What is the hybridization of the central iodine atom in the tetrachloroiodide anion, ICl4^-?

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

Answer: (C)

Verified step by step guidance

1. Step 1: Determine the Lewis structure of the ICl4^- ion. Iodine (I) is the central atom bonded to four chlorine (Cl) atoms, and the ion carries an extra negative charge.
2. Step 2: Count the total number of valence electrons. Iodine has 7 valence electrons, each chlorine has 7, and there is 1 extra electron due to the negative charge. Total valence electrons = 7 (I) + 4 × 7 (Cl) + 1 (charge) = 36 electrons.
3. Step 3: Assign electrons to bonds and lone pairs. Four single bonds between iodine and chlorine atoms use 8 electrons. Subtracting these from 36 leaves 28 electrons, which are placed as lone pairs on the chlorine atoms and the central iodine atom. Each chlorine gets 3 lone pairs (18 electrons), leaving 10 electrons (5 pairs) for iodine.
4. Step 4: Determine the number of electron pairs around iodine. There are 4 bonding pairs (to Cl) and 2 lone pairs on iodine, totaling 6 electron pairs.
5. Step 5: Use VSEPR theory to find the hybridization. Six electron pairs correspond to an octahedral electron geometry, which requires sp\^3d\^2 hybridization for the central iodine atom.