CHM 3411 – Problem Set 8
Due date: Friday, April 8th.
Do all of the following problems. Show your work.
1) We showed in class that for a Morse oscillator the dissociation energy, D e, is given by the expression
De = e2/(4exe)
(1.1)
For real diatomic molecules the above is expected to be approximately correct.
a) Based on the data below calculate the value for De. Compare the results to the experimental values for
De, and briefly comment on the agreement observed.
b) Find k, the force constant, for each of the molecules listed below. Give your values for k in units of N/m.
c) Estimate the value for e for 2H35Cl (the deuterated for of hydrogen chloride).
e (cm-1)
molecule
H35Cl
CO
N2
O2
2889.6
2169.5
2358.6
1580.2
exe (cm-1)
52.06
13.45
14.34
11.98
De (cm-1)
37190.
90680.
79890.
42040.
2) The NO2 molecule is non-linear, and has three vibrations.
vibrational energy, relative to the ground (0,0,0) state, is
Erel(v1,v2,v3) = E(v1,v2,v3) – E(0,0,0) = 1v1 + 2v2 + 3v3
In the harmonic oscillator approximation the
(2.1)
where 1 = 1358. cm-1, 2 = 757. cm-1, and 3 = 1666. cm-1.
a) Identify all of the vibrational states with relative vibrational energy E rel < 5000. cm-1.
b) Find the number of vibrational states with Erel(v1,v2,v3) in the following ranges of energy: 0-1000 cm-1,
1000-2000 cm-1, 2000-3000 cm-1, 3000-4000 cm-1, and 4000-5000 cm-1. This represents a rough measure of the
dependence of the density of vibrational states on energy.
c) Comment on the general trend observed in the density of vibrational states with energy.
3) The following transitions are observed in the infrared spectrum of N 2O (nitrous oxide). All transitions originate in
the ground (0,0,0) vibrational state of the molecule.
Band location (cm-1)
Relative Intensity
589.
1167.
1285.
1868.
2224.
2462.
2564.
2799.
3366.
3481.
strong
very weak
very strong
very weak
very strong
weak
weak
very weak
very weak
medium
a) Assign the spectrum, that is, give the quantum numbers (v1,v2,v3) for the final state for each observed
band. Note: You may identify the lower energy stretching vibration as v1 and the higher energy stretching vibration
as v3.
b) For each combination and overtone band give the experimental location of the band, the location
predicted using the harmonic oscillator approximation, and the difference between the two (predicted experimental).
Also do the following from Atkins.
Exercise 12.19b For 14N2, the fundamental and first overtone vibrational transitions occur at the following
energies v = 0  v = 1 (2345.15 cm-1); v = 0  v = 2 (4661.40 cm-1); v = 0  v = 3 (6983.73 cm-1). Assuming that
the vibrational energy is given by the expression
Ev = e (v + ½) - exe (v + ½)2
find the values for e and exe for N2 based on the above data. (Hint: Starting with the above expression for E v find
an expression for Ev – E0. Use the expression you derive to find values for e and exe).
EXTRA CREDIT – Starting with the Boltzmann distribution law as applied to the rotational states of a
diatomic molecule (in the rigid rotor approximation)
PJ/P0 = (2J+1) exp[-(J(J+1)B/kT]
derive an expression for Jmax, the value of J corresponding to the most populated level of a diatomic molecule. Use
your result to find the corresponding expression for T in terms of J max. Note that this represents a way to estimate the
approximate temperature of a gas containing a diatomic molecule based on its rotational spectrum (or the rotational
structure in the vibrational spectrum).
Solutions.
1)
a)
Molecule
De(exp)
(cm-1)
De(calc)
(cm-1)
HCl
CO
N2
O2
37190.
90680.
79890.
42040.
40100.
87490.
96980.
52110.
% diff = |De(calc) – De(exp)|/De(exp)
7.8 %
3.5 %
21.4 %
24.0 %
The value for De calculated using the approximation derived from a Morse potential is good to within ~
25% of the actual value. Not great agreement, but not a bad first approximation.
It would be interesting to test a few more molecules, since there appears to be much better agreement for the
heteronuclear diatomics than the homonuclear diatomics. This may simply be coincidence, but it would be
interesting to see.
b)
e = (1/2c) (k/)1/2
So
k = 42c2e2
Mass data are given in the Appendix of Atkins.
Molecule
H35Cl
CO
N2
O2
e (cm-1) (kg)
2889.6
2169.5
2358.6
1580.2
c)
k (N/m)
0.1627 x 10-26
1.139 x 10-26
1.163 x 10-26
1.328 x 10-26
482.
1902.
2296.
1177.
Since e ~ (1/)1/2
it follows that
e(2H35Cl)/e(1H35Cl) = (1H35Cl)/(2H35Cl)
or
e(2H35Cl) = e(1H35Cl) [(1H35Cl)/(2H35Cl)]1/2
Since
(2H35Cl) = 0.3162 x 10-26 kg
e(2H35Cl) = (2889.6 cm-1) [0.1627/0.3162]1/2 = 2072.8 cm-1
2)
All of the states for which Erel(v1,v2,v3) = E(v1,v2,v3) – E(0,0,0) < 5000. cm-1 are listed below
state
energy (cm-1)
state
energy (cm-1)
state
energy (cm-1)
000
001
002
003
010
011
012
020
021
022
030
031
0.
1666.
3332.
4998.
757.
2423.
4089.
1514.
3180.
4846.
2271.
3937.
040
041
050
060
100
101
102
110
111
120
121
3028.
4694.
3785.
4542.
1358.
3024.
4690.
2115.
3781.
2872.
4538.
130
140
200
201
210
220
230
300
310
3629.
4386.
2716.
4382.
3473.
4230.
4987.
4074.
4831.
If we group these by energy we get the following results
energy range (cm-1)
# vibrational states
0 – 1000
1000 – 2000
2000 – 3000
3000 – 4000
4000 – 5000
2
3
5
9
13
Notice that the number of states per energy interval (the density of states) increases as energy increases. For larger
polyatomic molecules, with more atoms, this increase is expected to be even more rapid.
3)
Band location
(exp, cm-1)
Assignment
589.
1167.
1285.
1868.
2224.
2462.
2564.
2799.
3366.
3481.
010
020
100
110
001
120
200
011
021
101
Band location
(pred, cm-1)
pred - exp
(cm-1)
1178.
11.
1874.
6.
2463.
2570.
2813.
3402.
3509.
1.
6.
14.
36.
28.
Relative Intensity
strong
very weak
very strong
very weak
very strong
weak
weak
very weak
very weak
medium
All transitions originate in the (0,0,0) state. The fundamentals are bolded and underlined. The lower energy stretch
is assumed to be v1, and the higher energy stretch is assumed to be v3. The reasoning behind the assignment is as
follows:
Bends are lower in energy than stretches, and so the first observed band is expected to be to the (0,1,0)
state. The intensity is consistent with an allowed fundamental band. The band at 1167. cm-1 is close to twice this
fundamental, and is very weak, consistent with its being the (0,2,0) overtone. There is no band close to three times
the bending fundamental.
The strong band at 1285. cm-1 is not an overtone of the bend, and so must be one of the stretching
fundamentals. It is thus the (1,0,0) band. The predicted location of the (1,1,0) combination band is close to the
1868. cm-1 band and so has been assigned to it. Two times the v1 fundamental is 2570. cm-1, close to a band at 2564.
cm-1, which would be forbidden and so weak, and so this band has been assigned to it.
The strong band at 2224. cm-1 is not an overtone or combination band, and so must be the third fundamental
band (0,0,1). Since the weak band at 2462. cm-1 cannot involve v3, the only good possibility is that it is the
forbidden (1,2,0) band.
The (1,0,1) combination band is predicted to appear at 3509. cm-1. There is a band at 3481. cm-1, and the
intensity is medium, consistent with a transition to an allowed combination. It has thus been assigned (1,0,1).
The only reasonable possibilty for the remaining band at 3366. cm-1 is the forbidden (0,2,1) combination
band, and so it has been assigned as that, even though the difference between the observed and predicted location is
~ 35 cm-1.
Notice that we used both the intensities and predicted band locations to make our assignments. Also notice
that anharmonicity becomes greater with greater energy, as indicated by the deviations between the predicted band
positions (based on the HO approximation) and the observed positions.
Exercise 12.19b
Using the expression that is given for Ev, we may say
Ev – E0 = [e (v + ½) - exe (v + ½)2 ] – [ (1/2) e – (1/4) exe ]
= e v - exe (v2 + v) = (e - exe) v - exe v2
There are many ways to proceed from this point. The most straightforward is to plot (E v – E0) vs v, and fit the data
to a second order polynomial. The second order term will be equal to - exe, and the first order term to e - exe.
v
Ev – E0 (cm-1)
0
1
2
3
0.00
2345.15
4661.40
6983.73
Using POLFIT to find the best fitting second order polynomial gives
Ev – E0 = (1.8 cm-1) + (2343.9 cm-1) v – (5.70 cm-1) v2
and so
exe = 5.70 cm-1
e - exe = 2343.9 cm-1 , and so e = (2343.9 + 5.70) cm-1 = 2349.6 cm-1
EXTRA CREDIT
PJ/P0 = (2J+1) exp[-(J(J+1)B/kT]
To find the maximum we use the first derivative test
d/dJ { (2J+1) exp[-(J(J+1)B/kT] }
= 2 exp[-(J(J+1)B/kT] + (2J+1) exp[-(J(J+1)B/kT] (-1) (B/kT) (2J+1) = 0
And so 0 = 2 – (2J+1)2 (B/kT)
(2J+1)2 = (2kT/B)
Jmax = (kT/2B)1/2 – ½
If we solve for T, we get
T  (B/2k) (2Jmax + 1)2