Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section B / 20

Place the trigonal axis along $z$, the axial ligands on $\pm z$ and the equatorial ligands in the $xy$ plane. The ordering is $e^{\prime \prime }(d_{xz},d_{yz})<e^{\prime }(d_{xy},d_{x^2-y^2})<a_1^{\prime }(d_{z^2})$. The $e^{\prime \prime }$ pair has nodal behaviour along both the equatorial ligand directions and the axial axis, minimizing direct ligand repulsion. The in-plane $e^{\prime }$ pair points into the equatorial region. The $d_{z^2}$ lobes point at both axial ligands and its torus also samples the equatorial ligands, giving the largest repulsion in the regular point-charge geometry.

Rotation by 120° and reflection in the equatorial plane preserve the geometry; they enforce the indicated doublet degeneracies. The ordering concerns a regular trigonal bipyramid with equivalent ligand-field assumptions, not every distorted five-coordinate complex.

For two independent identical ions at sufficiently high temperature, Curie's law gives $\chi T=2(N_A{\mu }_B^2/3k_B)g^{2}S(S+1)$. In the cgs units used here, $N_A{\mu }_B^2/3k_B\simeq 0.125\mathrm{c}{\mathrm{m}}^3\mathrm{K}\mathrm{mo}{\mathrm{l}}^{-1}$.

Fe(II), Co(II) and Ni(II) have $S=2$, $3/2$ and 1, respectively. The corresponding high-spin occupancies, ordered as $(e^{\prime \prime },e^{\prime },a_1^{\prime })$, areHund filling puts one electron in each orbital before additional pairing. Spin-only dimer values at $g=2$ would be 6.0, 3.75 and 2.0; the observed larger values anticipate orbital contributions. The spin assignments require the chemically appropriate high-spin states, rather than rounding the measured $\chi T$ directly to a spin-only integer.

Using the spin values above and accounting for two metal ions per formula unit,The deviations from 2 indicate spin-orbit coupling and incomplete quenching of orbital angular momentum. The partly filled $e^{\prime \prime }$ manifold in Fe and $e^{\prime }$ manifold in Ni permit orbital degeneracy in the ideal geometry. Co's high-spin orbital occupation is more orbitally balanced, but mixing with excited ligand-field states still produces a substantial effective $g$ shift. These fitted scalar values summarize a thermally averaged response; they do not determine the individual components of an anisotropic g tensor.

The radical's magnetic orbital has $\pi$ symmetry. Fe(II)'s partly occupied $e^{\prime \prime }$ set supplies a matching metal $\pi$ magnetic orbital, permitting electron transfer and predominantly antiferromagnetic metal–radical exchange. In Co(II) and Ni(II), that set is filled; their unpaired metal electrons occupy orbitals of different symmetry. Approximate orthogonality suppresses the usual antiferromagnetic kinetic-exchange pathway, favouring ferromagnetic metal–radical coupling through the remaining exchange contributions.

This orbital-symmetry argument assumes the intended idealized binding geometry. It predicts exchange signs, not exact magnitudes; distortion and additional orbital mixing can alter the competition.

Adopt $\mathcal{H}=-2J({\mathbf{S}}_1+{\mathbf{S}}_2)\cdot \mathbf{s}-2J^{\prime }{\mathbf{S}}_1\cdot {\mathbf{S}}_2$, with $S_1=S_2=1$ and ligand spin $s=1/2$. Let ${\mathbf{S}}_{12}={\mathbf{S}}_1+{\mathbf{S}}_2$ and $\mathbf{S}={\mathbf{S}}_{12}+\mathbf{s}$. The angular momentum addition identities giveThe five allowed multiplets areTheir degeneracies $2,2,4,4,6$ sum to 18, matching $3\times 3\times 2$. For dominant ferromagnetic $J$, the ground multiplet has $S_{12}=2$, $S=5/2$. A weak antiferromagnetic $J^{\prime }$ does not overcome the two strong metal–radical bonds.

A frustrated triangle has two ferromagnetic metal–radical bonds and an antiferromagnetic metal–metal bond. With the Hamiltonian convention above, take $J/k_B=+800$ K and $J^{\prime }/k_B=-8$ K. In the maximally aligned ground state the two metal–radical products are $1\times 1/2$ and the metal–metal product is $1\times 1$. ThereforeHere $F_C=-1$ denotes complete satisfaction of all exchange interactions; the small deviation $1+F_C=0.0198$ quantifies the weak frustrated contribution in this normalization. The numerical answer assumes antiferromagnetic metal–metal exchange. The printed magnitudes alone do not establish that sign. If the metal–metal exchange were instead ferromagnetic, all three bonds would be satisfied and $F_C=-1$. The ground-state scalar products also follow directly from $S_{12}=2$, $S=5/2$, so a separate classical approximation is unnecessary for this fully polarized multiplet.

For an axial spin Hamiltonian ${\mathcal{H}}_D=D{S}_z^2$ with $D/k_B=-25$ K, the minima are $m_S=\pm 5/2$. The highest levels along the reversal ladder are the $m_S=\pm 1/2$ doublet. Henceabout 1.25 kJ mol${}^{-1}$ or 12.9 meV per molecule. Using $|D|S^2$ would incorrectly put a half-integer spin at a nonexistent $m_S=0$ level. Transverse anisotropy and quantum tunnelling of magnetization can create faster relaxation pathways, so this axial ladder barrier need not equal an experimentally fitted effective barrier.

Frequency-dependent ac susceptibility probes relaxation: a peak in the out-of-phase component occurs approximately when $\omega \tau =1$. Measuring its displacement with frequency determines $\tau (T)$; an activated region may fit $\tau ={\tau }_0\exp (U_{\mathrm{eff}}/k_{B}T)$. A low-temperature departure can reveal tunnelling or other relaxation channels.

Low-temperature, sweep-rate-dependent magnetization hysteresis tests whether reversal is slow on the measurement timescale. A loop opening below the blocking temperature, variation with sweep rate and possible tunnelling steps provide complementary evidence. Isothermal remanent magnetization decay after field removal directly measures the associated relaxation time. A single open loop alone would not distinguish isolated-molecule blocking from collective ordering or impurities.