I have some more questions about this problem. In molecule PBr2F3 the two Br lie at equatoral places because they need bigger space than F. Also with XeF2, this molecule has 4 lone pairs, the lone pairs must have the bigger space than the others so why the geometry of XeF2 does not has V sharp but linear? I think with V sharp the angle of two of the lone pairs would be 180, bigger than 90 when XeF2 is linear. Please explain this for me. Thank you.

I think you are familiar with VSEPR theory, right?

Electron pairs tend to minimize the repulsions.

Electron-electron pair repulsion decrease:

l.p.-l.p. > l.p.-b.p. > b.p.-b.p. (l.p.=lone pair, b.p.=bond pair)

For XeF2, the electronic geometry is octahedral, 4 l.p. and 2 b.p..

What you need to do is to calculate their repulsions that finding the most stable configuration.

There are only 2 configurations in XeF2, either both F atoms are 180 degree away from each other or 90 degree separation.

For F-F 180 degree separation, there are:

4 x 90 degree l.p.-l.p. repulsions, 8 x 90 degree l.p.-b.p. repulsions...(no need to discuss 180 degree repulsions at this moment)

For F-F 90 degree separation, there are:

5 x 90 degree l.p.-l.p. repulsions (the main factor), 6 x 90 degree l.p.-b.p. repulsions and 1 x 90 degree b.p.-b.p. repulsion.

It is obvious that the electronic repulsion in 2nd one is higher. Therefore, the 1st one is prefer to the 2nd one.