How many lone pairs of electrons are present on the iodine atom in the Lewis dot structure of IF_4^+?

• A. 0 lone pairs
• B. 3 lone pairs
• C. 1 lone pair
• D. 2 lone pairs

Answer: (C)

Verified step by step guidance

1. Determine the total number of valence electrons available for the molecule IF_4^+. Iodine (I) has 7 valence electrons, each fluorine (F) has 7 valence electrons, and since the species has a +1 charge, subtract one electron from the total count. So, total valence electrons = 7 (I) + 4 × 7 (F) - 1 (charge).
2. Draw the skeletal structure with iodine as the central atom bonded to four fluorine atoms. Each I–F bond represents a pair of shared electrons (2 electrons per bond).
3. Subtract the electrons used in bonding from the total valence electrons to find the number of electrons remaining as lone pairs. Since there are 4 bonds, total bonding electrons = 4 × 2.
4. Assign the remaining electrons as lone pairs, first completing the octets of the fluorine atoms (each fluorine needs 6 more electrons as lone pairs to complete its octet).
5. After completing the fluorine octets, any leftover electrons are placed as lone pairs on the iodine atom. Count these lone pairs to find how many lone pairs are on iodine.