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# chapter09 - 402 Chapter 9 Molecular Structures...

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Chapter 9: Molecular Structures 402 End-of-Chapter Solutions for Chapter 9 Summary Problem Part 1 Answer: (a) see structure below; tetrahedral, angular, 109.5°, sp 3 (b) see structure below; octahedral, square planar, 90°, sp 3 d 2 (c) see structure below; octahedral, octahedral, 90°, sp 3 d 2 (d) see structure below; octahedral, square planar, 90°, sp 3 d 2 Strategy and Explanation: Write Lewis structures for a list of formulas and identify their electron geometry, the shape, and the hybridization of the central atom. Follow the systematic plan for Lewis structures given in the solution to Question 8.13, then determine the number of bonded atoms and lone pairs on the central atom, determine the designated type (AX n E m ) and use Table 9.1. Notice: It is not important to expand the octet of a central atom solely for the purposes of lowering its formal charge, because the shape of the molecule will not change. Hence, we will write Lewis structures that follow the octet rule unless the atom needs more than eight electrons. Use the electron geometry to determine the number of electron pairs around the central atom. Select a hybridization that accommodates the appropriate number of electron pairs. Use Tables 9.2 or 9.3 for reference. (a) BrF 2 (20 e ) The type is AX 2 E 2 , so the electron-pair geometry is tetrahedral and the molecular geometry is angular. The bond angle is slightly less than 109.5°. The Br atom must be sp 3 -hybridized, according to Table 9.2, to accommodate the two bonds and two lone pairs. F Br F . F Br F (b) BrF 4 (36 e ) The type is AX 4 E 2 , so the electron-pair geometry is octahedral and the molecular geometry is square planar. The bond angles are 90°. The Br atom The P atom must be sp 3 d 2 -hybridized, according to Table 9.3, to accommodate the four bonds and two lone pairs. . . .. .. . .. .. . . . . . . . . . . . . . . . F Br F F F F Br F F F . . (c) BrF 6 + (48 e ) The type is AX 6 E 0 , so electron-pair geometry and the molecular geometry is octahedral. The bond angles are 90°. The Br atom The P atom must be sp 3 d 2 - hybridized, according to Table 9.3, to accommodate the six bonds. . . .. .. .. .. . . . . . . . . . . . . . . F Br F F F F Br F F F F F F F (d) IBrCl 3 (36 e ) The type is AX 4 E 2 , so the electron-pair geometry is octahedral and the molecular geometry is square planar. The bond angles are 90°. The Br atom The P atom must be sp 3 d 2 -hybridized, according to Table 9.3, to accommodate the four bonds and two lone pairs. . . .. .. . .. .. . . . . . . . . . . . . . . . Cl I Br Cl Cl Cl I Cl Br Cl . .

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Chapter 9: Molecular Structures 403 Part 2 Answer: (a) C 14 H 18 O 5 N 2 (b) sp 3 (c) sp 3 (d) sp 3 (e) sp 2 (f) See structure below Strategy and Explanation: (a) To get the formula, count the number of each type of atom. Don’t forget the six carbon atoms and five hydrogen atoms in the ring. H N H C C N C C O C H H C H H C O O H O H C H C H H O C C C C C H H H H H H H .. .. . . . . . . . . . . . . . .. The formula is C 14 H 18 O 5 N 2 .
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chapter09 - 402 Chapter 9 Molecular Structures...

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