Biomolecular Spectroscopy

1
Chemistry 331
Lecture 7 Vibrational and Rotational
Spectroscopy
NC State University
2
The Dipole Moment Expansion
The permanent dipole moment of a
molecule oscillates about an equilibrium value as
the molecule vibrates. Thus, the dipole
moment depends on the nuclear coordinate
Q. where m is the dipole operator.
3
Rotational Transitions
Rotational transitions arise from the rotation of
the permanent dipole moment that can interact
with an electromagnetic field in the microwave
region of the spectrum.
4
The total wave function
The total wave function can be factored into an
electronic, a vibrational and a rotational wave
function.
5
Interaction with radiation
An oscillating electromagnetic field enters as
E0cos(wt) such that the angular frequency hw is
equal to a vibrational energy level difference
and the transition moment is
6
Interaction with radiation
The choice of cos(q) means that we
consider z-polarized microwave light. In
general we could consider x- or y-polarized as
well. x sin(q)cos(f) y sin(q)sin(f) z cos(q)
7
Vibrational transitions
Vibrational transitions arise because of the
oscillation of the molecule about its
equilibrium bond configuration. As the
molecule oscillates infrared radiation can
interact to alter the quantum state.
8
Vibrational transitions
As an example we can calculate the
transition moment between the state v 0 and v
1.
9
Vibrational transitions
Note that this result is a statement of the
vibrational selection rule. Within the
harmonic approximation transitions can only occur
between states separated by one quantum number
(Dv 1 or Dv -1). This general rule can be
seen by considering integrals of the type shown
in the previous slide.
10
Classical Vibration of a Diatomic
As was the case for rotation, we can consider a
simple model of a mass on a spring attached to a
wall of infinite mass and a diatomic molecule as
two simple examples. Mass on a spring
Diatomic
k
k
m2
m1
m
k is the force constant
Reduced mass
11
Harmonic approximation
At equilibrium
Assume terms higher than quadratic are zero By
definition
12
Classical approach to vibration
Solution is oscillatory Any energy is possible
Q
13
Classical vibrational motion

• A particle undergoes harmonic motion if it
  experiences a restoring force that is
  proportional to its displacement, x.
• F -kQ (k is a force constant)
• F -dV/dQ and V 1/2kQ2.
• The classical harmonic oscillator can also be
  written as
• Solutions have the form of sin(wt) or cos(wt)
  depending on the initial conditions.
• These solutions imply that

14
Classical potential function

• The potential is V 1/2kQ2, which is a parabolic
  function.
• This potential is called a harmonic potential.
  
• The force constant k has units of Newtons/meter
  (N/m) or Joules/meter2 (J/m2).
• The angular frequency w 2pn, n is the frequency
  in Hz.

15
Quantum approach to vibration
Solution is Gaussian Energy is quantized
v is the quantum number Allowed transitions v
v 1, v v - 1
Q
16
Vibrational wavefunctions

• Energy levels are given by Ev (v 1/2)hw
• Wavefunctions are Yv NvHve-y2/2
  where Hv is the Hermite polynomial
• Typical energies are of the order of 0 - 3000
  cm-1

17
Solutions to harmonic oscillator
The Hermite polynomials are derivatives of a
Gaussian, y a1/2Q
The normalization constant is
18
The zero point energy

• The lowest level is E0 1/2hw
• The lowest vibrational level is not zero in
  energy.
• This is consistent with the uncertainty
  principle. If atoms were completely still at
  absolute zero then we would know both their
  position and moment to arbitrary accuracy.
• The width of the wavefunction is related to
  positional uncertainty of an atom.
• We call E0 the zero point energy.

19
Polyatomic Molecules

• There are 3N total degrees of freedom in a
  molecule that contains N atoms.
• There are three translational degrees of freedom.
  These correspond to motion of the center of mass
  of the molecule.
• In a linear molecule there are two rotational
  degrees of freedom. In a non-linear molecule
  there are 3 rotational degrees of freedom.
• The remaining degrees of freedom are vibrational.

20
Normal modes - water
Asymmetric Stretch n3 3935 cm-1
Bend n2 1654 cm-1
Symmetric Stretch n1 3825 cm-1
There are 3 normal modes (3N - 6). All of them
are infrared active since all show a dipole
moment change in their motion. The harmonic
approximation can be applied to each normal mode.
21
Normal modes - CO2
Asymmetric stretch n3 2349 cm-1 (IR
active)
Symmetric stretch n1 2289 cm-1 (Raman
active)
Asymmetric stretch n3 2349 cm-1 (IR
active)
Bends n2 667 cm-1
(IR active)
There are 4 normal modes (3N - 5). Three of them
are infrared active since they show a dipole
moment change in their motion.
22
Vibrational Transition
23
Vibrational Transition
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Vibrational Selection Rule
v 2
v v 1
v 1
v v - 1
v 0
Q
25
Overtones of water
Even in water vapor n1 n3, but symmetries are
different, G1 ¹ G3. However, the third
overtone of mode 1 has the same symmetry as the
combination band G1 G1 G1 G1 G3 G3 . Strong
anharmonic coupling leads to strong overtones at
11,032 and 10,613 cm-1. These intense bands
give water and ice their blue color.
26
Frequency shift due to molecular interactions
Hydrogen bonding lowers O-H force constant and
H-O-H bending force constant.
vapor liquid n1 3825 3657 n2 1654 1595 n3
3935 3756
27
Question
Which expression is correct?
28
Question
Which expression is correct?
29
Question
How many normal modes of vibration does methane
have? A. 15 B. 12 C. 9 D. 6
30
Question
How many normal modes of vibration does methane
have? A. 15 B. 12 C. 9 D. 6
31
Rotation in two dimensions
The angular momentum is Jz pr.
Jz
r
p
m
Using the deBroglie relation p h/l we also have
a condition for quantization of angular motion Jz
hr/l.
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Classical Rotation
In a circular trajectory Jz pr and E Jz2/2I.
I is the moment of inertia. Mass in a circle I
mr2. Diatomic I mr2
m
m1
r
r
m2
Reduced mass
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The 2-D rotational hamiltonian

• The wavelength must be a whole number fraction of
  the circumference for the ends to match after
  each circuit.
• The condition 2pr/m l combined with the
  deBroglie relation leads to a quantized
  expression,Jz mh.
• The hamiltonian is

34
The 2-D rotational hamiltonian

• Solutions of the 2-D rotational hamiltonian are
  sine and cosine functions just like the particle
  in a box.
• Here the boundary condition is imposed by the
  circle and the fact that the wavefunction must
  not interfere with itself.
• The 2-D model is similar to condition in the Bohr
  model of the atom.

35
The 3-D rotational hamiltonian

• There are two quantum numbers
  J is the total angular momentum quantum
  number M is the z-component of the
  angular momentum
• The spherical harmonics called YJM are functions
  whose probability YJM2 has the well known shape
  of the s, p and d orbitals etc.
• J 0 is s , M 0
• J 1 is p , M -1, 0 , 1
• J 2 is d , M -2 , -1, 0 , 1, 2

36
Space quantization in 3D

• Solutions of the rotational Schrödinger equation
  have energies E h2 J(J 1)/2I
• Specification of the azimuthal quantum number mz
  implies that the angular momentum about the
  z-axis is Jz hmz.
• This implies a fixed orientation between the
  total angular momentum and the z component.
• The x and y components cannot be known due to the
  Uncertainty principle.

z
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Spherical harmonic for P0(cosq)

• Plot in polar coordinates represents Y002 where
  Y00(1/4p)1/2 .
• Solution corresponds to rotational quantum
  numbers J 0,
• M or Jz 0.
• Polynomial is valid for n³1 quantum numbers of
  hydrogen wavefunctions

38
Spherical harmonic for P1(cosq)

• Plot in polar coordinates represents Y112 where
  Y11(1/2)(3/2p)1/2 sinqeif.
• Solution corresponds to rotational quantum
  numbers J 1, Jz 1.
• Polynomial is valid for n³2 quantum numbers of
  hydrogen wavefunctions

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Spherical harmonic for P1(cosq)

• Plot in polar coordinates represents Y102 where
  Y10(1/2)(3/p)1/2 cosq with normalization.
• Solution corresponds to rotational quantum
  numbers J 1, Jz 0.
• Polynomial is valid for n³2 quantum numbers of
  hydrogen wavefunctions.

40
Spherical harmonic for P2(cosq)

• Plot in polar coordinates of Y222 where
  Y221/4(15/2p)1/2cos2qe2if
• Solution corresponds to rotational quantum
  numbers J 2, Jz 2.
• Polynomial is valid for n³3 quantum numbers of
  hydrogen wavefunctions

41
Spherical harmonic for P2(cosq)

• Plot in polar coordinates of Y212 where
  Y21(15/8p)1/2sinqcosqeif
• Solution corresponds to rotational quantum
  numbers J 2, Jz 1.
• Polynomial is valid for n³3 quantum numbers of
  hydrogen wavefunctions

42
Spherical harmonic for P2(cosq)

• Plot in polar coordinates of Y202 where
  Y201/4(5/p)1/2(3cos2q-1)
• Solution corresponds to rotational quantum
  numbers J 1, Jz 0.
• Polynomial is valid for n³3 quantum numbers of
  hydrogen wavefunctions

43
Rotational Wavefunctions
J 0
J 1
J 2
These are the spherical harmonics YJM, which are
solutions of the angular Schrodinger equation.
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The degeneracy of the solutions

• The solutions form a set of 2J 1 functions at
  each energy (the energies are E h2 J(J 1)/2I).
• A set of levels that are equal in energy is
  called a degenerate set.

.
J 3
J 2
J 1
J 0
45
Rotational Transitions

• Electromagnetic radiation can interact with a
  molecule to change the rotational state.
• Typical rotational transitions occur in the
  microwave region of the electromagnetic spectrum.
• There is a selection rule that states that the
  quantum number can change only by or - 1 for an
  allowed rotational transition (DJ 1).

J 2
J 1
J 0
46
Orthogonality of wavefunctions

• The rotational wavefunctions can be represented
  as the product of sines and cosines.
• Ignoring normalization we have
• s 1
• p cosq sinqcosf sinqsinf
• d 1/2(3cos2q - 1), cos2qcos2f , cos2qsin2f ,
  
• cosqsinqcosf , cosqsinqsinf
• The differential angular element is sinqdqdf/4p
  over
• the limits q 0 to p and f 0 to 2p.
• The angular wavefunctions are orthogonal.

47
Orthogonality of wavefunctions

• For the theta integrals we can use the
  substitution
• x cosq and dx sinqdq
• For example, for s and p-type rotational wave
  functions we have

48
Question

• Which of the following statements is true
• A. The number of z-projection of the quantum
  numbers is 2J1.
• B. The spacing between rotational energy levels
  increases as 2(J1).
• C. Rotational energy levels have a degeneracy of
  2J1.
• D. All of the above.

49
Question

• Which of the following statements is true
• A. The number of z-projection of the quantum
  numbers is 2J1.
• B. The spacing between rotational energy levels
  increases as 2(J1).
• C. Rotational energy levels have a degeneracy of
  2J1.
• D. All of the above.

50
Question

• The fact that rotational wave functions are
  orthogonal means that
• A. They have no overlap
• B. They are normalized
• C. They are linear functions
• D. None of the above

51
Question

• The fact that rotational wave functions are
  orthogonal means that
• A. They have no overlap
• B. They are normalized
• C. They are linear functions
• D. None of the above

52
The moment of inertia
The kinetic energy of a rotating body is
1/2Iw2. The moment of inertia is given by The
rigid rotor approximation assumes that molecules
do not distort under rotation. The types or rotor
are (with moments Ia , Ib , Ic) - Spherical
Three equal moments (CH4, SF6) (Note No
dipole moment) - Symmetric Two equal moments
(NH3, CH3CN) - Linear One moment (CO2, HCl,
HCN) (Note Dipole moment depends on
asymmetry) - Asymmetric Three unequal moments
(H2O)
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Pure rotational spectra

• A pure rotational spectrum is obtained
• by microwave absorption.
• The range in wavenumbers is from
• 0-200 cm-1.
• Rotational selection rules dictate that
• the change in quantum number must
• be DJ 1 and DMJ 0.
• A molecule must possess a ground
• state dipole moment in order to have
• a pure rotational spectrum.

54
Energy level spacing
Energy levels
Energy Differences of DJ 1
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The rotational constant
The spacing of rotational levels in spectra is
given by DE EJ1- EJ according to the
selection rule.
The line spacing is thus proportional to the
rotational constant


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A pure rotational spectrum

2B
A pure rotational spectrum is observed in
the microwave range of electromagnetic spectrum.
57
Question

• Which molecule found in the atmosphere has a
• pure rotational spectrum?
• Diatomic oxygen
• Diatomic nitrogen
• Water
• Carbon Dioxide

58
Question

• Which molecule found in the atmosphere has a
• pure rotational spectrum?
• Diatomic oxygen
• Diatomic nitrogen
• Water
• Carbon Dioxide

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A typical rovibrational spectrum
Note that the rotational spectrum is centered a
vibrational frequency