Given the neutral compound IF₃, what is the hybridization of the central iodine atom in its Lewis structure?

• A. sp³d
• B. sp³d²
• C. sp³
• D. sp²

Answer: (A)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons for the central atom (iodine) and the surrounding atoms (fluorine). Iodine is in group 17, so it has 7 valence electrons, and each fluorine atom also has 7 valence electrons.
2. Step 2: Calculate the total valence electrons in the molecule IF₃ by adding iodine's valence electrons to three times the valence electrons of fluorine atoms: \$7 + 3 \times 7 = 28$ electrons.
3. Step 3: Draw the Lewis structure by placing iodine in the center and connecting it to three fluorine atoms with single bonds. Each bond uses 2 electrons, so 6 electrons are used in bonding. Distribute the remaining electrons to satisfy the octet rule for fluorine atoms first, then place any leftover electrons on iodine as lone pairs.
4. Step 4: Count the regions of electron density (bonding pairs + lone pairs) around the central iodine atom. This total number determines the electron geometry and hybridization. For IF₃, iodine has 3 bonding pairs and 2 lone pairs, making 5 regions of electron density.
5. Step 5: Use the number of electron density regions to identify the hybridization. Five regions correspond to an \(sp^{3}d\) hybridization for the central iodine atom.