Lecture Presentation Chapter 9 Molecular Geometry and Bonding
![Lecture Presentation Chapter 9 Molecular Geometry and Bonding Theories 考題詳解 James F. Kirby Quinnipiac Lecture Presentation Chapter 9 Molecular Geometry and Bonding Theories 考題詳解 James F. Kirby Quinnipiac](https://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-1.jpg)
Lecture Presentation Chapter 9 Molecular Geometry and Bonding Theories 考題詳解 James F. Kirby Quinnipiac University Hamden, CT
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![31) The basis of the VSEPR model of molecular bonding is _____. A) hybrid 31) The basis of the VSEPR model of molecular bonding is _____. A) hybrid](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-27.jpg)
31) The basis of the VSEPR model of molecular bonding is _____. A) hybrid orbitals will form as necessary to, as closely as possible, achieve spherical symmetry B) regions of electron density on an atom will organize themselves so as to maximize s-character C) electron domains in the valence shell of an atom will arrange themselves so as to minimize repulsions D) atomic orbitals of the bonding atoms must overlap for a bond to form E) regions of electron density in the valence shell of an atom will arrange themselves so as to maximize overlap
![31) The basis of the VSEPR model of molecular bonding is _____. A) hybrid 31) The basis of the VSEPR model of molecular bonding is _____. A) hybrid](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-28.jpg)
31) The basis of the VSEPR model of molecular bonding is _____. A) hybrid orbitals will form as necessary to, as closely as possible, achieve spherical symmetry B) regions of electron density on an atom will organize themselves so as to maximize s-character C) electron domains in the valence shell of an atom will arrange themselves so as to minimize repulsions D) atomic orbitals of the bonding atoms must overlap for a bond to form E) regions of electron density in the valence shell of an atom will arrange themselves so as to maximize overlap Answer: C 分子結合的VSEPR模型的基礎是_______。 A)混成軌域要形成,是盡可能接近地實現球形對稱性(無此要求) B)原子上的電子密度區域將自身最大化s特性(無此要求) C)原子的價電子殼中的電子域將自身排列,以便最小化排斥(正確 ) D)原子軌域鍵結原子必須重疊以形成鍵(不用重疊 例如反鍵結軌 域) E)電子密度在一個原子的價電子殼中的區域將自己排列以便最大 化重疊(沒有特別說明 ) 課本 9. 1 p 384
![32) The electron-domain geometry of _____ is tetrahedral. A) PH 3 B) CBr 4 32) The electron-domain geometry of _____ is tetrahedral. A) PH 3 B) CBr 4](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-29.jpg)
32) The electron-domain geometry of _____ is tetrahedral. A) PH 3 B) CBr 4 C) Xe. F 4 D) CCl 2 Br 2 E) all of the above except Xe. F 4
![32) The electron-domain geometry of _____ is tetrahedral. A) PH 3 B) CBr 4 32) The electron-domain geometry of _____ is tetrahedral. A) PH 3 B) CBr 4](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-30.jpg)
32) The electron-domain geometry of _____ is tetrahedral. A) PH 3 B) CBr 4 C) Xe. F 4 D) CCl 2 Br 2 E) all of the above except Xe. F 4 Answer: E _____的電子域幾何是四 面體。 A) PH 3 CBr 4 B) C) Xe. F 4 CCl 2 Br 2 D) 課本 9. 2 p 391
![33) The central Xe atom in the Xe. F 4 molecule has _____ unbonded 33) The central Xe atom in the Xe. F 4 molecule has _____ unbonded](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-31.jpg)
33) The central Xe atom in the Xe. F 4 molecule has _____ unbonded electron pairs and _____ bonded electron pairs in its valence shell. A) 4, 0 B) 2, 4 C) 4, 1 D) 1, 4 E) 4, 2
![33) The central Xe atom in the Xe. F 4 molecule has _____ unbonded 33) The central Xe atom in the Xe. F 4 molecule has _____ unbonded](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-32.jpg)
33) The central Xe atom in the Xe. F 4 molecule has _____ unbonded electron pairs and _____ bonded electron pairs in its valence shell. A) 4, 0 B) 2, 4 C) 4, 1 D) 1, 4 E) 4, 2 Answer: B Xe. F 4分子中的中心Xe原子具有未鍵合的電子對和鍵在其 價鍵殼中的電子對。 課本 9. 2 p 388
![34) The molecular geometry of the PF 3 molecule is _____, and this molecule 34) The molecular geometry of the PF 3 molecule is _____, and this molecule](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-33.jpg)
34) The molecular geometry of the PF 3 molecule is _____, and this molecule is _____. A) trigonal planar, polar B) trigonal pyramidal, polar C) trigonal planar, nonpolar D) tetrahedral, unipolar E) trigonal pyramidal, nonpolar
![34) The molecular geometry of the PF 3 molecule is _____, and this molecule 34) The molecular geometry of the PF 3 molecule is _____, and this molecule](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-34.jpg)
34) The molecular geometry of the PF 3 molecule is _____, and this molecule is _____. A) trigonal planar, polar B) trigonal pyramidal, polar C) trigonal planar, nonpolar D) tetrahedral, unipolar E) trigonal pyramidal, nonpolar Answer: B PF 3分子的分子幾何是_____,這個分子 是_____。 A)三角平面,極性 B)三角錐體,極性 C)三角平面,非極性 D)四面體,單極 E)三角錐體,非極性 課本 9. 2 p 388
![35) The bond angles marked a, b, and c in the molecule below are 35) The bond angles marked a, b, and c in the molecule below are](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-35.jpg)
35) The bond angles marked a, b, and c in the molecule below are about __________, and _____, respectively A) 90°, 90° B) 120°, 90° C) 109. 5°, 120°, 109. 5° D) 109. 5°, 90°, 120° E) 120°, 109. 5°
![35) The bond angles marked a, b, and c in the molecule below are 35) The bond angles marked a, b, and c in the molecule below are](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-36.jpg)
35) The bond angles marked a, b, and c in the molecule below are about __________, and _____, respectively A) 90°, 90° B) 120°, 90° C) 109. 5°, 120°, 109. 5° D) 109. 5°, 90°, 120° E) 120°, 109. 5° Answer: C a. N (3個鍵結+一個孤對電子對) sp 4 109. 5 b. C (4個鍵結其一算三鍵) sp 3 120 c. C (4個鍵結 ) sp 4 109. 5 課本 9. 5 p 386
![36) The hybridizations of iodine in IF 3 and IF 5 are _____ and 36) The hybridizations of iodine in IF 3 and IF 5 are _____ and](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-37.jpg)
36) The hybridizations of iodine in IF 3 and IF 5 are _____ and _____, respectively. A) sp 3 d 2, sp 3 d 2 B) sp 3, sp 3 d C) sp 3 d, sp 3 D) sp 3 d 2, sp 3 d E) sp 3 d, sp 3 d 2 Answer: E
![36) The hybridizations of iodine in IF 3 and IF 5 are _____ and 36) The hybridizations of iodine in IF 3 and IF 5 are _____ and](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-38.jpg)
36) The hybridizations of iodine in IF 3 and IF 5 are _____ and _____, respectively. A) sp 3 d 2, sp 3 d 2 B) sp 3, sp 3 d C) sp 3 d, sp 3 D) sp 3 d 2, sp 3 d E) sp 3 d, sp 3 d 2 IF 3和IF 5中的碘的混成分別是_____和_____。 Answer: E (3個鍵結+2組未共用電子對)sp 3 d 2 (5個鍵結+1組未 課本 9. 6 p 402
![37) Based on molecular orbital theory, the bond orders of the H�� H bonds 37) Based on molecular orbital theory, the bond orders of the H�� H bonds](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-39.jpg)
37) Based on molecular orbital theory, the bond orders of the H�� H bonds in H 2, H 2+, and H 2 - are _____, respectively A) 1, 1/2, and 0 B) 1, 1/2, and 1/2 C) 1, 0, and 1/2 D) 1, 0, and 0 E) 1, 2, and 0
![37) Based on molecular orbital theory, the bond orders of the H�� H bonds 37) Based on molecular orbital theory, the bond orders of the H�� H bonds](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-40.jpg)
37) Based on molecular orbital theory, the bond orders of the H�� H bonds in H 2, H 2+, and H 2 - are _____, respectively A) 1, 1/2, and 0 B) 1, 1/2, and 1/2 C) 1, 0, and 1/2 D) 1, 0, and 0 E) 1, 2, and 0 Answer: B 基於分子軌道理論,H 2 +和H 2 - 中的 H – H 鍵的鍵序是分別 是? Bond order = 1/2 (1 -0) = 1/2 Bond order = 1/2 (2 -0) = 1 Bond order = 1/2 (2 -1) = 1/2 課本 9. 7 p 413
![38) A molecule has the formula AB 3 and the central atom is in 38) A molecule has the formula AB 3 and the central atom is in](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-41.jpg)
38) A molecule has the formula AB 3 and the central atom is in a different plane from the surrounding three atoms. Its molecular shape is _____. A) bent B) tetrahedral C) trigonal pyramidal D) T-shaped E) linear
![38) A molecule has the formula AB 3 and the central atom is in 38) A molecule has the formula AB 3 and the central atom is in](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-42.jpg)
38) A molecule has the formula AB 3 and the central atom is in a different plane from the surrounding three atoms. Its molecular shape is _____. A) bent B) tetrahedral C) trigonal pyramidal D) T-shaped E) linear Answer: C 分子具有式AB 3,並且中心原子在與周圍三個原 子不同的平面中。其分子形狀為_____。 課本 9. 2 p 386
![39) PCl 5 has _____ electron domains and a _____ molecular arrangement. A) 5, 39) PCl 5 has _____ electron domains and a _____ molecular arrangement. A) 5,](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-43.jpg)
39) PCl 5 has _____ electron domains and a _____ molecular arrangement. A) 5, square pyramidal B) 6, trigonal bipyramidal C) 6, tetrahedral D) 5, trigonal bipyramidal E) 6, seesaw
![39) PCl 5 has _____ electron domains and a _____ molecular arrangement. A) 5, 39) PCl 5 has _____ electron domains and a _____ molecular arrangement. A) 5,](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-44.jpg)
39) PCl 5 has _____ electron domains and a _____ molecular arrangement. A) 5, square pyramidal B) 6, trigonal bipyramidal C) 6, tetrahedral D) 5, trigonal bipyramidal E) 6, seesaw Answer: D PCl 5具有_____電子和______分子排列。 P和5個氯共用電子, 構型trigonal bipyramidal A) 5,正四棱錐 B) 6,三角雙錐體 C) 6,四面體 D) 5,三角雙錐體 E) 6,蹺蹺板 課本 9. 2 p 386
![40) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of 40) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-45.jpg)
40) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of these, _____ is/are polar. A) SF 2 only B) SF 4 only C) SF 2, SF 4, and SF 6 D) SF 6 only E) SF 2 and SF 4 only
![40) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of 40) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of](http://slidetodoc.com/presentation_image/54bcae3f882ea4f6a5b837580131a4e3/image-46.jpg)
40) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of these, _____ is/are polar. A) SF 2 only B) SF 4 only C) SF 2, SF 4, and SF 6 觀察到三種單硫化物:SF 2,SF 4和SF 6。 其中, D) SF 6 only E) SF 2 and SF 4 only _____是極性的。 Answer: E A) 僅SF 2 B) 僅SF 4 C) SF 2,SF 4和SF 6 D) 僅SF 6 E) 僅SF 2和SF 4 SF 6都抵銷掉沒有 極性 課本 9. 3 p 395
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