Take $[\mathrm{NiCl}{}_4]{}^{2-}$ as tetrahedral and $[\mathrm{Ni}(\mathrm{CN}){}_4]{}^{2-}$ as square planar. Their spin-only magnetic moments (BM) are respectively

Both contain Ni(II), $d^8$. Tetrahedral $e^4{t}_2^4$ has 2 unpaired electrons, so $\mu =\sqrt{2\times 4}=2.83$ BM. In the strong-field square planar complex all eight electrons are paired in the lower four d orbitals, so $\mu =0$.