Problem 1: Determine the geometry of XeOF4

Step 1: Count valence electrons on Xe Xe has 8 valence electrons

Step 2: Count bonds XeOF4 has: 1 Xe=O bond + 4 Xe-F bonds = 5 bond pairs total (The double bond to O uses 2 electrons from Xe in terms of sigma framework)

Step 3: Calculate lone pairs on Xe Total valence $e^{-}$ on Xe = 8 Used in bonding = 5 (for 5 bonds, one pair per bond from Xe's perspective for sigma framework) Remaining = 8 - 5 = 3... but wait, Xe=O is a double bond. Recalculate using formal approach: Xe: 8 valence $e^{-}$. Forms 4 single bonds to F (uses 4$e^{-}$) + 1 double bond to O (uses 2$e^{-}$) = 6$e^{-}$ used. Remaining = 2$e^{-}$ = 1 lone pair.

Step 4: Total electron pairs = 5 bond pairs + 1 lone pair = 6 → sp3d2

Step 5: Apply VSEPR 6 electron pairs → Octahedral base. 1 lone pair in one axial position → 5 bond pairs remain. With O in equatorial (double bond prefers equatorial to minimize repulsion) and lone pair in the other axial position, XeOF4 is square pyramidal.