Which of the following interhalogen compounds has a Lewis structure that violates the octet rule by having more than eight electrons around the central atom?

• A. ClF
• B. BrF3
• C. ICl
• D. IF

Answer: (B)

Verified step by step guidance

1. Identify the central atom in each interhalogen compound. Typically, the less electronegative halogen is the central atom (e.g., Br in BrF3, I in IF, Cl in ClF and ICl).
2. Determine the total number of valence electrons available for bonding in each molecule by summing the valence electrons of the central atom and the surrounding halogen atoms.
3. Draw the Lewis structure for each compound, placing single bonds between the central atom and each surrounding atom, then distribute the remaining electrons as lone pairs to satisfy the octet rule where possible.
4. Check the number of electrons around the central atom in each Lewis structure. If the central atom has more than eight electrons (expanded octet), it violates the octet rule by having an expanded valence shell.
5. Compare the structures: BrF3 is known to have an expanded octet because bromine can accommodate more than eight electrons due to available d orbitals, while ClF, ICl, and IF typically obey the octet rule.