Question

PdCl42- has a weak field ligand, Cl-, so it should has a tetrahedral structure, but the experiment show that PdCl42- has a square planar structure. EXPLAIN why PdCl42-has a square planar structure and give me other six metal that have this behaviour.

Answer 1

Consider the splitting of the d orbitals in a generic d8 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 and on top of that, there are additional steric repulsions introduced by moving the ligands closer together.

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

• If ?E>P+S then the complex will be square planar;
• If ?E<P+S, then the complex will be tetrahedral.

The two main factors that responsible for the square planar geometry of Pd(II) complex are therefore:

1. The 4d orbitals of Pd2+ are more radially diffuse (i.e. bigger) and therefore form stronger overlap with the orbitals on Cl?. This leads to larger splitting of the d orbitals - or the relevant MOs and hence a larger ?E
2. Again because the d orbitals 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 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? and [AuCl4]^_