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In this answer @Jan says:

penta- and hexacoordination; tetracoordination with additional lone pairs and related: attempt to form as many normal bonds with p orbitals as possible; keep one lone pair in an s orbital if possible. Use remaining p orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. $90^{∘}$ going from 2e-2c bonds to 4e-3c bonds; $180^{∘}$ between a pair of coordinating atoms contributing to the same 4e-3c bond.)

Let us consider $\ce{PCl5}$. According to several books the hybridization in this compound is $\mathrm{sp^3d}$. But @Jan says that pure $d$ orbitals never take part in hybridization in case of main group elements. Okay. So, if we assume that there is no hybridization in $\ce{PCl5}$ then as Jan says we form "as many normal bonds" with $\ce{Cl}$ as possible using $p$ orbitals. Since, $\ce{P}$ has 3 p orbitals we use them to form mutually perpendicular bonds. Suppose they are the top, bottom and leftmost bonds as in the following diagram:

enter image description here

After this, we are still left with two $\ce{Cl}$ atoms and only one $s$ orbital. How can we place two $\ce{Cl}$ atoms in only one pure $s$ orbital ? This part makes no sense to me.

Now, let us consider the $\ce{ClF3}$ example that @Jan gives. If we follow Jan's method we keep one lone pair in a pure $s$ orbital , then the shape can never be trigonal bipyramidal like this:

enter image description here

This is because $s$ orbital is spherical in nature and does not look like a dumbbell shaped $p$ orbital.

Also, could someone please give me some examples in cases of "penta- and hexacoordination; tetracoordination with additional lone pairs", where $3c-4e$ bonds are formed using remaining $p$ orbitals when no hybridization occurs (as mentioned by @Jan) ?

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    I think they’re looking good =) – Jan Dec 17 '16 at 20:15
  • This is very broad as written, and it also reads more like a critique of another person's answer to a question than as a question itself. – Todd Minehardt Dec 17 '16 at 20:19
  • @ToddMinehardt Well, Jan himself/herself told me that this question is fine. And it simply asks for clarification. No one is criticizing anyone or anyone's answer here. –  Dec 17 '16 at 20:22
  • @Jan Do you also think that is very broad as Todd says? Should I edit it further ? (In my opinion if I trim this question further it will lose its essence ) –  Dec 17 '16 at 20:25
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    @ToddMinehardt Imho, the question is fine. The critique is on scientific level imho; they question the (very simplified) concept I gave in another answer asking for expansion. It was too much to include in the other answer but I think it’s good here. – Jan Dec 17 '16 at 20:31
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    In fact, I even think this can serve as a canonical dupe target for ‘My book says compound X uses d orbitals; here I read it doesn’t. Why/how?’ – Jan Dec 17 '16 at 20:33
  • One possibility, PCl3^2+ Cl- Cl-, P uses sp2 hybridization. – Rodriguez Dec 17 '16 at 20:38
  • Sounds like it's overflow from the other question - I trust your assessment, withdraw close vote. Thanks for clarifying. – Todd Minehardt Dec 17 '16 at 20:39
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    This has been asked (and explained) in one form or the other multiple times: http://chemistry.stackexchange.com/q/29142/4945 http://chemistry.stackexchange.com/q/18427/4945 http://chemistry.stackexchange.com/q/28692/4945 http://chemistry.stackexchange.com/q/6785/4945 Simply let go of the hybridisation concept, forget the VSEPR model and try to form a MO scheme with ligand group orbitals, the lone pairs will fall in place. An while I disagree with the name 'hypervalent' the wikipedia article is quite well explained. (cc @jan ) – Martin - マーチン Dec 17 '16 at 21:19

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