Which of the following best describes the Lewis dot structure for the iodine difluoride ion, IF2^-?

• A. Iodine is the central atom with two single bonds to fluorine atoms, three lone pairs on iodine, and a negative charge on the ion.
• B. Iodine is the central atom with two double bonds to fluorine atoms, two lone pairs on iodine, and a negative charge on the ion.
• C. Iodine is the central atom with two single bonds to fluorine atoms, one lone pair on iodine, and a positive charge on the ion.
• D. Fluorine is the central atom with single bonds to two iodine atoms, each iodine having three lone pairs.

Answer: (A)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons for the IF2^- ion. Iodine (I) has 7 valence electrons, each fluorine (F) has 7 valence electrons, and the negative charge adds 1 extra electron. So, total valence electrons = 7 (I) + 2 × 7 (F) + 1 (charge).
2. Step 2: Identify the central atom. Iodine is less electronegative than fluorine, so iodine will be the central atom bonded to two fluorine atoms.
3. Step 3: Draw single bonds between iodine and each fluorine atom. Each single bond accounts for 2 electrons, so subtract these bonding electrons from the total valence electrons.
4. Step 4: Distribute the remaining electrons as lone pairs to satisfy the octet rule for fluorine atoms first (each fluorine needs 8 electrons total, including bonding electrons). Then place any leftover electrons as lone pairs on iodine.
5. Step 5: Check the formal charges on each atom to ensure the most stable structure. The negative charge should be placed on the atom that can best accommodate it, which is iodine in this case, and confirm the number of lone pairs on iodine matches the electron count.