Chem 4502 Quantum Mechanics & Spectroscopy...

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Lecture 34: Sections 1-3 General trend in energy spacing: Electronic > vibrational > rotational >> translational and the corresponding regions of the electromagnetic spectrum; Rotational-vibrational spectra of gas phase molecules; Boltzmann distribution gives relative populations at thermal equilibrium. Office Hours: 3-5pm Tomorrow (Tuesday)

Chem 4502 Quantum Mechanics & Spectroscopy (Jason Goodpaster)

Chapter 13 - Spectroscopy

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Trends in Molecular Energy Level Spacings

Electronic > Vibrational > Rotational ( >> Translational )

50,000 cm-1

Engel p. 305

O2

0 v=0 v=1 v=2

v=0 v=1

v=3

O2 X3Σg

De 41,760 cm-1

(16O18O rot levels)

0 cm-1 787 cm-1

1556

3088

J=0 J=1

J=0 J=1

3 cm-1

3 cm-1

(Particle in a 3D Box)

Gas Phase

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Types of Quantized Energy Levels in Gas Phase Molecules

Electronic > Vibrational > Rotational ( >> Translational )

Nuclear Motion: 3 Types Born Oppenheimer Approximation:

Ignore nuclear motion

Vibration of the atoms in a molecule (molecule�s center of mass fixed) Rotation of the molecule about its (fixed) center of mass (�Rigid Rotator�) Translation of the molecule in space (motion of the center of mass)

What about atoms? No rotational or vibrational energy levels; just electronic and translational

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Molecular Spectroscopy and Regions of the Electromagnetic Spectrum

Electronic > Vibrational > Rotational

λ (m) 1 100 10,000 cm-1

(Gas Phase)

In the gas phase, vibrational transitions are accompanied by rotational transitions

Electronic transitions are generally accompanied by vibrational (and, in the gas phase, rotational) transitions

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Rigid Rotator: Diatomic Molecule

EJ (cm-1) = B J (J+1) ∼ ∼

Selection rule for absorption or emission of a photon: ∆J = ± 1

Fig. 5.10 p. 178

J

0

E ∼ (cm-1)

2B ∼

6B ∼

20B ∼

12B ∼

g

1 3

5

7

9

Degeneracy g = 2J+1

since m = 0, ±1, ..., ±J

B = h / 8 π2 I c

I = µr2

µ = m1 m2 / (m1+m2)

∼

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Ken Leopold�s Research Group: Microwave Spectroscopy of Gas Phase Complexes (trimethylamine trimethylborane, a Lewis acid-base complex)

convert 9492 MHz to cm-1:

1/ λ = ν / c

= 9 x 109 /s 3 x 1010 cm/s

= 0.3 cm-1

Range: 1.2 MHz = 0.00004 cm-1 Very high resolution

http://www.chem.umn.edu/groups/kleopold/

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Recall: Assuming that H79Br has a vibrational frequency of

∼2560 cm-1, predict its gas phase IR absorption spectrum.

Figure 13.2 p. 500

There is a gap centered at ∼2560 cm-1.

On either side, absorption lines are spaced by ∼17 cm-1.

These are due to rotational transitions (plus ∆v=1).

Vibrational Spectroscopy of Gas Phase Molecules

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Vibrational Spectroscopy of Gas Phase Molecules

Each vibrational transition is accompanied by a rotational transition.

Selection rules:

∆v = 1 (harmonic oscillator)

∆J = +1 �R branch�

-1 �P branch�

0 �Q branch� forbidden in HBr (1Σ electronic state) allowed for Π, ∆ etc. states

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Relative populations of molecules in various quantum states at thermal equilibrium

N2/N1 = (g2/g1) e -(E2-E1)/kT

where the 2 quantum states are labeled �1� and �2�

N2/N1 relative population of molecules in state 2 vs. state 1

g1, g2 degeneracies of states 1 and 2

E2-E1 energy difference between states 2 and 1

k Boltzmann constant: 0.695 cm-1/Kelvin

T temperature in Kelvin

E1

E2

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N2/N1 = (g2/g1) e -(E2-E1)/kT

v=0

v=1

What is the relative population of molecules in v=1 vs. v=0 for HCl at 298 Kelvin? (HCl vibrational frequency is 2885 cm-1).

2885 cm-1

Nv=1/Nv=0 = e -(2885 cm-1) / (207 cm-1)

= e -13.9 = 1 x 10-6

Answer: degeneracy of vibrational states = 1

kT = (0.695 cm-1/K)(298 K) = 207 cm-1

1 million times more molecules in v=0 than v=1.

In IR absorption spectrum, will see transitions from v=0 only.

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% Acetophenone molecules in low energy vibrational states at thermal equilibrium at 40 K (top) and 298 K (bottom).

In contrast to diatomics like HCl, polyatomic molecules with low frequency vibrational modes can have a relatively small population in the zero point vibrational level at room temperature.

Journal of Chemical Physics 75, 4758 (1981) 12


N2/N1 = (g2/g1) e -(E2-E1)/kT

J=0

J=3

What is the relative population of molecules in J=3 vs. J=0 for HCl at 298 Kelvin? (HCl rotational constant is 10.6 cm-1).

More molecules are in J=3 than in J=0. In absorption spectrum, will see transitions from many different J levels of the same v.

NJ=3/NJ=0 = (7/1) e - (EJ=3 - EJ=0) / kT

Answer: degeneracy of rotational states = 2J+1

e -127/207 = 3.8

0 EJ=3 = BJ(J+1) = 10.6 cm-1 (3)(4) = 127 cm-1

NJ=3/NJ=0 = (7/1)

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Figure 13.2 p. 500

Gas Phase H79Br Rotation-Vibration Spectrum

Labels show J in the lower v state → J in the higher v state

1→0

2→1

3→2

4→3

0→1

1→2

2→3

3→4

P Branch R Branch

R(0) R(1)

R(2) R(3)

P(1) P(2)

P(3) P(4)

Lines are labeled by initial value of J:

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Ene

rgy

≈

v=0

v=1

J

R(0) R(1)

We will use the �rigid� rotator approximation: assume that B (rotational constant) does not vary with v (vibrational state). Then:

R(2)

R(1) - R(0) = 2B ∼

R(2) - R(1) = 2B ∼

etc.

Lines in the R branch are

spaced by about 2B. ∼ hνo

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Ene

rgy

≈

v=0

v=1

J

P(1) P(2)

P(3)

Lines in the P branch are

also spaced by about 2B. ∼

2B ∼

P R

Actually the bond is longer, so B is smaller, in v=1 than v=0 (�vibration-rotation interaction�).

→ With increasing J, P spacings increase; R spacings decrease. 16

From the IR rotational-vibrational spectrum of a gas phase molecule, can measure:

bond length

vibrational force constant

Problem: For H81Br, lines in the above spectrum are spaced by about 17 cm-1. Calculate the equilibrium bond length.

ћ = 1.055 x 10-34 J•s 1 amu = 1.661x10-27 kg

Answer: 1.41 Å

∼ 2B ∼

P R

B = h / 8 π2 I c

I = µr2

µ = m1 m2 / (m1+m2)

∼

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Lecture 35: Chapter 13 Spectroscopy Chap. 13 Sect. 5 and Chap. 5 Section 3:

Vibrational energy levels for a Morse potential energy curve

Chap. 13 Sect. 9:

Vibrational modes of polyatomic molecules

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Quantum Mechanical Harmonic Oscillator

When the S. equation is solved exactly, the energy levels turn out to be:

Ev = (v+ ½) h νo v = 0, 1, 2, ...

Energy levels are equally spaced.

Spacing between adjacent levels is h νo .

To absorb or emit light, require: Ephoton = ∆Esystem = hνo

REVIEW

E0 = ½ hνo

Classical Oscillator Frequency

µ

Zero point vibrational energy

( the molecule is always vibrating at any T )

k vibrational force constant

µ reduced mass

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Quantum Mechanical Harmonic Oscillator

nodes v Energy

∆E = h νo

∆E = h νo

∆E = h νo

As v increases:

tunneling decreases

particles more likely to be found near classical turning points

Correspondence Principle

U=½ kx2

REVIEW

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Vibrational Normal Modes

To specify positions of N nuclei in a molecule requires 3N coordinates (called 3N �degrees of freedom�).

Vibrational Rotational Translational

Nuclear Motion: 3 Types

Translation of the molecule in space (motion of the center of mass) Rotation of the molecule about its (fixed) center of mass (�Rigid Rotator�) Vibrations of the nuclei (molecule�s center of mass fixed)

Degrees of Freedom (3N Total)

3

3 (nonlinear molecule) 2 (linear molecule)

3N-6 (nonlinear) 3N-5 (linear)

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Vibrational Normal Modes of Polyatomic Molecules

H2O

CO2 linear

nonlinear

N=3 (# atoms) 3N-6 = 3 normal modes

3N-5 = 4 normal modes

The normal modes are independent, so the total vibrational energy is the sum:

p. 521

( ) j

n

jjvib hvE

vib

ν∑=

+=1

21vj ½

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Comparison of the HO and the

(more realistic) Morse potential for a diatomic (pp. 163, 505)

The HO is a good model near Re (the equilibrium bond length), i.e., for low vibrational levels.

separated atoms

internuclear separation

Pote

ntia

l Ene

rgy

Morse potential (solid line)

Morse:

finite # of vibrational levels

energy spacing (and νo) decrease with increased v

HO (parabolic) potential (dashed line)

REVIEW

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Since the shape of the molecular potential energy curve does not depend on the # of neutrons in the nucleus, it is essentially unchanged upon isotopic substitution.

The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location.

Engel p. 135

Bond Dissociation Energy

Potential Energy Curve for HCl

E.g., for the diatomic molecule 1H35Cl, there are 24 vibrational levels.

So, k for HCl and DCl are (almost) the same, and νo for DCl is about √2 lower.

REVIEW

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Morse Potential Energy Curve

Parabolic potential energy curve (HO approx):

U(x) = ½ k x 2

where x = R - Re (= displacement from equilibrium bond length )

Morse potential energy curve (more realistic):

2)( )1( eRR

e eDhcV −−−= β

where De = depth of potential minimum (bond strength), cm-1

ωµ

β21

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

ehcDω (s-1) = (k / µ)1/2

cm-1

m-1

no units

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2)( )1( eRRe eD

hcV −−−= β

ωµ

β21

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

ehcD

What does this function look like?

At R = Re , have 20 )1( eDhcV

e −= = 0 This is the minimum, since elsewhere V will be positive.

As R → ∞, have 2)1( ∞−−= eDhcV

e = De

As R → 0, have 2)1( eRe eD

hcV β+−= V increases

large + number

De

Re

V(R

)

→ R

→

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When the Morse potential (instead of the parabolic potential)

is used in the Schrödinger equation for the 1-D oscillator:

2)( )1( eRReMorse eDhcV −−−= β

x

Resulting quantized energy levels relative to V=0, the potential energy minimum:

The vibrational energy levels get closer together as vibrational quantum number v increases.

�harmonic frequency�

�anharmonicity constant�, a small + number

G(v) = (v + ½) ωe - (v + ½)2 ωexe v = 0, 1, 2, ...

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G(v) = (v + ½) ωe - (v + ½)2 ωexe v = 0, 1, 2, ...

�Energies� relative to minimum of potential energy curve:

= (v + ½) ωe - (v2 + v + ¼) ωexe

Zero point vibrational energy (v=0) above potential minimum:

G(0) = (½) ωe - (¼) ωexe v = 0

So, relative to v=0, the vibrational energy levels are:

G0(v) = v ωe - v (v+1) ωexe v = 0, 1, 2, ...

Same as: Text p. 506 28


G0(v) = v ωe - v (v+1) ωexe

Problem: For the 3Σg ground state of O2, the vibrational constants* are: ωe = 1580.19 cm-1, ωexe = 11.98 cm-1

What is the zero point energy? Predict the energies of v=1, 2, 3 relative to v=0.

v

0

1

2

3

G0 = __ ωe - __ ωexe

0 0 1 2

2 6

3 12

Interval

2 ωexe

4 ωexe

6 ωexe

Each sequential vibrational interval decreases by 2 ωexe

G(0) = (½) ωe - (¼) ωexe

* Constants of Diatomic Molecules, G. Herzberg http://webbook.nist.gov/chemistry/form-ser.html

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Morse Potential Energy Curve G0(v) = v ωe - v (v+1) ωexe

Problem: For the 3Σg ground state of O2, the vibrational constants* are: ωe = 1580.19 cm-1, ωexe = 11.98 cm-1

What is the zero point energy? Predict the energies of v=1, 2, 3 relative to v=0.

G(0) = (½) ωe - (¼) ωexe

Answer:

Zero point energy: G(0) = (½) ωe - (¼) ωexe

ωe = 1580.19 cm-1

ωexe = 11.98 cm-1

v=1 G0(1) = ωe - 2 ωexe

G0(2) = 2ωe - 6 ωexe v=2

v=3 G0(3) = 3ωe - 12 ωexe

787 cm-1

1556 cm-1

3089 cm-1

4597 cm-1

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G(0) = (½) ωe - (¼) ωexe

We can also use the Morse vibrational parameters to estimate the bond dissociation energy:

ee

ee xD

ωω4

2

=

Problem: For the 3Σg ground state of O2, the vibrational constants are ωe = 1580.19 cm-1 and ωexe = 11.98 cm-1. Estimate the dissociation energy.

De

Re

V(R

)

→ R

( )98.11419.1580 2

=eD = 52,110 cm-1 = 6.46 eV

( 8066 cm-1 = 1 eV )

Actual value: 41,760 cm-1 (5.18 eV) Morse estimate of De based on only ωe and ωexe is typically too high.

At high v, higher order terms ( (v + ½)3, etc. ) become important.

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G(0) = (½) ωe - (¼) ωexe

We can also define Do , the bond dissociation energy relative to the zero point level:

ee

ee xD

ωω4

2

=

Problem: For the 3Σg ground state of O2, De is 41,760 cm-1, ωe = 1580.19 cm-1 and ωexe = 11.98 cm-1. What is D0?

D0 = De - G(0) De

Re

V(R

)

→ R

D0 v=0 -

Answer: D0 = De - G(0) = De - ((½) ωe - (¼) ωexe)

D0 = 41,760 cm-1 - 787 cm-1 = 40,970 cm-1

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Morse Potential Energy Curve G(0) = (½) ωe - (¼) ωexe

G0(v) = v ωe - v (v+1) ωexe

Problem: Draw a potential energy curve for the 3Σg ground state of O2 incorporating our results:

D0 = 40,970 cm-1

De = 41,760 cm-1 (5.18 eV)

Zero point energy: 787 cm-1

G0(1) = 1556 cm-1

G0(2) = 3089 cm-1

G0(3) = 4597 cm-1

The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location.

1.21 Å Engel p. 135 (HCl)

41,7

60 c

m-1

40,9

70 c

m-1

787

v=1 1556

v=2 3089

v=3 4597

(# vib. levels not correct) O2 3Σg

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Spectroscopy Lecture 36: Chap. 13 Sect. 6-7

Excited electronic states and electronic spectroscopy

Franck-Condon Principle

Fates of excited electronic states: emission of light (fluorescence, phosphorescence), photochemistry

Laser-induced fluorescence spectrum of gas phase I2 (p. chem lab 4511)

Chap. 10 pp. 393, 396; Chap. 3 problem 6: Excited states of linear polyenes (e.g., butadiene, hexatriene) and the Particle-in-a-Box model

Photoisomerization of retinal and the process of vision

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6 eV

Engel p. 305

0

The excited 1∆g and 1Σg states

also have the ground state ...(2pπg)2 configuration.

The spin-forbidden absorption from the X3Σg ground state to the b1Σg excited state (7900 Å, red) accounts for the blue color of liquid oxygen.

Excited �a� and �b� States of O2

Singlet O2 molecules are very reactive. They are deliberately prepared in photodynamic therapy (PDT) to treat diseases including cancer and macular degeneration.

REVIEW

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Excited Electronic �B� State of O2

These �Schumann Runge bands� of O2 absorb harmful UV sunlight (175-200 nm).

6 eV

Engel p. 305

0

In the B state, O2 has enough energy to photodissociate into two O atoms: essential for O + O2 → O3 (ozone) in stratosphere

(2sσg)2(2sσu)2(2pσg)2(2pπu)4(2pπg)2 X3Σg

(2sσg)2(2sσu)2(2pσg)2(2pπu)3(2pπg)3 B3Σu

The bond order decreases from 2 in the ground state to 1 in the excited state (πg ← πu transition).

X3Σg B3Σu

vib. freq. (cm-1) 1580 709 bond length (Å) 1.21 1.60

REVIEW

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(2pπu)4(2pπg)2 X3Σg

(2pπu)3(2pπg)3 B3Σu

Engel p. 305

0

Te 49,793 cm-1

ν00 ∼

49,358 cm-1

Excited electronic state energies (with respect to the ground state) can be expressed as ν00

∼

(also called T0), the separation between zero point levels,

or as Te (�equilibrium�), the separation between the potential energy minima (not directly observed).

Excited Electronic �B� State of O2

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Electronic Spectroscopy: Franck-Condon Principle

What is the �selection rule� that governs which vibrational states will be observed in electronic absorption and emission spectra?

Franck-Condon Principle: The electronic excitation is much faster than nuclear motion, so the electronic transition will be �vertical� (positions of nuclei initially unchanged).

As a result, if the 2 electronic states have similar equilibrium geometries, little vibrational excitation will be observed.

If their equilibrium geometries (bond lengths) are very different, a vibrational �progression� will be observed, as in the I2 emission spectrum.

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Laser-Induced Fluorescence Spectroscopy of Gas Phase I2 (Chem 4511 lab)

http://itl.chem.ufl.edu/4411L_f00/i2_lif/i2_lif1.html

Laser: 5145 Å (19,436 cm-1)

Absorption: v� = 43←v� = 0

(Spin forbidden transition observed due to spin-orbit coupling)

Emission:

v� = 43→v� = ?

Re = 2.67 Å

Re = 3.03 Å

orange

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Laser-induced fluorescence spectrum of gas phase I2 ( from Chem 4511 lab manual )

http://itl.chem.ufl.edu/4411L_f00/i2_lif/i2_lif0.html#intro

v' = 43 → v" = 1

v' = 43 → v" = 26

211

183 cm-1

excited electronic state ground electronic state

Can map out the detailed potential energy curve from this kind of data.

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In liquids, emission occurs from v=0 following rapid collisional vibrational relaxation in the excited electronic state.

Fig. 15.2 p. 594

Fates of Excited Electronic States:

Emission of Light (Fluorescence, Phosphorescence)

This can produce a shift between the maxima (λmax) of the absorption and fluorescence spectra.

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Ψn ∝ sin (nπx/a) Particle-in-a-Box

Wade, �Organic Chemistry�

π and π* Molecular Orbitals of Conjugated Linear Hydrocarbons

antibonding (π*)

⇂

↿ hν

⇌

⇌

⇌

Ephoton = ΔEsystem ≈ E3-E2

MO�s: Chap. 10 page 396

butadiene, C4H6 Same symmetry, # of nodes

As for PIB, energy increases with # of nodes.

cis-trans photoisomerization

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Fates of Excited Electronic States: Photochemistry Example: cis → trans photoisomerization around a double C=C bond upon the absorption of a visible photon by visual pigment,

11-cis retinal

all-trans retinal

Atkins and de Paula, P. Chem. for the Life Sciences, p. 587

the protein rhodopsin, in rods and cones on the retina, anchors the retinal molecule

hν π* ← π excitation

Photoisomerization takes about 200 femtoseconds,

and initiates the events of vision.

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