What is the molecular geometry of IF_5?

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

Answer: (B)

Verified step by step guidance

1. Step 1: Determine the Lewis structure of IF_5 by counting the total valence electrons. Iodine (I) has 7 valence electrons, and each fluorine (F) has 7 valence electrons. So, total valence electrons = 7 (I) + 5 × 7 (F) = 42 electrons.
2. Step 2: Draw the skeletal structure with iodine as the central atom bonded to five fluorine atoms. Use single bonds between iodine and each fluorine, which accounts for 10 electrons (5 bonds × 2 electrons each).
3. Step 3: Distribute the remaining electrons to complete the octets of the fluorine atoms first. Each fluorine needs 6 more electrons to complete its octet (since each already has 2 from the bond). This uses 30 electrons (5 fluorines × 6 electrons).
4. Step 4: Place any leftover electrons on the central iodine atom. After bonding and completing fluorine octets, iodine will have one lone pair of electrons remaining.
5. Step 5: Use the VSEPR theory to determine the molecular geometry. With 5 bonding pairs and 1 lone pair around iodine, the electron geometry is octahedral, but the molecular geometry (shape) is square pyramidal.