Why can #PF_3# form so many zerovalent, homoleptic metal complexes?

1 Answer
anor277
Apr 13, 2017

#PF_3# is an excellent, so-called #pi-"acid"#..........

Explanation:

#PF_3# forms a number of formally zerovalent metal complexes, for which the analogous (say) carbonyl complex is unknown. For example, #Pd(PF_3)_4# is known; #Pd(CO)_4# is UNKNOWN. The common rationale is that #PF_3# is a better #pi-"acid"# than #CO#, and effectively accepts electron density from the metal centre.

AS another example, consider the reaction of #"bisbenzene chromium"# with #PF_3#:

#Cr(eta^6-C_6H_6)_2 + 6PF_3 rarr Cr(PF_3)_6 + 2C_6H_6#

........The #"bis-benzene chromium"# complex is fairly difficult to make (certainly more so than #"ferrocene"#); formation of the homoleptic #PF_3# complex is facile by substitution, and this arguably reflects the #pi-"acidity"# of #PF_3#. And while #NiCO_4# is well-known, #PdCO_4# (so far as I know!) is STILL unknown. #Pd(PF_3)_4# is certainly known.

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