Physical Chemistry Third Edition

1012 24 Magnetic Resonance Spectroscopy

a particular nucleus the coupling constant for that nucleus and that orbital will be

negligibly small. Sincesorbitals are the only hydrogen-like orbitals that are nonzero

at the nucleus, the coupling constant at a nucleus is sometimes said to be a measure of

the “scharacter” at that nucleus of the orbital containing the unpaired electron.

An ESR spectrometer uses microwave radiation with wavelengths roughly equal

to 1 cm and a magnetic field near 1 T. This magnetic field can easily be attained with

a permanent magnet. The microwaves are conducted by wave guides to the sample

chamber, which is a cavity with conducting walls in which standing electromagnetic

waves can occur. Absorption by the sample is detected by its effect on these standing

waves. Since a particular cavity can support standing waves of only a few frequen-

cies, the frequency of the radiation is kept fixed and the applied magnetic field is

varied. Absorption will occur when the magnetic field affecting the unpaired elec-

tron equalsBres, the value such that Eq. (24.3-3) is satisfied for the frequency being

used:

B 0 +BinternalBresB 0 +

∑n

j 1

ajMIj

hν

gβe

(24.3-5)

A single substance can produce several values ofB 0 at which resonance occurs,

because different nuclear spin states will be found in different molecules in the sample

and because the coupling constants at different nuclei can be different from each other.

In the hydrogen molecule ion, H+ 2 , the electron couples equally with the two protons.

The molecule could be in a state with both proton spins up, in either of two states with

one proton spin up and one down, or in a state with both proton spins down. Since the

sum of theMIvalues can equal 1, 0, or−1, we obtain a spectrum with three lines,

where each line is produced by a different set of molecules. The states are nearly equally

populated and the middle line is twice as intense as the other two, because there are

two states with one spin up and one spin down.

EXAMPLE24.10

Assume that the benzene negative ion, C 6 H− 6 , has its unpaired electron in a delocalized

orbital with equal magnitude at each of the protons. Predict the ESR spectrum.

Solution

We assume that all of the coupling constants are equal. There will be one line for each value

of the sum of theMIvalues. This sum can equal 3, 2, 1, 0,−1,−2, or−3, so that seven

lines occur in the spectrum, with a splitting between two adjacent lines equal to the coupling

constant. For a field of 0.500 T, the difference in energy between two nuclear spin states with

values of the sum of theMI’s differing by unity is given by Eq. (24.2-19):

∆EmaggNβNBz(5.5857)(5. 0508 × 10 −^27 JT−^1 )(0.500 T)

 1. 411 × 10 −^26 J

At 300.0 K the ratio of the populations of two such states is

e−∆Emag/kBTexp

(

−

1. 411 × 10 −^26 J

(1. 3807 × 10 −^23 JK−^1 )(300.0K)

)

 0. 99999659