Chapter10ISM

Chapter10ISM - CHAPTER 10 CHEMICAL BONDING II: MOLECULAR...

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CHAPTER 10 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS Problem Categories Biological : 10.88, 10.103. 10.112. Conceptual : 10.21, 10.22, 10.23, 10.24, 10.41, 10.49, 10.50, 10.53, 10.57, 10.63, 10.64, 10.65, 10.66, 10.67, 10.68, 10.71, 10.72, 10.75, 10.77, 10.79, 10.82, 10.84, 10.85, 10.87, 10.90, 10.94, 10.96, 10.97, 10.99, 10.101, 10.105. Descriptive : 10.19, 10.20. Environmental : 10.86, 10.100. Organic : 10.23, 10.24, 10.38, 10.41, 10.43, 10.44, 10.63, 10.64, 10.65, 10.83, 10.84, 10.85, 10.92, 10.102, 10.109, 10.110. Difficulty Level Easy : 10.19, 10.20, 10.23, 10.40, 10.42, 10.50, 10.52, 10.53, 10.64, 10.78, 10.89, 10.91. Medium : 10.7, 10.8, 10.9, 10.10, 10.11, 10.13, 10.14, 10.22, 10.24, 10.31, 10.32, 10.33, 10.34, 10.35, 10.36, 10.37, 10.38, 10.41, 10.42, 10.49, 10.51, 10.54, 10.55, 10.57, 10.58, 10.59, 10.60, 10.65, 10.67, 10.68, 10.69, 10.70, 10.72, 10.73, 10.75, 10.77, 10.80, 10.81, 10.82, 10.84, 10.87, 10.88, 10.90, 10.92, 10.93, 10.94, 10.95, 10.96, 10.98, 10.99, 10.100, 10.101, 10.103. Difficult : 10.12, 10.21, 10.39, 10.43, 10.44, 10.56, 10.63, 10.66, 10.71, 10.74, 10.76, 10.79, 10.83, 10.85, 10.86, 10.97, 10.102, 10.104, 10.105, 10.106, 10.107, 10.108, 10.109, 10.110, 10.111, 10.112, 10.113, 10.114, 10.115. 10.7 (a) The Lewis structure of PCl 3 is shown below. Since in the VSEPR method the number of bonding pairs and lone pairs of electrons around the central atom (phosphorus, in this case) is important in determining the structure, the lone pairs of electrons around the chlorine atoms have been omitted for simplicity. There are three bonds and one lone electron pair around the central atom, phosphorus, which makes this an AB 3 E case. The information in Table 10.2 shows that the structure is a trigonal pyramid like ammonia. What would be the structure of the molecule if there were no lone pairs and only three bonds? (b) The Lewis structure of CHCl 3 is shown below. There are four bonds and no lone pairs around carbon which makes this an AB 4 case. The molecule should be tetrahedral like methane (Table 10.1). P Cl C H
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CHAPTER 10: CHEMICAL BONDING II 270 (c) The Lewis structure of SiH 4 is shown below. Like part (b), it is a tetrahedral AB 4 molecule. (d) The Lewis structure of TeCl 4 is shown below. There are four bonds and one lone pair which make this an AB 4 E case. Consulting Table 10.2 shows that the structure should be that of a distorted tetrahedron like SF 4 . Are TeCl 4 and SF 4 isoelectronic? Should isoelectronic molecules have similar VSEPR structures? 10.8 Strategy: The sequence of steps in determining molecular geometry is as follows: draw Lewis ⎯⎯→ find arrangement of find arrangement determine geometry structure electrons pairs of bonding pairs based on bonding pairs Solution: Lewis structure Electron pairs Electron Lone pairs Geometry on central atom arrangement (a) 3 trigonal planar 0 trigonal planar, AB 3 (b) 2 linear 0 linear, AB 2 (c) 4 tetrahedral 0 tetrahedral, AB 4 10.9 Lewis Structure e pair arrangement geometry (a) C Br Br Br Br tetrahedral tetrahedral (b) B Cl Cl Cl trigonal planar trigonal planar Si H H H H Te Cl Al Zn Cl Zn 2
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CHAPTER 10: CHEMICAL BONDING II 271 (c) N F F F tetrahedral trigonal pyramidal (d) Se HH tetrahedral bent (e) NO O trigonal planar bent 10.10 We use the following sequence of steps to determine the geometry of the molecules. draw Lewis ⎯⎯→ find arrangement of find arrangement determine geometry structure electrons pairs of bonding pairs based on bonding pairs (a) Looking at the Lewis structure we find 4 pairs of electrons around the central atom. The electron pair arrangement is tetrahedral. Since there are no lone pairs on the central atom, the geometry is also tetrahedral .
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Chapter10ISM - CHAPTER 10 CHEMICAL BONDING II: MOLECULAR...

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