CHEM 342. Spring 2002. Problem Set #8. Mortimer Chapters 19, 20. Answers.
Note: Please note that B 
h
8 2 cI
when B is expressed in cm1; and B 
h
8 2 I
when
B is expressed in Hz.
Rotating Diatomic Molecules
1. For a diatomic molecule in a 1  electronic state, we observe a microwave transition
from J = 1 to J = 2 in the presence of an electric field. How many lines will appear in the
spectrum?
The selection rules for rotational transitions in a 1  electronic state are J  1 and
M J  0,1 , where there are g J  2 J  1 values of M J (i.e. M J  0,1,...  J ). The following
transitions are allowed:
J = 1, MJ = +1 to J = 2, MJ = +2, +1, 0
J = 1, MJ = 0
to J = 2, MJ = +1, 0, 1
J = 1, MJ = 1 to J = 2, MJ = 0, 1, 2
This is a total of nine lines.
2. Which of the following diatomic molecules have a rotational microwave spectrum: IF,
O 2 , KCl, Cl 2 .
A pure rotational spectrum will be observed only for those molecules that contain a
permanent dipole moment. Therefore, spectra will be observed only for IF and KCl.
3. Calculate the bond length of
  1.14 10 26 kg .
h
h
B 2  2
8 cI 8 cR 2
R
R
12
C 16O using B  1.9302 cm 1 and the reduced mass is
h
8 cB
2
8 3.0 10
2
6.626 10 34 J s
8


m/s 1.14 10  26 kg 193.02 m 1

R  1.13 10 10 m
Vibrating Diatomic Molecules
4. Which of the following vibrational transitions will be observed for a diatomic molecule
(treated as a harmonic oscillator): v = 1 to v = 3; v = 2 to v = 3; v = 5 to v = 4.
The selection rules for vibrational translations is v  1. Therefore, the allowed
transitions are v = 2 to v = 3; and v = 5 to v = 4.
CHEM 342. Spring 2002. Problem Set #8. Mortimer Chapters 19, 20. Answers.
5. Calculate the frequency of the J = 3 to J = 4 transition in the pure rotational absorption
spectrum of 14 N 16O . The equilibrium bond length is 115 pm. Assume no centrifugal distortion.
The mass of a nitrogen atom is 14.003 amu; the mass of an oxygen atom is 15.995 amu; and the
conversion factor is 1.6605 10 27 kg/amu .
 m m 
I  R 2   N O  R 2
 m N  mO 
 14.003 amu 15.995 amu  
I 
1.6605 10  27 kg/amu 1.15 10 10 m

 14.003 amu  15.995 amu 



2
I  1.64 10  46 kg m 2
B
B
h
8 2 cI
6.626 10 34 J s
8 3.0 10
2
The frequency is
  2B cJ  1

8


m/s 1.64 10
 46
kg m
2

 170.6 m 1  1.706 cm 1

  2 1.706 cm 1 3.0 10 8 m/s 3  1  4.09 1011 s 1
Rotation of Polyatomic Molecules
6. Identify the molecules that will exhibit a pure rotational absorption microwave
spectrum: N 2 O, NO2 , CClF3 , NF3 , SF6 , CH 4 , CO2 .
The molecules that have a permanent dipole moment will have a rotational spectrum:
N 2 O, NO2 , CClF3 , NF3 .
7. What information about the molecular geometry for N 2 O can be determined from
knowing that a pure rotational absorption spectrum is observed for this molecule?
There are four possible arrangements for the atoms in N 2 O : linear N-N-O; linear N-O-N;
bent N-N-O; and bent N-O-N. The linear N-O-N hypothesis can be eliminated as a possibility
because this molecule would not have a permanent dipole moment and would not exhibit a pure
rotational spectrum.
8. The moment of inertia about an axis perpendicular to the principal axis ( I  ) for NH 3
is 2.82  10 47 kg m 2 . There are different types of rigid rotors: linear, spherical top, prolate
symmetric top, oblate symmetric top, asymmetric top. Which type of rotor is NH 3 ? Calculate
the separation (expressed in cm 1 ) of the pure rotational spectrum lines for NH 3 . Hint: The
moment of inertia about the principal axis is given by I ll  2m H R 2 1  cos  , where the mass of
a hydrogen atom = mH = 1.6735 10 27 kg ; the N-H bond length = R = 1.014 10 10 m ; and the
bond angle is 106.78. The moment of inertia about the principal axis is I ll .
CHEM 342. Spring 2002. Problem Set #8. Mortimer Chapters 19, 20. Answers.
In order to determine if NH 3 is an oblate or a prolate symmetric rotor, we need to
compare I  and I ll (the moment of inertia about the principal axis).
I ll  2m H R 2 1  cos 


I ll  2 1.6735 10  27 kg 1.014 10 10 m
 1  cos 106.78
2
I ll  4.435 10  47 kg m 2
I ll is greater than I  (which equals 2.82  10 47 kg m 2 ). Therefore, NH 3 is an oblate
symmetric top.
B
B
h
8 cI 
2
6.626 10 34 J s
8 3.0 10
2
8

m/s 2.82 10
 47
kg m
2

 992 m 1  9.92 cm 1
The rotational line separations are
  2 B J  1  2 B J  2 B  2 9.92 cm 1  19.8 cm 1


9. The molecule CClF3 is a prolate symmetrical top with A  0.1908 cm 1 and
B  0.1111 cm 1 . Calculate the energy corresponding to J = 2 and K = 1.
E J , K  J J  1B hc  K 2 A  B hc







E J , K  22  1 0.1111 cm 1   12 0.1908 cm 1  0.1111 cm 1 100 cm/m  6.63 10 34 J s 3.0 10 8 m/s
E J , K  1.48  10 23 Joules
Vibration of Polyatomic Molecules
10. Consider the vibrational mode that corresponds to the uniform expansion of the
benzene ring. Is it infrared active?
This vibration does not change the molecular dipole moment. Therefore, the mode is
infrared inactive.
Raman Spectroscopy
11. Explain the difference between Stokes and anti-Stokes lines in Raman Spectroscopy.
Anti-Stokes lines correspond to transitions from higher to lower energy levels. In this
case, the molecule makes a transition with J  2 and the scattered photon emerges with
increased energy (and therefore higher frequency than the incident radiation).
Spectral lines corresponding to transitions from a lower to a higher molecular energy
levels are Stokes lines. The molecules makes a transition with J  2 and the lines appear at
lower frequency than the incident radiation.
