CHM 3411 – Problem Set 10
Due date: Friday, April 22nd (noon, my office). Note - The final exam is Friday, April 29th, at 9:45am, in GC 280.
It will be comprehensive.
Do all of the following problems. Show your work.
1) Predict the esr (electron spin resonance) spectra for the following systems. In each case include the effects of
coupling of electron spin to nuclear spin (There is a table of nuclear spins in the appendix of Atkins. Assume the
most common isotope for each element unless otherwise stated.). Give the peak locations (relative to the location of
the peak in the absence of spin-spin coupling) and relative peak intensities. Unless otherwise indicated, give the
spin-spin coupling in units of a, the coupling constant.
a) A CH4+ ion.
b) A C6H6+ ion.
c) An NO molecule.
2) 33S has a nuclear spin I = 3/2 and a nuclear g factor gI = 0.4289.
a) Find the energy levels for the various MI states of 33S in the presence of a magnetic field of magnitude B
= 16.0 T. Give the energies relative to the energy of the spinning nuclei in the absence of an external magnetic field,
in units of cm-1.
b) At what frequency, in MHz, will transitions occur between states of the 33S nuclei? Recall that the
selection rule for allowed transitions is MI =  1.
3) The nmr spectrometer in the instrumentation lab at FIU is a 600. MHz machine. This means that resonance
occurs for protons (1H) at  = 600. MHz = 6.00 x 108 s-1.
a) What is the value for the magnetic field strength, B 0, used in the above machine? Give your answer in
units of T (Tesla).
b) At what frequency (in MHz) will the following nuclei be in respnance for the same magnetic field
strength found in a? Do your calculation for 13C, 19F, and 31P.
4) For a 19F nucleus in a magnetic field of strength B = 18.6 T find the fraction of nuclei in the lower energy spin
state at T = 300.0 K, and at T = 195.0 K. Give your answers to 6 significant figures (I ask this because to a good
first approximation your fraction will be f = 0.50.) In each case, also calculate the number of "excess spins" in the
lower energy state per 106 nuclei.
Solutions.
1)
a) There are four equivalent hydrogen nuclei that the electron can couple to. Based on Pascal’s triangle this
should give a set of five peaks, with relative intensities 1 : 4 : 6 : 4 : 1.
b) There are six equivalent hydrogen nuclei that the electron can couple to. Based on Pascal’s triangle this
should give a set of seven peaks, with relative intensities 1 : 6 : 15 : 20 : 15 : 6 : 1.
c) The only nucleus for the electron to couple to is the 14N nucleus, which is I = 1 (and so M I = 1, 0, -1). So there
should be three peaks of equal intensity.
2)
a) E = - gINB0mI = - (0.4289) (5.051 x 10-27 J/T) (16.0 T) mI = - (1.745 x 10-3 cm-1) mI
hc
(6.626 x 10 -34 J.s) (2.998 x 1010 cm/s)
mI = - 3/2
So
E = 0.002617 cm-1
E = 0.000872 cm-1
mI = - ½
mI = + ½
E = - 0.000872 cm-1
mI = + 3/2
E = - 0.002617 cm-1
b) E = h = gINB0 , and so
 = gINB0 = (0.4289) (5.051 x 10-27 J/T) (16.0 T) = 5.23 x 107 s-1 = 52.3 MHz
h
(6.626 x 10 -34 J.s)
3)
a) Using an equation derived in problem 2, we may say
h = gINB0
and so
B0 =
h = (6.626 x 10-34 J.s) (6.00 x 108 s-1) = 14.09 T
gIN
(5.5857) (5.051 x 10-27 J/T)
b) Since at a particular value of magnetic field strength  ~ gI, and since resonance occurs for protons at
600. MHz, we may say that for other nuclei the resonance frequency is
res(X) = [gI(X)/gI(1H)] (600. MHz) , where X is the nucleus of interest.
Nucleus
gI
res (MHz)
1
5.5857
1.4046
600.0
150.9
5.2567
2.2634
564.7
243.1
H
C
13
19
31
F
P
Notice that 19F has a resonance frequency close to that for 1H. That is why in conventional nmr spectrometers it was
often possible to do both 19F and 1H nmr measurements on the same machine, as discussed in class.
4)
P-1/2 = exp(-E/kT) , where E (in cm-1) = gINB0 , k = 0.69503 cm-1/K
P1/2
hc
P-1/2 = P1/2 exp(-E/kT)
f(mI = ½) =
Since
E =
P1/2
P1/2 + P-1/2
=
1
1 + exp(-E/kT)
(5.2567) (5.051 x 10-27 J/T) (18.6 T) = 0.02486 cm-1
(6.626 x 10-34 J.s) (2.998 x 1010 cm/s)
f(mI = ½) =
1
1 + exp(- 0.03577/T)
, where T is the temperature in K.
The number of “excess spins” per 106 nuclei will be
“excess spins” = 2 [ 106 f(mI = ½) - 500000]
T(K)
f(mI = ½)
300.
195.
0.500030
0.500046
excess spins
59.6
91.7
Notice that cooling the sample to T = 195. K (dry ice temperature) increases the number of excess spins by
~ 50%, and so increases the signal intensity by the same amount.