What is the electron geometry of the IF_4^+ ion?

• A. Square planar
• B. Tetrahedral
• C. Octahedral
• D. Trigonal bipyramidal

Answer: (C)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons for the IF_4^+ ion. Iodine (I) has 7 valence electrons, each fluorine (F) has 7 valence electrons, and since the ion has a +1 charge, subtract one electron from the total count.
2. Step 2: Draw the Lewis structure of IF_4^+. Place iodine as the central atom and connect it to four fluorine atoms with single bonds. Distribute the remaining electrons to satisfy the octet rule for fluorine atoms and place any leftover electrons on iodine.
3. Step 3: Count the regions of electron density (bonding and lone pairs) around the central iodine atom. Each bond counts as one region, and each lone pair counts as one region.
4. Step 4: Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to determine the electron geometry based on the number of electron density regions. For example, 6 regions correspond to an octahedral electron geometry.
5. Step 5: Identify the electron geometry of IF_4^+ based on the number of electron density regions found. Remember that electron geometry considers both bonding pairs and lone pairs, while molecular geometry considers only bonding pairs.