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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.

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    $\begingroup$ You'd need lone pairs to be different. $\endgroup$ – Mithoron Sep 14 '15 at 18:17
  • $\begingroup$ Please avoid using Latex in titles due to searching issues $\endgroup$ – bon Sep 14 '15 at 20:19
  • $\begingroup$ There should be a bonding argument for one case that doesn’t fit with the other if I am informed correctly. But then again, $\ce{AsF5}$ pseudorotates, doesn’t it? $\endgroup$ – Jan Sep 16 '15 at 17:37
  • $\begingroup$ $\ce{AsF5}$ has a trigonal bipyramid structure with the axial bonds slightly longer than the equatorial bonds. The $\ce{As}$ atom is $\ce{sp^2}$ hybridized and hypervalent bonding is involved in the two axial bonds. Take a look at this earlier description of the hypervalent bonding in $\ce{PCL5}$ to understand what is going on here. $\endgroup$ – ron Aug 6 '17 at 23:36
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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.

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    $\begingroup$ While hybridisation is a very common (but by far too simple) concept and the question was asked in terms of VSEPR theory, it is essential to point out, that there is little to no involvement of the d orbitals in the bonding of that molecule. $\endgroup$ – Martin - マーチン Aug 2 '16 at 5:07
  • $\begingroup$ @Nij Please make sure you use mhchem for any occurrences of chemical elements, don't just leave them in plain maths mode. $\endgroup$ – Martin - マーチン Aug 2 '16 at 6:07

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