# Why is antimony pentafluoride a stronger Lewis acid than phosphorus pentafluoride or arsenic pentafluoride?

How can these relative reactivities as a Lewis acids be rationalized:

$$\ce{SbF5 > AsF5 > PF5}$$

One simple argument could be the size of the $\ce{Sb}$-center: Because it is bigger than for example a $\ce{P}$-center it can accept more ligands. I couldn't find a more precise explanation or rationalisation. My knowledge of the MO diagram of hypercoordinative compounds is really basic, so maybe this could provide - energy differences of the LUMO's of the several compounds - a neat access?

• I have not heard of a universally applicable Lewis acidity strength scale; so that statement is only partially correct. With those compounds, usually the fluoride affinity is compared. Antimony is (iirc) strongly in favour here because of the formation of oligomeric anions. Because of their size, the charge density is much lower than in other cases, i.e. the negative charge is dispersed better, making any counter cation extremely weakly coordinated and much better available for reaction. – Martin - マーチン Aug 6 '18 at 12:14
• Caution on the term hypervalent compounds. This term is deprecated and leads to more confusion, especially when looking at MO diagrams, since those usually involve d-orbitals. The involvement of those in bonding has been disproved. The preferable term is hyper-coordination, and the bonds tend to have a large ionic contribution. – Martin - マーチン Aug 6 '18 at 12:17
• @Martin-マーチン Thanks, I changed my "hypervalent" into "hypercoordinative" – Nilsfrank99 Aug 6 '18 at 12:37
• might be of interest: pubs.acs.org/doi/abs/10.1021/ed073p701 – Martin - マーチン Aug 6 '18 at 14:23