Spin state change during crystallisation for [NiCl2(PPh3)2]

Question

$\ce{NiCl2(PPh3)2}$ is a complex that is borderline between tetrahedral(high spin) and square planar(no unpaired electron, high splitting parameter) because $\ce{Cl-}$ is a weak field ligand and $\ce{PPh3}$ is a strong field ligand. Apparently, the tetrahedral and square planar states of the compound are interconvertible. The tetrahedral high spin state is blue, and produced directly by reacting hydrated nickel chloride and triphenylphosphine in alcohol. But when the complex is crystallised out from a cholrinated solvent like dicholoromethane, it converts to the red square planar complex.

Is there any theoretical explanation as to

1) Why the square planar form is favoured in the crystalline state?
2) What role does the chlorinated solvent play in isomerising the complex to its sq. planar form?

And are there any other examples like this?

Answer 1

I made both isomers as a PhD student many years ago. The red diamagnetic form crystallises out when a concentrated CH2Cl2 solution is cooled in a dry ice-acetone bath, and you can quickly filter off the red product. But on leaving the solid at room temperature, it reverts in a couple of days at room temperature to the blue, paramagnetic form. So it's a thermal equilibrium.