2. Lecture 19: Magnetic properties and the Nephelauxetic effect sample south thermometer Gouy Tube electromagnet balance north connection to balance left: the Gouy balance for determining the magnetic susceptibility of materials

3. Magnetic properties Magnetic susceptibility (μ) and the spin-only formula. Materials that are diamagnetic are repelled by a magnetic field, whereas paramagnetic substances are attracted into a magnetic field, i.e. show magnetic susceptibility. The spinning of unpaired electrons in paramagnetic complexes of d-block metal ions creates a magnetic field, and these spinning electrons are in effect small magnets. The magnetic susceptibility, μ, due to the spinning of the electrons is given by the spin-only formula: μ(spin-only) = n(n + 2) Where n = number of unpaired electrons.

4. The spin-only formula applies reasonably well to metal ions from the first row of transition metals: (units = μ B,, Bohr-magnetons) Metal iond n configuration μ eff ( spin only ) μ eff ( observed ) Ca 2+, Sc 3+ d 0 00 Ti 3+ d 1 1.731.7-1.8 V 3+ d 2 2.832.8-3.1 V 2+, Cr 3+ d 3 3.873.7-3.9 Cr 2+, Mn 3+ d 4 4.904.8-4.9 Mn 2+, Fe 3+ d 5 5.925.7-6.0 Fe 2+, Co 3+ d 6 4.905.0-5.6 Co 2+ d 7 3.874.3-5.2 Ni 2+ d 8 2.832.9-3.9 Cu 2+ d 9 1.731.9-2.1 Zn 2+, Ga 3+ d 10 00 Magnetic properties

5. Example: What is the magnetic susceptibility of [CoF 6 ] 3-, assuming that the spin-only formula will apply: [CoF 6 ] 3- is high spin Co(III). (you should know this). High-spin Co(III) is d 6 with four unpaired electrons, so n = 4. We have μ eff = n(n + 2) =4.90 μ B egeg t 2g energy high spin d 6 Co(III)

6. For the first-row d-block metal ions the main contribution to magnetic susceptibility is from electron spin. However, there is also an orbital contribution from the motion of unpaired electrons from one d-orbital to another. This motion constitutes an electric current, and so creates a magnetic field (see next slide). The extent to which the orbital contribution adds to the overall magnetic moment is controlled by the spin-orbit coupling constant, λ. The overall value of μ eff is related to μ(spin-only) by: μ eff =μ(spin-only)(1 - αλ/Δ oct ) Spin and Orbital contributions to Magnetic susceptibility

7. Diagrammatic representation of spin and orbital contributions to μ eff spin contribution – electrons are orbital contribution - electrons spinning creating an electric move from one orbital to current and hence a magnetic another creating a current and field hence a magnetic field d-orbitals spinning electrons

8. μ eff =μ(spin-only)(1 - αλ/Δ oct ) In the above equation, λ is the spin-orbit coupling constant, and α is a constant that depends on the ground term: For an A ground state, α = 4. and for an E ground state, α = 2. Δ oct is the CF splitting. Some values of λ are: Ti 3+ V 3+ Cr 3+ Mn 3+ Fe 2+ Co 2+ Ni 2+ Cu 2+ λ,cm -1 155 105 90 88 -102 -177 -315 -830 Spin and Orbital contributions to Magnetic susceptibility

9. Example: Given that the value of the spin-orbit coupling constant λ, is -316 cm -1 for Ni 2+, and Δ oct is 8500 cm -1, calculate μ eff for [Ni(H 2 O) 6 ] 2+. (Note: for an A ground state α = 4, and for an E ground state α = 2). High-spin Ni 2+ = d 8 = A ground state, so α = 4. n = 2, so μ(spin only) = (2(2+2)) 0.5 = 2.83 μ B μ eff = μ(spin only )(1 - (-316 cm -1 x (4/8500 cm -1 ))) =2.83 μ B x 1.149 =3.25 μ B Spin and Orbital contributions to Magnetic susceptibility

10. The value of λ is negligible for very light atoms, but increases with increasing atomic weight, so that for heavier d-block elements, and for f-block elements, the orbital contribution is considerable. For 2 nd and 3 rd row d- block elements, λ is an order of magnitude larger than for the first-row analogues. Most 2 nd and 3 rd row d-block elements are low-spin and therefore are diamagnetic or have only one or two unpaired electrons, but even so, the value of μ eff is much lower than expected from the spin- only formula. (Note: the only high-spin complex from the 2 nd and 3 rd row d-block elements is [PdF 6 ] 4- and PdF 2 ). Spin and Orbital contributions to Magnetic susceptibility

11. In a normal paramagnetic material, the atoms containing the unpaired electrons are magnetically dilute, and so the unpaired electrons in one atom are not aligned with those in other atoms. However, in ferromagnetic materials, such as metallic iron, or iron oxides such as magnetite (Fe 3 O 4 ), where the paramagnetic iron atoms are very close together, they can create an internal magnetic field strong enough that all the centers remain aligned: Ferromagnetism: a) b) a)paramagnetic, magnetically dilute in e.g. [Fe(H 2 O) 6 ]Cl 2. b)ferromagnetic, as in metallic Fe or some Fe oxides. Fe atoms unpaired electrons separated by diamagnetic atoms unpaired electrons aligned in their own common magnetic field unpaired electrons oriented randomly

12. Antiferromagnetism: Here the spins on the unpaired electrons become aligned in opposite directions so that the μ eff approaches zero, in contrast to ferromagnetism, where μ eff becomes very large. An example of anti- ferromagnetism is found in MnO. antiferromagnetism electron spins in opposite directions in alternate metal atoms

13. The spectrochemical series indicates how Δ varies for any metal ion as the ligand sets are changed along the series I - < Br - < Cl - < F - < H 2 O < NH 3 < CN -. In the same way, the manner in which the spin-pairing energy P varies is called the nephelauxetic series. For any one metal ion P varies as: F - > H 2 O > NH 3 > Cl - > CN - > Br - > I - The term nephelauxetic means ‘cloud expanding’. The idea is that the more covalent the M-L bonding, the more the unpaired electrons of the metal are spread out over the ligand, and the lower is the energy required to spin- pair these electrons. The Nephelauxetic Effect: Note: F - has largest P values

14. The nephelauxetic series indicates that the spin-pairing energy is greatest for fluoro complexes, and least for iodo complexes. The result of this is that fluoro complexes are the ones most likely to be high-spin. For Cl -, Br -, and I - complexes, the small values of Δ are offset by the very small values of P, so that for all second and third row d- block ions, the chloro, bromo, and iodo complexes are low- spin. Thus, Pd in PdF 2 is high-spin, surrounded by six bridging fluorides, but Pd in PdCl 2 is low-spin, with a polymeric structure: The Nephelauxetic Effect: bridging chloride low-spin d 8 square-planar palladium(II)

15. Δ gets larger down groups, as in the [M(NH 3 ) 6 ] 3+ complexes: Co(III), 22,900 cm -1 ; Rh(III), 27000 cm -1 ; Ir(III), 32,000 cm -1. This means that virtually all complexes of second and third row d-block metal ions are low-spin, except, as mentioned earlier, fluoro complexes of Pd(II), such as [PdF 6 ] 4- and PdF 2. Because of the large values of Δ for Co(III), all its complexes are also low-spin, except for fluoro complexes such as [CoF 6 ] 3-. Fluoride has the combination of a very large value of P, coupled with a moderate value of Δ, that means that for any one metal ion, the fluoro complexes are the most likely to be high-spin. In contrast, for the cyano complexes, the high value of Δ and modest value of P mean that its complexes are always low-spin. The Nephelauxetic Effect:

16. Distribution of high- and low-spin complexes of the d-block metal ions: 2 nd and 3 rd row are all low-spin except for PdF 2 and [PdF 6 ] 4- 1 st row tend to be high-spin except for CN - complexes Co(III) is big exception – all low-spin except for [CoF 6 ] 3-

17. Empirical prediction of P values: Because of the regularity with which metal ions follow the nephelauxetic series, it is possible to use the equation below to predict P values: P = P o ( 1 - h.k ) where P is the spin-pairing energy of the complex, P o is the spin-pairing energy of the gas-phase ion, and h and k are parameters belonging to the ligands and metal ions respectively, as seen in the following Table:

18. Metal Ion k Ligands h Co(III)0.356 Br - 2.3 Rh(III)0.286 Cl - 2.0 Co(II)0.246 CN - 2.0 Fe(III)0.243 en1.5 Cr(III)0.216 NH 3 1.4 Ni(II)0.126 H 2 O1.0 Mn(II)0.076 F - 0.8 Empirical prediction of P values:

19. The h and k values of Jǿrgensen for two 9-ane-N 3 ligands and Co(II) are 1.5 and 0.24 respectively, and the value of P o in the gas-phase for Co 2+ is 18,300 cm -1, with Δ for [Co(9-ane-N 3 ) 2 ] 2+ being 13,300 cm -1. Would the latter complex be high-spin or low-spin? To calculate P for [Co(9-ane-N 3 ) 2 ] 2+ : P = P o (1 - (1.5 x.24)) = 18,300 x 0.64 = 11,712 cm -1 P = 11,712 cm -1 is less than Δ = 13,300 cm -1, so the complex would be low-spin. Example:

20. The value of P in the gas-phase for Co 2+ is 18,300 cm -1, while Δ for [Co(9-ane-S 3 ) 2 ] 2+ is 13,200 cm -1. Would the latter complex be high-spin or low-spin? Calculate the magnetic moment for [Co(9-ane-S 3 ) 2 ] 2+ using the spin- only formula. Would there be anything unusual about the structure of this complex in relation to the Co-S bond lengths? P = 18,300(1 – 0.24 x 1.5) = 11,712 cm -1. Δ at 13,200 cm -1 for [Co(9-ane-S 3 ) 2 ] 2+ is larger than P, so complex is low-spin. CFSE = 13,200(6 x 0.4 – 1 x 0.6) = 23,760 cm -1. Low-spin d 7 would be Jahn-Teller distorted, so would be unusual with four short and two long Co-S bonds (see next slide). μ eff = (1(1+2)) 0.5 = 1.73 μ B egeg t 2g energy

21. Structure of Jahn-Teller distorted [Co(9-ane-S 3 ) 2 ] 2+ (see previous problem) Structure of [Co(9-ane-S 3 ) 2 ] 2+ (CCD: LAFDOM) longer axial Co-S bonds of 2.43 Å Co shorter in-plane Co-S bonds of 2.25 Å S S S S S S