Which of the following best represents the Lewis dot structure for ICl_5?

• A. Chlorine at the center with five single bonds to iodine atoms.
• B. Iodine at the center with five double bonds to chlorine atoms.
• C. Iodine at the center with five single bonds to chlorine atoms and one lone pair on iodine.
• D. Iodine at the center with five single bonds to chlorine atoms and no lone pairs on iodine.

Answer: (C)

Verified step by step guidance

1. Identify the central atom by considering electronegativity and bonding capacity. Iodine (I) is less electronegative than chlorine (Cl), so iodine will be the central atom in ICl_5.
2. Determine the total number of valence electrons available. Iodine has 7 valence electrons, and each chlorine has 7 valence electrons. For ICl_5, total valence electrons = 7 (I) + 5 × 7 (Cl) = 42 electrons.
3. Draw single bonds between the central iodine atom and each of the five chlorine atoms. Each single bond uses 2 electrons, so 5 bonds use 10 electrons.
4. Distribute the remaining electrons to satisfy the octet rule for the chlorine atoms first. Each chlorine needs 6 more electrons (3 lone pairs) to complete its octet, so 5 chlorines × 6 electrons = 30 electrons used.
5. Assign any leftover electrons to the central iodine atom as lone pairs. After bonding and completing chlorine octets, iodine will have 2 electrons left, which form one lone pair on iodine, consistent with the expanded octet possible for iodine.