Chemistry 2 - University of Sydney Faculty of Science · PDF fileChemistry 2 Lecture 6...
Transcript of Chemistry 2 - University of Sydney Faculty of Science · PDF fileChemistry 2 Lecture 6...
Chemistry 2Chemistry 2
Lecture 6
Vibrational Spectroscopy
Assumed knowledgeLight 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( h t )
Learning outcomes(photons).
• 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 f th l l t t f ib ti di t i l lfor the energy level structure of a vibrating diatomic molecule.
Revision: h l i SThe Electromagnetic Spectrum
Revision: Light as a EM fieldRevision: Light as a EM field
Revision: The Electromagnetic Spectrum
Spectral range
λ (nm) ν (Hz) (cm‐1) Energy (kJ/mol)
Spectroscopyν~
Radio ~1 x 109 ~108 ~0.03 ~10‐8 NMR/ESR
Microwave ~100,000 ~1012 ~30 ~10‐2 Rotational,
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 CoreX ray <100 >10 >100,000 >10 Core electronic
Revision: Fundamental equations
Quantity Symbol SI Unit Common Unit
Energy E J kJ/molνhE Energy E J kJ/mol
Frequency ν s−1 or Hz
s−1 or Hz
νhE =Hz
Wavelength λ m nm or μmλν c=
Wavenumber m−1 cm−1
cν
λν == 1~ ν~
Constant Symbol Valuecλ Constant Symbol Value
Speed of light c 3.00 x 108 m/s
Planck constant h 6.626 x 10‐34 Js
Interaction of light with matter
Light can interact with matter in many different ways. We will explore three of these in this course:in this course:
1. Absorption
2 Emission
now
no2. Emission
3. Scattering
now
later
Others include: stimulated emission, multiphoton processes, coherent spectroscopies
Classical absorption of lightClassical 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 resonancewith the field then energy can be absorbed. gy
Alternatively, an oscillating charge can emit radiation with frequency in resonance with the original oscillationfrequency 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 lightClassical absorption of lightTelevision
RadioChannel Frequency
(MHz)FM Broadcast Band88.0 ‐ 108.0 MHz with 200 kHz channel
Radio
2 69.75
7 187.75
88.0 108.0 MHz with 200 kHz channel spacing
AM B d B d9 201.75
10 214.75
AM Broadcast Band526.5 ‐ 1606.5 kHz with 9 kHz channel spacing. 10 214.75
28 532.75
p g
Q: If the optimal design of an antenna is ½ wavelength, what sizeQ: 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.
Classical absorption of lightClassical absorption of light
What about a molecule as an antenna?What about a molecule as an antenna?
H t ill ti h i l l ?How can we get oscillating charges in a molecule?
+ -+-
Classical absorption of lightClassical absorption of light
-+ Rotating a permanent
+dipole causes an oscillation of charge+
-
Classical absorption of lightClassical absorption of light
+ -Vibrating a permanent dipole causes an oscillation of charge
+ -
Classical absorption of lightClassical absorption of light
Moving electrons in a
j = 3
gmolecule can change the dipole of the
l l j = 2molecule. j = 2
j = 1⎟⎠⎞
⎜⎝⎛+=
2cos2
πβαε jj ⎟⎠
⎜⎝ 6
β jj
j = 0
The “resonance condition” and the EM spectrum
Th i l i d f l l i b 10 (10 11 ) ThThe 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 g g p y ppure rotational spectroscopy to lie?
11
=> λ = 3x108 / 1011 = 3x10‐3 m
ν = 1011 Hz
The “resonance condition” and the EM
Vib i l i k i h i f d i
spectrumVibrational 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.q y
Take mid‐IR to be 3000 cm‐1 (CH stretch)ν = c × ν 3x1010 × 3000 = 9 x 1013 (say 1014 s‐1)
~ν = c × ν 3x1010 × 3000 = 9 x 1013 (say 1014 s 1)Therefore a molecule vibrates in about 10 fs (or 10‐14 s)
The Quantum Harmonic Oscillator
Solution of Schrödinger equation:
⎟⎞
⎜⎛ +=
1n
khε ν k1=But:⎟
⎠⎜⎝
+22
nn μπε
μπ2But
Oscillation (vibrational) frequencyn = 0,1,2,…
νε hnn )( 21+= S.I. unitsn )(
ω)v()v( 21+=G G(v) & ω in cm-1
Need to know!ω)v()v( 2+G ( )
G(v) is the “energy” from the bottom of the well and ω is the “harmonic frequency”
“Anharmonic oscillator” (A.H.O.)Anharmonic oscillator (A.H.O.)
harmonic
anharmonic
Molecules dissociatesMolecules dissociates
The Morse potential as an i ti f AHOapproximation of an AHO
[ ]2[ ]2)(1)r( erre eDV −−−= β
V (r
)where2
122 ⎤⎡ cμπ2
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
ee hD
cμπωβDDe
You DO NOT need to know or use these equations
rre
The Schrödinger equationfor a Morse potentialfor a Morse potential
22 d ⎞⎛ h)()()(
2 2
22
xxrVdx
dnnn ψεψ
μ=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−
h
μ ⎠⎝
where ( )2)(1)( erreDrV −−= βwhere ( )1)( e eDrV −=
The energy solution is:The energy solution is:
( ) ( ) ( )ehhE nn
2211
νν +−+= ( ) ( )e
e DhE
4nn 2
12
1v ν +−+=
Dissociation energy
The important equations!
all in wavenumber (cm‐1) units:
( ) ( ) ee D
G4
+v+v=)v(2
22
12
1 ωω( ) ( )
ee D4
)( 22
( ) ( )211( ) ( ) 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
V (r)
Harmonic frequency
Anharmonicityconstant
( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
V
( ) ( ) eee 22)(
Harmonic Anharmonic
term term
wexe is usually positive, so vibrationale e y p ,levels get closer together
k1 k1
The force constant for the AHO is the same as for the harmonic case:
rμπν k
2
1= μπ
ωk
c2
1=or
Different levels of approximation…ff f pp• General AHO:
( ) ( ) ( )32( ) ( ) ( ) ...vvv)v(3
212
21
21 −+++−+= eeeee yxG ωωω
• Morse oscillator:
( ) ( )G211)( ( ) ( ) eee xG ωω 2
21
21 vv)v( +−+=
• Harmonic oscillator:
( )ω21v)v( +=G ( )ω2v)v( +=G
The level of approximation to use depends on:pp p
i) the information you have
ii) the information you need
The important equations!
• In wavenumber (cm‐1) units:
( ) ( ) eeG +v+v=)v(
22
21
21 ω
ω( ) ( )e
e DG
4vv)v( 22 ω
( ) ( )211( ) ( ) eee xG ωω 22
12
1 +v+v=)v(
2The ones you DO have to
know how to useee
ee x
Dωω
4
2
=eexω4
Dissociation energy
eeD
ω 2
= V (r
)
eee xω4
This is the energy from the bottom of the
V
DThis is the energy from the bottom of the well to the dissociation limit.
D0
But we know about zero point energy DeBut we know about zero‐point energy, therefore slightly less energy is required to break the bond.
D0 = De – G(0)
G(0)We can estimate the bond dissociation energy from spectroscopic
rG(0)
(Zero‐point energy)
energy from spectroscopic measurements!
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 di l
vibrationaldi tstate statedipole coordinate
• Selection rules tell us when this integral is zero
Harmonic Oscillator Selection Rules
selection rule:Δv=±1
Δv=+1: absorptionΔ 1 i iΔv=-1: emission
If an oscillator has only one frequency associated with it, then it can only interact with radiationit can only interact with radiation of that frequency.
Selection rules limit the number of allowed transitions
Thermal populationp p
At normal temperatures, only the lowest vibrational statethe 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(Their intensity is strongly
temp. dependent)
Much of IR spectroscopy can understood from just h l f h h i illthese two results of the quantum harmonic oscillator:
E = (v+½)hν and Δv = ±1
For example:
1 0v = 1 ← 0
More problems with harmonic model….
1.6
CO
p
What are these?
1.2
1.4
m-1 on
e)
amen
tal)
COCO
What are these?
1.0
1.2
1 tone
)
4260
cm
(firs
t ove
rto
m-1,
(fun
da
nce
0 .6
0.8
6352
cm
-1
econ
d ov
ert
2143
c
Abs
orba
n
0 2
0.4
(se
x 10
A
0.0
0.2x 100
2000 4000 6000
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=2
v=3These are almost 1 : 2 : 3
hich s ggests transitions
v=1
which suggests transitions might be
v=00 → 10 → 20 → 3
Anharmonic oscillator (A.H.O.) l ti lselection rules:
H.O.There are none!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.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
Typical Exam Question• Consider the infrared absorption spectrum of CO below.
a) From the wavenumber measurements on the spectrum, assign the
yp
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.
Fundamental:Fundamental:2143 = G(1) – G(0),
Overtone:4260 = G(2) – G(0)
Using the spectra to get information…g p g f
( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
G(1)−G(0) = [(1.5)ωe – (1.5)2 ωexe] − [(0.5)ωe – (0.5)2 ωexe]
( ) ( ) eee 22
2143 = ωe – 2ωexe …(1)
G(2)−G(0) = [(2.5)ωe – (2.5)2 ωexe] − [(0.5)we – (0.5)2 ωexe]
4260 = 2ωe – 6ωexe …(2)e e e ( )
Two simultaneous equations (simple to solve)
→ ωe = 2169 cm‐1, and ωexe = 13 cm‐1
Using the spectra to get information…
1-2
cm50090)2169(
==D V (r
)
eDω 2
= cm500,90134
=×
=eD
0 )0(−= GDD
V
Dee
e xD
ω4=
1-
0
cm 400,891080500,90
)0(
=−=
GDD e D0
89,400 cm‐1 = 1069 kJ/mol
c.f. exp. value: 1080 kJ/mol
De
c f e p a ue 080 J/ o
Why the difference?R b M i ill h
G(0)
Remember Morse is still an approx. to the true intermolecular potential. Still 2% error is pretty good for just 2
rG(0)
(Zero‐point energy)
measurements!
Equations to know how to useEquations to know how to use…
μπν k
2
1=μ ≡
m1 × m2
m1 + m2
ω)v()v( 21+=G
μπ21 2
( ) ( )2
ω)v()v( 2+=G
( ) ( ) eee xG ωω 22
12
1 vv)v( +−+=
ee x
Dωω
4
2=)0(0 GDD e −=
eexω4
SummarySummary
• Light interacts with vibrating dipoles. It may set a vibrating dipole in motion. Light may be emitted by a vibrating dipole
Th ib ti l l l t t f di t i l l• 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 p yrange repulsion. It is a better representation of the energy levels observed in a real molecule.
• Anharmonic oscillators exhibit overtone transitions.
Next lectureNext lecture
• The vibrational spectroscopy of polyatomic molecules.
Practice Questions1. Which of the following diatomic molecules will exhibit an infrared
spectrum?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) ) ) y
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:oscillator usually:a) increase b) decrease c) stay constant
Practice Questions
6. For a Morse oscillator the observed dissociation energy, D0, is related to h ilib i ib i l f d h h i i b hthe 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)ωea) ωe / ωe e b) [ωe / ωe e] G(0) c) ( ½)ωe ( ½) ωe e d) ( 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 quantumb) 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 quantume) The classical HO does allow the bond to break, whereas the quantum HO does.
Practice Questions
8. Which of the following statements correctly describe features of the ill d l l i d i h i ?quantum Morse oscillator and energy levels associated with it?
a) The vibrational energy levels get more closely spaced with increasing v.c eas gb) 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.) p y p