Final answer:
The hybridization of iodine in IF₃ is sp³d, due to the combination of 3 bonds and 2 lone pairs, while the hybridization in IF₅ is sp³d², caused by 5 bonding pairs and no lone pairs, making the answer (c) IF₃: sp³d, IF₅: sp³d².
Step-by-step explanation:
The question is asking about the hybridization of iodine in the molecules IF₃ (iodine trifluoride) and IF₅ (iodine pentafluoride). When determining the hybridization of a central atom, we must take into account the number of electron pairs (bonding and non-bonding) around the central atom.
For IF₃, iodine has 7 valence electrons and forms 3 bonds with fluorine atoms, leading to 2 lone pairs of electrons. This accounts for a total of 5 electron pairs, which corresponds to sp³d hybridization, suggesting a trigonal bipyramidal molecular geometry with a T-shaped structure as illustrated in the Figure 8.18 information provided.
For IF₅, iodine again has 7 valence electrons but forms 5 bonds with fluorine atoms, and no lone pairs. This situation leads to 5 bonding pairs and requires sp³d² hybridization, which supports an octahedral structure as indicated by the Figure 8.20.
Therefore, the correct answer is c. IF₃: sp³d, IF₅: sp³d².