What is the hybridization of the central iodine atom in ICl_5?

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

Answer: (D)

Verified step by step guidance

1. Identify the central atom in the molecule ICl_5, which is iodine (I).
2. Determine the number of valence electrons on the central iodine atom. Iodine is in group 17, so it has 7 valence electrons.
3. Count the number of atoms bonded to iodine and any lone pairs on iodine. ICl_5 has 5 chlorine atoms bonded to iodine, and since iodine has 7 valence electrons, after forming 5 bonds, 2 electrons remain as one lone pair.
4. Calculate the steric number, which is the sum of bonded atoms and lone pairs around iodine: 5 (bonded atoms) + 1 (lone pair) = 6.
5. Use the steric number to determine the hybridization. A steric number of 6 corresponds to an sp^3d^2 hybridization, which matches the correct answer.