Final answer:
Phosphorus trifluoride (PF3) is generally a stronger ligand for forming bonds with central metal atoms than phosphine (PH3) due to PF3 being a weaker σ-donor but a stronger π-acceptor, potentially allowing for stronger metal-ligand bonding in transition metal complexes.
Step-by-step explanation:
In considering the ligand properties of phosphine PH3 and phosphorus trifluoride PF3, and which forms a stronger bond with a central metal atom, it is important to consider their electronic effects as σ-donors and π-acceptors. Phosphorus trifluoride (PF3) has fluorine atoms that are more electronegative compared to the hydrogen atoms in PH3, which can withdraw electron density from the phosphorus atom, making PF3 a weaker σ-donor but a stronger π-acceptor.
This is because the more electronegative fluorine atoms in PF3 hold the bonding electrons closer and create a more positive phosphorous center that can accept electron density via π-backbonding more effectively from the central metal in a transition metal complex.
Moreover, as the phosphorus in PF3 can utilize sp³d hybridization to extend its valency, it potentially become part of a more complex geometry with a metal center. Referring to the nature of ligands and their effects on metal complexes, the strong π-acid character of PF3 often results in larger crystal field splitting energies in transition metal complexes, indicating stronger metal-ligand bonding.
Thus, PF3 is generally considered a stronger ligand for forming metal complexes than PH3, particularly with transition metals.