Geometry of AsF5 molecule

Question

Which of the following is the best description of the arrangement 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 ligands, and the 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.

Answer 1

We can predict the trigonal pyramidal geometry from electrostatic considerations. To wit, since arsenic is a metalloid its bonds with fluorine are quite polar, and we could expect the negatively charged fluorine atoms to arrange themselves in a manner that minimizes their mutual repulsion around the essentially spherical* arsenic atom. Thus a trigonal bipyramid.

*because arsenic has no directed lone pairs in the pentafluoride.

Answer 2

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 electrons 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 the absence of lone pairs.

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