Download chm3411t03-2009soln

CHM 3411
Third Hour Exam
April 9th, 2010
There are five problems on this exam. Do all of the problems. Show your work.
1. (15 points) Cyclic C5H5 has been considered as a possible intermediate in the formation of polycyclic
aromatic hydrocarbons (PAHs). The sigma bonding structure for C5H5 is shown below, along with a
numbering scheme for the carbon atoms.
a) Write the secular determinant for pi bonding in C5H5.
b) Diagonalization of the secular determinant in part a gives the following energy levels for the pi
molecular orbitals in C5H5.
 + 2 (1 level)
 + 0.62  (2 levels)
 - 1.62  (2 levels)
Based on this information give the total pi-bonding energy for C5H5.
c) What is S, the total spin, for the ground electronic state of C 5H5?
d) At what energy will the lowest energy electronic transition in C5H5 take place? Give your
energy in terms of  and/or .
2. (20 points) Astrophysicists are interested in the pure rotational spectrum of the HS radical as a means of
detecting interstellar clouds from which new stars may form. In the mid-1990s Klisch and coworkers
measured the microwave spectrum of 1H32S and found that the J = 0  1 transition occurs at an energy E =
18.919 cm-1.
a) What is the value for the rotational constant, B, for 1H32S? Give your answer in units of cm-1.
b) What is the value for r, the bond distance, in 1H32S? Give your answer in units of nm.
c) There is a small amount of 2H32S in interstellar space. At what energy would the J = 0  1
transition in 2H32S be expected to occur? Give your answer in units of cm-1.
Isotopic masses are as follows
m(1H) = 1.0078 amu
m(2H) = 2.0140 amu
m(32S) = 31.9721 amu
3. (18 points) Consider the following molecules
1,3,5 trifluorobenzene (C6H3F3)
trichloromethane (CHCl3)
formaldehyde (H2CO)
sulfur hexafluoride (SF6)
a) Which of the above molecules (if any) will exhibit a pure rotational absorption spectrum?
b) Which of the above molecules (if any) will exhibit a pure rotational Raman spectrum?
c) For 1,3,5 trifluorobenzene only, list all of the symmetry elements present in the molecule.
d) For formasldehyde only, give the total number of normal mode vibrations.
4. (32 points) 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)
The following data are for the diatomic molecule 127I19F, and may be used in this problem
e = 610.26 cm-1
exe = 3.14 cm-1
B0 = 0.2788 cm-1
B1 = 0.2769 cm-1
m(19F) = 18.9984 amu
m(127I) = 126.904 amu
a) What is k, the force constant, for the bond in 127I19F? Give your answer in units of N/m.
b) Estimate the value for D, the dissociation energy for 127I19F. Give your answer in units of cm-1.
c) At what energy, in cm-1, will the R(5) transition be observed in 127I19F?
d) An argon ion laser operating at a wavelength  = 514.22 nm is used to study the vibrational
Raman spectrum of 127I19F. At what wavelength (in nm) will the Stokes transition (v" = 0  v' = 1) occur?
Ignore rotational structure.
5. (15 points) The ground electronic state and first excited electronic state in SrS are both 1+ states. The
following two transitions are observed in the electronic absorption spectrum of SrS (note that v" is the
vibrational quantum number for molecules in the ground electronic state, and v' is the vibrational quantum
number for molecules in the first excited electronic state).
v" = 0  v' = 0 13908. cm-1
v" = 0  v' = 1 14247. cm-1
In a separate experiment, the vibrational constant for molecules in the ground electronic state is found to be
e" = 388. cm-1.
a) Are the above transitions allowed or forbidden transitions?
b) Based on the above data, estimate the values for e' and Te' for the excited electronic state in
SrS.
Solutions
1)
a)
-E

0
0

b)

-E

0
0
0

-E

0
0
0

-E


0
0

-E
________
________
 - 1.62 
______
_______
 + 0.62 
______
 + 2.00 
Epi = 2 ( + 2.00 ) + 3 ( + 0.62 ) = 5.00  + 5.86 
c)
S = 1/2 (see above)
d) There are two possible transitions that could be lowest in energy
 + 2.00    + 0.62 
E = - 1.38 
 + 0.62    - 1.62 
E = - 2.24 
The first of these is lower in energy and so the lowest energy electronic transition occurs at E = - 1.38 .
E = E(J=1) - E(J=0) = 2B - 0 = 2B
2)
a)
So
2B = 18.919 cm-1 ;
b)
and so
B = 9.460 cm-1
B = /4cr2
r = (/4cB)1/2
(1H32S) = (1.0078 amu) (31.9721 amu)
1 kg
= 1.622 x 10-27 kg
(1.0078 + 31.9721) amu 6.022 x 10 26 amu
So
r = { (1.055 x 10-34 J.s)/[ 4 (2.998 x 1010 cm/s) (1.622 x 10-27 kg) (9.460 cm-1)] }1/2
= 1.351 x 10-10 m = 0.1351 nm
c) Since B ~ 1/
Then
B(2H32S) = (1H32S)
B(1H32S)
(2H32S)
B(2H32S) = B(1H32S) (1H32S)
(2H32S)
But
(2H32S) = (2.0140 amu) (31.9721 amu)
1 kg
= 3.146 x 10-27 kg
26
(2.0140 + 31.9721) amu
6.022 x 10 amu
And so B(2H32S) = (9.460 cm-1) (1.622 x 10-27 kg) = 4.88 cm-1
(3.146 x 10-27 kg)
So J = 0  1 occurs at 2 B(2H32S) = 2 (4.88 cm-1) = 9.76 cm-1
3)
a) CHCl3, H2CO
b) CHCl3, H2CO, CH3F3
c) E, C3, 3 C2, h, 3 v
d) # vibrations = 3N - 6 = 3(4) - 6 = 6
4)
a)
 = [1/(2c)] (k/)1/2
k = 42c2 2
So
(127I19F) = (126.904 amu) (18.9984 amu)
1 kg
(126.904 + 18.9984) amu
6.022 x 1026 amu
= 2.744 x 10-26 kg
k = 42 (2.998 x 1010 cm/s)2 (610.26 cm-1)2 (2.774 x 10-26 kg) = 363. N/m
b)
D  e2/(4 exe) = (610.26 cm-1)2/[4(3.14 cm-1)] = 29650. cm-1
c) For the R(5) transition we have v"=0,J"=5  v'=1,J'=6
E(v=1,J=6) = 3/2 (610.26 cm-1) + 9/4 (3.14 cm-1) + (6) (7) (0.2769 cm-1) = 919.95 cm-1
E(v=0,J=5) = 1/2 (610.26 cm-1) + 1/4 (3.14 cm-1) + (5) (6) (0.2788 cm-1) = 312.71 cm-1
So
E = 919.95 cm-1 - 312.71 cm-1 = 607.25 cm-1
d) It is easiest to see what is happening using an energy diagram
So
E(laser) = E(Raman scatter) + E
E(Raman scatter) = E(laser) - E
E = E(v'=1) - E(v"=0)
E(laser) =
1
514.22 nm
107 nm =
1 cm
19446.9 cm-1
E = [3/2 (610.26 cm-1) + 9/4 (3.14 cm-1)] - [1/2 (610.26 cm-1) + 1/4 (3.14 cm-1)] = 603.98 cm-1
E(Raman scatter) = 19446.9 cm-1 - 603.98 cm-1 = 18842.9 cm-1
(Raman scatter) = 1 =
1
E 18842.9 cm-1
107 nm = 530.70 nm
1 cm
5)
a) The electronic part of the transition is allowed, and since there is no vibrational selection rule
for transitions between different electronic states, both transitions are allowed
b) Again an energy diagram will help
Based on the above diagram the following relationships will be correct (in the harmonic oscillator
approximation)
E(v"=0,v'=1) - E(v"=0,v'=0) = ' (vibrational constant for upper electronic state)
E(v"=0,v'=0) = Te' + ('/2) - ("/2)
And so Te' = E(v"=0,v'=0) + ("/2) - ('/2)
So
' = 14247. cm-1 - 13908. cm-1 = 339. cm-1
Te' = 13908. cm-1 + (388. cm-1/2) - (339. cm-1/2) = 13932. cm-1