What is the electron geometry of the ion I3−?

• A. Bent
• B. Tetrahedral
• C. Linear
• D. Trigonal planar

Answer: (C)

Verified step by step guidance

1. Identify the central atom in the ion I3⁻, which is the iodine atom in the middle of the three iodine atoms.
2. Determine the total number of valence electrons for the ion: each iodine atom contributes 7 valence electrons, and the extra negative charge adds 1 more electron, so total valence electrons = 3 × 7 + 1 = 22 electrons.
3. Draw the Lewis structure by placing a single bond between the central iodine and each terminal iodine, then distribute the remaining electrons to satisfy the octet rule, keeping in mind the extra lone pairs on the central iodine.
4. Count the regions of electron density (bonding and lone pairs) around the central iodine atom. For I3⁻, there are 2 bonding pairs and 3 lone pairs, totaling 5 regions of electron density.
5. Use the VSEPR theory to determine the electron geometry based on 5 regions of electron density, which corresponds to a trigonal bipyramidal electron geometry; however, the molecular shape considering only atoms is linear because the 3 lone pairs occupy equatorial positions, forcing the bonded atoms into a linear arrangement.