1 + Ĥ2 + 1

R (33)

Ĥ1 = −1

2∇21 −

1

r1A − 1

r1B (34)

Ĥ2 = −12∇22 −

1

r2A− 1

r2B(35)

Like the particle in a box, we can solve this by using products of single-electron MOs as our wavefunctions,and the resultant energies are the sums of the single-electron energies!

|ψi = |σ1s1i |σ1s2i ∝ (|1s1Ai + |1s1Bi) (|1s2Ai + |1s2Bi) , E ψ = E (σ1s1) + E (σ1s2) (36)When we make molecular orbitals that are products of linear combinations of atomic orbitals, they are calledLCAO-MOs.

The one catch in doing this is the Pauli exclusion principle: wavefunctions must be antisymmetric withrespect to exchange of identical fermions. Recall that electrons are fermions (spin 1/2): we represent the twopossible spin states as |αi and |β i (each is a projection of the angular momentum on the electron’s own axis).We need to make sure that when we make a product of single-electron wavefunctions, it’s antisymmetricwith respect to the electrons. Our wavefunctions should be in the form |ψi |σi, where |ψi depends only onspatial coordinates and |σi depends only on spin. If our wavefunction is |σ1s1i |σ1s2i, where the subscriptsindicate the electron number, then we could try giving each electron an opposite spin:

|ψi = |σ1s1α1i |σ1s2β 2i (37)But here it is clear that if we swap the labels on the electrons (change all subscript 1 to 2 and vice-versa),we don’t get the negative of the same wavefunction!

p̂12 |ψi = |σ1s2α2i |σ1s1β 1i 6= − |ψi . (38)Instead, we can take a linear combination of spin functions such that

|ψi = |σ1s1i |σ1s2i (|α1β 2i − |α2β 1i) (39)Then if we apply the permutation p̂12

p̂12 |ψi = |σ1s2i |σ1s1i (|α2β 1i − |α1β 2i)= − |σ1s1i |σ1s2i (|α1β 2i − |α2β 1i)= − |ψi (40)

we can see that our wavefunction is appropriately antisymmetrized (though it’s not normalized; need factorof 1

√ 2).

The general method for doing this is by use of a Slater determinant (just like a multi-electron atom): eachcolumn contains the product of one orbital and spin function for each possible labeling of the electrons.

For our system above (two electrons in |σ1si, one with spin α, and the other with spin β ), we would write(removing the 1s for clarity)

|ψi = 1√ 2!

|σ(1)i |α(1)i |σ(1)i |β (1)i|σ2i |α(2)i |σ(2)i |β (2)i , (41)

where the number in parenthesis is the “number” of the electron. The Slater determinant therefore containsall of the possible ways that we could assign numbers to the electrons in the system.

A few things to know about Slater determinants:

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◦ If you swap two columns, the determinant changes sign (this is why it satisfies the exclusion principle!).◦ If any two columns in a determinant are the same, the determinant goes to 0. This is why the Pauli

exclusion principle only allows one electron of each spin to occupy a single orbital, which you learnedway back in Gen Chem. It also directly leads to the “Aufbau” principle: when populating molecular

orbitals, we fill up orbitals in order of increasing energy with two electrons each.◦ The energy of a Slater determinant is simply the sum of the orbitals that make it up (i.e., add up the

energy of each electron in the MO diagram).◦ The wavefunction described by a Slater determinant is normalized if the individual orbitals that make

it up are normalized.◦ Two Slater determinants are orthogonal if they diff er in any single orbital.

Given a MO diagram, it is easy to generate a Slater Determinant for the ground state using general chemistryrules: the aufbau principle and the Pauli exclusion principle. Each orbital can accommodate two electrons:one of each spin, and electrons are placed into orbitals from lowest to highest energy. A system with N electrons in orbitals |φ1i, |φ2i, etc., will generate an N × N dimensional Slater determinant, and the generalform is (if N is even):

|ψi = 1√ N !

|φ1(1)i |α(1)i |φ1(1)i |β(1)i |φ2(1)i |α(1)i |φ2(1)i |β(1)i · · ·φN/2(1)

|α(1)i

φN/2(1)

|β(1)i

|φ1(2)i |α(2)i |φ1(2)i |β(2)i |φ2(2)i |α(2)i |φ2(2)i |β(2)i · · ·φN/2(2)

|α(2)i φN/2(2)

|β(2)i

..

....

..

....

..

....

..

.|φ1(N )i |α(N )i |φ1(N )i |β(N )i |φ2(N )i |α(N )i |φ2(N )i |β(N )i · · ·

φN/2(N )

|α(N )i φN/2(N )

|β(N )i

(42)

So we’ve found that by making the non-interacting electron approximation, we can write the Hamiltonian interms of products of single-electron wavefunctions. But we won’t stop there. Using the variational method,we can compute the average interaction energy and add that in to improve the energies of the wavefunctionswe calculate. Although we won’t show it here, doing this also makes changes to our wavefunctions as well,but since the corrections are small, they just slightly tweak the shapes of the orbitals (qualitatively, they

don’t change much). For a general polyatomic molecule, the electronic Hamiltonian (i.e., the Hamiltonianunder the Born-Oppenheimer approximation) is:

Ĥ = −12

Xi

∇2i −Xi,k

Z krik

+Xk,qq


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It’s this last term that causes all of the difficulty with the quantum mechanics of molecules.2. Choose a set of basis functions to describe each atom, such as hydrogenic wavefunctions centered on

each atom A, B , . . . (instead of |1ski, etc., this time we’ll call them |n,l,m; ki).3. Calculate the Hamiltonian matrix for the single-electron problem (elements are hn,l,m; k| Ĥ0+V (R) |n0, l0, m0; k

V (R) is the constant nuclear repulsion potential).4. Diagonalize the single-electron Hamiltonian matrix to calculate LCAOs that make up the molecular

orbitals and their energies. The number of MOs that comes out is equal to the number of atomicorbitals that was put in.

5. Those are just the solutions to the (imaginary) single-electron problem. In reality, we need to findthe best LCAOs that take into account electron-electron repulsion as well. We can start this processby writing out a Slater determinant for our system, which gives us a product basis of spatial-spinLCAO-MOs.

6. With our Slater determinant |ψi (which is a product of LCAOs |φni), we calculate the average interac-tion energy for our wavefunction hψ| Ĥ1 |ψi, which evaluates to terms that are like our earlier Coulomband exchange integrals but with spin and spatial components (won’t derive this here because it is verytedious, but it makes use of the fact that the single-electron wavefunctions are orthonormal):

hĤ1i = 12

Xi,ji6=j

J̃ ij − 12

Xi,ji6=j

K̃ ij (47)

J̃ ij = hσi(1)σj(2)| hφi(1)φj(2)| 1r12

|φi(1)φj(2)i |σ1(1)σj(2)i (48)

K̃ ij = hσi(1)σj(2)| hφi(1)φj(2)| 1r12

|φj(1)φi(2)i |σj(1)σi(2)i , (49)

where |σi(1)σj(2)i are the spins of electrons 1 and 2 in orbitals |σii and |σji7. The total energy, therefore, is the sum of the single electron energies and the interaction terms:

hĤi =Xi

E i + 1

2

Xi,ji6=j

J̃ ij − 12

Xi,ji6=j

K̃ ij . (50)

8. However, we’re not done! We now need to recalculate our wavefunction to minimize this energy. We

use hĤi as an operator, and find the solutions to the eff ective Hamiltonian equation ˆhHi |ψi = |ψi.Essentially, this adds off -diagonal terms to the Hamiltonian that we just diagonalized. The problemwith this is that these off -diagonal elements themselves are defined in terms of LCAO-MOs, so if thespatial MOs change, the values of the off -diagonal elements change too! Simply re-diagonalizing thematrix isn’t enough; after diagonalization, the off -diagonal elements must be recalculated using the newMOs, and they won’t be zero (but they should be smaller). Therefore, the process must be repeatediteratively until the off -diagonal elements are still very close to zero after diagonalizing and recalculatingthem. This is an iterative procedure called a Hartree-Fock Self-Consistent Field calculation. Essentially,this means that we’re trying to find the best possible product of single-electron wavefunctions thatminimize the energy of the system and produces a self-consistent inter-electron repulsion/exchangepotential. For any real system, this is way too hard to do by hand!

9. When this calculation converges, the energies and wavefunctions are said to be at the “Hartree-Focklimit.” The drawback to these solutions is that, as single-electron wavefunctions, there is no wayto account for dynamic electron correlation (coming from those rij terms), only the average mutualrepulsion/exchange. Physically, electrons just don’t tend to get close to one another; they “like” tostay far away. If we just look at all the positions where we could find each electron individually, thenoverlap those electron densities and average their repulsion/exchange energy, we will overestimate therepulsion! The very presence of the other electron stops the first electron from moving into regions of space that are close to the other electron, but if we calculate their positions independently, nothing stopsthem from getting that close. Because products of single-electron wavefunctions cannot dynamically



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account for electron correlation, they will always overestimate the electronic energy: the gap betweenthe Hartree-Fock energy and the “true” energy is called the “correlation energy,” There are post-Hartree-Fock methods, such as Configuration Interaction, that can make progress toward closing thecorrelation energy gap, but we will not discuss them in this course other than to note that they exist.

10. Now, we repeat this whole procedure at many diff erent values of Rkq so that we know the electronicenergy at many diff erent nuclear configurations (this is called calculating a potential energy surface,because we’re calculating the potential energy that the nuclei experience from the electrons as a functionof the nuclear coordinates). The minimum energy that we find on the PES is the equilibrium geometry,but there may be multiple local minima on the surface for diff erent conformers, isomers, and such.

In practice, calculations are performed on computers, because there are numerical methods available forsolving integrals and eigenvalue problems. It’s important to have a general understanding of how theseprocesses work so that you can understand the power and limitations of these computational approaches.All calculations begin with the selection of a basis set, like we were doing above. For our MO theorycalculations on H+2 , our basis set was a linear combination of hydrogen atom wavefunctions centered on eachatom. We found that there were three types of integrals that we needed to evaluate: S AB = h1sA| 1sBi,K AB = h1sA| 1rB + 1R |1sBi, and J AA = h1sA| 1rB + 1R |1sAi. How does a computer evaluate integrals likethese if the analytical solution is not known? Recall the definition of an integral as a Riemann sum:Z b

a

f (x)dx = lim∆x→0

bXi=a

f (xi)∆x. (51)

That is, if we define an evenly-spaced grid of points (called a quadrature grid) that spans the distance froma to b, evaluate the function at each point, multiply by the grid spacing, and add up the results, we havean approximation of the integral that becomes exact as the number of points in the grid approaches ∞. Adouble integral can be calculated as:

Z ba

Z dc

f (x, y)dxdy = lim∆x→0∆y→0

bXi=a

dXj=c

f (xi, yj)∆x∆y. (52)

Now the number of calculations that needs to be done is the product of the number of quadrature pointsbetween a and b and that between c and d. In the limit that the grid sizes are equal with number of pointsn, t 2D integral requires n2 calculations (and a m-dimensional integral requires nm calculations!). Now,this isn’t the only way to estimate the value of an integral: I’ve described the “left point rule” with evensampling. There are more advanced methods (trapezoidal rule, Simpson’s rule, adaptive sampling), andimplementations of these methods are widely available in a variety of programming languages (for Python,see NumPy, for C++, see the GNU Scientific Library).



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Diff erent methods of approximating a one-dimensional integral as a finite Riemann sum.

We find that there is a tradeoff between accuracy and computational time: to get more accurate results,we have to evaluate the function at more grid points, which takes a longer time. The next issue we face isthat not all of the integrals in our hydrogen atom wavefunctions have nice, definite limits: the integrationregion for the radial portion of the wavefunction goes from 0 to ∞. As a practical matter, our grid justwon’t cover the full domain of the radial functions, so we have to choose some point in the calculation to cutit off . Fortunately, as r increases, past a certain point the value of the radial function approaches 0, so thecontributions of those grid points to the sum will be very small. We can choose a cutoff point such that the

error is small compared to the grid spacing error. But again, the larger the region we choose to integrateover, the more grid points we need to get the same grid spacing ∆x, and so the longer the calculation willtake.

What kinds of integration do w need to do with molecular calculations? Getting back to our S , J , and K integrals, each is a three-dimensional integral because the basis functions are functions of r, θ, and φ. Upuntil this point, we’ve always been using hydrogenic orbitals as our basis functions. However, the hydrogenatom wavefunctions are not the best choice for heavier atoms: the more nuclear charge, the closer electronstend to be held to the nucleus. To capture this behacior, so-called Slater-type orbitals were used aroundthe time computers became available to serve as basis functions for arbitrary atoms (your book gives thesein terms of spherical harmonics; most general calculations are done in Cartesian coordinates except for veryspecial cases, because most molecules aren’t spherically symmetric):

φ

STO

abc (x,y,z; ζ )

= N e−ζ r

xa

yb

zc

, (53)where N is a normalization constant, a, b and c are integers controlling angular momentum (L = a + b + c),r2 = x2 + y2 + z2, and ζ is a term that can be optimized to account for how “tight” the radial part of the wavefunction is. In eff ect, ζ accounts for the increased nuclear charge and screening eff ects: there arerecommended tabulated values of ζ for many atoms at many values of the n quantum number. The largerthe value of ζ , the narrower the wavefunction. The problem is that products of e−ζ r functions at diff erentpositions (as we get in overlap and exchange integrals) are cumbersome to calculate; they must be numericallyevaluated. However, by using Gaussian-type orbitals (GTOs), such multicenter integrals become trivial toevaluate: the solutions can be found analytically, eliminating a fair portion of the numerical integration



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required. The GTOs have the form (again in Cartesian coordinates):

φGTOabc (x,y,z; α)

= N e−αr

2

xaybzc, (54)

where α is the shape term that can be optimized, and the other variables have the same meaning as theircorresponding values in STOs.

The only problem with GTOs is that they don’t have quite the right shape close to the nucleus: theexponential term goes as −αr2 rather than −αr. This can be overcome by using a linear combination of GTOs to approximate an STO:

ψSTO-NGi E = N Xn

dinφGTO(αin) (55)

where N is the number of GTOs being used to approximate the STO, i is the type of STO being approximated(eg 1s), din are the optimized expansion coefficients for each GTO, and αin are the optimized α terms foreach GTO. The d and α terms are varied to maximize overlap with the corresponding STO; tabulations of

these are available, though they’re not difficult to calculate.

STO-NG basis functions for the 1s STO.

Almost all modern electronic structure calculations are performed using basis sets built from combinationsof GTOs, and there are many, many types of basis sets out there. Here’s some terminology that is used asa shorthand for knowing which basis functions are employed in a particular calculation:

◦ The “minimal” basis set consists of one basis function (usually itself a sum of optimized GTOs) foreach orbital from each atom. For instance, a H atom would just have a minimal basis of a single 1s

orbital, while a 2nd row atom (C, N, O, etc.) would have 1s, 2s, 2 px, 2 py, and 2 pz orbitals. If aminimal basis set is used for each atom in the molecule, the basis is called a “Zeta” basis Z.

◦ Larger basis sets can be denoted by a multiple of “Zeta” (double-Zeta = DZ, triple Zeta = TZ, QZ,5Z, etc.). In these sets, each atomic orbital is described by sums of functions with diff erent shapeparameters (ζ terms). This broadens out the shape of the orbital and lets it describe more diff useshapes. In a DZ basis, 2 STOs with diff erent ζ values would be used to represent the H atom 1sorbital, and usually each STO is instead expressed as a combination of GTOs.

◦ Often, core electrons are fairly well localized and we don’t need the extra flexibility that the multipleZeta basis sets give. Then we use “split-valence” basis sets, where fewer orbitals are used for core



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electrons than valence electrons. The popular 6-31G basis functions from the program Gaussian aresplit-valence double-Zeta functions: the core electrons are treated as a single orbital that is a sum of 6 GTOs, and the valence electrons by two orbitals: one that is the sum of 3 GTOs and the other asingle GTO (with a more diff use form than the other). This means that each heavy atom contributes

a basis of 9 orbital functions (each of which may actually be multiple GTOs).◦ Asterisks or the lower-case letter p designate additional polarization terms that are added by mixing

in basis functions with higher angular momentum. The 6-31G∗ basis set has d character mixed in withthe p orbitals.

◦ Increasingly nowadays you will see basis sets with cc in them (e.g. aug-cc-pVTZ). This means thatthe coefficients for the GTOs are optimized to match the results of wavefunctions derived from post-Hartree-Fock calculations, and are used almost exclusively in such advanced calculations.

Most calculations you will see employ some type of standard basis set, where all of the adjustable parametersin the basis functions are fixed to predetermined values and only the relative contributions of each orbitalare adjusted. For any particular problem, it is likely that the results of the calculation would be improvedif these terms were optimized as well, but this is an extremely nonlinear optimization problem that wouldoverwhelm even the most advanced computers now.

In the end, the accuracy of a calculation depends on the basis set chosen, and there are tradeo ff s to be made.For the best answer, the basis set must be flexible enough to fit the shape of the “true” wavefunction, whichcan be accomplished by adding additional functions with diff erent shape parameters or polarization propertiesto the set. However, the more functions that are added, the longer it takes for the calculation to complete.Calculations involving systems with hydrogen bonding, van-der-Waals interactions, or electronegative atomswill benefit from adding diff use and polarized functions to the basis, and calculations on anions eff ectivelymandate use of such functions. The farther down the periodic table the atoms are, the more functions needto be included in the basis, and the more difficult the calculations will be.

An alternative method for solving the Schrödinger equation is through Density Functional Theory (DFT).The conceptual picture is based on the Hohenberg-Kohn theorem, which eff ectively shows that the electrondensity (itself a function of space and time) can be used to uniquely specify the ground state of a system in

the Born-Oppenheimer approximation. That is, it can be shown that after some manipulation, the energycan be expressed as,E (ρ) = E K + E P ;e,N + E P ;e,e + E XC (ρ) (56)

where ρ is the electron density, E K is the kinetic energy, E P ;e,N and E P ;e,e are the potential energies fromelectron-nucleus and electron-electron interaction, and E XC is the exchange-correlation energy, which takesinto account eff ects arising from electron spin and correlation. The first three terms are all classical in nature,while it is entirely unknown how to calculate the final term. The electron density is expressed in terms of a basis of normalized orbitals (typically GTOs). By applying the variational principle to the energy andelectron density, we arrive at the Kohn-Sham equations:

−1

2∇2 −

N Xj

Z jrj1

+

Z ρ(r2)

r12+ V XC (r1)

ψ(r1) = ψ(r1). (57)

You can see this is analogous to the Hartree-Fock equations (it is another way of converting the Hamiltonianinto a noninteracting problem), except for the exchange-correlation potential. This is in turn expressed asthe functional derivative of the exchange-correlation energy; i.e., like the Hartree-Fock system, the potentialis defined in terms of the electron density that gives rise to it. However, since the functional is unknown,an assumption must be made, and there are a variety of approximate functionals out there (the mostcommonly-used functional in chemistry is called B3LYP). In the end, given a density functional like B3LYP,the Kohn-Sham equation is solved iteratively.

Because the electron density can be expressed as a function of spatial coordinates, the dimensionality of a



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DFT calculation is much lower than for a Hartree-Fock (or higher-level) ab initio calculation, and it cantherefore be applied to much larger systems. DFT can be and is often used to calculate the propertiesof proteins, solid state materials, and other large molecules, though it has trouble with certain types of intermolecular interactions and dispersion. Compared with high-level ab initio methods, the accuracy of

DFT is relatively poor, but it does calculate bond lengths to within a few percent, vibrational energies towithin 10-20%, and reaction energetics that are at least qualitatively reasonable. While the initial formalismfor DFT could only be applied to ground states of nondegenerate systems, it has been expended to coverdegenerate systems and, with the advent of Time-Dependent DFT (TDDFT), it can even be used to calculateexcited state properties. Nearly any calculation on a system involving more than, 10-20 atoms will be donewith DFT because to treat them accurately with ab initio methods is computationally infeasible at present.

4 � Further approximations in MO theory: Hybridization and Hückel the-ory

4.1 | Hybridization

Full-scale electronic structure calculations are very nice, but it is useful to have more approximate andqualitative theories that let us describe the general structure of a molecule without getting bogged downin the details. One common approximation that organic chemists really like is to explain bonds in termsof hybrid valence orbitals. Since the valence orbitals in an atom are similar in energy, we can take linearcombinations of these valence orbitals to help explain structure. Before we start, you MUST understandthat hybridization is a lie! Much like “resonance structures,” it’s an imaginary concept that

you can use to rationalize why some molecules have certain shapes, but it does NOT provide

the correct explanation for bonding!

Take, for instance, the linear molecule BeH2. The Be atom has an electron configuration 1s22s2, so its

electron distribution is nominally spherically symmetric (in the approximation that its electron distribution

can be described by H-atom orbitals). However, the 2 p orbitals are close by in energy. If we let the hydrogenatoms lie along the z axis, then it makes some intuitive sense that there could be overlap between thehydrogen 1s orbitals and the Be 2 pz orbital in addition to the 2s orbital. That means the linear combinationc1 |H, 1si + c2 |Be, 2si + c3 |Be, 2 pzi is likely to be a good representation of the interaction between eachindividual H atom and the Be atom. We can form so-called sp-hybrid orbitals and normalize them:

φ±sp = 1√ 2

(|2si ± |2 pzi) (58)

Note that |φ+i builds wavefunction amplitude constructively in the +z direction, and vice-versa for |φ−iThus, when we bring in the 1s orbitals from the hydrogen atoms, we would expect the two Be −H bondingorbitals to be on opposing sides:

|ψ1

i =

1

√ 2(φ+s p+ |1s, H 1i) (59)

|ψ2i = 1√ 2

(φ−s p+ |1s, H 2i), (60)

and their energies would be expected to be equal (likewise, there would be two degenerate antibondingorbitals with subtraction of the H atom orbitals).

Note that the choice of axis is entirely arbitrary. We could have decided that the atoms lie along the x axisinstead, then our hybrid orbitals would involve |2 pxi instead. Fundamentally, the atom doesn’t care which

p orbital we call x, y , or z ; it only cares that there are three p orbitals that are mutually orthogonal.



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Now let’s go larger and look at the trigonal planar BH3 molecule. We can also construct hybrid orbitalshere; let all of the atoms lie in the x, y plane like in this figure:

Coordinate system for BH3 sp2 hybrid orbitals, showing the B |2si, |2 pxi, and |2 pyi orbitals, and the target hybrid orbitals.

We would like to make linear combinations of the form |ψni = c1n |2si + c2n |2 pxi + c3n |2 pyi. What shouldthe coefficients be? Knowing that the molecule has equal bond angles and bond lengths, we can figureout the coefficients purely by symmetry and normalization arguments. Since the 2s orbital is sphericallysymmetric, it must contribute equally to all three hybrid orbitals. This means that c211 + c

212 + c

213 = 1,

so all c1n coefficients arep

1/3. Next, we see that the |2 pxi orbital’s positive lobe is oriented directlyalong one of the bonds, and the |2 pyi orbital is completely orthogonal to that bond. Therefore, all thecontribution from the p orbitals for that hybrid orbital (|ψ1i) must come from |2 pxi (i.e., c13 = 0), andbecause hψ1| ψ1i = 1, c21 =

p 2/3. Looking now at the remaining

p 1/3 of |2 pxi, we see that it should be

distributed equally between |ψ2i and |ψ3i, and that it needs a negative sign in order to contribute positivelyin those directions. This means c22 = c23 = −p 1/6. Finally, we know that |2 pyi contributes only to |ψ2i and|ψ3i, the contributions must be equal in magnitude but opposite in sign, and therefore c23 = −c33 = p 1/2.Our results are called sp2 hybrid orbitals since they’re formed by a linear combination of one s and two porbitals:

|ψ1i =r

1

3 |2si +

r 2

3 |2 pxi (61)

|ψ2i =r

1

3 |2si −

r 1

6 |2 pxi +

r 1

2 |2 pyi (62)

|ψ3i =r

1

3 |2si −

r 1

6 |2 pxi −

r 1

2 |2 pyi . (63)

We could do more examples, but the upshot is that knowing the molecular geometry allows us to rationalizethe electronic structure of the bonding orbitals in terms of hybridization of atomic orbitals on a central atom.It is important to remember that hybridization is a lie! The hybridization picture does not allow usto make predictions of molecular structure from first principles; it gives a misleading picture of the energeticsof molecular orbitals, and it even makes wrong predictions about the electron densities of molecules! Its onlyreal use is in rationalizing the structures and total ground state energies of organic molecules, but since theorganic chemists rely on this picture so heavily, we’re almost forced to teach it.

Why is hybridization wrong? Let’s consider our BeH2 example again: we said that |ψ1i and |ψ2i would bemolecular orbitals representing the two Be−H bonds, and that the two orbitals would be equal in energy.



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Shown below are the LCAO-MOs and their relative energies (not shown is the non-bonding 1 s orbital fromBe).

LCAO-MOs for BeH2

in relative energy order. Note that the +z axis goes to the left in this figure. The + and − signsindicate the sign of the wavefunction in that lobe.

As it turns out, the LCAOs that describe the two Be−H bonds look more like |1sH1i + |2sBei + |1sH2i and|1sH1i+|2 pz,Bei− |1sH2i, and they are not equal in energy. In fact, they don’t even look like individual Be −Hbonds at all; rather, they are full molecular orbitals that

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CHE 110B: Handout 110a Review