Exam 3 Review

Shervin Fatehi

Contents

1 Introduction 2

2 The Hamiltonian for diatomic molecules 2

3 Molecular orbital theory 4

4 MO theory predictions for bond order and magnetism 5

5 Molecular term symbols 6

6 Huckel theory 8

7 The harmonic oscillator / rigid rotor model 11

8 The electromagnetic spectrum 13

9 Molecular dipole moments 13

10 Selection rules for rovibrational transitions 14

11 The rovibrational absorption spectrum 16

12 Selection rules for rovibronic transitions 17

13 Corrections to the harmonic oscillator / rigid rotor model 20

14 Isotope effects 21

15 Good luck! 22

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1 Introduction

The material to be covered on exam 3 lies at the heart of chemistry. Overthe last month or so, we’ve constructed molecules, studied their structure,and considered how we can probe that structure experimentally. The topicsto be addressed in this review, then, fall under two basic umbrellas: molec-ular orbital theory (constructing molecules) and spectroscopy of diatomics(probing their structure). Spectroscopy of polyatomic molecules will be cov-ered for the remainder of the course, as will NMR.

There are a few things we discussed in this portion of the course that we’renot going to review here. Specifically, we’re not going to address valencebond theory or hybrid orbitals, since they’re outside the scope of the exam— as was stated in class by Profs. Miller and Yang. We’re going to omitProf. Yang’s discussions of nonadiabatic processes (curve crossing) and ofthe origins of the London dispersion interaction for similar reasons.

If you find this review unhelpful, well, we’re sorry! Drop us a line or lookback to McQuarrie & Simon. Wei’s notes on diatomics present some of thesame material with a little bit of a different style, so check there, too.

With that said, let’s begin.

2 The Hamiltonian for diatomic molecules

In order to study molecules, we need to simplify the Hamiltonian

H = − ~2

2mα

∑α

∇2α−

~2

2me

∑i

∇2i−

e2

4πε0

∑i

∑α

Zα

riα+

e2

4πε0

∑i<j

1rij

+e2

4πε0

∑α<β

ZαZβ

Rαβ

where α labels nuclei, i labels electrons, sums of form i < j are over pairsof particles of the same type, and rij = |~ri − ~rj | and similar expressions aredistances between particles. The first two terms are the kinetic energy of thenuclei and electrons, respectively, while the rest denote attractions betweenelectrons and nuclei, interelectronic repulsions, and internuclear repulsions.

Since we are studying diatomic molecules, this Hamiltonian simplifies quitea bit from the get-go, giving us

H = − ~2

2mα

B∑α=A

∇2α−

~2

2me

N∑i=1

∇2i−

e2

4πε0

N∑i=1

B∑α=A

Zα

riα+

e2

4πε0

∑i<j

1rij

+e2

4πε0ZAZB

RAB

2

This expression is still quite daunting, so we rewrite the Hamiltonian as

H = − ~2

2M∇2

R + Hel(R)

where translational motion of the system has been neglected, the formerterm is the kinetic energy associated with relative motion between the nu-clei of reduced massM , and the latter term is a Hamiltonian for the electronsfor a given value of R ≡ RAB, the internuclear separation.

We expect the nuclei in our molecule to be moving significantly more slowlythan the electrons at any given moment. After all, even the hydrogen nu-cleus, which is just a proton, is 1000 times heavier than an electron. As such,it might not be that bad of an approximation to assume that the nuclei moveso slowly that they are essentially fixed with respect to the electrons. Wemay therefore solve the Schrodinger equation associated with Hel(R),

Hel(R)ψeln (r;R) = Eel

n (R)ψeln (r;R)

where the semicolon separating the electronic positions r ≡ ri and R ismeant to stress the fact that the wavefunction is only parametrically depen-dent on the latter.

Having done so, we choose a product form of the overall wavefunction,Ψ(r,R) = ψel

n (r;R)χ(R), where χ(R) is a nuclear wavefunction. The Born-Oppenheimer approximation consists of asserting

− ~2

2M∇2

Rψeln (r;R) = 0

such that

HΨ(r,R) =[− ~2

2M∇2

R + Hel(R)]ψel

n (r;R)χ(R)

=[− ~2

2M∇2

R + Eeln (R)

]ψel

n (r;R)χ(R)

In other words, we may use the electronic energy as the potential functionfor which to solve the nuclear Schrodinger equation[

− ~2

2M∇2

R + Eeln (R)

]χ(R) = Eχ(R)

We’ll have more to say about the nuclear wavefunction later, but let’s goahead and examine how we can can approximately solve the electronicSchrodinger equation.

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3 Molecular orbital theory

We’d like to tackle the problem of finding solutions to the electronic Schrodingerequation

Helψeln (r;R) = Eel

n (R)ψeln (r;R)

with Hamiltonian

Hel = − ~2

2me

N∑i=1

∇2i −

e2

4πε0

N∑i=1

B∑α=A

Zα

riα+

e2

4πε0

∑i<j

1rij

+e2

4πε0ZAZB

RAB

This may be rewritten as a sum of 1-electron Hamiltonians with two nucleus-electron interactions, hi = − ~2

2me∇2

i − e2

4πε0

∑Bα=A

Zαriα

, and various repulsions:

Hel = hi +e2

4πε0

∑i<j

1rij

+e2

4πε0ZAZB

RAB

Any time we have a sum of 1-electron Hamiltonians with some extra termshanging out, we should think of doing a variational treatment of our prob-lem. In this case, we will choose to let our variational wavefunctions bethose appropriate for a single electron swimming around in a sea of nuclei— linear combinations of atomic orbitals of the same energy.1

The canonical example of MO theory is the H2 molecule. For H2, ourvariational wavefunctions will be ψ± = cA1sA ± cB1sB. These lead, aftersome struggle, to the secular determinant∣∣∣∣∣ HAA − E HAB − ESAB

HAB − ESAB HBB − E

∣∣∣∣∣ = 0

which, when solved, yields the energies (which aren’t so germane to our dis-cussion here) and the wavefunctions.

One of the two wavefunctions has been stabilized with respect to a 1s on itsown, while the other has been destabilized. They are, respectively,

ψ+ = ψb =1√

2(1 + SAB)(1sA + 1sB)

1It was shown on problem set 8 that orbitals of same type — or even just orbitals withclose to the same energy — will mix best. So 1s orbitals mix with other 1s’s, 2s orbitalswith other 2s’s, 2pz’s with other 2pz’s, etc. Of course, this isn’t rigorously and strictlytrue. For example, 2s and 2pz might also mix a little bit, although not as much.

4

andψ− = ψab =

1√2(1− SAB)

(1sA − 1sB)

Now, these are single electron wavefunctions. As usual, though, we will treattheir properly antisymmetrized products with appropriate spin functions asour basic approximation to the full multielectron wavefunction. Similarreasoning and similar work — and we know that this particular sectionis highly abbreviated — leads eventually to the LCAO-MO orbital energydiagrams for diatomics given in Figure 9-13 on p. 340 of M&S. These maybe filled up with electrons pairwise, as per the Pauli exclusion principle.

4 MO theory predictions for bond order and mag-netism

Aside from spectroscopic predictions, which we’ll be examining later, molec-ular orbital theory allows us to predict the bond order of a given moleculeas well as its magnetic properties.

The bond order is a quantity defined as

bond order =12

[(# of electrons in bonding orbitals)

− (# of electrons in antibonding orbitals)]

It turns out that bond order predictions tend to be more or less in agree-ment with our expectations from Lewis structures, once resonance is takeninto account. On the other hand, one can determine the bond order evenfor systems with odd numbers of electrons, which doesn’t make much sensein the Lewis picture.

The magnetic properties of molecules are pretty simple to determine usingMO theory. To do so, we simply look at the MO diagram of the moleculeand check to see whether it contains any unpaired electrons. If all of theelectrons are paired, the molecule will be diamagnetic, and it will be weaklyrepelled by a magnetic field. If one or more electrons are unpaired, themolecule will be paramagnetic, and it will be attracted by a magnetic field.An example of the former is N2, while an example of the latter is O2, whichhas an unpaired electron in each of its 1π∗u orbitals.

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5 Molecular term symbols

Molecular term symbols are a bit of a different beast from their atomiccousins, but they present no special challenges once the general procedurefor assigning them has been grasped. This is ironically a bit difficult to ex-plain without a concrete example, so let’s lay out some nomenclatural stuffand then look at a specific molecule.

The anatomy of a molecular term symbol 2S+1 |ML|mirrorinversion is as follows:

• The spin multiplicity, 2S + 1, which is the same as it was for atoms.Hurray for stasis!

• The orbital angular momentum of an MO will be given by 0 if it is aσ orbital and either 1 or −1 if it is one of a pair of π orbitals.2 Theabsolute value of the total orbital angular momentum, |ML|, is used toassign the term a letter. Unlike in the atomic case, we “go all Greekon it,” as David memorably commented, and assign

|ML| =⇒ Greek letter0 =⇒ Σ1 =⇒ Π2 =⇒ ∆

and so on.3

• The inversion symmetry subscript may be either g or u, for the Germanwords gerade and ungerade. It may only be applied to homonucleardiatomics, though, as heteronuclear diatomics don’t have an appro-priate center of inversion in the first place. Polyatomics are assignedterm symbols in even more Byzantine, group theoretic ways . . .

• The mirror symmetry superscript may be either + or −. It may onlybe applied to terms of Σ type.

So how are these determined? Let’s look at a simple example, the borondimer B2. It has the electron configuration (in SCF-LCAO-MO notation)

B2 : KK(2σg)2(2σ∗u)2(1πu)2

2δ orbitals have ±2 units of angular momentum, etc.3We guess a Φ would be next, but we don’t honestly know.

6

where the 1πu level is doubly degenerate and can therefore hold four elec-trons. The microstates associated with this electron configuration are shownin Table 1.

Having found them, we make another table — Table 2, naturally — slotting

State 1πu− 1πu+ ML MS

a −2 0b 0 1c 0 0d 0 0e 0 −1f 2 0

Table 1: Microstates for the boron dimer’s ground state configuration. 1πu+and 1πu− are just convenient ways of noting the fact that one of the twowill have orbital angular momentum of +1 and the other of −1.

states into entries according to their values of ML and MS . By inspectingthis table, we find that we have a 1∆, 3Σ, and 1Σ term corresponding tothis electron configuration.

Now let’s assign inversion subscripts to these states. This is done by taking

MS→ML

↓ −1 0 +1-2 a0 e c, d b2 f

Table 2: Microstates for the boron dimer’s ground state configuration. 1πu+and 1πu− are just convenient ways of noting the fact that one of the twowill have orbital angular momentum of +1 and the other of −1.

the direct product of the subscripts for all of the electrons in the system,using the rules

g ⊗ g = g

g ⊗ u = u

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u ⊗ g = u

u ⊗ u = g

One thing that should be immediately obvious is that any orbital filled with2 electrons will contribute a g to the proceedings and can therefore be ig-nored for our purposes. Applying that reasoning shows that we need onlyconsider the u⊗u from the outermost electrons, and, thus, all of these statesare gerade: we have 1∆g, 3Σg, 1Σg. Pleasingly, this will be true in general:the inversion symmetry subscript for one term corresponding to a given elec-tron configuration will be that for any other terms as well.

Finally, we assign mirror symmetry superscripts + and −, which are a littletrickier; they are only assigned to Σ terms, as the result for others can beambiguous. For this purpose we are interested in the symmetry of the 1πu

orbitals across a mirror plane containing the internuclear axis z. For conve-nience we can choose either the xz or yz plane. Now, one of the 1πu orbitalscomes primarily from combining the px orbitals, while the other comes pri-marily from combining the py’s. This means that one of the orbitals willhave + mirror symmetry, and the other will have − symmetry. Since thereis one electron in each of these orbitals for our Σ terms, the symmetry willbe − overall.

We’re done! Boron dimer has three term symbols, 1∆g, 3Σ−g , and 1Σ−g . Ac-cording to Hund’s rules, which apply equally well to molecular as to atomicterm symbols, they will be ordered energetically as 3Σ−g < 1∆g <

1Σ−g .

6 Huckel theory

Huckel theory is an approximate molecular orbital theory principally usedto treat systems of π electrons.4 As with so many of our other methods, itis based on setting up and solving a secular determinant.

In the context of using Huckel theory to examine an extended π-electronsystem, as we did on problem set 9, we can imagine that all of the σ bondsin our molecule have been formed. We would like to form π MO’s from thepz orbitals that remain on our atoms such that they will be of the lowest

4It may also be applied to other types of systems. For example, it can be used totreat systems of σ electrons in which exchange between nuclei is going to be the principalinteraction. It can also be modified so that some energies and exchange interactions aredistinct from others, as would be desirable in a heteroatomic π system like pyrrole.

8

possible energy in the effective field of the σ electrons. In other words, we’dlike to find the optimal linear combinations of pz’s,

N∑i=1

cipz,i

where i labels the atoms. This is a classically-framed variational problem,and it leads us to the general secular equation∣∣∣∣∣∣∣∣∣∣∣∣

H11 − ES11 H12 − ES12 · · · H1N − ES1N

H21 − ES21 H22 − ES22 · · · H2N − ES2N

......

. . ....

HN1 − ESN1 HN2 − ESN2 · · · HNN − ESNN

∣∣∣∣∣∣∣∣∣∣∣∣= 0

where everything is defined as usual — Hii is the energy of an electron inorbital pz,i, Hij = Hji is the exchange energy between orbitals pz,i and pz,j ,and Sij = Sji is the overlap between orbitals pz,i and pz,j .

Huckel theory consists of a set of approximations to the terms describedabove. Specifically:

• Sij = δij . The overlap between pz orbitals on different atoms is as-sumed to be negligible.

• Hii = α. All of the energies are assumed to be equal, since the orbitalsare all sitting in the same effective field from the σ electrons.

• Hij = β for |i − j| = 1, and it’s 0 otherwise. All of the exchangeenergies are set to β for exchange between neighboring orbitals, andexchange between orbitals further away from each other is neglected.

Our secular equation for a linear π-system therefore becomes∣∣∣∣∣∣∣∣∣∣∣∣

α− E β 0 · · · · · ·

β α− E β · · · · · ·

0 β α− E β · · ·...

.... . . . . . . . .

∣∣∣∣∣∣∣∣∣∣∣∣= 0

where entries have been added to help stress the fact that the matrix will betridiagonal and symmetric. Cyclic π-systems will have additional nonzero

9

entries of β for any pair of neighboring atoms ij = ji.

A typical simplification is to pull β out of each column of the determinantin order to get a new equation∣∣∣∣∣∣∣∣∣∣∣∣

x 1 0 · · · · · ·

1 x 1 · · · · · ·

0 1 x 1 · · ·...

.... . . . . . . . .

∣∣∣∣∣∣∣∣∣∣∣∣= 0

where x ≡ α−Eβ . Solving for x and rewriting in terms of α and β makes it

harder to flub the math.

The secular equation can be solved in general using the properties of de-terminants, specifically the fact that they may be expanded in terms of theentries of any row or column and their cofactors. For example, a 3×3 matrixmay be expanded in the following way:∣∣∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣∣∣ = a11

∣∣∣∣∣ a22 a23

a32 a33

∣∣∣∣∣− a12

∣∣∣∣∣ a21 a23

a31 a33

∣∣∣∣∣ + a13

∣∣∣∣∣ a21 a22

a31 a32

∣∣∣∣∣= a11(a22a33 − a23a32)− a12(a21a33 − a23a31) + a13(a21a32 − a22a31)= a11a22a33 − a11a23a32 − a12a21a33 + a12a23a31 + a13a21a32 − a13a22a31

which is in agreement with other methods of expanding determinants. Ofcourse, we could have chosen another row or a column if we chose — knowinghow to do this is important, since sometimes there will be a row or columnwith an appealing small number of nonzero entries.

After the determinant is expanded out, one needs to find the roots λiof the resulting polynomial. We promise that any polynomial you’ll see ona test will be factorizable at least into a product of first- and second-orderpolynomials (that is, polynomials like x − a or ax2 + bx + c). The corre-sponding energies, Ei, may then be found using the definition of x.

Something to watch out for: the exchange energy β is always negative! Keepthis in mind when constructing energy diagrams, because the fact α + β isa lower energy than α− β will matter. Don’t forget that two electrons can

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populate each level according to Pauli exclusion, either!

One can also use the roots to find the corresponding coefficients of theorbitals and, thereby, the wavefunctions for a given system. This is difficultfor all but the most trivial systems, so don’t worry about it. If you wantedto do it, though, you could try to solve the matrix equation

x− λi 1 0 · · · · · ·

1 x− λi 1 · · · · · ·

0 1 x− λi 1 · · ·...

.... . . . . . . . .

c1,i

c2,i

c3,i

...

= 0

for each of the λi.

7 The harmonic oscillator / rigid rotor model

Now that we’ve examined how one can solve for the electronic structure of amolecule, we need to think about the nuclei again. The nuclear Schrodingerequation is [

− ~2

2M∇2

R + Eeln (R)

]χ(R) = Eχ(R)

This is just a three-dimensional wave equation with a radial potential! Assuch, we already know that we can separate the radial and angular motion,and we can write the nuclear wavefunction as χ(R) = φ(R)YMJ

J (θ, φ), wherethe latter are just spherical harmonics with angular quantum number J andmagnetic quantum number MJ .5 This leaves us with the radial equation[

− ~2

2Md2

dR2+

~2

2MR2J(J + 1) + Eel

n (R)]φ(R) = Eφ(R)

In general, Eeln is not going to be a nice, exactly solvable potential. Even if

it were, we would have the centrifugal potential butting in and mucking itup. As a result, we choose to make the simplest possible approximation tothe potential.

5Note that this means that the rotational wavefunctions will be the same regardless ofwhat electronic state we’re in.

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• For the purposes of the rotational coupling to the vibrational motion(as represented by the centrifugal potential term), we fix the inter-nuclear separation to be R = Re, where Re minimizes the electronicpotential energy.

dEeln

dR

∣∣∣∣R=Re

= 0

• We approximate the electronic potential energy by its second-orderTaylor expansion around Re, such that

Eeln (R) ≈ Eel

n (Re) +dEel

n

dR

∣∣∣∣R=Re

(R−Re) +12

d2Eeln

dR2

∣∣∣∣R=Re

(R−Re)2

= Eeln (Re) +

12

d2Eeln

dR2

∣∣∣∣R=Re

(R−Re)2

= Eeln (Re) +

12k(R−Re)2

where we’ve simply defined the force constant in writing the secondequality.

This work gets us to the following form of the radial equation:[− ~2

2Md2

dR2+

~2

2MRe2J(J + 1) + Eel

n (Re) +12k(R−Re)2

]φ(R) = Eφ(R)

The middle two terms in the brackets are just constants now, while the outertwo are simply the Hamiltonian for a harmonic oscillator. So we concludethat the functions φ(R) will be harmonic oscillator wavefunctions. As aresult, we find the following final form for the wavefunction of our moleculein a state described by electronic quantum number n, vibrational quantumnumber v, and rotational quantum numbers J and MJ ,

ΨnvJMJ(r,R, θ, φ) = ψel

n (r;R)φv(R)YMJJ (θ, φ)

which will have an energy

EnvJ = Eeln (Re) + ~ωe

(v +

12

)+BJ(J + 1) v, J ∈ 0, 1, 2, . . .

where ωe =√

kM , the rotational constant B = ~2

2MRe2 , and the specific values

of n aren’t known without solving a specific problem.

12

Phew! With this model for diatomics — the harmonic oscillator / rigidrotor model (HORR) — firmly (re-)established, let’s have a look at itsspectroscopy. In order to do that, we’ll need to think a little bit aboutelectromagnetic radiation and dipole moments.

8 The electromagnetic spectrum

In doing molecular spectroscopy we bombard molecules with light. Through-out the course we’ve stressed that quantum mechanical systems will onlyabsorb or emit light with a frequency that corresponds to the energy gap be-tween any two of its energy levels, according to the formula ∆E = hν = ~ω.Table 3 provides a quick rundown of which wavelengths and frequencies ofelectromagnetic radiation correspond to which kinds of molecular transition.

Radiation ∼ λ ∼ hcλ (cm−1) ∼ hc

λ (eV) TransitionRadio 10 m 10−3 10−7 NMR

Microwave 1 cm 1 10−4 rotational

Far IR 0.1 cm 10 10−3torsional,

floppy vibrationalIR 10−3 cm 1000 0.1 vibrational

Visible 400 - 800 nm 20000 1 - 3electronic

(large molecules)

UV 100 - 200 nm 80000 4 - 12electronic

(small molecules)

X-ray 0.1 nm 108 104electronic

(inner shell)

Table 3: Portions of the electromagnetic spectrum relevant to spectroscopy.

9 Molecular dipole moments

The molecular dipole moment is a vector quantity dependent on the chargedistribution within a given molecule. Specifically, it’s the sum

~µ(r,R) = −∑

i

e~ri +∑α

eZα~Rα

where e is the unit charge, Zα is the charge on nucleus α and ~ri and ~Rα arethe positions of electron i and nucleus α, respectively. Not all molecules have

13

a permanent dipole moment — homonuclear diatomics and other symmet-rical molecules (like methane) being a good example — but for those thatdo, the dipole moment describes the separation of two effective charges,one negative and one positive. Even some molecules that don’t have per-manent dipole moments — like methane, again — can have induced dipoles.

Through arguments very similar to those given in our solutions to problemset 7, we can determine that the probability amplitude of a spectroscopictransition from some molecular state a to some other state b will be⟨

b∣∣∣~µ(r,R) · ~E

∣∣∣a⟩where ~E is the electric field associated with the light. Since we may chooseour coordinate system such that ~E is directed along the z-axis, this is equiv-alent to writing

〈b|µz(r,R) cos θ|a〉

where θ is the angle between the vectors and we assume that the angularcoordinates are oriented identically the same for r and R.

It also turns out that the intensity I of a transition from a to b is proportionalto its probability amplitude’s modulus squared (that is, the probability),

I(a→ b) ∝ P (a→ b) = |〈b|µz(r,R) cos θ|a〉|2

Taken together, these results allow us to predict selection rules for spec-troscopic transitions as well as the relative intensities of different lines inmolecular spectra.

10 Selection rules for rovibrational transitions

To obtain selection rules for a rovibrational transition — one between statesa = n, v, J,MJ and b = n, v′, J ′,M ′

J — we need to evaluate the expec-tation value 〈b|µz(r,R) cos θ|a〉, which can also be written in integral formas∫

dRR2

∫dΩ

∫drr2 ψel∗

n (r;R)φ∗v′(R)YM ′J∗

J ′ (θ, φ)µz(r,R) cos θψeln (r;R)φv(R)YMJ

J (θ, φ)

where dΩ = sin θdθdφ. Collecting terms, we have[∫dRR2 φ∗v′(R)

(∫drr2 ψel∗

n (r;R)µz(r,R)ψeln (r;R)

)φv(R)

]×

[∫dΩ Y

M ′J∗

J ′ (θ, φ) cos θYMJJ (θ, φ)

]14

We call the result of the integral over r, whatever it may be, µz(R) —there’s no way we can know what it is without having worked out somespecific problem. This leaves us with a product of two integrals to evaluate:[∫

dRR2 φ∗v′(R)µz(R)φv(R)]×

[∫dΩ Y

M ′J∗

J ′ (θ, φ) cos θYMJJ (θ, φ)

]The latter integral may be solved quite straightforwardly once we realizethat cos θ ∝ Y 0

1 (θ, φ). We also note that Y 01 (θ, φ)YMJ

J (θ, φ) ∝ YMJJ+1(θ, φ).

These facts allow us to write the second integral in two different ways andthen use the orthonormality of the spherical harmonics to see:∫

dΩ YM ′

J∗J ′+1 (θ, φ)YMJ

J (θ, φ) or∫

dΩ YM ′

J∗J ′ (θ, φ)YMJ

J+1(θ, φ)

= δJ ′+1,JδM ′J ,MJ

or = δJ ′,J+1δM ′J ,MJ

Since the Kronecker delta is only nonzero when its subscripts are equal, weobtain the rotational selection rules

∆J = J ′ − J = ±1∆MJ = M ′

J −MJ = 0

Getting the vibrational selection rule is a little trickier, since we don’t knowwhat µz(R) is. To bypass this, we’ll approximate the dipole moment by itsfirst order Taylor expansion around the equilibrium bond length, µz(R) ≈µz(Re)+µ′z(Re)(R−Re). This should be a sensible choice, since we anticipatethe the dipole moment will grow vary linearly in that region anyway.6 Thismakes our vibrational integral∫

dRR2 φ∗v′(R)µz(Re)φv(R) +∫

dRR2 φ∗v′(R)µ′z(Re)(R−Re)φv(R)

Pulling out the constants and recalling that the harmonic oscillator wave-functions are orthonormal, we find

µz(Re)δv′,v + µ′z(Re)∫

dRR2 φ∗v′(R)(R−Re)φv(R)

The first term will only contribute in the case of a pure rotational transition— not something terribly interesting, since it’d have the same spectrum asthe rigid rotor — so let’s look at the second. Similar to the rotational case,

6We expect it to drop off and eventually go to 0 as we pull the molecule apart, butthat’s certainly not the case here.

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we can notice here that (R − Re) ∝ φ1(R). Furthermore, φ1(R)φv(R) ∝φv+1(R). This allows us to write down∫

dRR2 φ∗v′+1(R)φv(R) or∫

dRR2 φ∗v′(R)φv+1(R)

= δv′+1,v or = δv′,v+1

which gives us the vibrational selection rule

∆v = v′ − v = ±1

Let’s summarize our results here: at least within the confines of the HORR,it is only possible to have rovibrational transitions that are described by

a = n, v, J,MJ → b = n, v ± 1, J ± 1,MJI(a→ b) ∝ µ′z

2(Re)

where it’s important to remember that µ′z(Re) is the dipole derivative.

11 The rovibrational absorption spectrum

Given our selection rules, let’s see what the possible absorption energiesare for a rovibrational transition. Since rotational energies are substantiallysmaller than vibrational energies, we must choose the v → v+ 1 vibrationaltransition, but either of ∆J = ±1 is acceptable. We therefore have

∆E− = En,v+1,J−1 − En,v+1,J

= Eeln (Re) + ~ωe

(v +

32

)+B(J − 1)J − Eel

n (Re)− ~ωe

(v +

12

)−BJ(J + 1)

= ~ωe +B(J2 − J − J2 − J)= ~ωe − 2BJ J ∈ 1, 2, 3, . . .

∆E+ = En,v+1,J+1 − En,v+1,J

= Eeln (Re) + ~ωe

(v +

32

)+B(J + 1)(J + 2)− Eel

n (Re)− ~ωe

(v +

12

)−BJ(J + 1)

= ~ωe +B(J2 + 3J + 2− J2 − J)= ~ωe + 2B(J + 1) J ∈ 0, 1, 2, . . .

This means that we will the spectrum will be composed of lines spacedequally around the vibrational energy ~ωe in intervals of 2B. Their inten-sities will be determined partially by the dipole derivative and partially by

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Figure 1: Simulated rovibrational spectrum of carbon monoxide 12C16Oat room temperature, presented as both a series of sticks (left panel) andwith some line-broadening (right panel). The y-axis is in arbitrary units ofintensity, while the x-axis is in wavenumbers. In each image, the so-calledP branch (associated with ∆J = −1) is at left, the R branch (associatedwith ∆J = +1) is at right, and the Q branch (associated with ∆J = 0) isimplied on the boundary between P and Q.

the populations in different rotational levels. This contribution has to dowith degeneracy and temperature, such that I ∝ µ′z(Re)g(J)e−βEJ , whereg(J) = 2J + 1 is the degeneracy and the exponential Boltzmann factor isdependent on the inverse temperature β = 1

kBT .

Figure 1 shows our predicted stick spectrum and a simulated “actual” spec-trum for carbon monoxide.7

12 Selection rules for rovibronic transitions

To obtain selection rules for a rovibronic transition — one between statesa = n, v, J,MJ and b = n′, v′, J ′,M ′

J — we once again need to evaluatethe expectation value 〈b|µz(r,R) cos θ|a〉, which we previously wrote as∫

dRR2

∫dΩ

∫drr2 ψel∗

n′ (r;R)φn′∗v′ (R)YM ′

J∗J ′ (θ, φ)µz(r,R) cos θψel

n (r;R)φnv (R)YMJ

J (θ, φ)

7I made it using a fun Java applet created by Adam Bridgeman at the Universityof Hull. Find it at http://hull.ac.uk/chemistry/spectroscopy/rovibrational.php?res=high.In his notation, the difference between the corrected and rigid rotor rotational constants,∆B = Bv −Be, is called α in the applet. Make sure to set α = 0 if you want to scope outan HORR spectrum.

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where dΩ = sin θdθdφ.

Note that we have labeled the vibrational states with their correspondingelectronic quantum numbers. This is for a very good reason — they’re go-ing to be different, since the electronic potentials they’re sitting on are, ingeneral, going to be differently shaped and shifted in radius. The rotationalwavefunctions, however, will be exactly the same.

Collecting terms, we once again have[∫dRR2 φn′∗

v′ (R)(∫

drr2 ψel∗n′ (r;R)µz(r,R)ψel

n (r;R))φn

v (R)]×

[∫dΩ Y

M ′J∗

J ′ (θ, φ) cos θYMJJ (θ, φ)

]The second term will lead us to exactly the same rotational selection rulesas before, ∆J = ±1 and ∆MJ = 0. The first term, however, will take ussomewhere new. Let’s focus on it, calling the result of the integral µz(R) asbefore: ∫

dRR2 φn′∗v′ (R)µz(R)φn

v (R)

Our usual trick here would be to take a Taylor expansion to first order.However, even the zeroth order Taylor expansion will give us non-trivialinformation here. This is true, once again, because the vibrational levelsassociated with different electronic energy curves are going to have non-orthonormal wavefunctions. What we do, then, is replace µz(R) with µz(R),where R is the average of the equilibrium bond lengths on electronic curvesn and n′; this is called the Condon approximation, and it gives us

µz(R)∫

dRR2 φn′∗v′ (R)φn

v (R) = µz(R)⟨φn′

v′

∣∣∣φnv

⟩In other words, the probability amplitude of a transition n, v → n′, v′will be determined by the overlap between the vibrational states on the twoelectronic curves! This is represented pictorially, including a couple of fa-vorable transitions, in Figure 2.

How can we understand this equation and this picture? It’s actually prettystraightforward. Light moves incredibly quickly — one might say it movesat the speed of light. This means that the process of exciting an electronspectroscopically is it very fast, and so we expect that the nuclei won’t havetime to move very far in the meantime. It therefore makes sense that theoverlap between the spot on the initial curve where the electrons were most

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Figure 2: A figure shamelessly stolen (under license) from Wikipedia’s articleon the Franck-Condon principle. Their E0 corresponds to our En, their E1

to our En′ , their v′′ to our v, and their v′ to our . . . v′.

likely to be found (the tail end of the blue arrow in the picture) and the spoton the final curve above it (the head end of the blue arrow in the picture) iswhat we shoud compute to determine the likelihood of the transition. Thisis the Franck-Condon principle.

One other thing: the maxima in the higher wavefunctions are typicallyfound at the sides of the potential, near what are called the “classical turn-ing points.” They’ve earned this appellation because the molecule is veryclose to being either as compressed or as stretched as it is allowed to bewhen in those regions. As a result, the forces opposing its further motionin are very large, and it may be very close to standing still. Eventually itwill stop, and the force begins to drive its motion in the other direction.Therefore, the probability of finding the particle around the turning pointsis typically large.

Let’s sum up again: within the confines of the HORR and the Franck-

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Condon approximation, it is only possible to have rovibronic transitionsthat are described by

a = n, v, J,MJ → b = n′, v′, J ± 1,MJ

I(a→ b) ∝ µ′z2(Re)

∣∣∣µz(R)⟨φn′

v′

∣∣∣φnv

⟩∣∣∣2

13 Corrections to the harmonic oscillator / rigidrotor model

There are several corrections that are commonly made to the HORR modelto make it more realistic. They are:

• Anharmonicity.Since the electronic potential is anharmonic, the vibrational levels willno longer be evenly spaced; in fact, their spacing needs to decrease as vincreases. This is typically corrected with an empirical anharmonicitycorrection xe such that

Ev → Ev = ~ωe

(v +

12

)− ~ωexe

(v +

12

)2

• Rotation-vibration coupling.We know that higher vibrational states are more likely to be foundat higher values of R. As such, it seems appropriate to let B varyaccording to the average R for the vibrational state being occupied:

B → Bv =~2

2M

⟨φv

∣∣∣∣ 1R2

∣∣∣∣φv

⟩In practice, Bv is typically well-approximated by the simpler expres-sion B −∆B

(v + 1

2

)such that

EJ → ErotvJ = BJ(J + 1)−∆BJ(J + 1)

(v +

12

)• Centrifugal distortion.

As J grows, the system will stretch out under the influence of thecentrifugal potential. This leads to a lowering of the rotational levels.Empirically this is associated with a parameter D such that

ErotvJ → Erot

vJ = BJ(J + 1)−∆BJ(J + 1)(v +

12

)+D

(J(J + 1)

)2

20

The expression for the total energy of the system that results from thesecorrections is

EnvJ = Eeln (Re) + ~ωe

(v +

32

)− ~ωexe

(v +

12

)2

+B(J − 1)J

−∆BJ(J + 1)(v +

12

)+D

(J(J + 1)

)2

Beautiful, isn’t it? Even so, general corrections to the HORR take the formof terms in a Dunham expansion

∞∑n=0

∞∑m=0

Cnm

(v +

12

)n (J(J + 1)

)m

Dunham expansion is an empirical method of correction, although explicitforms of the coefficients Cnm may be derived.

14 Isotope effects

Some elements exist in several forms, each having a different number ofneutrons. Hydrogen, for example, may have no neutrons (1H, or protium,which is typically just called “hydrogen” since it’s the most abundant), oneneutron (D or 2H, called deuterium), and 2 neutrons (T or 3H, called tri-tium). We can therefore imagine taking a molecule that contains hydrogenand replacing it with one of the heavier isotopes. If there’s a change in theenergy associated with a given kind of molecular motion when we make thisreplacement, there is said to be an isotope effect.

Mathematically, replacing an element with one of its isotopes is just a changein M , the reduced mass of the nuclei. This is basically the only effect from achemical standpoint, as well, since nuclear structure is what’s being changed.For a diatomic AB,

M =mAmB

mA +mB

If mB is much larger than mA, a good approximation is to say that M ≈ mA.In that case, replacing A with an isotope A′ will lead to a change in M suchthat M ′ =

mA′mA

M .

There are basically three sets of energies that we are able to probe spec-troscopically: electronic energy, vibrational energy, and rotational energy.

21

Let’s see which of these display isotope effects.

Within the context of the Born-Oppenheimer approximation, electrons onlyinteract with nuclei Coulombically, with nuclear charges and positions de-termining the details of the interaction. The motion of the nuclei, associatedwith the operator KR = − ~2

2M∇2R, isn’t part of the electronic Hamiltonian,

and it has no effect on the electronic wavefunctions. The electronic energies,therefore, will stay the same if we change M .

Vibrational and rotational energies, on the other hand, will change. Theexpressions for these energies are (assuming an HORR situation)

Ev = ~ωe

(v +

12

)v ∈ 0, 1, 2, . . .

EJ = BJ(J + 1) J ∈ 0, 1, 2, . . .

The reason these energies will change is that both ωe and B are dependenton M . Specifically, if M →M ′, they will change such that

ωe → ω′e =

√M

M ′ωe

B → B′ =M

M ′B

15 Good luck!

We hope you all do well on this exam; we certainly believe it to be possi-ble. Regardless, have a lovely Thanksgiving . . . or at least a lovely break, ifturkey, turducken, or even tofurkey isn’t your thing.

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