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9.1 Calculate the separation, in magnetic field units of G, of two EPR absorptions with g-values of 2.0023 and 2.0071 using (i) an X-band instrument operating at 9.2500 GHz and (ii) a Q-band instrument operating at 33.600 GHz. Which instrument provides the greater resolution?
Eqn 9.3 is required and needs to be rearranged
g |
X-band (9.2500 GHz) |
Q-band (33.600 GHz) |
2.0023 |
3300.6 G |
11989.1 G |
2.0071 |
3292.7 G |
11960.5 G |
difference |
7.9 G |
28.6 G |
Therefore, experiments at Q-band frequencies provide the greater separation, and hence resolution between the spectral features.
9.2 Predict the g-value for ·ArF.
λ is negative for Ar, just as it is for Kr and Xe, as the Ar 3p orbital is more than half-filled. With S = ½, λ = −ζ/2S, so using the data in Table 9.1 the value of λ for Ar is −940 cm−1. A simple ratio of the ζ values for Ar (−940 cm−1) and Kr (−3480 cm−1) multiplied by the g-shift for ·KrF (Δg = 0.0427) gives Δg = 0.0115 and a g-value of 2.0138 for ·ArF. As expected the g-value increases from ·ArF (2.0138) through ·KrF (2.0450) to ·XeF (2.0769).
9.3 Interpret the X-band EPR spectrum shown in Figure 9.9 obtained from a compound containing an element with a 9.5% abundant I = 3/2 isotope, assume all other isotopes have I = 0.
The intense central feature at 3300 G in Figure 9.3ST is due to the I = 0 nuclei. The four weaker satellite features at 3150, 3250, 3350 and 3450 G are due to hyperfine coupling, A, of 100 G between the unpaired electron and the nucleus with I = 3/2 using the 2nI + 1 rule. The 9.5% abundance of this isotope is spread out amongst the four features, so each one is fairly weak to give an intensity ratio of 2.4:2.4:90.5:2.4:2.4. This is the pattern expected for chromium complexes as 53Cr has I = 3/2 with 9.5% abundance and all the other isotopes (50Cr, 52Cr and 54Cr) have I = 0. This behaviour is analogous to the coupling observed in NMR spectra for dilute spin nuclei.
Figure 9.3 ST Schematic X-band EPR spectrum of a compound containing one 53Cr nucleus.
9.4 Figure 9.11 shows isotropic X-band (9.25 GHz) EPR spectrum of ●PF2 radicals produced by g-irradiation of PF3 in solid C2F6. Measure, and explain the hyperfine coupling patterns. How would the spectra of ●NF2 and ●AsF2 be different?
(31P, I = 1/2, 100% abundant; 19F, I = 1/2, 100% abundant, 14N, I = 1, 99.6%; 75As, I = 3/2, 100%))
The g-values and hyperfine coupling constants for ●PF2 are shown in Figure 9.4ST(a). In ●PF2 the spin density could be on either or both of 31P and 19F both of which have I = 1/2. The spectrum in Figure 9.4ST(a) is a doublet of triplets, with the doublet due to coupling to one 31P nucleus with Aiso = 85 G, and the triplet from coupling to two equivalent 19F nuclei with Aiso = 33 G. Therefore, there is spin density on both the 31P and 19F. The Aiso(31P) of 85 G is about two and a half times that of Aiso(19F) of 33 G, and this simple analysis indicates that there is much higher spin density on the 31P than the 19F nuclei. However, the atomic hyperfine interaction data available in the Online Resource Centre indicate that the isotropic hyperfine constant (Ans) for unit population for 19F (17640 x 10−4 cm−1) is about four times that of 31P (4438 x 10−4 cm−1), and this more detailed analysis implies that the s orbital spin density on the phosphorus is about ten times that on the fluorine.
The assignment of the spectral features in ●PF2 was not without controversy (see references).
Figure 9.4ST(a) Isotropic X-band (9.25 GHz) EPR spectra of ●PF2. (Data from W. Nelson, G. Jackel and W. Gordy, J. Chem. Phys. 52 4572 (1970), A. J. Colussi, J. R. Morton, K. F. Preston and R. W. Fessenden, J. Chem. Phys. 61 1247 (1974))
The spectra of ●NF2 and ●AsF2 will be different because of the different I values of N and As. For ●NF2 the hyperfine coupling to the two 19F nuclei will result in a 1:2:1 triplet, but coupling to the single I = 1 14N nucleus will result in a 1:1:1 triplet, so the spectrum will consist of a triplet of triplets. The literature value for Aiso(14N) is 17 G and that of Aiso(19F) is 60 G, and the simulated isotropic spectrum with these parameters is shown in Figure 9.4ST(b) (C. A. McDowell, H. Nakajima and P. Raghunathan, Can. J. Chem. 48 805 (1970))
Figure 9.4ST(b) Isotropic X-band (9.25 GHz) EPR spectra of ●NF2. (Data from C. A. McDowell, H. Nakajima and P. Raghunathan, Can. J. Chem. 48 805 (1970))
Aiso(14N) of 17 G is now smaller than that of Aiso(19F) of 60 G, and would indicate that the s orbital spin density on the N is less than that of the phosphorus in ●PF2. However, the isotropic atomic hyperfine constant for unit population of an N 2s orbital (Ans), (see Online Resource Centre for table of data) is ca. 650 G, whereas Ans for the P 3s orbital is ca. 4750 G, and using these values the spin density of 2.6% on the nitrogen in ●NF2 is slightly larger than that of 1.8% for phosphorus in ●PF2.
For ●AsF2, hyperfine coupling to the two fluorine nuclei will again give rise to a 1:2:1 triplet, but coupling to the single I = 3/2 75As nucleus will give rise to a 1:1:1:1 quartet, so the spectrum will consist of either a 1:1:1:1 quartet of 1:2:1 triplets, or a 1:2:1 triplet of 1:1:1:1 quartets as shown in Figure 9.4ST(c).
As the isotropic atomic hyperfine constant (Ans) of 75As is similar to that of 31P a quartet of triplets is expected, as shown in the lower spectrum.