Q.How does the splitting of the $d$-orbitals in a tetrahedral crystal field differ from that in an octahedral field, both in the order of the two orbital sets and in the magnitude of the splitting energy $\Delta_t$ compared with $\Delta_o$? Why are tetrahedral complexes almost always high-spin?

[!TLDR] Tetrahedral splitting is reversed ($e$ lower, $t_2$ higher) and much smaller ($\Delta_t\approx\tfrac{4}{9}\Delta_o$) — too small for pairing to ever be worthwhile. [!ANSWER] $e$ (lower, $-0.6\Delta_t$) and $t_2$ (higher, $+0.4\Delta_t$) …

In a tetrahedral complex, four ligands approach between the Cartesian axes rather than along them, which reverses the ordering seen in the octahedral case: the $e$ set ($d_{z^2}, d_{x^2-y^2}$, pointing along the axes) now points away from the ligands and is lowered in energy, by $0.6\Delta_t$, while the $t_2$ set ($d_{xy}, d_{yz}, d_{xz}$, pointing between the axes) points more nearly toward the ligands and is raised, by $0.4\Delta_t$ — exactly opposite to the octahedral $t_{2g}$/$e_g$ ordering. In addition, the overall magnitude $\Delta_t$ is intrinsically much smaller than $\Delta_o$ for the same metal and ligands, roughly $\Delta_t \approx \tfrac{4}{9}\Delta_o$, because there are fewer ligands (four instead of six) and none of them lies directly along a d-orbital's axis of maximum repulsion. Because $\Delta_t$ is essentially always smaller than the pairing energy $P$, regardless of how strong-field the ligand is, it is never energetically worthwhile for a tetrahedral complex's electrons to pair up early to stay …

Compare which orbital set points toward versus away from the four tetrahedrally-arranged ligands (reversed from the octahedral case) and note the smaller geometric/ligand …

Assuming a strong-field ligand could still produce a low-spin tetrahedral complex, by analogy with the octahedral case — in practice $\Delta_t$ never gets large enough (even with $\text{CN}^-$ or $\text …