Number of valence electrons in Wilkinson’s catalyst RhCl[P(C6H5)3]3?

Consider (OC)5MnCl There are two ways of counting e⁻s: one is to consider it as a classical Werner type cmplx: Mn^+ + :Cl^- the other is to split the Mn:Cl bond homolytically: Mn• [Mn(0)] + •Cl. The later method is used by experts in the field but some texts use the old method which can be confusing.

They behave like covalently bonded cmpds.

So this is Rh(0) group 9, and has 9 valence e⁻s; •Cl donates one e⁻ and the three triphenylphosphine ligands (:PPh3) donate 6 (3×2) e⁻s for a total of 16. The formal oxdn state of Rh is however Rh(I). Rh can flip between "14", 16 and 18 e⁻ configurations which explains why it is the most famous "organometallic" cmpd after ferrocene (the structure of which Wilkinson first explained). I received lectures from Wilkinson as an undergrad just after his group discovered the catalyst (I told him to stick with it as it had potential!!).

confident you may because of fact in case you the two have 8 valence electrons you may no longer bond that's: toddler, we could be chemically bonded because of fact at the same time you and that i've got 8 valence electrons