chemistry 2 · in spectroscopy, we tend to use the letter “vee” to indicate the quantum number...
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Chemistry 2
Lecture 6
Vibrational Spectroscopy
Learning outcomes
Light behaves like an oscillating electromagnetic field. The electric field interacts with charges. Two charges separated in space represent a dipole moment which can interact with an electric field. Energy can only be taken or added to the electric field in units of hν(photons).
Assumed knowledge
• Be able to manipulate and use the key equations given in the green box at the end of the lecture.
•Utilize the harmonic oscillator and anharmonic oscillator as a model for the energy level structure of a vibrating diatomic molecule.
Revision: The Electromagnetic Spectrum
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Revision: Light as a EM field
Revision: The Electromagnetic Spectrum
Spectral range
λ (nm) ν (Hz) (cm‐1) Energy (kJ/mol)
Spectroscopy
Radio ~1 x 109 ~108 ~0.03 ~10‐8 NMR/ESR
Microwave ~100,000 ~1012 ~30 ~10‐2 Rotational
ν~
,
Infrared ~1000 ~1014 ~3,000 ~103 Vibrational
Visible 400‐750 4‐6 x 1014 14,000‐25,000
1 – 3x105 Electronic
Ultraviolet 100‐400 ~1015 ~40,000 ~5x105 Electronic
X ray <100 >1016 >100,000 >106 Core electronic
Quantity Symbol SI Unit Common Unit
Energy E J kJ/mol
Frequency ν s−1 or Hz
s−1 or Hz
Revision: Fundamental equations
νhE =Hz
Wavelength λ m nm or μm
Wavenumber m−1 cm−1
λν c=
cν
λν == 1~ ν~
Constant Symbol Value
Speed of light c 3.00 x 108 m/s
Planck constant h 6.626 x 10‐34 Js
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Interaction of light with matter
Light can interact with matter in many different ways. We will explore three of these in this course:
1. Absorption2. Emission3. Scattering
now
Others include: stimulated emission, multiphoton processes, coherent spectroscopies
now
later
Classical absorption of light
B li ht i ill ti l t ti fi ld itBecause light is an oscillating electromagnetic field, it can cause charges to oscillate. If the charge can oscillate in resonance with the field then energy can be absorbed.
Alternatively, an oscillating charge can emit radiation with frequency in resonance with the original oscillation.
For example, in a TV antenna, the oscillating EM field broadcast by the transmitter causes the electrons in your antenna to oscillate at the same frequency.
Classical absorption of light
Channel Frequency (MHz)
2 69.75
7 187.75
FM Broadcast Band88.0 ‐ 108.0 MHz with 200 kHz channel spacing
AM B d t B d
TelevisionRadio
9 201.75
10 214.75
28 532.75
AM Broadcast Band526.5 ‐ 1606.5 kHz with 9 kHz channel spacing.
Q: If the optimal design of an antenna is ½ wavelength, what size should your TV antenna be to pick up Ch 9? What about SBS (Ch28)?
A: ν = c/λ. 532.75×106 = 2.9979×108/λ, λ=1.78m.
Best antenna is 0.89m for SBS.
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Classical absorption of light
What about a molecule as an antenna?
How can we get oscillating charges in a molecule?
+ -+-
Classical absorption of light
-+
+-
Rotating a permanent dipole causes an oscillation of charge
Classical absorption of light
+ -
+ -
Vibrating a permanent dipole causes an oscillation of charge
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Classical absorption of light
Moving electrons in a molecule can change the dipole of the
l l j = 2
j = 3
molecule.
j = 1
j = 0
j = 2
⎟⎠⎞
⎜⎝⎛+=
62cos2 πβαε jj
The “resonance condition” and the EM spectrum
The rotational period of a molecule is about 10 ps (10‐11 s). The charge is therefore oscillating every 10‐11 s or 1011 times per second. In what range of the electromagmetic spectrum would you expect pure rotational spectroscopy to lie?
11
=> λ = 3x108 / 1011 = 3x10‐3 m
ν = 1011 Hz
Vibrational spectroscopy is known to occur in the infrared region of the spectrum. Calculate the frequency of the oscillating dipole and hence determine how quickly molecules vibrate.
Take mid‐IR to be 3000 cm‐1 (CH stretch)ν = c × ν 3x1010 × 3000 = 9 x 1013 (say 1014 s‐1)~
The “resonance condition” and the EM spectrum
ν = c × ν 3x1010 × 3000 = 9 x 1013 (say 1014 s 1)Therefore a molecule vibrates in about 10 fs (or 10‐14 s)
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The Quantum Harmonic OscillatorSolution of Schrödinger equation:
⎟⎠⎞
⎜⎝⎛ +=
21
2nkh
n μπε μπ
ν k21
=But:
νε hnn )( 21+= S.I. units
ω)v()v( 21+=G G(v) & ω in cm-1
G(v) is the “energy” from the bottom of the well and ω is the “harmonic frequency”
Oscillation (vibrational) frequencyn = 0,1,2,…
Need to know!
harmonic
“Anharmonic oscillator” (A.H.O.)
anharmonic
Molecules dissociates
The Morse potential as an approximation of an AHO
[ ]2)(1)r( erre eDV −−−= β
V (r
)where2
122 ⎤⎡ cμπ
r
2
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
ee hD
cμπωβ
re
De
You DO NOT need to know or use these equations
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The Schrödinger equationfor a Morse potential
)()()(2 2
22
xxrVdxd
nnn ψεψμ
=⎟⎟⎠
⎞⎜⎜⎝
⎛+−
h
where ( )2)(1)( erreDrV −−= β
Dissociation energy
where ( )1)( e eDrV −=
The energy solution is:
( ) ( ) ( )e
ee D
hhE4
nn2
22
12
1v
νν +−+=
The important equations!
( ) ( )e
ee D
G4
+v+v=)v(22
21
21 ω
ω
( ) ( )211
all in wavenumber (cm‐1) units:
( ) ( ) eee xG ωω 22
12
1 +v+v=)v(
This one you DO have to know how to use
In spectroscopy, we tend to use the letter “vee” to indicate the quantum number for vibration. The vibration frequency is indicated by ωe in cm-1. When solving the general quantum mechanical problem we used the letter n, to minimize confusion with “nu”, the vibrational frequency in s-1.
The Morse energy levels
( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
V (r)
Harmonic Anharmonic
Harmonic frequency
Anharmonicityconstant
r
term term
wexe is usually positive, so vibrationallevels get closer together
μπν k
21
= μπω
kc2
1=
The force constant for the AHO is the same as for the harmonic case:
or
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Different levels of approximation…• General AHO:
( ) ( ) ( ) ...vvv)v( 32
122
12
1 −+++−+= eeeee yxG ωωω
• Morse oscillator:
( ) ( )G 211)( ( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
• Harmonic oscillator:
( )ω21v)v( +=G
The level of approximation to use depends on:i) the information you haveii) the information you need
The important equations!
( ) ( )e
ee D
G4
+v+v=)v(22
21
21 ω
ω
( ) ( )211
• In wavenumber (cm‐1) units:
( ) ( ) eee xG ωω 22
12
1 +v+v=)v(
The ones you DO have to know how to use
ee
ee x
Dωω
4
2=
ee
ee x
Dωω
4
2=
This is the energy from the bottom of the well to the dissociation limit.
V (r
)
D0
But we know about zero point energy
Dissociation energy
r
De
G(0)
(Zero‐point energy)
D0 = De – G(0)
But we know about zero‐point energy, therefore slightly less energy is required to break the bond.
We can estimate the bond dissociation energy from spectroscopic measurements!
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Selection rules• All forms of spectroscopy have a set of selection rules that limit the number of allowed transitions.
• Selection rules arise from the resonance condition, which may be expressed as a transition dipole moment:
0)( )()( 1*221 ≠= ∫ drrrr ψψ μμ
Upper state
lower state
molecular dipole
vibrationalcoordinate
• Selection rules tell us when this integral is zero
selection rule:Δv=±1
Δv=+1: absorptionΔ 1 i i
Harmonic Oscillator Selection Rules
Δv=-1: emission
Selection rules limit the number of allowed transitions
If an oscillator has only one frequency associated with it, then it can only interact with radiation of that frequency.
Thermal populationAt normal temperatures, only the lowest vibrational state(v =0 ) is usually populated, therefore, only the first transition is typically seen.
Transitions arising from v≠0 are called “hot bands”(Their intensity is strongly
temp. dependent)
Much of IR spectroscopy can understood from just these two results of the quantum harmonic oscillator:
E = (v+½)hν and Δv = ±1
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For example:
v = 1 ← 0
1.0
1.2
1.4
1.6
1 tone
)4260
cm
-1
(firs
t ove
rtone
)
m-1,
(fund
amen
tal)
CO
nce
More problems with harmonic model….
COWhat are these?
2000 4000 6000
0.0
0.2
0.4
0.6
0.8
6352
cm
-1
(sec
ond
over
t
2143
c
x 100x 10
Abso
rban
W avenum ber (cm -1)
What are the new peaks?
• Three peaks…i. 2143 cm‐1
ii. 4260 cm‐1
iii. 6352 cm‐1
v 3
v=4
v=0
v=1
v=2
v=3These are almost 1 : 2 : 3
which suggests transitions might be
0 → 10 → 20 → 3
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Anharmonic oscillator (A.H.O.) selection rules:
H.O.There are none!
But!...Harmonic and anharmonicmodels are very similar at low
A.H.O.
energy, so selection rules of AHO converge on HO as the anharmonicity becomes less:
A.H.O. selection rule:Δv=±1,±2, ±3
Intensity gets weaker and weaker (typically 10×weaker for each)
Anharmonic oscillator (A.H.O.)
A.H.O. selection rule:Δv= ±1,±2, ±3
Δv = 1 : fundamentalΔv = 1 : fundamentalΔv = 2 : first overtoneΔv = 3 : second overtone, etc
• Consider the infrared absorption spectrum of CO below.a) From the wavenumber measurements on the spectrum, assign the
spectrum, hence determine the harmonic frequency, ωe (in cm‐1) and the anharmonicity constant ωexe (in cm‐1).
b) Estimate the bond dissociation, D0, for this molecule. There is no absorption below 2000 cm‐1.
Typical Exam Question
Fundamental:2143 = G(1) – G(0),
Overtone:4260 = G(2) – G(0)
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Using the spectra to get information…
G(1)−G(0) = [(1.5)ωe – (1.5)2 ωexe] − [(0.5)ωe – (0.5)2 ωexe]
2143 = ωe – 2ωexe …(1)
( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
G(2)−G(0) = [(2.5)ωe – (2.5)2 ωexe] − [(0.5)we – (0.5)2 ωexe]
4260 = 2ωe – 6ωexe …(2)
Two simultaneous equations (simple to solve)
→ ωe = 2169 cm‐1, and ωexe = 13 cm‐1
Using the spectra to get information…
1-2
cm 500,90134
)2169(=
×=eD
1-0
cm 400,891080500,90
)0(
=−=
−= GDD e
V (r
)
D0
ee
ee x
Dωω
4
2=
89,400 cm‐1 = 1069 kJ/mol
c.f. exp. value: 1080 kJ/mol
r
De
G(0)
(Zero‐point energy)
Why the difference?Remember Morse is still an approx. to the true intermolecular potential. Still 2% error is pretty good for just 2 measurements!
Equations to know how to use…
ω)v()v( 21+=G
μπν k
21
=μ ≡m1 × m2m1 + m2
( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
ω)v()v( 2+=G
ee
ee x
Dωω
4
2=)0(0 GDD e −=
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Summary
• Light interacts with vibrating dipoles. It may set a vibrating dipole in motion. Light may be emitted by a vibrating dipole
• The vibrational energy level structure of a diatomic molecule may be represented by a harmonic oscillator, giving rise to quantization and spectroscopic selection rules.quantization and spectroscopic selection rules.
• The anharmonic oscillator represented by the Morse potential correctly describes bond dissociation and short range repulsion. It is a better representation of the energy levels observed in a real molecule.
• Anharmonic oscillators exhibit overtone transitions.
Next lecture
• The vibrational spectroscopy of polyatomic molecules.
Practice Questions1. Which of the following diatomic molecules will exhibit an infrared
spectrum?a) HBr b) H2 c) CO d) I2
2. An unknown diatomic oxide has a harmonic vibrational frequency of ω = 1904 cm−1 and a force constant of 1607 N m −1. Identify the molecule.a) CO b) BrO c) NO d) 13CO
3 A th i th ib ti l l l i f h i3. As the energy increases, the vibrational level spacing for a harmonic oscillator is:a) increases b) decreases c) stays constant
4. As the energy increases, the vibrational level spacings for a Morse oscillator usually:a) increase b) decrease c) stay constant
5. As the energy increases, the vibrational level spacings for an anharmonicoscillator usually:a) increase b) decrease c) stay constant
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Practice Questions6. For a Morse oscillator the observed dissociation energy, D0, is related to
the equilibrium vibrational frequency and the anharmonicity by the following expression:a) ωe
2/4ωexe b) [ωe2/4ωexe]‐G(0) c) (v+½)ωe+(v+½)2ωexe d) (v+½)ωe
7. Which of the following statements about the classical and quantum harmonic oscillator (HO) are true (more than one possible answer here)?( ) ( p )a) The classical HO frequency is continuous, whereas the quantum frequency is discrete.b) The classical HO has continuous energy levels, whereas the quantum HO levels are discrete.c) The classical HO depends on the force constant, but the quantum HO does not.d) The classical HO may have zero energy, but the quantum HO may not.e) The classical HO does allow the bond to break, whereas the quantum HO does.
Practice Questions8. Which of the following statements correctly describe features of the
quantum Morse oscillator and energy levels associated with it?a) The vibrational energy levels get more closely spaced with increasing v.b) The vibrational energy levels approach a continuum as the dissociation energy is approached.) Th M ill t d HO l th t lc) The Morse oscillator and HO are nearly the same at very low vd) The Morse potential is steeper than the HO for r < ree) The Morse potential exactly describes the interatomic potential.