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Section 4 M l l Vib ti l S t
Section 4 M l l Vib ti l S tMolecular Vibrational Spectroscopy
(lectures 6‐8 ish)Molecular Vibrational Spectroscopy
(lectures 6‐8 ish)( )( )
Quantum theoryPreviously: Quantum V l
Q yof atoms / molecules
Previously: QMechanics
Valence
Molecular Vibrational SpectroscopyDiatomic molecules (harmonic oscillator revision)Beyond the Harmonic OscillatorRotational structure in vibrational spectraPolyatomic molecules: Normal modesSymmetry of normal modesSymmetry of vibrational levelsCombination differencesCombination bands
Molecular l
Molecular lEnergy LevelsEnergy Levels
i.e., typically ΔEel >> ΔEvib >> ΔErot
Different electronic states (electronic arrangements)(electronic arrangements)
λΔ ≈≈E 2 x 104 – 105 cm‐1
500 – 100 nm
102 – 5 x 103 cm‐1
100 μm – 2 μm
3 – 300 GHz (0.1 – 10 cm‐1)
Transitions at λVis – UV
100 μm 2 μminfrared
10 cm – 1 mmmicrowave
4.1 Vibrational Energy Levels: The Harmonic oscillator4.1 Vibrational Energy Levels: The Harmonic oscillatorR0
Obeys Hooke’s Law, FF k x= −m1
m2
R0
Where kF is the Force constant (Nm‐1) and
∴The potential energy
R0
Where kF is the Force constant (Nm ) and x the displacement from equilibrium
( )2
0 2x
Fk xV x Fdx= − =∫
∴The potential energy, μ
x 2
Classical frequency of oscillation:
V=1/2 kx2
q y
12
Fvib
kν
π μ=
μ
Robert Hooke1635 ‐1703
4.2 Harmonic Oscillator Eigenvalues4.2 Harmonic Oscillator EigenvaluesEigenvalues:
( ) ‐11 1/cm vib FkG ν+
gThe vibrational terms
V=1/2 kx2
( ) 112 /cm
2vib F
v e eG vc c
ω ωπ μ
= + = =
v is the vibrational quantum number = 0, 1, 2,..q , , ,ωe the vibrational constant (in cm‐1),kF is the Force Constant,μ is the reduced mass
92 eω ω μ is the reduced mass
52 eω3
72 eω
eω
n.b., existence of zero point energy G0 = ωe/2 ≠ 0
no classical analogue and arises due 12 eω
32 eω
to Heisenberg Uncertainty
Equally spaced non‐degenerate energy levels
For HCl (1Σ+): kF = 510 Nm‐1 ωe = 2990.9 cm‐1 vibrational frequency ~ 9 x 1013 HzCO: kF = 1902 Nm‐1 ωe = 2169.8 cm‐1 vibrational frequency ~ 6.5 x 1013 Hz
4.2 Harmonic Oscillator Wavefunctions4.2 Harmonic Oscillator Wavefunctions
Th i f i f h H i O illThe eigenfunctions of the Harmonic Oscillator are products of Hermite polynomials and Gaussian functions: 1
⎛ ⎞⎛ ⎞v = 3
( ) ( )2 4
2
Fv v v
mkyx N H y exp y x .ψ⎛ ⎞⎛ ⎞−
= = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠v = 2
v = 1
v = 0
Note: Penetration into classically forbidden regionsProbability distributions becomeProbability distributions become more classical at high v
4.3 Harmonic Oscillator Selection Rules (preamble)4.3 Harmonic Oscillator Selection Rules (preamble)
22 2d221
2
22
2ˆ kx
dxd
mH +
−= 2
21
2
2
21ˆ q
dqdH +−=simplifies in scaled coordinates
and units to become
b l
⎟⎞
⎜⎛ 2q
q is a vibrationalcoordinate
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
2exp0
qψ is the lowest energy eigenfunction: 120 0ψ ψ=
We can obtain all other eigenfunctionsusing the ladder operators: ⎟⎟
⎠
⎞⎜⎜⎝
⎛−=+
dqdqQ
21ˆ
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
dqdqQ
21ˆ
⎠⎝ ⎠⎝ q
1v vψ ψ+
+⇒ 1v vψ ψ
−⇒With the properties:
qQQ 2ˆˆ =++and
⎫⎧ if '1Finally, of course all vibrational eigenfunctionsare orthonormal ⎭
⎬⎫
⎩⎨⎧
≠=
==∫∞+
∞−′ vvif
vvifdq vvvv '0
'1'δψψ
4.3 Harmonic Oscillator Selection Rules4.3 Harmonic Oscillator Selection Rules
∫∞
∞−
∗′′ = dqR vvvv ψμψ ˆConsider the Transition Dipole
Moment for a v – v’ transition where ∑−=i
ireˆμ̂
We can expand the dipole moment operator about q = 0:
⎞⎛⎞⎛ 2 ˆ1ˆ dd …+⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
===
2
02
2
00
ˆ21ˆˆˆ q
dqdq
dqd
qqq
μμμμ
Consider each term in the TDM in turn:0
∫∫∞
∞−
∗′=
∞
∞− =∗′′ == dqdqR vvqvqvvv ψψμψμψ 00 ˆˆ
0
Hence this term is identically zero for any vibrational transition
+⎟⎟⎞
⎜⎜⎛
+= qd ˆˆˆ μμμ04.3 Harmonic Oscillator Selection Rules4.3 Harmonic Oscillator Selection Rules
…+⎟⎟⎠
⎜⎜⎝
+==
= qdq q
q0
0μμ
0vv v' v
ˆdR q dqdqμ ψ ψ
∞∗
′ −∞
⎛ ⎞= ≠⎜ ⎟⎝ ⎠
∫For an allowed transition (Rvv’ ≠0) we require:
0qdq ∞=⎝ ⎠∫
2 ∫∞
0ˆ
0
≠⎟⎟⎠
⎞⎜⎜⎝
⎛
=qdqdμ1.
and2. 0≠∫
∞
∞−
∗′ dqq vv ψψ
( )∫∫∞ +∗∞ ∗ + dqQQdqq ψψψψ ˆˆ1
i.e., a dipole moment that changes during the vibration
( )∫∫ ∞−′
∞−′ += dqQQdqq vvvv ψψψψ
2
i e ∫ ∫∞ ∞ ∗+∗ ≠≠ 0ˆ0ˆ dqQordqQ ψψψψi.e., ∫ ∫∞− ∞−
′′ ≠≠ 00 dqQordqQ vvvv ψψψψ
1+=′ vv 1−=′ vvGross selection rule
i.e., Δv = ± 1
Specific selection rule
4.4 The (very simple) spectrum of a harmonic oscillator4.4 The (very simple) spectrum of a harmonic oscillator
( )12v eG v ω= +Vibrational Terms:
Δv = ± 19ω
Selection Rule:
52 eω
2 eω72 eω
eω
Transitions at: 1v v eG Gν ω+
= − =
12 eω
2 e32 eω
i e only one line in the spectrumi.e., only one line in the spectrum
b I b ti l t iti f 0 i t t hi h Tn.b. In absorption, only see transitions from v =0 since, except at very high T, this is the only level populated:
n hcg ω⎧ ⎫⎪ ⎪1
e.g., if ωe = 1000 cm‐1, T = 300 K1 1
0 0
exp 0.008v e
v
n hcgn g kT
ω=
=
⎧ ⎫−⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
4.5 Anharmonicity4.5 Anharmonicity
But of course molecular bonds are not harmonic oscillatorsBut, of course, molecular bonds are not harmonic oscillators
Dissociation limit (e.g., isolated atoms)
We can break bonds (pull them apart)We can break bonds (pull them apart)
We can identify (finite) dissociation energy for doing soenergy for doing so
As we stretch a bond the bond itself gets weaker (the restoring force is reduced)
4.5 Accounting for anharmonicity: The Morse Oscillator4.5 Accounting for anharmonicity: The Morse Oscillator
( ) ( )( )2
1V R D R Rβ⎡ ⎤l ( ) ( )( )1e eV R D exp R Rβ⎡ ⎤= − − −⎣ ⎦Morse Potential:
SHO a good approx. at small x (low v) n.b., the experimental dissociation energy, D0 = De ‐ZPE
4.5 Accounting for anharmonicity: The Morse Oscillator4.5 Accounting for anharmonicity: The Morse Oscillator
( ) ( )( )2
1V R D R Rβ⎡ ⎤l
vmax21 1
2 2v e e eG v v xω ω= + − +
( ) ( )( )1e eV R D exp R Rβ⎡ ⎤= − − −⎣ ⎦Morse Potential:
Energy Eigenvalues:a 2 2
0 1 2 3 ma
v e
x
e e
vv , , , ,= …
D0 De
xe is the (dimensionless) anharmonicity constant: 4
ee
e
xDω
=
Specific Selection rule: ( )1 2 3 weakerv , , ,Δ = ± ± ±
Transitions: ( )( )
fundamental 0 1 21 overtone 0 2 2 6
e e est
e e e
xx
ν ω ων ω ω
→ = −
→ = −( ) e e e
SHO a good approx. at small x (low v) n.b., the experimental dissociation energy, D0 = De ‐ZPE
HCl (1Σ+): ωe = 2990.9 cm-1, ωe xe = 52.8 cm‐1, (xe = 0.018), De = 35486 cm‐1, D0 = 34092 cm‐1
4.6 Rotational fine structure: I. Diatomics4.6 Rotational fine structure: I. Diatomics
Associated with a vibrational transition
HF
Associated with a vibrational transition
we observe rotational transitions:
Combined vibration rotation terms:S v G v F+J JCombined vibration‐rotation terms:S v , G v F= +J J
e g or an anharmonic rigid rotor:i.e., Etot = Evib + Erot
e.g., or an anharmonic rigid rotor:
( ) ( ) ( ) ( )21 12 2 1e e e vS v , v v x Bω ω= + − + + +J J J
transitions: ( ) ( )S v , S v ,ν ′ ′ ′′ ′′= −J Jvibrational rotational
transitions: ( ) ( ), ,subject to combined selection rules
All ΔJ = +1 transitions give rise to an R- branch andthe ΔJ = ‐1 transitions comprise a P‐ branch
Recall rotational selection rule ΔJ = ±1:
4.7 Rotation Vibration Spectra4.7 Rotation Vibration Spectra
e.g., in the fundamental band:
( ) 2R e e exν ω ω′′ = −J( )( )( ) ( )1 0 1 2 1B B′′ ′′ ′′ ′′+ + + − +J J J J
( ) 2P e e exν ω ω′′ = −J( )( )( ) ( )1 0 1 1
P e e e
B B′′ ′′ ′′ ′′+ − − +J J J J
Assuming; BBB ~~~01 ==
Spacing in each branch = B~2P‐branchR‐branch
Central gap = R(0) –P(1) =
xωωω 2B~4
eee xωωω 20 −=
*Q‐branch (ΔJ=0) absent unless additional angularQ branch (ΔJ 0) absent unless additional angular momentum present (e.g., NO 2Π) or a degenerate bending mode is involved (see polyatomics later)
e.g., in the fundamental band of HCl
35note offset of H35Cl and H37Cl lines due to different reduced mass (and hence different ωeand B constants)
Also, note bunching of the lines in the R‐branchlines in the R branch
?~~ BB
4.8 Vibrational dependence of rotational constants4.8 Vibrational dependence of rotational constants
?01 BB =
21~B ∝ but what exactly do we mean by
How good is the assumption
2Rμbut what exactly do we mean by 1/R2 in a vibrating molecule?
We actually mean <1/R2>We actually mean <1/R >
<1/R2> decreases with v and therefore:
0 1 1v >> >
12v e α= − +
Under high resolution we can determine the rotational constant in each vibrational level using a method of combination differencesusing a method of combination differences.
4.9 Combination Differences4.9 Combination Differences
Use transitions with common upper or lower levels to determine rotational constants for each vibrational level:
e.g., R(J) and P(J) originate at v =0, J;1
( ) ( ) ( )R 1Pν ν− = +J J J
1→
Likewise the difference of R(J‐1) and P(J+1), which both terminate at v =1, J, gives:
0 ( ) ( ) ( )R 0Pν ν− − + = +J J J
0→
4.10 Band heads in Vibrational Spectra4.10 Band heads in Vibrational Spectra
C id th R b h i li ( ) 2′′JConsider the R‐branch in a linear molecule:
( )( )( ) ( )( )21 0 1 0
2
1 1R e e ex
B B B B
ν ω ω′′ = −
′′ ′′+ + + + − +
J
J J
‐ve
At high enough J the wavenumberseparation of adjacent transitions canseparation of adjacent transitions can change sign.
This leads to a bandhead in R branch:This leads to a bandhead in R‐branch:
Diatomics have only one vibrational mode
4.11 Beyond Diatomics: Normal Modes in Polyatomic Molecules4.11 Beyond Diatomics: Normal Modes in Polyatomic Molecules
Diatomics have only one vibrational mode. In polyatomics we define Normal Modes as independent, harmonic vibrations which:
1. leave the centre of mass unmoved;2. involve all atoms moving in phase (coherent motion);3. transform as an irreducible representation of the molecular point group
N isolated atoms have 3N degrees of freedom (x y z translations for each)
I. Number of Normal Modes:I. Number of Normal Modes:
N atoms in a molecule have 3N degrees of freedom arranged amongst translations, vibrations and rotations:
N isolated atoms have 3N degrees of freedom (x, y, z translations for each)
vibrations and rotations:
Molecule: Linear Non-linear
no. translations = 3 3
no. rotations = 2 3
∴ no. vibrations = 3N-5 3N-6
4.12 Visualising normal modes4.12 Visualising normal modes
H OH2O
Symmetric
CO2
Symmetric ν1Symmetric
stretch
Asymmetric
ystretch
ν3
Bend
stretch
A t iDegenerate
B dν2Asymmetric
stretchBend 2
4.13 Symmetry of normal mode vibrations4.13 Symmetry of normal mode vibrations
A normal mode must transform as an irreducible representation of the point group of the molecule.
Once we can picture the mode itself this is usually easyOnce we can picture the mode itself this is usually easy:
e.g., water, C2v
C2v E C2 σxz σyz
A 1 1 1 1
E C2 σxz σyz
A1 1 1 1 1
A2 1 1 ‐1 ‐1
B 1 ‐1 1 ‐1
1 1 1 1 A1Mode 1Symmetric stretch
B1 1 1 1 1
B2 1 ‐1 ‐1 11 1 1 1 A1Mode 2
1 -1 -1 1 B2
Bend
M d 3 2Mode 3Asymmetric stretch
4.14 Symmetry of normal modes: Linear molecules (D∞h, C∞v)4.14 Symmetry of normal modes: Linear molecules (D∞h, C∞v)
CO2 D∞h2 ∞h
HCN CHCN C∞v
acetylene D∞h
As well as the normal modes the vibrational wavefunctions have distinct
4.15 Symmetry of vibrational wavefunctions4.15 Symmetry of vibrational wavefunctions
As well as the normal modes, the vibrational wavefunctions have distinct symmetry which can be useful in determining selection rules. These are different for different v levels.
I. Ground levels (v = 0 levels)I. Ground levels (v = 0 levels)
⎞⎛ 2
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
2exp
2
0qψAs we have seen, ψ0 has the form
The sign ofΨ0 is independent of the sign of q:
→It follows that Ψ0 transforms as the totally symmetric irreducible representation of the relevant point group
i.e., A1 in C2vΣg
+ in D∞hΣ+ in C∞v
2 21 d q qˆ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −
II. First excited vibrational levels (v = 1 levels)II. First excited vibrational levels (v = 1 levels)
1 0
1 22 22
d q qQ̂ q exp qexpdq
ψ ψ+⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −
∝ = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
ψ1 has the same symmetry as the normal coordinate q
→ ψ1 transforms as the same symmetry as the normal mode
ψ0 ψ1Normal
So, in the case of water:
Mode 1 A1
ψ0 ψ1mode
A1 A1Mode 1
Mode 2
A1
A1
A1
A1
A1
A1Mode 2
Mode 3
A1
B
A1
A
A1
BMode 3 B2 A1 B2
III. Higher vibrational levelsIII. Higher vibrational levels
212 qQ̂ ∝= +ψψ
For non‐degenerate modes:
→ transforms as tot. symmetry IR
323
12
qqQ̂
+∝= +ψψ
ψψ f y y
→ transforms as same symmetry as the normal mode
(degenerate modes, E, T, etc., are more complex)
H2O
4.16 Vibrational Selection Rules in Polyatomic Molecules4.16 Vibrational Selection Rules in Polyatomic Molecules
As normal modes are harmonic, the selection rules are:
The dipole moment must change during the vibrationΔ ±1Δvi = ±1
For polyatomics it can be hard to see if the former is true. Clearly though it relates to the transition dipole moment,p
∫∞
∞−
∗′′ = dqR vvvv ψμψ ˆ a vector with x, y, z components of:
∫∞
∞−
∗′′ = dxR vxvxvv ψμψ ˆ ∫
∞
∞−
∗′′ = dyR vyvyvv ψμψ ˆ ∫
∞
∞−
∗′′ = dzR vzvzvv ψμψ ˆ
Each of these is a scalar and thus in order for Rv,v’ ≠ 0 at least one of these components must transform as the totally symmetric IR.
i.e., see if any of the direct products v'v XΓ⊗Γ⊗Γ μ
Γ⊗Γ⊗Γ v'v YΓ⊗Γ⊗Γ μ
v'v ZΓ⊗Γ⊗Γ μ
transform as the totally symmetric irrep
C id th f d t l (0→1) b d
4.16 Example of Vibrational Selection Rules4.16 Example of Vibrational Selection Rules
Consider the fundamental (0→1) band in mode 3 (B2) of H2O (C2v):
0ψ transforms as the totally symmetric IR, i.e., 1A
1ψ transforms as the normal mode, i.e., 2B1ψ , , 2BThe components of transform, in C2v as: μ̂
2
1
B
BX
=Γ
=Γμ transforms as x, y, z read IRs from characterμ̂
1
2
A
B
Z
Y
=Γ
=Γ
μ
μ
The components ˆ' Γ⊗Γ⊗Γ become
read IRs from character table:
212112:ABBABBABBABBX
⊗⊗⊗=⊗=⊗⊗μ
The components vz,y,xˆ'v Γ⊗Γ⊗Γ μ become
The y‐component transforms as A1, so the transition is “dipole‐allowed with transition
212112
122122
::
BABAABABBABB
X
Y
=⊗=⊗⊗=⊗=⊗⊗
μμ p
dipole along y”.
4.17 General Vibrational Selection Rules4.17 General Vibrational Selection Rules
The fundamental transition is allowed provided the symmetry of the normal mode (and thus v = 1) is the same as x, y, or z. In this case the mode is infra‐red active.
If the transition moment is parallel to the symmetry (z‐) axis the mode is called a parallel mode.f h i i i di l h i h d i ll dIf the transition moment is perpendicular to the symmetry axis the mode is called a perpendicular mode.
4.18 Combination Bands4.18 Combination Bands
Just as anharmonicity permits changes in vibrational quantum number Δv > ±1, it also allows for simultaneous changes in more than one normal mode.
Such transitions all break the harmonic oscillator selection rule but we can again use symmetry to determine if such combination bands are allowed:
e.g., Consider the v1=0, v3=0 → v1=1, v3=1 combination band in H2O(i.e., simultaneous excitation of both symmetric and asymmetric stretches)v =0 transforms as A symmetryv =0 transforms as A1 symmetry.v1=1 level transforms as A1,v3= 1 level transforms as B2. H h 1 1 l l f BBA ⊗Hence, the v1=1, v3=1 level transforms as 221 BBA =⊗
112212 AABBAB =⊗⊗=⊗Γ⊗ 112212 AABBABY
⊗⊗⊗Γ⊗ μ
The transition is symmetry allowed with transition dipole along y. (Despite breaking the harmonic oscillator selection rule)
4.19 Rotational Fine Structure4.19 Rotational Fine Structure
1. Linear Molecules1. Linear Molecules1. Linear Molecules1. Linear Molecules
2 types of bands:2 types of bands: Σ−Σ (parallel bands) and Σ−Π (perpendicular bands)ΔJ = ±1 (P,R, branches) ΔJ = 0, ±1 (Q, P, R branches)
( ) ( ) ( )1el vibQ B Bν−
′ ′′= + − +J J J
Q‐branch (ΔJ=0) arises from the additional angular momentum from the degenerate bending modemomentum from the degenerate bending mode
Σ+−Σ+ infrared band in HCN(asymmetric stretch fundamental) Πu−Σg
+ overtone band in HC≡CH
A. Parallel transitions:ΔJ = 0 ±1 ΔK = 0
2. Symmetric tops2. Symmetric topsB. Perpendicular transitions:
ΔJ= 0, ±1 ΔK = ± 1 ΔK=ΔJ = 0, ±1 ΔK = 0 , ΔK=
‐1+1
+1‐1
‐1+1
+1