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F10HW12-CH10 - HOMEWORK WEEK 12 CHAPTER 10 10.2 10.7 10.8...

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HOMEWORK WEEK 12 CHAPTER 10 10.2 The arrangements are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. 10.7 sp 3 hybridization is shown by an atom with 4 groups of electrons, i.e. with a tetrahedral electron arrangement. The angle is 109.5 ° . 10.8 A sigma bond has a cylindrical shape about the bond axis. The maximum electron density is on the bond axis. A pi bond has a distribution of electrons above and below the bond axis. There is no electron density in the pi bond on the bond axis. The pi bond is much more reactive than the sigma bond; for instance, Br 2 reacts with the C=C pi bond but not with the C-C sigma bond. Each sigma bond is mainly localized between the relevant nuclei, but pi bonds often extend over several atoms (“delocalization”). 10.9 In ethylene, C 2 H 4 , the orbital diagram on each carbon atom may be described as follows: C atom (ground state) 1 s 2 s 2 p 1 s C atom (hybridized) 1 s C atom (in C 2 H 4 ) σ bonds π bond 2 p 2 p sp 2 sp 2 Energy An sp 2 hybrid orbital on one carbon atom overlaps a similar hybrid orbital on the other carbon atom to form a σ bond. The remaining hybrid orbitals on the two carbon atoms overlap 1 s orbitals from the hydrogen atoms to form four C-H bonds. The unhybridized 2 p orbital on one carbon atom overlaps the unhybridized 2 p orbital on the other carbon atom to form a π bond. The σ and π bonds together constitute a double bond. The pi bond requires that the overlapping 2p orbitals on the C atoms have their lobes pointing in the same directions. This requires that the region of the molecule around the C atoms (i.e. the whole molecule!) is planar. See Fig. 10.26! 010. The answer is b, H 2 S. This is because this molecule is bent. All the other species are very symmetrical, i.e. linear (CO 2 and BeF 2 ), tetrahedral (CH 4 ) or octahedral (SF 6 ), so that all the bond dipoles cancel.
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10.23 In order to solve this problem, draw the Lewis structure for each of the listed molecules.
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