So determine the electron pair of molecular geometries for PF5.

Any time I want to determine the geometries of a molecule,

I first need to get the Lewis structure of that molecule.

So I'm first going to figure out the number of valence electrons I have to

work with here.

So phosphorous has 5.

And I have 5 fluorines, which each have 7 and

that gives me a total of 40 electrons that are going to be in Lewis structure.

I start by putting the Phosphorous in the center,

one because it has a lower electro-negativity than fluorine, and

we know that fluorine is always going to be a terminal atom.

So I'm going to draw my skeletal structure all single bonds.

Now I've used up ten electrons.

Now I'm going to start filling in my octets on each of my fluorines, and

see how many electrons I have used after I've done that.