Question 7.EP.8: Determine the shape of the following molecules using VSEPR t...
Determine the shape of the following molecules using VSEPR theory: (a) SF_{4}, (b) BrF_{5}
Strategy
As always, we start by drawing the Lewis structures. Then count the number of electron pairs around the central atom, and determine the spatial arrangement of electrons pairs, consulting Table 7.3 as necessary. Place the lone pairs in positions where the electron repulsions are minimized, and describe the resulting geometric arrangement of the atoms.
| Table 7.3 | ||||||||
| Each of the geometrical arrangements shown in the table minimizes the electron pair repulsions for the indicated number of electron pairs. To visualize the shapes of molecules, it is essential that you have a sound mental picture of each of these geometries. | ||||||||
| Number of Electron Pairs |
Geometric Name | Bond Angles | Diagram | |||||
| 2 | Linear | 180° |  |
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| 3 | Trigonal planar | 120° |  |
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| 4 | Tetrahedral | 109.5° |  |
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| 5 | Trigonal bipyramidal | 120°, 90° |  |
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| 6 | Octahedral | 90°, 180° |  |
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(a) The Lewis structure of SF_{4} is as shown.
There are five pairs of electrons around sulfur, so the shape with minimum interaction is a trigonal bipyramid. Unlike the previous example, though, here the central sulfur atom has one lone pair. The five positions of a trigonal bipyramid are not equivalent, so we need to determine the best location for the lone pair. Electron pairs in the planar triangle (equatorial positions) have two near neighbors at an angle of 120° and two at 90°, whereas the axial positions have three near neighbors, each at an angle of 90°. Thus there is “more space” for electron pairs in the equatorial positions. Because lone pair–bond pair repulsion is larger than bond pair–bond pair repulsion, this means that the lone pair should occupy an equatorial position. This reduces the overall repulsive interactions. (Only the two closest bonding pairs are 90° away, whereas if the lone pair was in an axial position, there would be three bonding pairs 90° away.)
The resulting shape is usually described as a seesaw. If you don’t see why, try looking at a model.
(b) The Lewis structure of BrF_{5} is as shown.
There are six electron pairs around the central bromine atom, so the distribution of electrons is based on an octahedron. A lone pair occupies one of the six positions, so we must consider where to place it. But because an octahedron is completely symmetrical and all of its vertices are equivalent, the lone pair can occupy any corner. The resulting shape of the molecule is a square pyramid.
Discussion
For many students, the confusing part of this type of problem lies in the appearance that a lone pair of electrons is considered for part of the problem but not for all of it. There are two things to remember: (1) The shape of a molecule is determined by the positions of its atoms. This is a result of the ways we measure shapes. Because nuclei are so much more massive than electrons, our experiments detect them more readily, and thus we consider that the shape is defined by the positions of the nuclei. (2) Whether or not it is involved in bonding, every electron pair in the Lewis structure constitutes a region in space where there is a relatively high concentration of negative charge. The key concept of VSEPR theory is that these negatively charged regions will repel each other and therefore must be arranged as far apart as possible. So the electrons dictate the shape of a molecule, but they are not directly measured. That’s why we have a process where we appear to consider them at some stages but not at others.