ch. 8 hw

ch. 8 hw - of the simplest stable hydrocarbon would be CH 2...

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Buttler 25. Write a Lewis formula for each of the following species. Indicate the number of electron groups and the electronic and molecular or ionic geometries. b. SO 2 S three electron groups; electronic geometry is trigonal planar; molecular geometry is angular O O a. BF 3 F three electron groups; both geometries are trigonal planar B F F d. SiCl 4 Cl four electron groups; both geometries are tetrahedral Cl Si Cl Cl 29. Pick the member of each pair that you would expect to have the smaller bond angles, if different, and explain why. a. SO 2 because O has less valence electrons than F, so they won’t be as repulsive to each other d. NF 3 because it contains more atoms, so they’re obviously going to have to be closer together 47. Briefly summarize the reasoning by which we might have predicted that the formula
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Unformatted text preview: of the simplest stable hydrocarbon would be CH 2 , if we did not consider hybridization. I wouldnt have predicted that because that would only give it six valence electrons. Would this species satisfy the octet rule? No. 54. Predict the hybridization at each carbon atom in each of the following molecules. a. acetone: 1 sp 2 and two sp 3 s b. glycine: 1 sp 3 and one sp 2 c. nitrobenzene: 6 sp 2 s d. chloroprene: 4 sp 2 s e. 4-penten-1-yne: 2 sp 2 s, 1 sp 3 , 2 sp s 65. Write the Lewis formulas and predict the hybrid orbitals and the shapes of these polyatomic ions and covalent molecules: (next page) b. BF 3 F sp 2 hybridization, trigonal planar B F F c. SbF 6-F F sp 3 d 2 hybridization, octahedral F Sb F F F...
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This note was uploaded on 04/21/2008 for the course CM 1113 taught by Professor Hutchinson during the Fall '07 term at Lipscomb.

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ch. 8 hw - of the simplest stable hydrocarbon would be CH 2...

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