How does VSEPR theory explain the formation of different bond angles in PCl₅?

We know, that in the exited state, electrons of Phosphorus could go to higher states and form $\ce{dsp^3}$ hybridization. Therefore it will create five equal energy states of electrons. But my question is that, why are the angles of the bonds in $\ce{PCl5}$ 90 and 120 degree instead of equally distributed? How does VSEPR theory explain the difference in the angles?


Why do the five equal bonding pairs of electrons have a trigonal
 bipyramidal arrangement, with two different internal bond angles -
 120° in the plane and 90° between the plane and the axial atoms?




There are only five Platonic solids: 


1. tetrahedron,
2. octahedron,
3. cube,
4. dodecahedron, and
5. isocahedron.


Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane.  

Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages [*Min-Energy Configurations of Electrons On A Sphere*](http://www.mathpages.com/home/kmath005/kmath005.htm) ([Stable link via The Internet Archive](https://web.archive.org/web/20180411111522/http://www.mathpages.com/home/kmath005/kmath005.htm)).


Also there are not five equal energy states as demonstrated in the lecture notes by Marcel Schlaf and Kathryn Preuss from the University of Guelph:
[Chapter 5.5 (continued 2) *MO Theory ICl4, PCl5*. (pdf)](http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf) ([Stable pdf link via The Internet Archive](https://web.archive.org/web/20170516230216/http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf))


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Here's a proof that you can't place 5 ligands equidistant from each other and from the central atom in a **nonplanar** molecule:


1. Place two points anywhere you like on the surface of a sphere around the central atom.
They will define a great circle (an "equator") that cuts the sphere in half.
2. You have 3 points left. You can't place them all off the "equator" without putting two in one hemisphere and 1 in the other. *You cannot make all the points equidistant.* The only way you can balance the number of points in each hemisphere is to put one in one hemisphere, one in the other, and one on the "equator" with the first two you placed. But the distance between the equatorial points will not be the same as all of the distances to points in the hemisphere.


Why is it trigonal pyramidal, though? The best we can do is make subsets of the ligands equidistant. Place three equidistant points on the equator, and two at the "poles". Imagine a linear shape (the axial positions) passing at right angles through a trigonal planar shape (the equatorial positions). You have a trigonal bipyramidal shape.


The dsp3 hybridization is a futile attempt to explain 5 orbitals around the central atom by mixing the central atom's orbitals alone. In fact back in the early 1990's it was shown that this is a poor explanation (source: E. Magnusson, Hypercoordinate molecules of second-row elements: d functions or d orbitals? *J. Am. Chem. Soc.* **1990,** *112* (22), 7940–7951. [DOI: 10.1021/ja00178a014](https://doi.org/10.1021/ja00178a014).)


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