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Physical Chemistry (II)
Lecture 18
CHEM 3172-80
Lecturer: Hanning Chen, Ph.D.04/03/2017
Molecular Vibrational Spectra
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Quiz 17
5 minutes
Please stop writing when the timer stops !
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Three Types of Molecular Motion
Translation Vibration Rotation
O
O
O
O
: center of massChange? Yes
d d
d : inter-particle distance
Change? No
d
: center of massChange? No
d : inter-particle distance
Change? Yes
: center of massChange? No
d : inter-particle distance
Change? Noone-dimensional box quantum tunneling rigid rotorone-dimensional
harmonic oscillator(hard to study due to frequent
molecular collisions)
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the Vibrations of Diatomic Molecules
equilibrium state: R = R0 : equilibrium bond length
vibrational state: R ' = R0 + ΔRlinear spring
Hooke’s law: F = −kΔRThe spring always tends to move the particle towards the equilibrium position.
k ↑ : stiff spring k ↓ : floppy spring
A B
vibrational energy: Evib R( ) = Evib (R0 )+∂Evib
∂R⎛⎝⎜
⎞⎠⎟ R=R0
R − R0( ) + 12
∂2Evib
∂R2⎛⎝⎜
⎞⎠⎟ R=R0
R − R0( )2 + ...
: bond distortion
0Evib R( ) ≈ 1
2∂2Evib
∂R2⎛⎝⎜
⎞⎠⎟ R=R0
ΔR2 = 12
− ∂F∂R
⎛⎝⎜
⎞⎠⎟ R=R0
ΔR2 = 12kΔR2harmonic approximation:
(Taylor expansion)
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Quantum Harmonic Oscillator
system’s Hamiltonian:
A B
H = T + V = − !
2
2mA
d 2
dRA2 −!2
2mB
d 2
dRB2 +
12k(RA − RB )
2
effective mass: meff =mAmB
mA +mBif mA ≫ mB meff ≈ mB
seems like only B is moving
bond distortion: ΔR = RA − RB
rewritten Hamiltonian:
Heff = Teff + Veff = − !
2
2meff
d 2
dΔR2+ 12kΔR2
two-particle system:
effective one-particle harmonic oscillator
vibrational energy: Ev = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω
nv : vibrational quantum number non − negative integerquantized
ω = kmeff
: characteristic frequency
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Ladder of Energy Levels
nv = 0
nv = 1
nv = 2
nv = 3
nv = 4
equally spaced energy levels !
E nv = 0( ) = 1
2!ω zero point
energy
a quantum oscillator can NOT be entirely frozen !
Heinsberg uncertainty principle
Δx→ 0Δp→ 0×
ΔpΔx ≥ !
2
Ev = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω
ΔE = !ω
ΔE = !ω
ΔE = !ω
ΔE = !ωnv = 0→ nv = 1 ΔE = !ω observednv = 0→ nv = 2 ΔE = 2!ω nevernv = 0→ nv = 3 ΔE = 3!ω nevernv = 1→ nv = 2 ΔE = !ω observednv = 1→ nv = 3 ΔE = 2!ω nevernv = 2→ nv = 3 ΔE = !ω
×
observed
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Selection Rules for Vibrational TransitionsFermi’s Golden Rule:
Ti→ f =
2π!
ψ i"r ψ f
2ρ f density of final states
transition dipole moment !r : polarization direction of incident light
harmonic oscillator: ϕv = NvHv y( )e− y2 /2 y = ΔR
α,α = !2
meff k⎛
⎝⎜⎞
⎠⎟
14
ϕv ' = Nv 'Hv ' y( )e− y2 /2Hv : Hermite polynomials
Decomposition of total wavefunction: ψ =ϕvϕe ϕe ΔR( ) : electronic wavefunction
ϕe ΔR( ) =ϕe(Re )+dϕe
dR (R=Re )ΔR + ...variation of electronic wavefunction:
ψ i!r ψ f = ϕe(Re )
!r ϕe(Re ) ϕv ϕv ' +∂ ϕe(R)
!r ϕe(R)∂R
ϕv!r ϕv ' = ∂µ
∂R⎛⎝⎜
⎞⎠⎟ ϕv
!r ϕv ' ≠ 00Hermitian Hamiltonian
molecular dipole for allowed
transitions
(Born-Oppenheimer approximation)
δ vv '
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Allowed Vibrational Transitions
µif = ψ i
!r ψ f = ∂µ∂R
⎛⎝⎜
⎞⎠⎟ ϕv
!r ϕv ' ≠ 0transition dipole moment:
1. ϕv!r ϕv ' ≠ 0
NvNv ' Hv y( ) !rHv ' y( )∫ e− y
2
dr ≠ 0
only satisfied when Δv = v − v ' = ±1A vibrational transition can ONLY occur between two neighboring vibrational states !
2. ∂µ∂R
⎛⎝⎜
⎞⎠⎟ ≠ 0
A vibrational transition is ONLY possible when changing molecule dipole moment !
!E
initial final initial
µi = 0
A
A A
A
finalµ f = 0 ΔH = 0
NO photon is absorbed in order to conserve the energyH =
!E i!µ
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Anharmonicity of Molecular Vibration
H2
potential energy profile
experimental data
harmonic approximation
OK
terrible Evib R( ) = 12
∂2Evib
∂R2⎛⎝⎜
⎞⎠⎟ R=R0
ΔR2 + 16
∂3Evib
∂R3⎛⎝⎜
⎞⎠⎟ R=R0
ΔR3 + 124
∂4Evib
∂R4⎛⎝⎜
⎞⎠⎟ R=R0
ΔR4 + ...
higher-order terms are neededwhen ΔR is large
F ≠ −kΔR the restoring force is no longer linear !
Morse potential:
VMorse = De(1− e−αΔR )2 α =
meffω2
De
Morse parameter
for a perfect harmonic oscillator, α → 0,De →∞
E nv( ) = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω − nv +
12
⎛⎝⎜
⎞⎠⎟2
xe!ω
corrected energy levels:
xe : anharmonicity constant
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Vibrational OvertonesWith the help of anharmonicity:A vibrational transition is NO LONGER limited between two neighboring vibrational states !
ψ v!r ψ v ' ≠ 0 even when Δv = v − v ' ≠ ±1
ΔEnv→nv+2= 2!ω − (4nv − 6)xe!ω
Vibrational overtones:
ΔEnv→nv+3= 3!ω − (6nv −12)xe!ω ΔEnv→nv+4
= 4!ω − (8nv − 20)xe!ω
overtone intensity drops substantially with increasing Δv
ΔEnv→nv+1= !ω − (2nv − 2)xe!ω
why ΔE1→2 < ΔE0→1 ? redshift ?
ΔE0→1 = !ω + 2xe!ω ΔE1→2 = !ω
difference maker
greater redshifts for vibrational overtones
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the Birge-Sponer Plot
nv = 0
nv = 1
nv = 2
nv = 3
nv = 4
ΔE = !ω + 2xe!ω
ΔE = !ω
ΔE = !ω − 2xe!ω
ΔE = !ω − 4xe!ω ΔE = !ω − 8xe!ω
nv = 5 E nv( ) = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω − nv +
12
⎛⎝⎜
⎞⎠⎟2
xe!ω
D0 = E0→1 + E1→2 + E2→3 + ...= Ei→i+1i=1
∞
∑Dissociation energy of a chemical bond:
assuming all vibrational transitions are detectable:
In most experiments, only the first several transitions can be observed
linear assumption:
ΔEnv→nv+1= !ω + 2xe!ω( )− 2!ω xenv
y axis interception: !ω + 2xe!ω
x axis interception: 1+ 12xe
Sshade =
(2xe +1)2
4xe!ω ≈ D0
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Vibration-Rotation SpectraCoupling between molecular vibration and molecular rotation
~ 10 TetraHertz (1013 / s) ~ 100 GigaHertz (1011 / s)In general, molecular vibration is ~100 times faster than molecular rotation
ΔEvib >> ΔErot
molecular vibrational energy gap is much much greater than that of molecular rotation !
Combined energy term:
Etotal = Evib + Erot = v + 1
2⎛⎝⎜
⎞⎠⎟ !ω + BJ(J +1)
hω >> Bv : vibrational quantum number J : rotational quantum number
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Spectral Branchesselection rules for rotational spectra:
ΔJ = 0,±1ΔJ = −1: P branch
ΔJ = +1: R branch
ΔJ = 0 : Q branch
EP = ΔEvib + ΔErot = !ω − 2BJ
ER = ΔEvib + ΔErot = !ω + 2B(J +1)
EQ = ΔEvib + ΔErot = !ωSeparation of the three branches yields
B : rotational constant
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Combinational Differences
vibrational excitation
nv
BI 'A BI ''↑
A
the rotational “constant” B is actually a function of vibrational quantum number,
I ' ≠ I ''
EP = !ω − (B '+ B '')J + (B ''− B)J 2Corrected spectral differences: B '(B '') for the initial(final) vibrational state
ER = !ω + (B '+ B '') J +1( ) + (B ''− B) J +1( )2
EQ = !ω + B ''− B '( )J(J +1)Combinational differences:
ER J( )− EP J( ) = 4B'' (J + 12)
starting from the same rotational state
ER J −1( )− EP J +1( ) = 4B '(J + 12)
ending at the same rotational state
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Energy Diagram of Combinational Differences
B '
B ''E R
J ()
E PJ ()
E PJ+1
()
E RJ−1
()
ER J( )− EP J( ) = 4B'' (J + 12)
ER J −1( )− EP J +1( ) = 4B '(J + 12)
J
JJ −1
J +1
J −1
J +1
For example, in HCl
Bv=0' : 10.440 cm−1 Bv=1
' : 10.136 cm−1
apparently due to the bond elongation
the reduction of B was also observed in DCl
Bv=0' : 5.392 cm−1 Bv=1
' : 5.280 cm−1
~ 3% reduction
~ 2% reduction
the bond distortion is less prominent in the heavier DCl
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Review of Homework 6Review of Homework 1712.7 The microwave spectrum of 16 O12 CS gave absorption lines (in GHz) as follows:
J 1 2 3 4
32S 24.32592 36.48882 48.65164 60.81408
34S 23.73233 47.46240
Use the expressions for moments of inertia in Table 12.1 and assume that the bond lengths are unchanges by substitution; calculate the CO and CS bond lengths in OCS.
RCO = R1 and RCS = R2
I = mOmS
mOCS
R1 + R2( )2 + mC mOR12 +mSR2
2( )mOCS
!v = 2B J +1( ) = "2πcI
J +1( )
I = !2πc!v
J +1( )
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Review of Homework 6Review of Homework 17
I(16O12C32S)=1.38 ×10−45kgm2 I(16O12C34S)=1.42 ×10−45kgm2
I = mOmS
mOCS
R1 + R2( )2 + mC mOR12 +mSR2
2( )mOCS
R1 = RCO = 1.16Å R2 = RCS = 1.56Å
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Homework 18
Reading assignment: Chapters 12.8, 12.9, 12.10, 12.11 and 12.12
Homework assignment: Exercises 12.20 Problems 12.9
Homework assignments must be turned in by 5:00 PM, April 4th, Tuesday
to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall