Question 7.EP.9: Use VSEPR theory to determine the shape of the NOF molecule.
Use VSEPR theory to determine the shape of the NOF molecule.
Strategy
Once again, we start by drawing the Lewis structures. Then count the number of regions of electron density around the central atom, remembering to count any double or triple bonds as a single region. Determine the spatial arrangement of electrons pairs, consulting Tables 7.2 as needed. Place any lone pairs in positions where the electron repulsions are minimized, and describe the resulting geometric arrangement of the atoms.
| The five common orbital hybridization schemes are shown. The names for these hybrid orbitals are derived from the type and number of orbitals that combine to form them. Thus sp² hybrids result from combining an s orbital with a pair of p orbitals. The orbital geometries shown in the left-hand column give rise to the common molecular shapes described in the next section. | |||||
| Orbitals Combined | Hybridization | Orbital Geometry | |||
| s, p | sp |  |
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| s, p, p | sp² |  |
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| s, p, p, p | sp³ |  |
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| s, p, p, p, d | sp³d |  |
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| s, p, p, p, d, d | sp³d² |  |
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The choice of central atom is perhaps less obvious here than in some cases we have looked at. But nitrogen is the least electronegative atom in this molecule, so it will be central. Starting from an F—N—O skeleton, the Lewis structure is as shown below.
\overset{..}{\underset{..}{O}} \xlongequal[]{}\ddot{N}—\overset{..}{\underset{..}{F}:}
\ There are three regions of electron density around the nitrogen atom, so the shape with minimum interaction is trigonal planar. The central nitrogen atom has one lone pair. In the trigonal planar geometry, all three positions are equivalent, so it does not matter which one the lone pair occupies. The molecule will be bent, with an O—N—F bond angle of approximately 120°. Because lone pairs provide stronger repulsion, the actual bond angle should be slightly less than 120°.
Discussion
Much like the situation we saw for lone pairs, double bonds can cause confusion for many students. When counting the number of valence electrons to draw the Lewis structure, we must be sure to count the double bond as four electrons. (And a triple bond, if one were present, would count as six electrons.) But in determining the shape, we treat the double (or triple) bond as a single region of electron density. It may help if you realize that both of the bonds in the double bond must “point in the same direction” because they connect the same pair of atoms.