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Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?

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We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

  1. Planar-tetrahedral equilibria. Compounds such as $\ce{[NiBr2(PEtPh2)2]}$ mentioned above as well as a number of sec-alkylsalicylaldiminato derivatives (i.e. Me in Fig. 27.6b replaced by a sec-alkyl group) dissolve in non-coordinating solvents such as chloroform or toluene to give solutions whose spectra and magnetic properties are temperature-dependent and indicate the presence of an equilibrium mixture of diamagnetic planar and paramagnetic tetrahedral molecules.
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    $\begingroup$ Can it be justified by saying that due to steric repulsions of the bulky PPh3, to minimise repulsions, the equilibrium is on the side of tetrahedral form? $\endgroup$ – user226375 May 19 at 13:52
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Dichlorobis(triphenylphosphine)nickel(II), or $\ce{NiCl2[P(C6H5)3]2}$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both weak field ($\ce{Cl−}$) and strong field ($\ce{PPh3}$) ligands comprise $\ce{NiCl2(PPh3)2}$, hence this compound is borderline between the two geometries.

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Steric effects also affect the equilibrium; larger ligands favoring the less crowded tetrahedral geometry.[1]

Reference

  1. Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.).
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