 Solution

Ni(II) is $d^8$. A square-planar complex can keep its strongly antibonding in-plane orbital empty and has a paired, $S=0$ configuration. A tetrahedral $d^8$ arrangement has two unpaired electrons, $S=1$.

The important change here is the substituent on the imine nitrogen: Me, Et, then tert-butyl. The phenolic ortho-tert-butyl substituent is common to all three ligands. The small N-Me groups permit coplanar N${}_2$O${}_2$ coordination in $\mathrm{Ni}({\mathrm{L}}_{\mathrm{a}}{)}_2$. N-tert-butyl crowds that arrangement and favors twisting the two chelate planes toward a distorted tetrahedron in $\mathrm{Ni}({\mathrm{L}}_{\mathrm{c}}{)}_2$. N-Et gives a closer energetic competition, so $\mathrm{Ni}({\mathrm{L}}_{\mathrm{b}}{)}_2$ accesses both structures. The observed equilibrium reflects ligand-field stabilization, pairing, steric strain and temperature-dependent entropy; it is not a change in nickel oxidation state.