Why is the shape of $\ce{PCl5}$ trigonal bipyramidal? All the $\ce{Cl}$ atoms must be at equal distances forming a star like shape. However, in the actual shape, the distance between the $\ce{Cl}$ atoms in the equatorial position is less than the ones in the axial position.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Have a look at this question, it might already contain an answer. $\endgroup$ – Martin - マーチン Jun 21 '16 at 14:26
  • $\begingroup$ see the last para of the answer to this question. It basically covers up your question. $\endgroup$ – Nilay Ghosh Jun 21 '16 at 18:25
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    $\begingroup$ I can’t believe that this is not a dupe! =O $\endgroup$ – Jan Jun 21 '16 at 20:19

The short answer is stated on the Wikipedia page for trigonal bipyramidal:

… there is no geometrical arrangement with five terminal atoms in equivalent positions.

This website from the School of Mathematics and Statistics at UNSW explains why distributing points on a sphere is tough and that perfect distribution is possible only in certain cases $(n=1,2,4,6,8,12,\&\ 20)$. These cases have been known for a long time — they correspond to the platonic solids (plus a point and a line segment). Even trigonal planar does not represent an even distribution on the sphere.

Trigonal bipyamidal then just happens to be the minimum energy conformation. The repulsive interactions between the chlorine atoms cannot be optimized but they can be minimized.

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    $\begingroup$ One could say that the positions are indeed optimized - they just aren't all the same positions geometrically. $\endgroup$ – Jon Custer Jun 21 '16 at 14:53

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The geometry of $sp^3d" hybridisation, i.e: trigonal bipyramidal happens because of unequal hybridization of atomic orbitals.

The hybrid orbitals of a trigonal bipyramidal structure consist of two sets namely axial and equatorial orbitals. The two axial orbitals are made by inter-mixing of $p$ and $d$ atomic orbitals whereas the three equatorial orbitals are formed by intermixing of one $s$ and two $p$ orbitals.

Since the equatorial orbitals have a higher percentage of $s$ character compared to the axial orbitals, they are shorter than axial orbitals.

This is why you have two set of distances in $PCl_5$. This is NOT true when there are more than 1 atom except for one specific case which I will add to the answer if requested (did not add it here to avoid complication since the OP appears to be new to the topic)

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    $\begingroup$ This is a little bit backwards as orbital hybridization is a mathematical construct developed to rationalize observed geometry with the shapes and orientations of the canonical hydrogenic atomic orbitals. It is more appropriate to say that $\ce{PCl5}$ is $sp^3 d$ hybridized because it has trigonal planar geometry. $\endgroup$ – Ben Norris Jun 21 '16 at 15:27
  • $\begingroup$ $PCl_5$ is trigonal bipyramidal. Of course, hybridization is a mathematical construct developed to explain observed geometry but I believe that the OP isn't through with the basics yet. So I thought an answer based on VSEPR theory would be sufficient. $\endgroup$ – Yashas Jun 21 '16 at 15:30
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    $\begingroup$ Most, if not all, trigonal bipyramidal structures I know, including $\ce{PCl5}$, have no contribution of d-orbitals. $-1$. $\endgroup$ – Jan Jun 21 '16 at 20:11
  • $\begingroup$ huh? You can't form 5 bonds with just s and p orbitals. $PCl_5$ is a hypervalent molecule. $\endgroup$ – Yashas Jun 22 '16 at 6:18
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Yashas Jun 22 '16 at 6:18

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