Past exam of the condensed matter physics major option course of the University of Oxford / 2019 / 4 / b

For the regular octahedron, repulsion from the six oxygen ligands raises the two $e_g$ orbitals, $d_{z^2}$ and $d_{x^2-y^2}$, above the three $t_{2g}$ orbitals, $d_{xy},d_{xz},d_{yz}$. The latter point between the metal-ligand axes and experience less direct repulsion. The crystal field splitting is therefore a lower triplet and an upper doublet.

For the $3d^2$ V${}^{3+}$ ion, Hund's rules put the two electrons into different $t_{2g}$ orbitals with parallel spins, giving $S=1$. There are three choices for the unoccupied $t_{2g}$ orbital, so the orbital ground state is triply degenerate. Its spin degeneracy is separate. This orbital degeneracy makes a symmetry-lowering Jahn-Teller effect possible.

Take the elongation axis as $z$. The axial oxygen ligands move farther away while the equatorial ones move closer, at fixed mean bond length. Orbitals containing $z$ are stabilized relative to their in-plane partners. Thus $d_{xz},d_{yz}$ move below $d_{xy}$ within $t_{2g}$, while $d_{z^2}$ moves below $d_{x^2-y^2}$ within $e_g$. The schematic level ordering appears in the figure below; the octahedral gap remains open for a small distortion.

The two parallel-spin electrons occupy $d_{xz}$ and $d_{yz}$, one each. Filling the entire lower doublet gives a unique orbital configuration, so its orbital degeneracy is one and the spin is still $S=1$. This is the orbitally quenched configuration relevant to the later spin Hamiltonian.

The energetic preference can be made explicit in a traceless linear splitting model. For an elongation coordinate $Q>0$, write the $t_{2g}$ levels relative to their barycentre as $(-aQ,-aQ,2aQ)$ with $a>0$. Their two-electron ground energy decreases by $2aQ$. For an equal-magnitude compression, the levels are $(-2aQ,aQ,aQ)$ and the two-electron energy decreases only by $aQ$. With the same quadratic elastic cost, minimizing $-2aQ+k{Q}^2/2$ gives four times the stabilization of minimizing $-aQ+k{Q}^2/2$. Thus the small-distortion Jahn-Teller effect favors elongation for this high-spin $d^2$ filling. The argument assumes the linear coupling and harmonic elastic model, rather than claiming that symmetry alone fixes all real-material distortions.

For axial compression along $z$, the ordering is reversed: $d_{xy}$ lies below the $d_{xz},d_{yz}$ doublet, and $d_{x^2-y^2}$ lies below $d_{z^2}$ within $e_g$. One electron occupies $d_{xy}$ and the other occupies either $d_{xz}$ or $d_{yz}$, with parallel spins in the weak-distortion regime. The orbital degeneracy is therefore two, with $S=1$.

The retained orbital degeneracy allows further symmetry lowering and explains why this compressed arrangement does not provide the fully quenched equilibrium of the elongated case in the specified simple model.