How to explain shape of molecules in penta and hexa coordination if hybridization involving d-orbitals (in main block) is considered incorrect?

In [this answer](https://chemistry.stackexchange.com/questions/64613/what-is-dragos-rule-does-it-really-exist/64636#64636) @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](https://i.stack.imgur.com/vovyq.png)](https://i.stack.imgur.com/vovyq.png)


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](https://i.stack.imgur.com/KnUhbs.gif)](https://i.stack.imgur.com/KnUhbs.gif)


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) ? 


