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Vibrational Spectroscopy

Harmonic Oscillator

Potential Energy

V x

( )

1k x2 x R Re 2

= = − where Re – equilibrium bond length

Selection Rules

The dipole moment of a molecule can be expanded as a function of x = R – Re.

( ) ( )

2 2 3 3 2 3 0 0 0 1 1 x 0 x x x x 2 x 6 x     ∂µ ∂ µ ∂ µ   µ = µ + + + + ∂ ∂ ∂       L

( )

0 0

µ = µ – dipole momentum at equilibrium, permanent dipole moment.

0 x

∂µ   

  – change in dipole moment along vibrational coordinate

– effect of stretching, compression, bending, etc… on dipole moment.

Recall that transition is possible only when transition dipole matrix element is nonzero, that is f µ ≠ˆ i 0

For the dipole operator in the transition dipole matrix element, substitute the series approximation above. 2 3 2 3 0 2 3 0 0 0 0 0 1 1 ˆ f i f x x x i f x i x 2 x 6 x x     ∂µ ∂ µ ∂ µ ∂µ     µ = µ + + + + ≈ µ + ∂ ∂ ∂ ∂       L   0 0 0 0 0 0 f x i f i f x i f i f x i x x x ∂µ ∂µ ∂µ       µ + = µ + = µ + ∂ ∂ ∂      

A transition implies that the final state must be different from the initial state f i =0 Thus f x i ≠0 for a transition to occur.

Also, 0 0 x ∂µ   ≠

  for a transition to occur. This requirement means that the vibration that is

being excited must cause the dipole moment of the molecule to change. This requirement is true for transitions in infrared vibrational spectroscopy.

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Recall that

( )

2 2 N H e ξ − ν ν ν = ξ where 1 4 2 k x µ  

ξ =  h Careful: µ is reduced mass.

( )

( )

2 1 2 1 2 4 2 4 2 2 f f i i 0 0 1 2 2 0 0 f i f x i N H e N H e d x x k k f i f i x k x k ∞ ξ ξ −∞  ∂µ ∂µ     µ = = ξ ξ ξ ξ ∂ ∂ µ µ           ∂µ ∂µ ω     = ξ = ξ ∂ µ ∂      

h h h h Recall that H H 1 1H 1 1 1 1 2 2 ν ν− ν+ ξ = ν + ⇒ ξ ν = ν ν − + ν + 0 0 0 1 f i f i f i i 1 f i 1 x k x k 2 1 i f i 1 f i 1 x k 2 ∂µ ω ∂µ ω       µ = ξ = − + + ∂ ∂       ∂µ ω     = − + + ∂     h h h Or perhaps stated a little more clearly.

i 0 1 f i f i 1 f i 1 x k 2 ∂µ ω     µ = ν − + + ∂     h

The first integral is nonzero only when ν = νf i 1 ⇒ ∆ν = −1 The second integral is nonzero only when ν = νf i 1+ ⇒ ∆ν = +1

Therefore the selection rule for a harmonic oscillator in vibrational spectroscopy is ∆ν = ±1

Anharmonic Oscillator

General Potential Energy

The actual potential energy (not idealized) can be written as a Taylor series about the point R = Re (i.e., x = 0).

( )

( )

2 2 3 3 4 4 2 3 4 0 0 0 0 V 1 V 1 V 1 V V x V 0 x x x x x 2 x 6 x 24 x       ∂ ∂ ∂ ∂   = + + + + + ∂ ∂ ∂ ∂         L

The origin of the potential energy can be set arbitrarily (and sensibly) to zero. That is, V(0) = 0 by definition.

Also, since the slope of the potential energy curve at Re is zero

0 V 0 x ∂   =   

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Thus the first nonzero coefficient of the potential energy series above is 2 2 0 1 V 2 x ∂     

This coefficient can be related to the force constant.

2 2 2 2 0 0 1 V 1 V k k 2 x 2 x ∂  ∂  = ⇒ =        

The next nonzero coefficient of the potential energy series causes the oscillator to be

anharmonic. The nonzero coefficients can be manipulated to yield to yield the cubic anharmonicity constant, χeνe, the quartic anharmonicity constant, yeνe, etc, …

Energy levels of the anharmonic oscillator.

2 3 e e e e e 1 1 1 E h y 2 2 2 ν = ν ν + − χ ν ν +  + ν ν +  +       L

Note: The energy levels for the harmonic oscillator are evenly spaced. (∆ = νE h e) However the energy levels for the anharmonic oscillator are not evenly spaced. (∆ < νE h e). The energy levels for the anharmonic oscillator get closer together as the quantum number ν increases.

Morse Potential

Electronic potential energy curve is not purely quadratic, i.e., V x

( )

1kx2 2

A better approximation of an actual potential energy curve is the Morse potential.

( )

(

ax

)

2 e V x =D 1 e− − where 1 2 2 e a 2D µω  =  

  and De is the potential well depth.

V(x) x harmonic (quadratic) anharmonic (Morse) D0 De D0 – dissociation energy e 0 1 D D 2

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Using the Morse potential, the spectroscopic term becomes

( )

e e e 2 1 1 G 2 2     ν = ν ν + − χ ν ν +     % % where e e e h 4D ν χ =

This expression is an exact solution for the Morse potential.

Worth emphasizing is that the Morse potential is only an approximation (albeit, a good one).

If the above spectroscopic term, cannot reproduce the experimental spectra, then a more accurate potential energy function is needed. (Usually based on Taylor series expansion.) When the polynomial found from the Taylor expansion is used, a more general spectroscopic term can be written.

( )

e e e 2 e e 3 e e 4 1 1 1 1 G y z 2 2 2 2         ν = ν ν + − χ ν ν + + ν ν + + ν ν + +         % % % % L

Selection Rules of the Anharmonic Oscillator

Nonzero transition dipole matrix elements of the anharmonic oscillator have two origins 1.) Anharmonic wavefunctions are combinations of HO wavefunctions.

2.) Dipole can have quadratic (cubic, etc…) dependence i.e., 2 2 0 0 x ∂ µ    thus f µ i and 2 f x i , f x i , 3 f x i , etc… 4

The resultant selection rules are complicated, but transitions where ∆ν is a single digit are all allowed. That is, ∆ ν = ± ± ± ± ±1, 2, 3, 4, 5,K

Consequences of Anharmonicity

1.) Transitions where ∆ ν >1 are allowed - Such transitions are called overtones. - 1 ← 0 transition is called the fundamental.

2.) More than one vibration can be excited with a single transition. - Such transitions are combination bands.

- Consider water with its three modes of vibration.

O H H O H H symmetric stretch - νs O H H antisymmetric stretch - νa O H H bending - νb O H H O H H

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- The total vibrational wavefunction is product of individual vibrational wavefunctions.

s a b s a b

ν ν ν = ν ν ν

000 - vibrational ground state

100 ← 000 - fundamental for symmetric stretching mode 010 ← 000 - fundamental for antisymmetric stretching mode 001 ← 000 - fundamental for bending mode

- Examples of combinations bands would be

201 ← 000 110 ← 000

312 ← 221 001 ← 211

Miscellaneous

A hot band is an absorptive transition from an excited state.

Vibration – Rotation Spectra

Analysis for the HCl/DCl experiment is part of this section. The discussion won’t be repeated here. Refer to the lab handout.

Combination Differences

The technique of combination differences is used to find the rotational constant for an excited state.

The analysis is similar (but not identical) to the analysis for the band spectrum of nitrogen. Consider the simplest term for rotovibrational spectroscopy

- harmonic oscillator and rigid rotor only.

( )

e

(

)

1 S , J B J J 1 2   ν = ν ν + + +   %

However, the rotational constant changes for each vibrational state. We can find the rotational constant for the excited vibrational state by taking the difference of specific spectral lines to remove the influence of the rotation of the ground vibrational state.

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Consider a portion of a molecular energy level diagram. J+1 J J-1 J+1 J J-1 νP(J+1) νR(J-1) νP(J) νR(J) ν = 1 ν = 0

( ) (

) (

)

(

)

(

)

(

) (

)

P e 1 e 0 2 e 1 0 1 0 J S 1,J 1 S 0,J 3 1 B J 1 J B J J 1 2 2 B B J B B J ν = ν = − − ν =   = ν + − − ν + +   = ν − + + − % % % %

( ) (

) (

)

(

)(

)

(

)

(

)(

) (

)(

)

R e 1 e 0 2 e 1 0 1 0 J S 1,J 1 S 0,J 3 1 B J 1 J 2 B J J 1 2 2 B B J 1 B B J 1 ν = ν = + − ν =   = ν + + + − ν + +   = ν − + + + − + % % % %

B1 – rotational constant of the ν = 1 state

B0 – rotational constant of the ν = 0 state

Bν – general rotational constant (has dependence on ν) Remember P branch ⇒ ∆ = −J 1

Remember R branch ⇒ ∆ = +J 1

Since the frequencies of the spectral lines depend on B1 and B0, one can take combination

differences of the appropriate lines to eliminate either B1 or B0, thus solving for the other

rotational constant.

(

)

(

)

R P 0 1 J 1 J 1 4B J 2   ν − − ν + = +   % %

( )

( )

R P 1 1 J J 4B J 2   ν − ν = +   % %

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Polyatomic Vibrational Spectroscopy

Normal Vibrational Modes

Vibrations in a polyatomic molecule involve movement of all the atoms in a molecule. - In contrast, to only two atoms vibrating at a time

Modes of vibration that are independent of each other (i.e., don’t affect each other) are called

normal mode vibrations.

For a molecule with N atoms, 3N total modes of motion exist. 3 translational modes

3 rotational modes

2 modes for linear molecule 3N-6 vibrational modes

3N-5 modes for linear molecule Symmetry of Polyatomic Vibrations

To find symmetry of a vibration, attach vectors to atoms to indicate atomic movements. The symmetry label of the vibration is found by noting how these vectors transform when they are operated upon by symmetry operators.

Example: Consider “puckering” mode of BF3.

B

F F

F

above plane

below plane

BF3 belongs to the D3h symmetry group

D3h E σh 2 C3 2 S3 3 C2′ 3 σv A1′ 1 1 1 1 1 1 A2′ 1 1 1 1 -1 -1 A1″ 1 -1 1 -1 1 -1 A2″ 1 -1 1 -1 -1 1 E′ 2 2 -1 -1 0 0 E″ 2 -2 -1 1 0 0

Note the effect of symmetry operations on vectors. E ⇒ vectors same ⇒χ = 1 σh ⇒ vectors reversed ⇒χ = -1 C3 ⇒ vectors same ⇒χ = 1 S3 ⇒ vectors reversed ⇒χ = -1 C2′⇒ vectors reversed ⇒χ = -1 σv ⇒ vectors same ⇒χ = 1

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Raman Spectroscopy

Raman Scattering

Raman spectra come from fundamentally different process than “normal” absorption/emission spectra.

For Raman spectra, light is scattered by molecule, not absorbed or emitted.

Such scattering of light can be considered as a collision between the photon and the molecule.

1.) If the collision is elastic (elastic scattering), the frequency of the scattered photon is the same as the incident photon

- Process is known as Rayleigh scattering.

2.) If the collision is inelastic (inelastic scattering), the frequency of the scattered photon is different than the incident photon. That is, the photon has gained or lost energy.

- Process is known as Raman scattering.

Vibrational Raman Spectroscopy

An incident photon has an electric field that can interact with the charge distribution of a target molecule.

The applied electric field deforms the electron cloud and the effect of the deformation is proportional to the strength of the applied electric field.

- In other words, the dipole moment of the molecule changes in response to the electric field. 0 E µ = µ +α α - polarizability ν = 4 ν = 3 ν = 2 ν = 1 ν = 0 ground electronic state excited electronic state virtual electronic state R ay le ig h S to k es an ti -S to k

es The deformed electron cloud can return to its original state to yield Rayleigh

scattering.

Or the deformed electron cloud can return to another state.

1.) If the final state has higher energy than the initial state, the scattering event (and subsequent spectral line) is known as a Stokes shift (or Stokes scattering).

2.) If the final state has lower energy than the initial state, the scattering event (and subsequent spectral line) is known as an anti-Stokes shift (or anti-Stokes scattering).

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Selection Rules 2 2 0 0 0 2 0 0 0 0 0 0 0 0 0 0 1 ˆ f i f E i f x x E i x 2 x f x E i f i f E i f x E i x x 0 0 E f x i E f x i x x  ∂α ∂ α  µ = µ + α = µ + α +  +   +         ∂α  ∂α ≈ µ + α + = µ + α + ∂ ∂       ∂α ∂α     = + + = ∂ ∂     L

Thus for vibrational Raman spectroscopy, the selection rule is ∆ν = ±1 for the harmonic oscillator. For the anharmonic oscillator, the selection rule are complicated as for vibrational absorption spectroscopy. ∆ ν = ± ± ± ± ±1, 2, 3, 4, 5,K

Also important to note is that not all vibrations can participate in Raman scattering. Only those vibrations where the polarizability of the molecule changes during the course of a vibration will have a Raman spectrum (i.e.,

0 0 x ∂α   ≠    )

Rotational Raman Spectroscopy

Remember that an applied electric field deforms the electron cloud of a molecule.

As the molecule rotates, the molecular dipole changes because of the applied electric field.

- + - + - + - + - +

The induced dipole changes at double the rate of rotation. - The induced dipole changes after half of a rotation.

The consequence of the changing dipole is that the selection rule for rigid-rotor Raman spectroscopy is ∆ = ±J 2.

S branch

Q branch (Rayleigh) O branch

References

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