What is the hybridization of the central iodine atom in IF_5?

• A. sp^2
• B. sp^3
• C. sp^3d^2
• D. sp^3d

Answer: (C)

Verified step by step guidance

1. Identify the central atom and its valence electrons. In IF_5, iodine (I) is the central atom. Iodine is in group 17, so it has 7 valence electrons.
2. Count the number of atoms bonded to the central atom and any lone pairs on it. IF_5 has 5 fluorine atoms bonded to iodine, and since iodine has 7 valence electrons, after forming 5 bonds, it has 2 electrons left as a lone pair.
3. Determine the steric number, which is the sum of bonded atoms and lone pairs around the central atom. Here, steric number = 5 (bonded atoms) + 1 (lone pair) = 6.
4. Use the steric number to find the hybridization. A steric number of 6 corresponds to an octahedral electron geometry and an sp^3d^2 hybridization.
5. Conclude that the central iodine atom in IF_5 is sp^3d^2 hybridized due to 5 bonding pairs and 1 lone pair, which fits the octahedral electron domain geometry.