CHM 3400 – Problem Set 10
Due date: Wednesday, April 18th. The final exam is Monday, April 23rd, from 5:00pm to 7:00pm. It will be
comprehensive. The new material that we have covered since the third exam is Chapter 19, sections 1, 2, 6, 7, 9, 10,
and 11; Chapter 20 sections 5, 6; Chapter 21, sections 1, 2.
Do all of the following problems. Show your work.
"It's a particle; it's a wave." - Keanu Reeves
1) For each of the following list the total number of rotations and the total number of vibrations for the molecule.
a) HCN
b) CH2Cl2
c) HCCH
d) C6H5CH3
2) The observation of microwave transitions in interstellar space is one way in which molecules are identified. It is
also a method used to map out regions of space containing significant amounts of molecular species.
In a recent publication (Astrophys. J. 743, 94 (2011)), Rangwala and coworkers observed transitions
between rotational energy levels in the 12C16O molecule in the spectrum of the Arp-220 Galaxy. One transition was
observed at  = 461.1 GHz (1 GHz = 109 Hz; 1 Hz = 1 s-1).
a) The rotational constant for 12C16O is B = 1.923 cm-1. Based on this information and the other information
in the problem, assign the observed transition (that is, give the values for J for the two states between which the
transition is occurring).
b) Predict the next location for a rotational transition in the spectrum. Give the location in units of GHz.
3) The absorption coefficient for Br2(aq) is  = 160. L/mol.cm at  = 400. nm. What is the concentration of Br2(aq)
in a sample whose measured absorbance in a 1.00 cm cuvette, at  = 400. nm, is A = 0.4856? What is the value for
% transmittance of light at this wavelength for this sample?
4) The frequency at which a spinning nucleus will absorb electromagnetic radiation is given by equn 21.13 of Atkins
= NB/2
(4.1)
where N is the nuclear magnetogyric ratio and B is the magnetic field strength. Values for N are given in Table
21.2 of Atkins.
Find the value of , in MHz (1 MHz = 106 Hz; 1 Hz = 1 s-1) for the following nuclei when they are present
in a 12.0 T magnetic field.
a) 1H
b) 13C
c) 19F
Also do the following problem from Atkins:
19.27 The hydrogen halides have the following values for the vibrational constant
hydrogen halide
(cm-1)
HF
4141.3
HCl
2988.9
HBr
2649.7
HI
2309.5
Calculate the force constant for each bond. Does the observed trend in force constants match what you would expect
for the relative strengths of H-X bonds? Explain your answer.
Solutions.
1)
The general rule for the number of fundamental rotations and vibrations of a molecule consisting of n atoms
is as follows.
Type of molecule
So
# rotations
# vibrations
linear
2
3n – 5
non-linear
3
3n – 6
a) HCN is linear, n = 3, and so 2 rotations, 4 vibrations.
b) CH2Cl2 is non-linear, n = 5, and so 3 rotations, 9 vibrations
c) HCCH is linear, n = 4, and so 2 rotations, 7 vibrations
d) C6H5CH3 is non-linear, n = 15, and so 3 rotations, 39 vibrations
2) As discussed in class and on page 454 of Atkins, the rotational energy levels for a diatomic molecule are given by
the expression
E = J(J+1)B
(where E, B are in units of cm-1)
Because the selection rule for an allowed rotational transition is that the diatomic molecule be heteronuclear and that
J =  1, the photon energies required for allowed transitions are
E = 2(J+1)B
(where E, B are in units of cm-1)
where J is the quantum number of the lower energy rotational state (Fig 19.12, Atkins).
a) The molecule absorbs at at frequency  = 461.1 GHz = 461.1 x 109 s-1.
Since
c = 
E = (1/) = (/c) = (461.1 x 109 s-1)/(2.998 x 1010cm/s) = 15.380 cm-1
E = 2(J+1)B
(J+1) = E/2B
J = (E/2B) – 1 = [(15.380 cm-1)/2(1.923 cm-1)] – 1 = 2.999  3
The observed transition is therefore J = 3  4.
b) The next higher energy transition will be J = 4  5, at an energy
E = 2(J+1)B = 2(4+1)(1.923 cm-1) = 19.230 cm-1.
 = c/ = c(1/) = (2.998 x 1010cm/s)(19.230 cm-1) = 5.765 x 1011s-1 = 576.5 GHz
3) Beer’s law is
A = c, and so
c = A/() =
0.4856
= 3.04 x 10 -3 mol/L
.
(160. L/mol cm)(1.000 cm)
A = - log10(%T/100%)
10-A = %T/100%
%T = 10-A (100%) = 10-.4856 (100%) = 32.69 %
4)
= NB/2 , so
a)  = (26.752 x 107 T-1s-1)(12.0 T) = 5.109 x 108 s-1 = 510.9 MHz
2
b)  = (6.7272 x 107 T-1s-1)(12.0 T) = 1.285 x 108 s-1 = 128.5 MHz
2
c)  = (25.177 x 107 T-1s-1)(12.0 T) = 4.808 x 108 s-1 = 480.8 MHz
2
19.27) As shown in problem 5 of problem set 9
k = 42c22
We need to find  for each molecule, where (HX) = mHmX/(mH+mX). We will use the atomic masses from the
periodic table (which will introduce a small error, as we should use the masses of the most abundent isotopes). If we
give these masses in g/mol, then to get  in units of kg
(HX) = (10-3 kg/g) mHmX
NA
(mH+mX)
molecule
(kg)
HF
1.589 x 10-27
HCl
1.627 x 10-27
HBr
1.653 x 10-27
HI
1.659 x 10-27
For HF
k = 42c22 = 42(2.998 x 1010 cm/s)2(4143.3 cm-1)2(1.589 x 10-27 kg) = 968. N/m
For HCl
k = 42c22 = 42(2.998 x 1010 cm/s)2(2988.9 cm-1)2(1.627 x 10-27 kg) = 516 N/m
For HBr
k = 42c22 = 42(2.998 x 1010 cm/s)2(2649.7 cm-1)2(1.653 x 10-27 kg) = 412. N/m
For HI
k = 42c22 = 42(2.998 x 1010 cm/s)2(2309.5 cm-1)2(1.659 x 10-27 kg) = 314. N/m
All of the H-X bonds are single bonds. As we go to larger and larger halogen atoms, the size of the halogen
atom, and therefore the length of the covalent bond, increases. If the bond length increases we would expect the
bond strength to decrease, as is observed above.