Question

Why pdcl4 is not stable in palladium

Answer 1

Answer:

The geometry of the complex changes going from [NiCl4]2−[NiClX4]X2− to [PdCl4]2−[PdClX4]X2−. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference.

Consider the splitting of the dd orbitals in a generic d8d8 complex. If it were to adopt a square planar geometry, the electrons will be stabilised (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired, are now paired:

D4h vs Td

(source: Wikipedia)

and on top of that, there are additional steric repulsions introduced by moving the ligands closer together. (This is, to some extent, a throwback to VSEPR theory, which simply says that the ligands would rather adopt a tetrahedral arrangement because of their mutual repulsion.)

We can label these two factors as ΔEΔE (stabilisation derived from occupation of lower-energy orbitals) and P+SP+S (pairing energy + steric repulsions) respectively. One can see that:

If ΔE>P+SΔE>P+S, then the complex will be square planar;

If ΔE<P+SΔE<P+S, then the complex will be tetrahedral.

This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter ΔOΔO, also called 10 Dq10 Dq in older literature, plays the same role as ΔEΔE does above. (However, unlike ΔOΔO, there is no simple graphical way to represent ΔEΔE on a diagram since multiple orbitals are changed in energy.)