What are the expected bond angles in the ICl_4^+ ion?

• A. 90^ext{o}
• B. 109.5^ext{o}
• C. 120^ext{o}
• D. 180^ext{o}

Answer: (A)

Verified step by step guidance

1. Determine the central atom in the ICl_4^+ ion, which is iodine (I), and count the total number of valence electrons. Iodine has 7 valence electrons, each chlorine (Cl) has 7, and since the ion has a +1 charge, subtract one electron from the total count.
2. Calculate the total number of valence electrons: 7 (I) + 4 × 7 (Cl) - 1 (charge) = 34 electrons. Use these electrons to form bonds and lone pairs around the iodine atom.
3. Draw the Lewis structure by placing four chlorine atoms bonded to iodine and distribute the remaining electrons as lone pairs on iodine. Determine the number of lone pairs on iodine after bonding four chlorines.
4. Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the molecular geometry. With four bonded atoms and one lone pair on iodine, the electron geometry is trigonal bipyramidal, but the molecular shape is see-saw.
5. Identify the bond angles based on the molecular geometry. In a see-saw shape, bond angles around the central atom include approximately 90° and 120°, but the presence of the lone pair distorts these angles. For ICl_4^+, the expected bond angles are close to 90° due to the square planar arrangement after considering lone pair repulsions.