CHM 3411 – Problem Set 7
Due date: Friday, April 1st.
Do all of the following problems. Show your work.
1) The rotational constant for the molecule 1H81Br is B = 8.465 cm-1.
a) Find the number of molecules at equilibrium, at T = 300.0 K, that are in the J = 1, 2, 3, and 4 levels,
relative to the number in the J = 0 rotational level. Recall the expression for the Boltzmann factor as applied to
rotation in a diatomic molecule
PJ/P0 = (gJ/g0) exp[-(EJ – E0)/kT]
(1.1)
where PJ is the number of molecules in state J, relative to the number in J = 0 (P0), and gJ = (2J + 1) is the
degeneracy of the Jth rotational energy level (where g0 = 1). The value of the Boltzmann constant is k = 0.69503 cm1
/K.
b) Briefly explain the effect of the observed populations on the appearance of the pure rotational spectrum
of 1H81Br.
2) Classify the following molecules in one of the following four categories: linear rotor, symmetric rotor, asymmetric
rotor, or spherical rotor. Also indicate whether the molecule will exhibit a pure rotational spectrum.
a) CH2Cl2
d) NOF
b) N2O
e) C6H3Cl3 (1,3,5 trichlorobenzene)
c) C6H6
f) CCl4
3) To a good approximation the vibrational-rotational energy levels for a diatomic molecule are given by the
expression
Ev,J = e (v + 1/2) - exe (v + 1/2)2 + Bv J (J+1)
(3.1)
Note that B0  B1, because of the effect of vibrational excitation on the value for the rotational constant. Starting
with eq 3.1, find an expression for the location of the R-branch (J = + 1) transitions in a diatomic molecule.
Assume that the vibrational transition taking place is v = 0  v = 1. (You may check your result against equn
12.46, p 469 of Atkins).
4) The energy expression for a Morse oscillator is
Ev = e (v + 1/2) - exe (v + 1/2)2
(4.1)
Find the value for vmax, the maximum value for the vibrational quantum number, and De, the dissociation energy, for
a Morse oscillator. Hint: Since for a real molecule the vibrational energy must increase as v increases, you may find
vmax using the condition
dEv/d(v + ½) = 0
(4.2)
If you substitute the value of v you obtain you should get the value for D e, the dissociation energy. Note that vmax is
the largest integer value of v that is smaller than the value of v found by the above procedure.
Also do the following from Atkins.
Exercise 12.7b Given that the spacing of lines in the microwave spectrum of 35Cl19F is constant at 1.033
cm , calculate the moment of inertia and bond length of the molecule. Is the value for bond length r0 or re? Briefly
explain. Also predict the spacing of lines in the microwave spectrum of 37Cl19F. m(35Cl) = 34.9688 amn; m(37Cl) =
36.9651 amu; m(19F) = 18.9984 amu.
-1
Exercise 12.24b How many normal modes of vibration are there for the following molecules. a) C6H6, b)
C6H5CH3,(Atkins has a typo here) c) HC  C - C  CH.
Solutions.
1)
a) PJ/P0 = (gJ/g0) exp[-(EJ – E0)/kT]
Since
gJ = 2J + 1
g0 = 1
k = 0.69503 cm-1/K
B = 8.465 cm-1
T = 300.0 K
EJ – E0 = J(J+1)B – 0(1)B = J(J+1)B
then
PJ/P0 = (2J+1) exp[-(J(J+1)B/kT]
J
gJ/g0
EJ – E0
PJ/P0
0
1
2
1
3
5
0.0
16.93
50.79
1.000
2.766
3.919
3
4
5
7
9
11
101.58
169.30
253.95
4.301
3.996
3.254
(Note that I have included oe additional value for J in the above table).
b) Since to a first approximation the intensity of a transition in the pure rotational spectrum of a diatomic
molecule is proportional to population, the above indicates that at T = 300.0 K the most intense peak in the pure
rotational spectrum will be J = 3  4. If we were looking at the rotational structure of a vibrational transition, we
would expect P(3) to be the most intense P branch transition, and R(3) to be the most intense R branch transition.
2)
The general way of answering this question can be put in the form of a flow chart.
Is the molecule linear?

NO

YES – then a linear rotor
Does the molecule have the same symmetry
elements as CH4 or SF6?

NO

YES – then a spherical top rotor
Is the principal axis of rotation C3 or higher?

NO – then an asymmetric top rotor

YES – then a symmetric top rotor
a)
b)
c)
d)
e)
no spectrum)
f)
spectrum)
CH2Cl2 – Principal axis is C2, and so an asymmetric top rotor. (polar, so will have spectrum)
N2O – Linear so a linear rotor. (polar, so will have spectrum)
C6H6 – Principal axis is C6, and so a symmetric top rotor. (nonpolar, so no spectrum)
NOF – Nonlinear, only a C1 axis, and so an aymmetric top rotor. (polar, so will have spectrum)
C6H3Cl3 (1,3,5 trichlorobenzene) – Principal axis is C3, and so a symmetric top rotor. (nonpolar, so
CCl4 – Same elements of symmetry as CH4, and so a spherical top rotor. (nonpolar, so no
3) For the R-branch transition
E = E1,J+1 – E0,J
= [ 3/2 e – 9/4 exe + (J+1)(J+2)B1 ] - [ 1/2 e – 1/4 exe + (J)(J+1)B0 ]
= e – 2 exe + (J2 + 3J + 2)B1 – (J2 + J)B0
While this is a perfectly good way of expressing the result, we can play with the above to write it in terms of powers
of (J+1), as in Atkins
(J2 + 3J + 2)B1 = (J2 + 2J + 1)B1 + (J+1)B1 = (J+1)2B1 + (J+1)B1
(J2 + J)B0 = (J2 + 2J + 1)B0 – (J+1)B0 = (J+1)2B0 – (J+1)B0
Substituting gives the final result
E = e – 2 exe + (J2 + 3J + 2)B1 – (J2 + J)B0 = (e – 2 exe) + (J+1)(B1 + B0) + (J+1)2(B1 - B0)
Ev = e (v + 1/2) - exe (v + 1/2)2
4) Since
then dEv/d(v + ½) = d/d(v + ½) [ e (v + 1/2) - exe (v + 1/2)2 ]
= e – 2 exe (v + ½) = 0
and so (vmax + ½) = e/(2 exe)
vmax = [e/(2 exe)] – ½
In fact, since vmax must be an integer, vmax is the largest integer that is smaller than [e/(2 exe)] – ½ .
To get the value for De, we can substitute our value for vmax + ½ into the equation for Ev.
De = Evmax = e (vmax + ½) - exe (vmax + ½)2
= e [e/(2 exe)] - exe [e/(2 exe)]2
= [e2/(2 exe)] – [e2/(4 exe)] = [e2/(4 exe)]
Exercise 12.7b
The spacing between adjacent lines in the pure rotational spectrum of a molecule is equal to 2B.
so
B = (1.033 cm-1)/2 = 0.5165 cm-1
But B = h/82Ic
and so I = h/82Bc =
I = r2
 = mAmB/(mA+mB)
(6.626 x 10-34 J.s)
= 5.420 x 10-46 kg.m2
-1
10
8 (0.5165 cm ) (2.998 x 10 cm/s)
2
But
I = r2
r = (I/)1/2
(35Cl19F) = (34.9688 amu) (18.9984 amu)/[(34.9688 + 18.9984) amu]
= 12.3103 amu (1 kg/6.022 x 10 26 amu) = 2.044 x 10-26 kg
(37Cl19F) = (36.9651 amu) (18.9984 amu)/[(36.9651 + 18.9984) amu]
= 12.5489 amu (1 kg/6.022 x 10 26 amu) = 2.084 x 10-26 kg
Using the value for B for 35Cl19F gives
r = [(5.420 x 10-46 kg.m2)/(2.044 x 10-26 kg)]1/2 = 1.628 x 10-10 m = 0.1628 nm
While no information about the conditions of the experiment are given, it is reasonable to assume the
molecules are at equilibrium at room temperature, and so primarily in the v = 0 vibrational state. The above value
for r is thus r0 (and not re).
Since B ~ (1/) it follows (assuming r0 is the same for both 37Cl19F and 35Cl19F) that
B(37Cl19F) = (35Cl19F)
B(35Cl19F)
(37Cl19F)
or
B(37Cl19F) = B(35Cl19F) [(35Cl19F)/(35Cl19F)]
= (0.5165 cm-1) [(2.044/2.084)] = 0.5066 cm-1
and so the spacing between adjacent lines for the microwave spectrum of 37Cl19F should be
2B(37Cl19F) = 2 (0.5066 cm-1) = 1.013 cm-1
Exercise 12.24b
Recall that
# vibrations
= 3N – 5 (linear molecule
= 3N – 6 (nonlinear molecule) where N = # of atoms per molecule
a) C6H6 N = 12, nonlinear, and so 30 vibrations
b) C6H5CH3
N = 15, nonlinear, and so 39 vibrations
c) HC  C - C  CH
N = 6, linear, and so 13 vibrations