1. Revisit vibrational Spectroscopy
Frequency: Infrared
Near-IR cm-1
Mid-IR cm-1
Far-IR cm-1
Model:
Harmonic oscillator
Hamiltonian:
Wave function:
Energy level:
Transition energy:
Selection rule:
E= h

2. Anharmonic oscillator
Energy curve:
Harmonic oscillator
Anharmonic oscillator E 4 D0 3 De 2 1 1 h
at n=0, E0  0
Normal modes: 3N-6 (non-linear), 3N-5 (linear)

3. Revisit Rotational Spectroscopy
Frequency: Microwave
cm-1
Model:
Rigid rotor
Hamiltonian:
Wave function:
Energy level:
Degeneracy:
2J+1 values mJ = 0, 1, 2,.. J
Transition energy:
Selection rule:

4. Rotational angular momentum:
Rotational spectral lines:

5. Rovibrational Spectroscopy
Outlines
Operator, wavefunctions, energy
Rovibrational Energy of rigid diatomic molecules
Rotational and vibrational energy levels
Rovibrational transitions and selection rule
Transition energy of rovibration
Rovibrational spectrum
Intensity of a spectral line: Boltzmann distribution
Rovibration of non-rigid rotation

6. Operator, wavefunctions, energy
Rovibration = Rotation + Vibration
- Energy operator of rovibration motion :
Summation of H
- Wavefunctions of the system :
Product of 
Total energy of rovibration motion:
Erovib = Erot + Evibr

7. Rovibrational Energy of rigid diatomic molecules
Absorption of infrared region, molecules can change vibrational and rotational states because vibrational transitions can couple with rotational transitions to give rovibrational spectra.
By Treating as harmonic oscillator and rigid rotor, energy of vibration and rotation can expressed as:
n = 0, 1, 2, …
J = 0, 1, 2, …
J unit

8. From previous slide: n = 0, 1, 2, … J = 0, 1, 2, …
In unit of Wave number (cm-1)
n = 0, 1, 2, …
J = 0, 1, 2, …

9. Rotational and vibrational energy levels
A series of J rotational states in vibrational states.

10. n = + 1 Vibrational Transition
Rovibrational transitions and selection rule
Selection rules: n =  1 and J =  1
Absorption:
n = + 1 Vibrational Transition 
Rotational Transition
J = - 1
J = + 1
Typically at room temperature, only ground vibrational state populated but several rotational levels may be populated.

11. Transition energy of rovibration
Rotational Transition
Erot ;
Selection rules: J =  1
For J = + 1 :
Vibrational Transition
Evib ;
Selection rules: n = + 1
For J = - 1 :

12. n = + 1 Vibrational Transition
Rovibrational spectrum
n = + 1 Vibrational Transition 
Rotational Transition
J = - 1
J = + 1
Q branch 
P branch
(lower frequency) 
R branch
(higher frequency) 
cm-1 2B
Decreases as J increases
increases as J increases

13. Ex. Rovibrational transitions
J = - 1
J = + 1

14. Ex. Rovibrational transitions
J = - 1 (P branch)
J = + 1 (R branch)

15. Simulated 300. K infrared absorption spectrum and energy diagram for HCl.

16. Ex. Infrared absorption by HI gives rise to an R-branch from n = 0
Ex. Infrared absorption by HI gives rise to an R-branch from n = 0.What is the wavenumber of the line originating from the rotational state with J = 4? (Assuming HI as rigid rotor model where k =314 N m-1 and its bond length = 163 pm)
Absorption lines in R-branch region correspond to:
Calculate the vibrational transition energy at n=0
=1.66 10-27kg
= 6.92 1013 Hz
= 4.59  J

17. Calculate the vibrational transition energy at J=4
the wavenumber of the line originating from the rotational state with J = 4 is
= cm-1

18. Rovibrational spectrum of HCl

19. Intensity of a spectral line: Boltzmann distribution
Depends on the number of molecules in the energy level from which the transition originates.
Boltzmann distribution
A high-resolution spectrum is shown for CO in which the P and R branches are resolved into the individual rotational transitions.

20. Ex. Using the Boltzmann distribution equation to calculate the number of CO molecules in J = 2, 5, 9 and 18 relative to the ground state (J=0) at 300K. Giving the rotational constant = cm-1
kB = 1.38 X m2 kg s-2 K-1 J
2J+1
BJ(J+1)
NJ/N0 2 5
2.3019E-22
4.73 11
1.1509E-21
8.33 9 19
3.4528E-21
8.25 18 37
1.3121E-20
1.56

21. If we calculate NJ/N0 with various J values, we would obtain a graph below.
At 300K
At 700K 1
2.945
2.98 2
4.730
4.88 3
6.264
6.67 4
7.478
8.31 5
8.331
9.77 6
8.810
11.00 9
8.255
13.29 12
5.894
13.46 15
3.357
11.96 18
1.558
9.52 21
0.596
6.87 24
0.189
4.53 27
0.050
2.74 30
0.011
1.52 33
0.002
0.78

22. Rovibration of non-rigid rotation
Two effects:
Vibration-Rotation Coupling :
As the molecule vibrates more, bond stretches
The effective rotational constant
Be : the rotational constant for a rigid rotor
αe : the rotational-vibrational coupling constant
2) Centrifugal Distortion
As a molecule spins faster, the bond is pulled apart

23. Centrifugal Distortion : non-rigid rotation
When a molecule rotates faster and faster (J increases), the centrifugal force pulls the atoms apart. As a result, its bond length increases and its moment of inertia (I) increases, thus decreasing B
Centrifugal force
Rotational energy of non-rigid rotor:
rigid rotot
D = the centrifugal distortion constant

24. Real spectrum Ideal spectrum Rovibrational energy of non-rigid rotor:
As energy increases, the R-branch lines move closer and as energy decreases, the P-branch lines move farther apart. This is attributable to two phenomena: rotational-vibrational coupling and centrifugal distortion.

25. Ex. Using the rovibrational spectrum of HCl to answer the questions below (consider as non rigid molecule) A D E C F G
2906 B
2926 H
2865
Spectral region with J = -1 ……..
Spectral region with J = +1 ……..
R-branch ……..
P-branch ……..
Q-branch ……..
6) J=7  J=8 transition ……..
7) J=5  J=4 transition ……..
8) Spectral lines belong to 1H35Cl ……..
9) Spectral lines belong to 1H37Cl ……..
10) 𝐵 𝑒 = ………cm-1, 𝐷 = ………… cm-1

26. Transition Selection ruleJ
En,J
E0,0
E0,1
E1,0
E1,1
E1,2 2 1 1 1
Observed freq.
Transition Selection rule
n = 0  n =1
J = 0  J = 1
n = 1 J = +1
n = 0  n =1
J = 1  J = 0
n = 1 J = -1
n = 0  n =1
J = 1  J = 2
n = 1 J = +1

27. Rovibrational energy of non-rigid rotor:
2926
2865
2906

28. (1) - (2)
= 41
(3) - (1)
= 20

29. Ex. Using the rovibrational spectrum of HCl to answer the questions below
2906
2926 H F C G E D A B
Spectral region with J = -1 …A…..
Spectral region with J = +1 … B ..
R-branch … B ….
P-branch … A ..
Q-branch … D…..
6) J=7  J=8 transition … E …..
7) J=5  J=4 transition … C …..
8) Spectral lines belong to 1H35Cl … F ..
9) Spectral lines belong to 1H37Cl … G…..
10) 𝐵 𝑒 = …10.3 cm-1, 𝐷 = …0.02…… cm-1

31. Rigid rotor wavefunctions
The full wavefunctions
Reduce to
Consider only the angular part of the wavefunctions
Set the Schrödinger Equation equal to zero and

32. From the previous slide
Rearranging the differential equation separating the θ-dependent terms from the -dependent terms:
Only 
Only 
the Schrödinger equation can be solved using separation of variables.

33. Use the substitution method (similar to the previous one)
For the J=0 → J=2 transition,
Consider
Use the substitution method (similar to the previous one)
Replace x with  and integrate from 0 to , we get:
Do the same for

34. From the previous derivation:
For the J=0 → J=2 transition,
Thus:
From the previous derivation:
Therefore:
Thus, the J=0 → J=2 transition is forbidden.

35. Home Work 2
1. Spectral line spacing of rotational microwave spectrum of OH radical is 37.8 cm-1 . Determine the OH bond length (in pm unit) and moment of inertia (in kg m2)
mO = amu
Spectrum (cm-1)
mH = amu
37.8 cm-1
2. Use the bond length of diatomic molecules in Table to predict line spacing (in cm-1 unit) of rotational microwave spectrum.
molecule
bond length (pm) HF
91.7 HI
161
HCl
128
HBr
141

36. 3. Determine the bond length of these diatomic gases in Table and arrange them in order of increasing the bond length.
Bond length (pm) OH
37.80
ICl
0.11
ClF
1.03
AlH
12.60
4. Using the information in the Table to calculate the ratio between the transition energy of rotation from J=0 to J =1 and vibration from n=0 to n=1 for H2
Atomic mass
1.008 amu
Bond length of H2
74.14 pm
Force constant of H2
575 N/m