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

Apr 13, 2017

$P {F}_{3}$ is an excellent, so-called $\pi - \text{acid}$..........

#### Explanation:

$P {F}_{3}$ forms a number of formally zerovalent metal complexes, for which the analogous (say) carbonyl complex is unknown. For example, $P d {\left(P {F}_{3}\right)}_{4}$ is known; $P d {\left(C O\right)}_{4}$ is UNKNOWN. The common rationale is that $P {F}_{3}$ is a better $\pi - \text{acid}$ than $C O$, and effectively accepts electron density from the metal centre.

AS another example, consider the reaction of $\text{bisbenzene chromium}$ with $P {F}_{3}$:

$C r {\left({\eta}^{6} - {C}_{6} {H}_{6}\right)}_{2} + 6 P {F}_{3} \rightarrow C r {\left(P {F}_{3}\right)}_{6} + 2 {C}_{6} {H}_{6}$

........The $\text{bis-benzene chromium}$ complex is fairly difficult to make (certainly more so than $\text{ferrocene}$); formation of the homoleptic $P {F}_{3}$ complex is facile by substitution, and this arguably reflects the $\pi - \text{acidity}$ of $P {F}_{3}$. And while $N i C {O}_{4}$ is well-known, $P \mathrm{dC} {O}_{4}$ (so far as I know!) is STILL unknown. $P d {\left(P {F}_{3}\right)}_{4}$ is certainly known.