 Solution

For an octahedral complex, strong axial sigma donation raises the metal–ligand antibonding $e_g$-derived orbitals. A hydride is a very strong sigma donor: its occupied $1s$ orbital overlaps effectively along a metal–ligand axis and has no filled perpendicular p orbitals to act as a conventional pi donor. It can therefore produce a large octahedral splitting by raising $e_g$ relative to $t_{2g}$.

Triphenylphosphine also gives sigma donation, but its useful acceptor orbitals additionally receive metal pi backbonding. This lowers the metal $t_{2g}$-derived levels and enlarges their separation from $e_g$. The acceptors have substantial P–C antibonding character; invoking a dominant phosphorus $3d$ acceptor is unnecessary. Thus hydride is strong-field principally through sigma bonding, while PPh${}_3$ combines sigma donation with pi acceptance. The spectrochemical series describes the resulting splitting, not just a ranking of ligand basicity.