Geometry of AsF5 molecule Which of the following is the best description of the arragement of fluorine atoms around the arsenic atom in a molecule of $\ce{AsF5}$?
  (a) trigonal bipyramid
  (b) octahedron
  (c) tetrahedron
  (d) square pyramid
  (e) planar pentagon


The correct answer is (a). 

I know that the answer can't be (b), (c), or (e) since octahedron requires 6 ligands and tetrahedron requires 4 ligand, and planar pentagon is only observed in $\ce{[IF5]-}$ and $\ce{[XeF5]-}$. I recall that first row transition metals prefer trigonal bipyramidal configurations, thus trigonal bipyramidal is the more likely geometry, but I'm not sure why this is true?

Any help would be greatly appreciated.
There are no lone pairs in $\ce{As}$ after it forms a bond with each of the five $\ce{F}$.

Hence, the number of lone pairs of electron around the central atom ($\ce{As}$) is zero, and the number of sigma bonds the central atom forms is five (a single bond has one $\sigma$-bond only).

Hence the hybridization number of $\ce{As}$ in $\ce{AsF5}$ is $0+5=5$ and it follows $\ce{sp^3d}$ hybridization geometry with no change in actual shape because of absence of lone pairs.

So, it has a trigonal bipyramidal structure: two $\ce{F}$ ions in the axial positions and three others in equitorial positions.