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9.1. Predict the sign of Δg for Mo5+.
Mo5+ has a d1 configuration and as this is less than half full λ is positive (ca. 900 cm−1), so g < ge, and Δg will be negative. Values of 1.95 for g for Mo(V) are common, but they can extend to 1.75.
9.2. Sketch the first derivative spectrum of a radical containing one I = 1/2 nuclei with a hyperfine coupling constant A of 100 G, and one I = 3/2 nuclei with a hyperfine coupling constant of 20 G. Assume that g = 2.0023 and the operating frequency is 9.25 GHz.
Sketch the spectrum with the hyperfine coupling constants reversed.
Using Eqn 9.3, the magnetic field for g = 2.0023 at 9.25 GHz is 3300.6 G, so this will be the centre position of the multiplet. Coupling to one I = 1/2 nucleus will result in a doublet of equal intensity, and coupling to one I = 3/2 nucleus will cause a quartet (2nI + 1) with four lines of equal intensity. In the first case where the hyperfine coupling constant, A, from the I = 1/2 nucleus is greater than that of the I = 3/2 nucleus a doublet of quartets (of equal intensity) is observed. When the situation is reversed, a quartet of doublets is formed. The spectra are shown in Figure 9.2P.
Figure 9.2P. Isotropic X-band (9.25 GHz) EPR spectra for a radical with g = 2.0023 and A(I = 1/2) = 100 G and A(I = 3/2) = 20 G (top) and A(I = 3/2) = 100 G and A(I = 1/2) = 20 G (bottom)
9.3. For giso = 2.00, and Aiso = 200 G, calculate the hyperfine coupling constant in terms of wavenumber (cm−1) and MHz. (μB = 9.27401 x 10−24 J T−1 = 4.66858 x 10−5 cm−1 G−1).
For gz = 1.93 and Az = 170 G, and gx,y = 1.98 and Ax,y = 60 G, calculate the A⊥ and A|| hyperfine coupling constants in cm−1 and MHz.
Eqn 9.7 and 9.8 are required to convert the hyperfine coupling constants in G, to the field independent cm−1 and MHz values.
Eqn 9.7
Eqn 9.8
giso = 2.00, and Aiso = 200 G, 0.0187 cm−1 (normally expressed as 187 x 10−4 cm−1), 560 MHz.
gz = 1.93 and Az = 170 G, A|| = 0.0153 cm−1, 153 x 10−4 cm−1, 459 MHz
gx,y = 1.98 and Ax,y = 60 G, A⊥ = 0.00555 cm−1, 55.5 x 10−4 cm−1, 166 MHz
(A|| is equivalent to Az and A⊥ is equivalent to Ax,y.)
9.4. Sketch the isotropic X-band (9.2500 GHz) spectrum of ●GaH3− (g= 2.0055; A(69Ga)iso = 420.2 G, 60.4%; A(71Ga)iso = 534.1 G, 39.6%; and A(H)iso = 10 G.)
Ga has two isotopes, 69Ga, 60.4% abundant and 71Ga which is 39.6% abundant, both with I = 3/2. Hyperfine coupling to these will each give four lines of equal intensity. The hyperfine coupling constants of 420.2 G and 534.1 G reflect the different gyromagnetic ratios of the two nuclei. As the10 G Aiso(H) is much smaller than these, each of the four lines will be split into a binomial 1:3:3:1 quartet.
A g-value of 2.0055 at an operating frequency of 9.2500 GHz means that the multiplet is centred at 3295.3 G.
A simulated spectrum using the experimental values is shown in Figure 9.4P. The experimental spectrum (J. C. Brand and B. P. Roberts, Chem Comm 109 (1984)) has differential line broadening due to tumbling effects, and the quartets due to the hydrogen hyperfine coupling were poorly resolved.
Figure 9.4P Isotropic X-band (9.250 GHz) EPR spectrum of ●GaH3−. (Data adapted from J. C. Brand and B. P. Roberts, Chem Comm 109 (1984).
The atomic hyperfine interaction for 100% population of the Ga 4s orbital (Ans) for 69Ga is given as 4073 x 10−4 cm−1 in the table in the additional resources. This is equivalent to 4350 G at g = 2.0055. The experimental value of 420.2 G implies a 4s spin density of ca 10% in ●GaH3−. For the analogous ●AlH3− radical discussed in Example 9.4, the Ans value for Al is 1305 x 10−4 cm−1, and this corresponds to 1396 G at g= 2.0025. The experimental A(Al)iso value of 154.2 G indicates a 3s spin density of ca. 11%, which is very similar to that for ●GaH3−. In contrast for ●BH3−, with an Ans value of 909 G, the experimental value of A(B)iso of 19.9 G indicates a 2s spin density of ca. 2%. The higher s orbital spin density in the ●AlH3− and ●GaH3− radicals indicates that there is much greater s-p mixing in the ●AlH3−and ●GaH3− radicals than the ●BH3− radical. In planar units the s and p orbitals have different symmetry, but in pyramidal geometries the s and pz orbitals both have a1 symmetry and can therefore mix. Therefore, the greater spin density in ●AlH3− and ●GaH3− radicals is associated with pyramidal rather than planar geometries as for the ●BH3− radical.
9.5. The K-band (24.0619 MHz) isotropic EPR spectrum of the irradiation products of BF3 are shown in Figure 9.17. Due to the complexity of the spectrum from natural abundance boron, samples enriched in 10B and 11B (assume complete enrichment) were also prepared. Use these data to identify the radical formed. Natural boron: 10B, 20%, I = 3; 11B, 80%, I = 3/2.
The magnetic nuclei involved are: 10B, I = 3, ca. 20%; 11B, I = 3/2, ca. 80%; 19F, I = 1/2, 100%. Therefore, if the unpaired electron is coupled to one boron nucleus a seven line pattern of equal intensity is expected for 10B, and a four line pattern of equal intensity for 11B. In the spectrum of the 10B enriched sample there are seven lines of higher intensity than the remainder of the features. In the spectrum of the 11B enriched samples there are four lines of higher intensity. These observations indicate that the radical contains one boron atom. Hyperfine coupling to one 19F nucleus will result in doublets, coupling to two 19F nuclei will give binomial (1:2:1) triplets and coupling to three 19F will generate binomial (1:3:3:1) quartets. In the spectrum of 11B enriched sample there are four intense lines and eight of about half this intensity, and it is relatively easy to see that each of the four more intense features is the central component of a 1:2:1 triplet. The spectrum of the 10B enriched sample is more congested, but in addition to the seven more intense features, there are fourteen weaker ones of about half their intensity. As in the spectrum of the 11B sample, each of the more intense features is the central component of a 1:2:1 triplet. To identify which features belong together, the easiest way is to measure the distance between the first of the weaker features, and the first of the more intense ones, and use this separation to identify the third component, and also the members of the other triplets. The dotted lines in both the 10B and 11B enriched samples are the same length as Aiso(19F) is the same in both cases.
Figure 9.5P Isotropic K band (24.0619 GHz) spectra of irradiation products of BF3.
9.6. The isotropic X-band spectrum of bis(O,O'-diethyldithiophosphato)oxovanadium(IV) ([VO(dtp)2] Figure 9.18) in ether solutions shows a 24 line pattern at room temperature, but this changes to a 16 line pattern on cooling. Account for these observations. (Data from M. Sato, Y. Fujita and T. Kwan, Bull. Chem. Soc. Jpn. 46 3007 (1973), M. Sato, T. Katsu, Y. Fujita and T. Kwan, Chem. Pharm. Bull. 22 1393 (1974).)
The significant magnetic nuclei in [VO(dtp)2] shown in Figure 9.6P are 51V (I = 7/2, effectively 100% abundant) and 31P (I = 1/2, 100% abundant).
Figure 9.6P Structure of [VO(dtp)2].
Hyperfine coupling to one 51V will result in an eight line pattern of equal intensity. Superhyperfine coupling to one 31P nucleus will result in a doublet, and coupling to two 31P nuclei will give rise to a triplet. Therefore, if there is spin density only on the vanadium an eight line pattern will be observed, if there is spin density on the vanadium and one phosphorus, each of these lines will be split into a doublet, giving rise to a sixteen line pattern. If there is coupling to the vanadium and both phosphorus nuclei then the each of the original eight lines due to the vanadium hyperfine coupling will be split into triplets, giving a total of twenty four lines.
Therefore, at room temperature the hyperfine coupling of twenty four lines implies spin density on the central V as well as both of the 31P atoms. As the temperature is reduced, the sixteen line pattern indicateds that only one of the P atoms is contributing to the hyperfine pattern. This has been interpreted as one of the V-S bonds breaking, and being replaced by an ether, which is analogous to the situation observed for the stronger N donor bases such as pyridine.
(Data from M. Sato, Y. Fujita and T. Kwan, Bull. Chem. Soc. Jpn. 46 3007 (1973), M. Sato, T. Katsu, Y. Fujita and T. Kwan, Chem. Pharm. Bull. 22 1393 (1974).)
9.7. The isotropic X-band EPR spectrum of [Co(CO)2(Ph2C2)(P(OMe)3)] (prepared from the reduction of a dicobalt complex) consists of a sixteen line pattern of approximately equal intensity with g = 2.061, and two hyperfine coupling constants of 45.4 x 10−4 cm−1 and 166.3 x 10−4 cm−1. Explain these observations. (Data from L. V. Casagrande, T. Chen, P. H. Rieger, B. H. Robinson, J. Simpson and S. J. Visco, Inorg. Chem. 23 2019 (1984))
The magnetic nuclei are 59Co, I = 7/2, 100% and 31P, I = 1/2. Therefore, hyperfine coupling to one cobalt nucleus will result in an eight line pattern of equal intensity, and superhyperfine coupling to the phosphorus will result in all of these doubling up to give 16 lines of approximately equal intensity. While it might be thought that the larger of the two hyperfine couplings of 166.3 x 10−4 cm−1 would be due to the 59Co nucleus, this is in fact due to 31P, and the smaller value of 45.4 x 10−4 cm−1 is due to the 59Co, as the observed spectrum is a doublet of equal intensity octets.
9.8. Reaction of a large excess of NaCN with [VCl2Cp2] resulted in the formation of a green compound. Elemental analysis gave the following: %C 61.75, %H 4.40, N 12.0%. The IR spectrum contained intense bands at 2116 and 2111 cm−1 with corresponding bands in the Raman spectrum at 2116 and 2110 cm−1. The EPR spectrum in a water solution contained 8 lines of approximately equal intensity, with giso = 1.995 and Aiso = 60.8 G. When K13CN was employed for the synthesis, each of the lines in the EPR spectrum was split into a 1:2:1 triplet with Aiso of 12.8 G.
Use these data to identify the complex formed
The elemental analysis of the green compound gives an empirical formula of C6H5N1. However, the analysis only adds up to 78.1%. As the material is green it is not unreasonable to assume that the missing 21.9% is vanadium, which would give an empirical formula of C12H10N2V. Therefore, the green compound is likely to be [V(CN)2Cp2]. The two bands in the IR and Raman spectra at 2116 and 2111/2110 cm−1 are typical for νCN modes and show that the NC-V-CN units cannot be linear, confirming a tetrahedral geometry. (It does not rule out cis-square planar, but this is not an expected geometry for vanadium. Small differences between IR and Raman peaks for the same vibrational mode are not uncommon). The vanadium oxidation state in [V(CN)2Cp2] is V(IV), d1. 51V is I = 7/2 99.8%, therefore expect 2nI + 1 = 8 lines of equal intensity if the unpaired electron has spin density on the vanadium. When 13CN was introduced into the molecule, each of the eight lines split into a 1:2:1 triplet as 13C has I = 1/2, due to superhyperfine interaction thus indicating that the unpaired electron is delocalised onto the two cyanide ligands as well as the central vanadium indicating covalency in the vanadium-cyanide bonding.
(Data from : J. Honzíček, J. Vinklárek, Z. Černošek and I. Císařová, Magn. Reson. Chem. 45 508 (2007))