The structure of $$\text{IF}_7$$ is:

Iodine belongs to group 17, so the central atom brings $$7$$ valence electrons.

Each of the seven fluorine atoms contributes one electron to form a single $$\sigma$$-bond with iodine. Hence the total electron pairs around iodine are:

Bond pairs $$=7$$, Lone pairs $$=0$$.

According to VSEPR notation this is an $$\text{AX}_7$$ species.

For an $$\text{AX}_7$$ arrangement with zero lone pairs the minimum-repulsion geometry predicted by VSEPR is a pentagonal bipyramid (five F atoms in an equatorial pentagon and two F atoms in the axial positions).

Therefore the molecular shape of $$\text{IF}_7$$ is pentagonal bipyramid.

Option C which is: pentagonal bipyramid