why NiCl4 is tetrahedral but PtCl4 is square planar? co-ordination chemistry

The two main factors that differentiate the Ni(II) complex from the Pd(II) complex are therefore:

1. The 4d    orbitals of Pd2+$\mathrm{P}\mathrm{d}{\phantom{X}}^{2+}$ are more radially diffuse (i.e. bigger) and therefore form stronger overlap with the orbitals on Cl−$\mathrm{C}\mathrm{l}{\phantom{X}}^{-}$, as compared to the 3dorbitals of Ni2+$\mathrm{N}\mathrm{i}{\phantom{X}}^{2+}$. This leads to larger splitting of the d$\mathrm{d}$ orbitals - or the relevant MOs ^{_{(I'm too lazy to check their symmetry labels)}} - and hence a larger ΔE$\mathrm{\Delta }E$.
2. Again because the dorbitals are more diffuse, the pairing energy P is smaller in the palladium complex (it basically costs less energy to stuff them into the same orbital).

Together, these two factors ensure that practically all 4d and 5d d8 ML4$\mathrm{M}\mathrm{L}{\phantom{X}}_4$ complexes adopt a square planar geometry, even if the ligand is not a strong-field ligand. Other examples of such square planar complexes are [PtCl4]2−$[\mathrm{P}\mathrm{t}\mathrm{C}\mathrm{l}{\phantom{X}}_4]{\phantom{X}}^{2-}$ and [AuCl4]−$[\mathrm{A}\mathrm{u}\mathrm{C}\mathrm{l}{\phantom{X}}_4]{\phantom{X}}^{-}$.

In fact,  same factors also cause the octahedral complexes to be almost invariably low-spin. For example, the d5 complex [Fe(ox)3]3−$[\mathrm{F}\mathrm{e}(\mathrm{o}\mathrm{x}){\phantom{X}}_3]{\phantom{X}}^{3-}$ has five unpaired electrons, indicating a high-spin configuration, (t2g)3(eg)2$({\mathrm{t}}_{2\mathrm{g}}{)}^3({\mathrm{e}}_{\mathrm{g}}{)}^2$. However, [Ru(ox)3]3−$[\mathrm{R}\mathrm{u}(\mathrm{o}\mathrm{x}){\phantom{X}}_3]{\phantom{X}}^{3-}$ has one unpaired electron, which implies a low-spin configuration (t2g)5$({\mathrm{t}}_{2\mathrm{g}}{)}^5$.