What is the approximate bond angle between the chlorine atoms in iodine trichloride (ICl_3)?

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

Answer: (C)

Verified step by step guidance

1. Step 1: Determine the Lewis structure of iodine trichloride (ICl_3). Iodine (I) is the central atom bonded to three chlorine (Cl) atoms. Count the total valence electrons: Iodine has 7 valence electrons, each chlorine has 7, so total electrons = 7 + 3×7 = 28 electrons.
2. Step 2: Arrange the electrons to satisfy the octet rule for each atom. Place single bonds between iodine and each chlorine atom, using 6 electrons (3 bonds × 2 electrons each). Distribute the remaining electrons as lone pairs, first completing the octets of chlorine atoms, then placing any leftover electrons on iodine.
3. Step 3: Identify the number of lone pairs on the central iodine atom. After bonding and completing chlorine octets, iodine will have two lone pairs remaining, in addition to the three bonded pairs.
4. Step 4: Use VSEPR theory to determine the molecular geometry. With 3 bonded atoms and 2 lone pairs, the electron geometry is trigonal bipyramidal, but the molecular shape is T-shaped due to the lone pairs occupying equatorial positions.
5. Step 5: Determine the bond angles between the chlorine atoms. In a T-shaped molecule, the bond angles between the bonded atoms are approximately 90° and 180°, but the bond angle between the chlorine atoms bonded to iodine is close to 90°, which explains the correct answer.