Can anyone explain to me why IF5 is square pyrimidal and not trigonal bipyrimidal?
I think you are asking VSEPR-type molecular shape of iodine pentafluoride. Well, iodine can bind up to seven ligands to itself, as it has an electron configuration of s2p5, as fluorine does.
In the second step, we will try to draw its Lewis-type structure. The total valence electrons add up to 42. The molecule has five bonding electrons, so ten of the available will be used. This means 52 electrons will be needed to reach the "octet". The total non-bonding electron count is 32, which can be calculated with (total valence electrons)-(total bonding electrons).
As one fluorine atom has six non-bonding electrons, you will have to put an additional non-bonding electron pair onto iodine. This will make the molecule distorted from ideal geometry, i.e., trigonal bipyramid.
You cannot place the non-bonded pair onto one of axial positions, you must place it on one of the equatorial positions. This will give you a rotated type of square pyramid.
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