= 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 \b[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 $\mathrm{Ni(L_a)_2}$. N-tert-butyl crowds that arrangement and favors twisting the two chelate planes toward a distorted tetrahedron in $\mathrm{Ni(L_c)_2}$. N-Et gives a closer energetic competition, so $\mathrm{Ni(L_b)_2}$ 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>.
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