The hybridization of iodine orbitals in the compound ${\mathrm{IF}}_7$ is

${\mathrm{sp}}^3{\mathrm{d}}^3$ hybridization is involved.

Iodine makes 7 bonds with 7 fluorine atoms in this structure. As a result, hybridization, which is ${\mathrm{sp}}^3{\mathrm{d}}^3$, involves 7 orbitals.