What is the hybridization of the central atom (iodine) in the tetrafluoroiodide anion, IF4^-?

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

Answer: (B)

Verified step by step guidance

1. Determine the total number of valence electrons for the central atom (iodine) and the surrounding atoms. Iodine has 7 valence electrons, each fluorine has 7 valence electrons, and the anion has an extra electron due to the negative charge.
2. Calculate the total valence electrons: add iodine's 7 electrons, 4 fluorines × 7 electrons each, and 1 extra electron for the negative charge.
3. Draw the Lewis structure for IF4⁻ by placing iodine in the center, connecting it to four fluorine atoms with single bonds, and distributing the remaining electrons to satisfy the octet rule for fluorine and account for lone pairs on iodine.
4. Count the regions of electron density (bonding pairs and lone pairs) around the iodine atom. Each bond counts as one region, and each lone pair counts as one region.
5. Use the number of electron density regions to determine the hybridization of iodine. For example, 6 regions correspond to sp³d² hybridization.