Textbook Question
Consider the following XF4 ions: PF4-, BrF4-, ClF4+, and AlF4-. (b) For which of the ions will the electron-domain and molecular geometries be the same?
12. Molecular Shapes & Valence Bond Theory
Molecular Geometry
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PF3Br2 is a nonpolar molecule. Based on this information, what are the approximate bond angles for F–P–F, Br–P–Br, and F–P–Br in PF3Br2?
A
F–P–F: 120°, Br–P–Br: 180°, F–P–Br: 90°
B
F–P–F: 90°, Br–P–Br: 90°, F–P–Br: 120°
C
F–P–F: 180°, Br–P–Br: 120°, F–P–Br: 90°
D
F–P–F: 120°, Br–P–Br: 120°, F–P–Br: 90°
Verified step by step guidance1
Identify the molecular geometry of PF3Br2. Since PF3Br2 is a nonpolar molecule, it suggests a symmetrical shape. The central atom, phosphorus (P), is surrounded by five substituents: three fluorine (F) atoms and two bromine (Br) atoms.
Determine the electron pair geometry around the phosphorus atom. With five substituents, the electron pair geometry is trigonal bipyramidal.
In a trigonal bipyramidal geometry, there are two types of positions: axial and equatorial. The axial positions are 180° apart, and the equatorial positions are 120° apart.
Assign the more electronegative atoms (fluorine) to the equatorial positions to minimize repulsion, and the less electronegative atoms (bromine) to the axial positions.
Based on the geometry, the bond angles are: F–P–F at 120° (equatorial), Br–P–Br at 180° (axial), and F–P–Br at 90° (between axial and equatorial).
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