Question

The following complex of nickel, Ni(PPhs ),Cl2 is paramagnetic, whereas the palladium analog. Pd(PPhy)2Cl2 is diamagnetic, (a) Account for the differences in magnetism for the two complexes (b) Provide sketches of all possible isomers for both compounds

Answer 1

The given complexes:

Ni(PPh₃)₂Cl₂ - Tetrahedral Complex Pd(PPh₃)₂Cl₂ - Square planar Complex

In the given complexes ligands are the same but metals are different. Ni is 3d metal but Pd is 4d metal.

Condition for 3d metal forming a square planar complex:

d⁸ configuration metal with the strong field ligand. In the Ni(PPh₃)₂Cl₂ given complex both PPh₃ & Cl^- are not a strong field ligand so Ni form a tetrahedral complex with PPh₃ & Cl^- .

Ni(PPh₃)₂Cl₂ = e_g⁴,t_2g⁴ two unpaired electron present in t_2g orbital Ni(PPh₃)₂Cl₂ is paramagnetic.

Condition for 4d forming a square planar complex:

d⁸ configuration metal with either strong field or weak field ligands. So, Pd form square planar complex with PPh₃ & Cl^- ligands.

Pd(PPh₃)₂Cl₂ = Square plannar complex with no unpaired elctron (square plannar complexes always diamagetic). So,Pd(PPh₃)₂Cl₂ is diamagnetic.

2) Both Ni(PPh₃)₂Cl₂ , Pd(PPh₃)₂Cl₂ complex exist cis & trans isomerism.
Ni (PPh3) Cla a al pphz PPh F ppha Php cis-isomer Trans-isomer pd (PPh ₂/2 cl ch Pd - PPha a PPh₂ / pphz Phap cis - isomer Trans-isomer