The effects of nuclear spin
المؤلف:
Peter Atkins، Julio de Paula
المصدر:
ATKINS PHYSICAL CHEMISTRY
الجزء والصفحة:
ص551-552
2025-12-15
34
The effects of nuclear spin
Consider the effect on the EPR spectrum of a single H nucleus located somewhere in a radical. The proton spin is a source of magnetic field, and depending on the orientation of the nuclear spin, the field it generates adds to or subtracts from the applied field. The total local field is therefore
Bloc = B +amI mI =±
where a is the hyperfine coupling constant. Half the radicals in a sample have mI =±, so half resonate when the applied field satisfies the condition
The other half (which have mI =−) resonate when
Therefore, instead of a single line, the spectrum shows two lines of half the original intensity separated by a and centred on the field determined by g (Fig. 15.58). If the radical contains an 14N atom (I = 1), its EPR spectrum consists of three lines of equal intensity, because the 14N nucleus has three possible spin orientations, and each spin orientation is possessed by one-third of all the radicals in the sample. In general, a spin-I nucleus splits the spectrum into 2I + 1 hyperfine lines of equal intensity. When there are several magnetic nuclei present in the radical, each one contributes to the hyperfine structure. In the case of equivalent protons (for example, the two CH2 protons in the radical CH3CH2) some of the hyperfine lines are coincident. It is not hard to show that, if the radical contains N equivalent protons, then there are N + 1 hyperfine lines with a binomial intensity distribution (the intensity distribution given by Pascal’s triangle). The spectrum of the benzene radical anion in Fig. 15.55, which has seven lines with intensity ratio 1:6:15:20:15:6:1, is consistent with a radical containing six equivalent protons. More generally, if the radical contains N equivalent nuclei with spin quantum number I, then there are 2NI + 1 hyperfine lines with an intensity distribution based on a modified version of Pascal’s triangle as shown in the following Example.
Fig. 15.58 The hyperfine interaction between an electron and a spin-
nucleus results in four energy levels in place of the original two. As a result, the spectrum consists of two lines (of equal intensity) instead of one. The intensity distribution can be summarized by a simple stick diagram. The diagonal lines show the energies of the states as the applied field is increased, and resonance occurs when the separation of states matches the fixed energy of the microwave photon.
Fig. 15.59The analysis of the hyperfine structure of radicals containing one 14N nucleus (I=1) and two equivalent protons.
Fig. 15.60The analysis of the hyperfine structure of radicals containing three equivalent 14N nuclei.
The hyperfine structure of an EPR spectrum is a kind of fingerprint that helps to identify the radicals present in a sample. Moreover, because the magnitude of the splitting depends on the distribution of the unpaired electron near the magnetic nuclei present, the spectrum can be used to map the molecular orbital occupied by the unpaired electron. For example, because the hyperfine splitting in C6H6 −is 0.375 mT, and one proton is close to a C atom with one-sixth the unpaired electron spin density (because the electron is spread uniformly around the ring), the hyperfine splitting caused by a proton in the electron spin entirely confined to a single adjacent C atom should be 6 ×0.375 mT =2.25 mT. If in another aromatic radical we find a hyperfine splitting constant a, then the spin density, ρ, the probability that an unpaired electron is on the atom, can be calculated from the McConnell equation:
a=Q ρ
with Q=2.25 mT. In this equation, ρ is the spin density on a C atom and ais the hyperfine splitting observed for the H atom to which it is attached.
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