# 211chp9 - Chapter 9 Homework Solutions 16 a trigonal planar...

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Chapter 9 Homework Solutions 11/2/11 16. a) trigonal planar b) tetrahedral c) trigonal bipyramidal d) octahedral 21. a) linear, linear d) octahedral, octahedral b) tetrahedral, trigonal pyramidal e) tetrahedral, tetrahedral c) trigonal bipyramidal, seesaw f) linear, linear 22. a) tetrahedral, trigonal pyramidal d) tetrahedral, tetrahedral b) trigonal planar, trigonal planar e) trigonal bipyramidal, linear c) trigonal bipyramidal, T-shaped f) tetrahedral, bent 26. a) 109.5º, 120º b) 109.5º, 120º c) 109.5º, 109.5º d) 180º, 109.5º 36. polar: b, c, e nonpolar: a, d, f 38. The ortho and meta isomers would have non-zero dipole moments. Each chlorine pulls electron density towards itself from the carbon to which it is attached. In the para isomer, the chlorines are pulling the electron density in exactly opposite directions so the bond dipoles exactly offset. In the other 2 isomers, the dipoles are partly in the same direction and add. 40. a) b) c) 47. a) sp 2 b) sp 3 c) sp d) sp 3 d e) sp 3 d 2 48. a) sp 3 b) sp c) sp 2 d) sp 3 d e) sp 3 d 2 56. a) 120º, 120º, 109.5º b) sp 2 , sp 2 , sp 3 c) 21 58. All three possible Lewis structures are shown below. The hybridization on sulfur is sp 2 . S O: O .. .. .. .. :O: : S O: O .. .. .. .. :O: : S O: O .. .. .. .. :O: : Yes, SO 3 will exhibit delocalized bonding. The sulfur bonds to each oxygen with a bond via overlap of the sp 2 hybrid with an oxygen p orbital. In each resonance form the sulfur p orbital overlaps with a p orbital on a different oxygen.

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