Name: ________________________ Class: ___________________ Date: __________ID: A AP Chapter 9 Study Questions True/False Indicate whether the statement is true or false. 1. Possible shapes of AB3 molecules are linear, trigonal planar, and T-shaped. 5. XeF4 is a polar molecule. 6. Hybridization is the process of mixing atomic orbitals as atoms approach each other to form a bond. 2. Boron trifluoride has three bonding domains and its electron domain geometry is trigonal planar. 3. Electron domains for single bonds exert greater force on adjacent domains than the electron domains for multiple bonds. 7. Electrons in core orbitals contribute to atom bonding. 8. Nitrogen is colorless because the minimum energy to excite an electron is in the ultraviolet section of the spectrum. 4. The quantitative amount of charge separation in a diatomic molecule contributes to the dipole moment of that molecule. Multiple Choice Identify the choice that best completes the statement or answers the question. 9. For a molecule with the formula AB2 the molecular shape is __________. a. linear or bent b. linear or trigonal planar c. linear or T-shaped d. T-shaped e. trigonal planar 12. The electron-domain geometry and molecular geometry of iodine trichloride are __________ and __________, respectively. a. trigonal bipyramidal, trigonal planar b. tetrahedral, trigonal pyramidal c. trigonal bipyramidal, T-shaped d. octahedral, trigonal planar e. T-shaped, trigonal planar 10. According to VSEPR theory, if there are five electron domains in the valence shell of an atom, they will be arranged in a(n) __________ geometry. a. octahedral b. linear c. tetrahedral d. trigonal planar e. trigonal bipyramidal 13. The molecular geometry of __________ is square planar. a. CCl4 b. XeF4 c. PH3 d. XeF2 e. ICl3 14. The molecular geometry of the CS2 molecule is __________. a. linear b. bent c. tetrahedral d. trigonal planar e. T-shaped 11. According to VSEPR theory, if there are four electron domains in the valence shell of an atom, they will be arranged in a(n) __________ geometry. a. octahedral b. linear c. tetrahedral d. trigonal planar e. trigonal bipyramidal 15. The molecular geometry of the SiH2Cl2 molecule is __________. a. trigonal planar b. tetrahedral c. trigonal pyramidal d. octahedral e. T-shaped 1 a. which orbitals on bromine atoms overlap in the formation of the bond in Br2? a. The molecular geometry of the H3O+ ion is __________. one d. linear b.ion is approximately __________.5° c. The electron-domain geometry of the AsF6. The F-B-F bond angle in the BF3 molecule is __________. two b. sp b. tetrahedral 20. 120° d. trigonal planar d. The O-S-O bond angle in SO2 is slightly less than __________. The F-N-F bond angle in the NF3 molecule is slightly less than __________. The F-B-F bond angle in the BF2. 180° e. __________ s atomic orbital(s) and __________ p atomic orbital(s) must be mixed. trigonal planar e. tetrahedral c. 109. sp2d2 b. The hybrid orbitals used by the As atom for bonding are __________ orbitals. a. 90° d. bent d. The angles between sp2 orbitals are __________. 109.5° c.5° c. 90° b. 60° 31. sp3 d. bent b. two. trigonal bipyramidal e. sp3d d. 60° 25. 3p c. The hybridization of the carbon atom in carbon dioxide is __________. 109. trigonal planar d. three c. sp3d2 21. The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal. 180° e. sp3 c. two e. T-shaped 26. bent c. 4p e. a. octahedral 34. linear c. octahedral 28. sp2 23. tetrahedral 19. sp b. a. 60° 30. trigonal planar c. a.5° c. The hybridization of the central nitrogen atom is __________. 4s d. 120° 2 . 120° d. trigonal pyramidal d. tetrahedral e. trigonal bipyramidal 29. sp3d e. a. T-shaped 27. 90° b. sp b. sp2 c. According to valence bond theory. two. sp3d2 e.5° e. 90° b. one. a. The hybridization of orbitals on the central atom in a molecule is sp2. a. one. linear c. 3d 17. bent b. 180° e. sp3d e. a. a. trigonal pyramidal e. 180° e. sp3 d.ion is octahedral. a. 45° b. The molecular geometry of the PHCl2 molecule is __________. 109. 109. 180° c. 180° e. 90° b. a. The Cl-Si-Cl bond angle in the SiCl2F2 molecule is approximately __________. trigonal planar d. one.Name: ________________________ ID: A 16. The electron-domain geometry around this central atom is __________. The molecular geometry of the SF2 molecule is __________. a. In order to produce sp3 hybrid orbitals. The electron-domain geometry about this central atom is __________. sp3d e. trigonal pyramidal d. The hybridization of the central atom in the XeF4 molecule is __________. octahedral b. octahedral b.5° c. a. three 24. sp3 d. 3s b. sp2 c. 109. 60° 32. a. octahedral b. linear b. sp2 c. trigonal bipyramidal e. a. 60° 33. 90° b. trigonal pyramidal d. The molecular geometry of the PF4+ ion is __________. a. tetrahedral e. The molecular geometry of the CHCl3 molecule is __________. tetrahedral c. a. 120° d. trigonal planar c. 120° d. 120° d. tetrahedral e. sp3d2 22. The hybridization of orbitals on the central atom in a molecule is sp. sp3d2 18. 4 3 . a. 6. 0 37. sp2. The Lewis structure of carbon monoxide is given below. According to VSEPR theory. hybrid orbitals will form as necessary to. octahedral b. a. sp2 d. a. There is/are __________ π bond(s) in the molecule below. regions of electron density on an atom will organize themselves so as to maximize s-character b. 1 c. 4 and 3 d. 1 e. 18 39. sp3. sp e. electron domains in the valence shell of an atom will arrange themselves so as to minimize repulsions e. 2 a. There is/are __________ σ bond(s) in the molecule below. 4 e. sp2. 7 b. 36. 3 and 2 b. sp3 b. :C≡O: a. 13 e. 2 d. The total number of π bonds in the H–C≡C–C≡C–C≡N molecule is __________. 6 d. 6. 2 c. if there are three electron domains in the valence shell of an atom. a. 2 and 3 e. 9 e. There is/are __________ π bond(s) in the molecule below. a. achieve spherical symmetry a. tetrahedral d. they will be arranged in a(n) __________ geometry. 16 44. as closely as possible. 3 b. atomic orbitals of the bonding atoms must overlap for a bond to form d. 2. The hybridizations of the carbon and oxygen atoms in carbon monoxide are __________ and __________. There are __________ σ and __________ π bonds in the H–C≡C–H molecule. sp3 c. 6 e. 38. regions of electron density in the valence shell of an atom will arrange themselves so as to maximize overlap c. 12 41. 2 d. 4 c. trigonal planar e.Name: ________________________ ID: A 35. 0 b. linear c. ClF3 has "T-shaped" geometry. 0 b. 5 and 0 40. 3 e. 1 c. respectively. There are __________ non-bonding domains in this molecule. 6 c. 2 d. The basis of the VSEPR model of molecular bonding is __________. 4 c. There are __________ σ and __________ π bonds in the H2C=C=CH2 molecule. 43. sp. 1 b. 2. 2 b. a. 12 d. 4. trigonal bipyramidal a. sp2 42. sp. 3 and 4 c. 2 d. trigonal planar c. 90°. NCl3 d. trigonal planar. 120°. tetrahedral. a. a. 109. In counting the electron domains around the central atom in VSEPR theory. 109. a __________ is not included. double covalent bond e. 50.5° d. The O-C-O bond angle in the CO32.5°. respectively. The electron domain and molecular geometry of BrO2. 109. tetrahedral. core level electron pair d. 109. 109. seesaw 46. The molecular geometry of the BrO3. 47. All of these will have bond angles of 120°. __________ will have bond angles of 120°. 109. 90° e. __________. 120°. and c in the molecule below are about __________.5° e. ClF3 c. a. 109. all of the above except XeF4 48. a.5°.is _________. 120° d. 90°. tetrahedral e. CCl2Br2 d. octahedral e. T-shaped 51. 120° d.5°. nonbonding pair of electrons b. trigonal pyramidal. BCl3 e.5° c. single covalent bond c.ion is approximately __________. bent e. b.5°. 60° a. trigonal planar b. T-shaped 4 . 180° e. b.5°. 120°. a. 120°. The molecular geometry of the right-most carbon in the molecule below is __________. CBr4 b. Of the following species.5°. respectively. a. The electron-domain geometry of __________ is tetrahedral. 109. tetrahedral d. T-shaped 53. 90°. trigonal planar b. 109. 120°. and c in the molecule below are about __________. 109. 109. The bond angles marked a. 45. 90° b. 90° b. 120° 49. XeF4 e. trigonal pyramidal. The bond angles marked a. The molecular geometry of the left-most carbon atom in the molecule below is __________. octahedral e. triple covalent bond a. trigonal planar b. 180°. 109.5°. a.5° b.Name: ________________________ ID: A 52. 90°. trigonal bipyramidal c. PH3 c. 120° c. tetrahedral d. PH3 b. bent d.5°. and __________. trigonal planar c. 109.ion is __________. trigonal bipyramidal c. __________. trigonal pyramidal b. 90° c. 120°. linear d. 54. 90° a. and __________.5°. 6. bent 59. For molecules of the general formula ABn. not related e. 360° e. 180° e. 0 d. The central Xe atom in the XeF4 molecule has __________ unbonded electron pairs and __________ bonded electron pairs in its valence shell. 5. a. 5. 5 b. 0. 2 b. 180° d. According to VSEPR theory. never the same b. a. 2. 1 d. a. 90° b. 1. 120 b. only when A is carbon e. the octet rule is obeyed 57. The bond angle marked a in the following molecule is about __________. 60° 56. and c e. 4. a. 120° c. there are no lone pairs on the central atom b. only when A is boron or beryllium d. for any element A b.5° c. T-shaped e. they will be arranged such that the angles between the domains are __________. b and c 60. 6. 3. b only c. 3. a. there is more than one central atom c. A molecule has the formula AB3 and the central atom is in a different plane from the surrounding three atoms. 5. 109. 180° e. trigonal bipyramidal e. 4 c. 4 b.5° d. 360° b. 4 65. n is greater than four d.5° d. 109. a) a nonbonding pair of electrons b) a single bond c) a multiple bond 67. The central iodine atom in the ICl4. seesaw a. a. 1. 2 66. 1. 4. The bond angles in a trigonal planar molecule are __________ degrees. a. only when A is an element from the third period or below the third period c. trigonal pyramidal c. c only d. n is less than four e. a. 90° 5 . 180° c. n can be greater than four __________. 120° d. 4 c. Its molecular shape is __________. tetrahedral b. 4.5° c. 1 e. 90° 55. According to VSEPR theory. 3 d. if there are two electron domains on a central atom. 109. 2. a only b. The electron-domain geometry and the molecular geometry of a molecule of the general formula ABn are __________. 360° e. a. b. 45 e. trigonal bipyramidal b. if there are three electron domains on a central atom. 63. The central iodine atom in IF5 has __________ unbonded electron pairs and __________ bonded electron pairs in its valence shell. 4. 120° b. only when A is Xe 61. 90 d. a. sometimes the same d. 109. mirror images of one another a. An electron domain consists of __________. square pyramidal d. a. 4 64. they will be arranged such that the angles between the domains are __________. < 45 58. 6. 1 e. always the same c.Name: ________________________ ID: A 62. a. they will be arranged such that the angles between the domains are __________. linear d. 2 e. 5 c. a.5 c. 90° b.ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its valence shell. 109. if there are four electron domains on a central atom. tetrahedral c. 2. The electron-domain geometry and the molecular geometry of a molecule of the general formula ABn will always be the same if __________. 120° 68. PCl5 has __________ electron domains and a __________ molecular arrangement. According to VSEPR theory. 1. Consider the following species when answering the following questions: (i) PCl3 (ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SF6 For which of the molecules is the molecular geometry (shape) the same as the VSEPR electron domain arrangement (electron domain geometry)? a. only __________ is polar. (v) only 70. a. unipolar 82. trigonal pyramidal. NCl3 d. Of the molecules below. (v) 6 . Cl2 72. polar 75. (i) b. only __________ is polar. trigonal planar. polar d. SF2. bent. I2 d. Of the following molecules. and the molecule is __________. seesaw. Of the molecules below. a. linear. a. Of the following molecules. and this molecule is __________. polar b. BF3 c. a. SF4. SeF4 d. (ii) and (v) d. trigonal planar. Of these. HCl e. Cl2 71. tetrahedral. polar 76. 2 c. SF6 e. The molecular geometry of the BeCl2 molecule is __________. trigonal pyramidal. polar e. IF3 d. CH4 c. 1s on H and 4p on Br b. (iii) d. For molecules with only one central atom. nonpolar b. only __________ is nonpolar. (i) and (iii) c. (i) and (ii) b. a. trigonal pyramidal. 2s on H and 3p on Br 83. a. polar b. how many lone pairs on the central atom guarantees molecular polarity? a. CO2 b. The molecular geometry of the CHF3 molecule is __________. polar b. only __________ is polar. SF4. polar e. The molecular geometry of the BCl3 molecule is __________. NF3 c. SbF5 b. SF6 only e. According to valence bond theory. only __________ is polar. BeCl2 b. CCl4 b. a. nonpolar e. trigonal pyramidal. Three monosulfur fluorides are observed: SF2. (ii) c. 2s on H and 4p on Br e. PBr3 e. seesaw. SF2 only b. nonpolar c. (iv) and (v) e. SF2 and SF4 only c. The molecular geometry of the PF3 molecule is __________. Consider the following species when answering the following questions: (i) PCl3 (ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SF6 Which of the molecules has a see-saw shape? a. and SF6 81. trigonal planar. CH4 77. nonpolar d. 1s on H and 3p on Br d. nonpolar d. BrCl3 80. BeCl2 e. polar d. trigonal planar. H2O c. CCl4 b. nonpolar e. Of the molecules below. linear. tetrahedral. (iv) e. Of the molecules below. a. and this molecule is __________. only __________ is nonpolar. SiCl4 79. TeCl2 78. 1s on H and 4s on Br c. trigonal planar. AsH3 c. and SF6. SF4 only d. SiH2Cl2 e. nonpolar c. nonpolar c. trigonal pyramidal. polar 74. trigonal bipyramidal. BF3 b. a. and this molecule is __________. 1 or 3 73. CBr4 d. 1 or 2 d. a. bent. a.Name: ________________________ ID: A 69. BCl3 c. 3 e. tetrahedral. NH3 d. 1 b. which orbitals overlap in the formation of the bond in HBr? a. polar c. __________ is/are polar. a. a. sp3 d. sp3d d. __________ hybrid orbitals are used for bonding by Xe in the XeF4 molecule. sp3d. 90° 91. 4 e. sp3 e. The hybridization of the central carbon atom is __________. CO32. 3 c. sp3 c. The hybrid orbitals used for bonding by the sulfur atom in the SF4 molecule are __________ orbitals. a. sp3. sp2 c. respectively. 1 b. a. sp3 d. sp3d b. sp3d2. sp3d2 c. The hybrid orbitals used for bonding by Xe in the unstable XeF2 molecule are __________ orbitals. The electron-domain geometry of a carbon-centered compound is tetrahedral. 0 85. respectively. 2 b. 4 d. sp3. sp3d d. a. sp2. The hybridizations of nitrogen in NF3 and NH3 are __________ and __________. sp b. a. 2 c. 6 89. sp3d b. Of the following. The combination of two atomic orbitals results in the formation of __________ molecular orbitals. sp3d2 84. PH3 d. The sp3d2 atomic hybrid orbital set accommodates __________ electron domains. 5 e. PH3 b. PF5 94. sp3d e. sp2. 3 c. sp2d2 b. sp2 b. sp d. sp. sp2 98. sp2 c. The hybrid orbitals used by the As atom for bonding are __________ orbitals. sp3 d. sp3d2. a. sp 7 . sp2. sp3d e. The hybridization scheme for BeF2 is __________. The sp2 atomic hybrid orbital set accommodates __________ electron domains. 88. sp3 c. I3. sp2 b. sp3d2 97. sp b.5° c. sp3d. sp3d2 e. a. a. sp3d2. a. The hybridizations of iodine in IF3 and IF5 are __________ and __________. 3 d. sp 87. the central atom is sp3d2 hybridized only in __________. sp3 c. The C-O-H bond angle is __________. 5 e. Of the following. sp2 b. PCl5 b. 6 90.Name: ________________________ ID: A 92. 109. sp3d2 86.5° d. The electron-domain geometry of the AsF5 molecule is trigonal bipyramidal. sp3 c. 109. sp3d2 c. sp3d2 96.c. a. sp3d e. sp3. ICl3 d. a. sp3 a. sp3d e. sp3 d. sp3. 90° e. sp3 d. sp3d. sp3d2 d. sp3d e. a. sp b.e. sp3d2 e. sp. sp. a. sp3. BeF2 95. sp3d2.e. 4 d. sp3d e. sp2 c. Br3. 2 b. The hybridizations of bromine in BrF5 and of arsenic in AsF5 are __________ and __________. sp3d2 93. respectively. XeF4 c. only __________ has sp2 hybridization of the central atom. sp3d. The hybridization of the oxygen atom labeled y in the structure below is __________. 180° b. sp3d2. 4 109. 1 c. SO2 and SO3 d. (iii) only c. three sp3 hybrid orbitals d. There are __________ σ bonds and __________ π bonds in H3C-CH2-CH=CH-CH2-C≡CH. a. 104. A triatomic molecule cannot be linear if the hybridization of the central atoms is __________.d. a molecule must have __________. N3. 16. is stronger and shorter than a single bond b. (i) and (v) d. excited states of molecules b. "localized" bonding electrons are associated with __________. Which of the following molecules or ions will exhibit delocalized bonding? 103. two particular atoms c. a. 10. trigonal planar electron domain geometry 106. 3 e. sp2d or sp3d2 111. all of the π bonds in the molecule e. at least four atoms e. (i) and (ii) b. at least two π bonds b.b. at least three σ bonds d. and N3. a. one particular atom b. a. None of the above will exhibit delocalized bonding. consists of three shared electrons c. consists of one σ bond and two π bonds b. In a polyatomic molecule. NO2. a. five 110. is longer than a single bond 100. consists of one σ bond and one π bond c. NO2. Which of the following molecules or ions will exhibit delocalized bonding? 105. how many hybrid orbitals are formed? a. molecular shape c. six c. 13. consists of two π bonds and one π bond d. consists of two shared electron pairs e.and N3112. a. 3 c. two or more σ bonds in the molecule 101. SO2. and (v) e. sp2 c.e. (iii) only c.only e. 2 e. In order to exhibit delocalized π bonding. 2 d. a. hybridization e. (iv). imparts rigidity to a molecule d. at least two resonance structures c. 0 b. SO3. (v) only 108.only c. The blending of one s atomic orbital and two p atomic orbitals produces __________. all of the atoms in the molecule d. SO3 and SO32. (iv) and (v) e. NH4+ and N3. NH4+. (i) and (ii) b. 8 . covalent bonding d. four e. three sp hybrid orbitals b.Name: ________________________ ID: A Consider the following species when answering the following questions: (i) PCl3 (ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SF6 99. (iii). a. a. When three atomic orbitals are mixed to form hybrid orbitals. two sp3 hybrid orbitals e. 14. (iii) and (iv) d.b. 3 102. sp b. and SO32. (v) only 107. one b. All of the above answers are correct. SO32. In which of the molecules is the central atom hybridized? a. three sp2 hybrid orbitals SO2 SO3 SO32 a. two sp2 hybrid orbitals c. A typical triple bond __________. sp2 or sp3 e. There are __________ unhybridized p atomic orbitals in an sp-hybridized carbon atom. A typical double bond __________.only c. In which of the molecules does the central atom utilize d orbitals to form hybrid orbitals? a. multiple bonds NO2 NH4+ N3– a. 2 b. NO2-. 2 d. consists of six shared electron pairs e. three d. 12. sp3 d. Valence bond theory does not address the issue of __________. The hybridization of the carbon atom in carbon dioxide is __________. cones e. 117. The carbon-carbon σ bond in ethylene. sp hybrid orbitals b. p atomic orbitals 116. protein b. sp2 e. sp3d2 123. hybrid or atomic 114. sp b. a. three ionic bonds a. a. sigma 119. sp3 124. sp3 d. The π bond in ethylene.Name: ________________________ ID: A 120. Structural changes around a double bond in the __________ portion of the rhodopsin molecule trigger the chemical reactions that result in vision. H2C=CH2. The hybridization of the carbon atom labeled x in the molecule below is __________. sp3d2 122. one σ bond and two π bonds c. 113. sp b. 115. sigma. sp3d e. pi c. opsin c. pi. hybrid or atomic e. sp3 d. sp2 hybrid orbitals d. The hybridization of nitrogen in the H–C≡N: molecule is __________. The N-N bond in HNNH consists of __________. ionic. sp2 hybrid orbitals e. one σ bond and no π bonds a. s2p c. sp3 hybrid orbitals b. sp hybrid orbitals d. sp2 c. The hybridization of the oxygen atom labeled x in the structure below is __________. results from the overlap of __________. hybrid c. sp3d e. results from the overlap of __________. a. sp b. two σ bonds and one π bond d. sp2 c. one σ bond and one π bond b. a. hybrid. sp2 c. a. sigma. retinal d. a. three sigma bonds b. two sigma and one pi bond e. sigma b. a. one sigma and two pi bonds d. sp2d2 118. p atomic orbitals a. the σ bond results from overlap of __________ orbitals and the π bond(s) result from overlap of __________ orbitals. sp3d2 121. sp b. s3p d. Electrons in __________ bonds remain localized between two atoms. sp3d e. rods 9 . two σ bonds and two π bonds e. In a typical multiple bond. H2C=CH2. sp d. pi d. hybrid. s atomic orbitals e. A typical triple bond consists of __________. a. The hybridization of the terminal carbons in the H2C=C=CH2 molecule is __________. s atomic orbitals c. hybrid. The Lewis structure of carbon dioxide is given below. three pi bonds c. atomic. pi. sp2d e. hybrid or atomic. Electrons in __________ bonds can become delocalized between more than two atoms. sp3 hybrid orbitals c. a. hybrid d. sp3 d. sigma e. atomic b. sp3 b. sp2 c. Based on molecular orbital theory. and the __________ the bond energy. 0 b. 0. all electrons in the MO electron configuration of F2 are paired. c. there are two unpaired electrons in the MO electron configuration of O2 e. 0 b. F2 d. Li2 129. Based on molecular orbital theory. Molecular Orbital theory correctly predicts diamagnetism of fluorine gas. the F–F bond enthalpy is very low 127. respectively a. 1. and 0 b. 1. 2 d. the __________ the bond length. 0 b. smaller. a.are __________. a. C2 b. 3 e. 3 c. An antibonding π orbital contains a maximum of __________ electrons. 5 131. a. twelve 10 . empty. N2 c. 2 c. a. a. 3/2 e. b. 2 d. Based on molecular orbital theory. a. O2. 2. greater. one bonding molecular orbital and one antibonding molecular orbital 138. H2+. Molecular Orbital theory correctly predicts paramagnetism of oxygen gas. and 0 135. In molecular orbital theory. a. the bond order in F2 can be shown to be equal to 1. 3 e. smaller e. 1 d. greater. the σ1s orbital is __________ and the σ1s* orbital is __________ in the H2 molecule. d. two bonding molecular orbitals and one antibonding molecular orbital e. a. filled. and H2. b. 2 d. In comparing the same two atoms bonded together. filled. the __________ the bond order. 1 b. empty c. the only molecule in the list below that has unpaired electrons is __________. 1/2 133. there are more electrons in the bonding orbitals than in the antibonding orbitals. six e. the O–O bond distance is relatively short 126. 0. A molecular orbital can accommodate a maximum of __________ electron(s). the energy of the π2p MOs is higher than that of the σ2p MO e. 0 b. a. 2 136. half-filled. 1. This is because __________. the bond orders of the H–H bonds in H2. greater b. Based on molecular orbital theory. 1 c. 1. longer. the bond order of the N–N bond in the N22+ ion is __________. two bonding molecular orbitals c. 0 b. filled b. greater. 4 d. 8 137. a. greater. The bond order of any molecule containing equal numbers of bonding and antibonding electrons is __________. O2 e. Based on molecular orbital theory. one bonding molecular orbital and one hybrid orbital b. Based on molecular orbital theory. 3 e. 1/2 c. This is because __________. 1/2. 1/2 139. c. and 1/2 d. Based on molecular orbital theory. a. greater. the bond order of the H–H bond in the H2+ ion is __________. greater 134. the bond order of the N–N bond in the N2 molecule is __________. overlap of two s atomic orbitals produces __________. there are __________ unpaired electrons in the OF+ ion. 1 d. two c. the energy of the π2p MOs is higher than that of the σ2p MO d. a. a. 1 d. filled 128. 2 e. F2. According to MO theory. filled. and 0 c. the bond order of the Be–Be bond in the Be2 molecule is __________. greater. 1/2 130. 1 c. 1 c. 3 c. 1. 1/2. one b. the bond order in O2 can be shown to be equal to 2. and 1/2 e. greater d.Name: ________________________ ID: A 125. greater c. 4 132. half-filled d. greater. shorter. a. 6 e. there are more electrons in the bonding orbitals than in the antibonding orbitals. four d. filled e. 2 e. two bonding molecular orbitals and two antibonding molecular orbitals d. 0 b. adding electrons to a bonding MO or removing electrons from an antibonding MO c. According to MO theory. two bonding molecular orbitals and one antibonding molecular orbital e. one π+ MO and one σ* MO e. The bond order of a homonuclear diatomic molecule cannot be decreased by any means. the more metallic character of boron. __________. one π MO and one π* MO or one σ MO and one σ* MO d. F2. N2 c. N2 only c. O2 e. 4 146. two bonding molecular orbitals and two antibonding molecular orbitals d. carbon and nitrogen as compared to oxygen. three bonding molecular orbitals and three antibonding molecular orbitals 148. 2 d. and N2 e. and neon c. The more effectively two atomic orbitals overlap. octahedral 11 . tetrahedral d. a. trigonal planar c. Li2 142. only __________ appears to gain mass in a magnetic field. a. fluorine. a. and Ne2 d. the electron-domain geometry of the central atom in BF3 is __________. and N2 144. is different from the order in O2. According to MO theory. trigonal planar c. can accommodate more electrons than c. the lower will be the energy of the resulting bonding MO and the higher will be the energy of the resulting antibonding MO 141. C2. a. a. the fewer antibonding MOs will be produced by the combination e. less effective overlap of p orbitals in B2. adding electrons to any MO d. the higher will be the energies of both bonding and antibonding MOs that result d. one π MO and one σ MO c. The order of MO energies in B2. removing electrons from any MO e. Of the following. removing electrons from a bonding MO or adding electrons to an antibonding MO b. a. one σ MO. greater 2s-2p interaction in B2. 1 c. octahedral 145. the bond order of the C–C bond in the C2 molecule is __________. the electron-domain geometry of the central atom in SF2 is __________. and N2 (σ2p > π2p). linear b. B2 and O2 143. C2 b. __________ appear(s) to gain mass in a magnetic field. linear b. is always degenerate with 150. two π+ MOs. and Ne2 This is due to __________. The bond order of a homonuclear diatomic molecule can be decreased by __________. greater 2s-2p interaction in O2. two bonding molecular orbitals b. C2. a. O2 only b. a. Of the following. C2. 0 b. a. one π MO and one σ* MO b. F2 d.Name: ________________________ ID: A 140. the more bonding MOs will be produced by the combination b. is always lower in energy than b. overlap of two p atomic orbitals produces __________. Using the VSEPR model. one bonding molecular orbital and one antibonding molecular orbital c. 3 e. and Ne2 b. is always higher in energy than e. B2 N2 O2 147. B2 and N2 d. trigonal bipyramidal e. the higher will be the energy of the resulting bonding MO and the lower will be the energy of the resulting antibonding MO c. tetrahedral d. An antibonding MO __________ the corresponding bonding MO. trigonal bipyramidal e. F2. less effective overlap of p orbitals in O2. F2. a. N2 and O2 e. a. overlap of two p atomic orbitals produces __________. two π MOs. can accommodate fewer electrons than d. and one σ* MO 149. Using the VSEPR model. Based on molecular orbital theory. trigonal pyramidal 159. trigonal planar c. Each molecular orbital can accommodate. a. sp3 d. a. trigonal planar c. 167. What is the molecular geometry of a molecule that has three bonding and two non-bonding domains? 166. The hybrid orbital set used by the central atom in NO3.is __________. 165. Using the VSEPR model. sp b. tetrahedral d. a. linear b. the molecular geometry of the central atom in BCl3 is __________. trigonal bipyramidal e. the molecular geometry of the central atom in NCl3 is __________.Name: ________________________ ID: A 156. a. trigonal bipyramidal d. tetrahedral b. at most. Using the VSEPR model. a. tetrahedral d. a. trigonal planar c. The hybrid orbital set used by the central atom in KrF2 is __________. 169. a. Using the VSEPR model. trigonal pyramidal 161. a. Using the VSEPR model. sp3d2 154. sp2 c. linear b. a. bent e. Using the VSEPR model. tetrahedral d. 164. the molecular geometry of the central atom in XeF2 is __________. sp3d2 Completion Complete each statement. tetrahedral d. 162. trigonal planar c. sp2 c. The 1s hydrogen orbital overlaps with the __________ iodine orbital in HI. sp b. sp3d e. trigonal bipyramidal e. the electron-domain geometry of the central atom in BrF4. sp2 c. the molecular geometry of the central atom in CF4 is __________. Using the VSEPR model. What are the three bond angles in the trigonal bipyramidal structure? 168. bent e. tetrahedral d. square pyramidal 153. linear b. the molecular geometry of the central atom in SO2 is __________.is __________. trigonal pyramidal 151. This is called the __________. Using the VSEPR model. A covalent bond in which overlap regions lie above and below an internuclear axis is called a(n) __________. tetrahedral d. bent e. sp3 d. seesaw e. two electrons with their spins paired. sp3d e. the molecular geometry of the central atom in PF5 is __________. sp3d2 155. linear b.is __________. trigonal pyramidal 160. linear b. trigonal planar c. linear b. Using the VSEPR model. In molecular orbital theory the stability of a covalent body is related to its __________. tetrahedral d. bent e. a. octahedral 158. The sensation of vision results from a nerve impulse that is triggered by the separation of retinal from __________. linear b. trigonal planar c. trigonal pyramidal 152. In the valence shell of an atom there are six electron domains. a. trigonal planar c. octahedral 157. sp3d e. bent e. square planar c. 12 . sp3 d. 163. the electron-domain geometry of the central atom in ClF3 is __________. They will be arranged in a (an) __________ geometry. The hybrid orbital set used by the central atom in BF4. sp b. Essay 171. the stronger is the force of magnetic attraction. Three molecules have similar electron domains. but different molecular shapes. Why? 13 .Name: ________________________ ID: A 170. This is called __________. The more unpaired electrons in a species. 2 Sec.2 Sec.5 Sec.5 Sec.2 Sec. 9.6 MULTIPLE CHOICE 9.2 Sec. 32. 9. 22. 23.2 Sec. 9.6 Sec. 4.6 Sec. 9. 9. 21.2 Sec. 3. 9. 9.3 Sec.2 Sec. 9.2 Sec. 9.2 Sec. 24. 9.5 Sec.ID: A AP Chapter 9 Study Questions Answer Section TRUE/FALSE 1. 9.2 Sec. 14. 15. 13. 9. 9. 9. 9. 35. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: A E C C B A B C D B B C B C C B D D D B C A E D B E A E 1 . 31. 9. 9. 9. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: F T F T F T F T PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: 1 1 1 1 1 1 1 1 DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: 2 2 1 1 3 1 1 3 REF: REF: REF: REF: REF: REF: REF: REF: Sec. 6. 27. 9.2 Sec.5 Sec. 9.5 Sec. 9. 9. 9. 2. 34.3 Sec. 9. 9.8 PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: 1 1 1 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 1 1 1 1 2 2 1 1 1 1 REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: Sec. 9. 9.5 Sec.1 Sec. 9. 36. 30. 11. 5. 20. 33.2 Sec.3 Sec. 29.2 Sec.4 Sec. 17.1 Sec. 9. 10. 9. 7. 28. 18.5 Sec.2 Sec. 9. 26. 9.2 Sec.2 Sec. 9.5 Sec. 12. 9. 19. 25.2 Sec. 9. 9.5 Sec. 16. 8. 9.3 Sec.3 Sec. 51. 9.2 Sec. 75. 67. 9.2 Sec.3 .6 Sec.6 Sec. 61.2 Sec. 9. 48.2 Sec. 9. 9. 9. 78. 38.3 Sec. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: C C C D D D D C D C E C D A C A D C C E A B D E B D C A A B D B C B A C A B A C D C A D D PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: 1 1 1 1 2 1 2 4 4 1 2 2 1 2 1 1 2 2 1 1 1 1 1 1 1 1 1 1 2 2 3 2 2 2 1 1 1 2 1 2 1 1 1 1 1 REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: 2 Sec. 69. 66. 9. 56.2 Sec. 74.2 Sec.2 Sec.2 Sec. 9. 9. 9. 80. 9.2 Sec. 58. 9. 43. 53. 55. 9. 9.2 Sec. 45.3 Sec. 81. 71. 77. 39. 9.2 Sec. 9.3 Sec. 54. 47. 9. 9. 60.3 Sec. 9.3 Sec. 9.6 Sec. 9.2 Sec.2 Sec. 42. 9. 52. 59. 9. 9. 9.6 Sec. 50. 79.2 Sec.2 Sec. 73.2 Sec. 9. 9. 9.2 Sec. 76. 9. 9.3 Sec.2 Sec.3 Sec. 46. 9. 9. 9. 57.2 Sec.2 Sec.2 Sec. 9. 9. 9. 63. 65.2 Sec. 9. 9.2 Sec.3 Sec.2 Sec.2 Sec.2 Sec. 49. 9. 64.6 Sec. 62.2 Sec. 9.3 Sec. 9. 41. 9. 70. 9. 40. 68.2 Sec. 9. 44.ID: A 37. 72. 98. 9. 9.ID: A 82. 117.5 Sec. 96.5 Sec. 111. 9. 9. 87.6 Sec.5 Sec. 9.6 Sec.6 Sec. 9.5 Sec. 9.6 Sec. 106. 9. 9.5 Sec. 125.6 Sec. 9.5 Sec. 99. 102. 89. 9.6 Sec.6 Sec. 86. 9.5 Sec.3 | Sec.5 Sec. 9. 93. 9. 9. 95.5 Sec. 9. 109. 9.5 Sec. 92. 88. 9. 105. 9. 83.4 Sec. 101. 9. 9. 100. 116.6 Sec.5 Sec. 9. 9. 9.5 Sec. 112.6 Sec. 85. 9.7 Sec. 114.6 Sec. 9.5 Sec.5 Sec. 9. 94. 9. 9. 120. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: A C B C B B E B D B D D C A C D D D D C C E D A E A B E C B B A C E C A B A B B C B C A A PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: 1 3 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 2 1 1 1 4 1 1 1 1 1 1 1 1 2 1 1 1 2 2 3 3 4 REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: 3 Sec. 9. 9.7 . 124. 103. 9.5 Sec. 9. 9.5 Sec.6 Sec. 9.6 Sec. 126. 110.5 Sec.6 Sec.5 Sec. 122. 9.6 Sec. 9.5 Sec. 115. 9. 9. 97. 104. 90. 84.5 Sec.6 Sec. 9.6 Sec. 9. 123. 9. 9.5 Sec. 118.5 Sec. 113. 9. 108. 91.6 Sec.6 Sec.5 Sec. 9. 107. 119. 121.6 Sec. 9. 149. 9.7 Sec. 9.7 | Sec.8 Sec. 158. 143. 134. 9.2 Sec. 139. 9. 9. 161. 153.8 Sec. 146.8 Sec. 136.8 Sec. ANS: octahedral PTS: 1 4 Sec.2 COMPLETION 162. 9. 129. 9. 9. 155.8 Sec. 133.7 | Sec. 142. 147. 9. 145.7 Sec.8 Sec.7 Sec. 9.8 Sec. 130.2 Sec. 9. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: B D B B E B D C D D D D A C D E B C D E A D B C D E A B C D E C B C D PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: PTS: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: DIF: 4 4 4 3 3 3 4 4 3 3 4 4 4 4 4 4 3 3 3 4 4 5 1 1 1 1 1 1 1 1 1 1 1 1 1 REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: REF: DIF: 4 REF: Sec. 152. 9.8 Sec. 148.8 Sec. ANS: T-shaped PTS: 1 163. 9.7 Sec. 9. 9. 9. 150. 9. 9.5 Sec.ID: A 127.2 Sec. 154.7 Sec. 9.2 Sec. 9.8 Sec.1 Sec. 9. 141. 140. 9.2 Sec. 9.2 Sec.5 . 9.2 Sec. 132.8 Sec.2 Sec. 9. 128. 9. 160. 138. 9. 9.8 Sec.5 Sec. 9. 9. 137. 9. 144. 9. 9. 9. 159. 9.2 DIF: 5 REF: Sec. 9. 9.8 Sec. 131. 9.8 Sec. 135.7 Sec.8 Sec. 156.2 Sec. 157.8 Sec. 9. 151. 2 PTS: 1 166. ANS: bond order DIF: 2 REF: Sec. ANS: paramagnetism 4 REF: Sec.ID: A 164.7 PTS: 1 DIF: 170. ANS: 5p DIF: 4 REF: Sec. 180° PTS: 1 165. ANS: 90°.6 PTS: 1 DIF: 169. ANS: Pauli principle 2 REF: Sec. 9.6 PTS: 1 168.8 1 REF: Sec. 9. ANS: π bond DIF: 2 REF: Sec. 9. 9.5 PTS: 1 167.2 5 . ANS: opsin DIF: 2 REF: Sec. 9.8 PTS: 1 DIF: ESSAY 171. 120°. 9. 9. 9. ANS: different numbers of non-bonding domains PTS: 1 DIF: 4 REF: Sec.