Chapter 9 - University of Dayton Academic Webserver

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$MATH 219 - Differential Equations Quiz 3 Name Solutions Date ...$

294 .. . F.. . . . O. . Chapter 9: Molecular Structures . . . F S . . . F. . . . F. 120¡ F O 90¡ F S F (c) BrF 5 (42 e – ) The type is AX 5E 1, so the electron-pair geometry is octahedral and all .. . .. . . . . F. . . . . . . . . . .. .. . F the F–Br–F angles are 90°. F F F F F F Br . . F F Br F 24. Adapt the method given in the answers to Question 22. (a) SF 6 (48 e – ) The type is AX 6E 0, so the electron-pair geometry is octahedral and all . . . . . .. . .. F .. . . S . . F. . .. .. . . . . . . F F F the F–S–F angles are 90°. F F F F F (b) XeF 2 (22 e – ) The type is AX 2E 3, so the electron-pair geometry is triangular . . . F . .. . . Xe . .. F . . S F F 90¡ 90¡ bipyramidal, and the F–Xe–F angle is 180°. (c) ClF 2 – (22 e – ) The type is AX 2E 3, so the electron-pair geometry is triangular . . . F . .. .. . Cl .. F . . F Xe F 180¡ bipyramidal, and the F–Cl–F angle is 180°. 25. NO 2 molecule and NO 2 – ion differ by only one electron. NO2 has 17 electrons and NO 2 – has 18 electrons. Their Lewis structures are quite similar: .. . .. .. .. . F Cl F 180¡ . . . . . . . . . . O N O O N O .
Chapter 9: Molecular Structures 295 According to VESPR, both of these are triangular planar (AX 2E 1), where the O–N–O angle is approximately 120°. However, the single unpaired electron in NO 2 molecule is not as good at repelling other electrons as a full pair would be. When a nonbonded pair of electrons is repelling the bonding pairs, it pushes them closer together making a narrower angle. Hence, we will predict that the O–N–O angle in NO 2 – ion is slightly smaller than the angle in NO 2 molecule. 26. ClF 2 + cation and ClF2 – anion differ by two electrons. ClF2 + has 20 electrons and ClF2 – has 22 electrons. Use the VSEPR model on each of them. ClF 2 + cation has type AX2E 2..The electron-pair geometry is tetrahedral, and the F–Cl–F angle is 109.5° as shown below: . . . . F . . . .. .. . Cl F + F Cl 109.5¡ ClF 2 – anion has type AX2E 3. The electron-pair geometry is triangular bipyramidal, and the F–Cl–F angle is 180° as shown below: . . . F . .. .. . Cl .. F . . According to the VESPR model, we predict that the F–Cl–F angle in a ClF 2 – anion is a greater angle than the angle in a ClF 2 + cation. Hybridization and Multiple Bonds 27. (a) One s and three p orbitals combine to make sp 3 hybrid orbitals. (b) One s, three p, and two d orbitals combine to make sp 3 d 2 hybrid orbitals. (c) One s and two p orbitals combine to make sp 2 hybrid orbitals. 28. (a) When four atomic orbitals (one s and three p orbitals) combine to make hybrid orbitals, four sp 3 hybrid orbitals are formed. (b) When six atomic orbitals (one s, three p, and two d orbitals) combine to make hybrid orbitals, six sp 3 d 2 hybrid orbitals are formed. (c) When three atomic orbitals (one s and two p orbitals) combine to make hybrid orbitals, three sp 2 hybrid orbitals are formed. 29. Write a Lewis structure for HCCl 3 and use VSEPR to determine the molecular geometry as described in the answer to Question 13. Use Table 9.2 to determine the hybridization of the central atom using the electronpair geometry. The molecule is AX 4E 0 type, so its electron-pair geometry and its molecular geometry are tetrahedral. F Cl F F 180¡
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