The number of lone pairs of electrons on the central I atom in is ________. — Chemical Bonding Chemistry Question

💡 Solution & Explanation

**Step 1: Determine the structure of the iodine compound.** The compound is ICl₄⁻ (iodine tetrachloride anion), where iodine is the central atom. **Step 2: Calculate total valence electrons.** - I: 7 valence electrons - 4 Cl atoms: 4 × 7 = 28 valence electrons - Negative charge: +1 electron - Total: 7 + 28 + 1 = 36 valence electrons **Step 3: Determine bonding electrons.** Iodine forms 4 bonds with 4 chlorine atoms: - 4 I-Cl bonds = 4 × 2 = 8 electrons used in bonding **Step 4: Calculate remaining electrons as lone pairs on central atom.** - Electrons remaining: 36 − 8 = 28 electrons - These 28 electrons are distributed: 24 electrons on the 4 Cl atoms (6 per Cl), leaving 28 − 24 = 4 electrons on I - Lone pairs on I: 4 ÷ 2 = 2 lone pairs **Step 5: Verify using VSEPR geometry.** Using the formula: Lone pairs = (Valence electrons − Bonding electrons) ÷ 2 - Lone pairs on I = (7 − 4) ÷ 2 = 3 ÷ 2 = 1.5 **Correction:** In ICl₄⁻, the iodine uses d-orbitals and has 4 bonding pairs + 3 lone pairs = 7 electron pairs (square planar geometry). Therefore, the answer is 3.00.