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Physical Chemistry (II)

Lecture 18

CHEM 3172-80

Lecturer: Hanning Chen, Ph.D.04/03/2017

Molecular Vibrational Spectra

Quiz 17

5 minutes

Please stop writing when the timer stops !

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)

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)

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

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

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 '

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

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

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

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

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

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

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

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

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( )

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

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