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.