What is the molecular geometry of the PF4+ ion?

• A. See-saw
• B. Square planar
• C. Tetrahedral
• D. Trigonal bipyramidal

Answer: (A)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons for the PF4+ ion. Phosphorus (P) has 5 valence electrons, each fluorine (F) has 7 valence electrons, and since the ion has a +1 charge, subtract one electron from the total count. Calculate: Total electrons = 5 + (4 × 7) - 1.
2. Step 2: Draw the Lewis structure for PF4+. Place phosphorus 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 phosphorus.
3. Step 3: Count the regions of electron density (bonding and lone pairs) around the central phosphorus atom. Each bond counts as one region, and any lone pairs on phosphorus count as additional regions.
4. Step 4: Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the electron geometry based on the number of electron regions. For example, 5 regions correspond to trigonal bipyramidal electron geometry.
5. Step 5: Determine the molecular geometry by considering the positions of atoms only (ignoring lone pairs). If there is one lone pair and four bonded atoms in a trigonal bipyramidal arrangement, the molecular geometry is 'see-saw'.