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9 Molecular Geometry Solutions to Exercises (b) PF₃, 26 valence e⁻, 13 pr, 1 nonbonding pair on P, influences molecular shape : F: (c) HBr, 8 valence e⁻, pr, 3 nonbonding pairs on Br, no effect on molecular shape because Br is not "central" H-Br: (d) HCN, 10 valence e⁻, 5 e⁻ pr, 0 nonbonding pairs on C, no effect on molecular shape H-CEN: (e) SO₂, 18 valence e⁻, 9e⁻ pr, 1 nonbonding pair on S, influences molecular shape 9.20 Draw the Lewis structure of each molecule. If it has nonbonding electron pairs on the central atom, decide whether they will cause to bond angles to deviate from ideal values for the particular electron domain geometry. (a) H₂S, 8 valence e⁻, 4 e⁻ pr, tetrahedral electron domain geometry with 2 nonbonding electron pairs on S will cause the bond angle to deviate from ideal 109.5° angles H-S-H (b) 24 valence e⁻, 12 pr, trigonal planar electron domain geometry with zero nonbonding pairs on B. We confidently predict 120° angles. (c) 14 valence e⁻, 7 pr, tetrahedral electron domain geometry with zero nonbonding pairs on C. We confidently predict 109.5° angles. H H (d) CBr₄, 32 valence e⁻, 16 pr, tetrahedral electron domain geometry with zero nonbonding pairs on C. We confidently predict 109.5° angles. :Br: :Br Br: : Br: (e) TeBr₄, 34 valence e⁻, 17 pr, trigonal bipyramidal electron domain geometry with one nonbonding pair on Te. The structure is similar to shown in Sample Exercise 9.2. The bond angles will deviate from ideal values, but perhaps not as much as in (Structure follows.) 234