Hybridisation =‌1∕2 (Number of valence electron + number of monovalent atoms other than central atom + negative charge − positive charge) ‌=‌
1
2
(7+2+1) ‌=‌
10
2
‌=5≅sp3d.
Also it can be determined by sum of number of lone pairs and number of σ-bonds central metal atom will form. As I is further linked to two iodine atoms and number of lone pair on each will be 3 . ∴3+2=5≅sp3d hybridised. ∴3+2=5≅sp3d hybridised.