What is the value of the bond angle in ICl_2^-?

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

Answer: (B)

Verified step by step guidance

1. Identify the central atom in the molecule ICl_2^-. Here, iodine (I) is the central atom bonded to two chlorine (Cl) atoms.
2. Determine the total number of valence electrons around the central iodine atom. Iodine has 7 valence electrons, each chlorine has 7, and there is one extra electron due to the negative charge, so total valence electrons = 7 (I) + 2 × 7 (Cl) + 1 (charge) = 22 electrons.
3. Draw the Lewis structure for ICl_2^-. Place the two Cl atoms bonded to I, and distribute the remaining electrons to satisfy the octet rule, noting that iodine can have an expanded octet. This will result in 3 lone pairs on iodine and 2 bonding pairs with chlorine atoms.
4. Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to determine the molecular geometry. With 2 bonding pairs and 3 lone pairs, the electron geometry is trigonal bipyramidal, but the molecular shape is linear because the lone pairs occupy equatorial positions and repel the bonding pairs to be 180° apart.
5. Conclude that the bond angle between the two chlorine atoms in ICl_2^- is 180°, corresponding to a linear molecular geometry.