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9 Molecular Geometry Solutions to Exercises (c) 12 valence (d) HCOOH, 18 valence e- H C C H H 3σ, 1π 4σ, 1π H H 9.93 (a) H C C C H H H 3(4) + 3(6) + 6(1) = 36 18 e- pr (b) There are 11 σ and 1 π bonds. (c) The C=O on the right-hand C atom is shortest. For the same bonded atoms, in this case C and O, the greater the bond order, the shorter the bond. (d, e) The right-most C has three e- domains, so the hybridization is sp²; bond angles about this C atom are approximately 120°. The middle and left-hand C atoms both have four e- domains, are sp³ hybridized, and have bond angles of approximately 109°. 9.94 F P B F F F F F µ = 1.03D BF₃ is a trigonal planar molecule with the central atom symmetrically surrounded by the three F atoms (Figure 9.12). The individual B-F bond dipoles cancel, and the molecule has a net dipole moment of zero. PF₃ has tetrahedral electron-domain geometry with one of the positions in the tetrahedron occupied by a nonbonding electron pair. The individual P-F bond dipoles do not cancel and the presence of a nonbonding electron pair ensures an asymmetrical electron distribution; the molecule is polar. 9.95 (a) Square pyramidal (b) Yes, there is one nonbonding electron domain on A. If there were only five. bonding domains, the shape would be trigonal bipyramidal. With five bonding and one nonbonding electron domains, the molecule has octahedral domain geometry. (c) Yes. If the B atoms are halogens, each will have three nonbonding electron pairs; there are five bonding pairs, and A has one nonbonded pair, for a total of [5(3) + 5 + 1] = 21 e- pairs and 42 electrons in the Lewis structure. If the five halogens contribute 35 e-, A must contribute seven valence electrons. A is also a halogen. 260