= Solution
{c}
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 $-2aQ+kQ^2/2$ gives four times the stabilization of minimizing $-aQ+kQ^2/2$. 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.
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