[Enantiomer]

Hi, I'm going over a practice test and I found a question asking for VSPER structures, which is generally easy enough but I found a bit of a road block, (I think):

PF₄^-- Which I believe to have seesaw structure, (4 BP/1LP)
This is the one that has me mildly unsure
PF₄⁺-  Which I'm guessing would form a tetrahedral?  It's hard for me to tell if it changes shape, is this wrong or right?  And if even if it does change shape is this due to a lesser repulsion force from the loss of the lone pair?
Thanks for your time

[cth]

PF₄^-- Which I believe to have seesaw structure

It is simpler than that: the P atom is in the centre and fluorine atoms around. But it should be distorted.

PF₄⁺-  Which I'm guessing would form a tetrahedral?

Yes. It isn't distorted.


Why is that?
* First, think of PF₃: there are 3 fluorine atoms and one lone pair around the P atom.
* Now, add F⁺ cation to PF₃ to form PF₄⁺. The fluorine atom has lost one electron to form the ion F⁺, so it has 6 valence electrons left: 3 lone pairs and one empty 2p orbital (which previously contained the electron that had been kicked off). What do you think the F⁺ cation will do when it meet with PF₃? It will take over the lone pair located on the P atom to form a covalent bond (so called dative bond). So, in PF₄⁺, the P atom has 4 fluorines around and no lone pair anymore it is tetrahedral.
* Now, add F^- anion to PF₃ to form PF₄^-. The fluorine atom has gained one electron to form F^-, so it has 8 valence electrons. Its octet is filled. What do you think would happen when it meets with PF₃? The fluorine won't take over the lone pair on the P atom. Instead, it is the P atom that will accept one lone pair coming from F^- and put it in one of its 3d orbitals that are empty. So, the P atom will have 4 fluorine atoms around and will still have its own lone pair  5 "objects" around the P atom, it won't be tetrahedral.