Chapter 09 - Exam practice questions PDF

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Chemistry: The Central Science, 14e (Brown, et al.) Chapter 9 Molecular Geometry and Bonding Theories 9.1 Multiple-Choice Questions 1) The basis of the VSEPR model of molecular bonding is ________. A) regions of electron density on an atom will organize themselves so as to maximize s-character B) regions of electron density in the valence shell of an atom will arrange themselves so as to maximize overlap C) atomic orbitals of the bonding atoms must overlap for a bond to form D) electron domains in the valence shell of an atom will arrange themselves so as to minimize repulsions E) hybrid orbitals will form as necessary to, as closely as possible, achieve spherical symmetry Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 2) In counting the electron domains around the central atom in VSEPR theory, a ________ is not included. A) nonbonding pair of electrons B) single covalent bond C) core level electron pair D) double covalent bond E) triple covalent bond Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 3) The electron-domain geometry of ________ is tetrahedral. A) CBr4 B) PH3 C) CCl2Br2 D) XeF4 E) all of the above except XeF4 Answer: E Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2

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4) Of the following species, ________ will have bond angles of 120°. A) PH3 B) ClF3 C) NCl3 D) BCl3 E) All of these will have bond angles of 120°. Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 5) The molecular geometry of the BrO3- ion is ________. A) trigonal pyramidal B) trigonal planar C) bent D) tetrahedral E) T-shaped Answer: A Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 6) The molecular geometry of the left-most carbon atom in the molecule below is ________.

A) trigonal planar B) trigonal bipyramidal C) tetrahedral D) octahedral E) T-shaped Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2

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7) The molecular geometry of the right-most carbon in the molecule below is ________.

A) trigonal planar B) trigonal bipyramidal C) tetrahedral D) octahedral E) T-shaped Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 8) The bond angles marked a, b, and c in the molecule below are about ________, ________, and ________, respectively.

A) 90°, 90°, 90° B) 120°, 120°, 90° C) 120°, 120°, 109.5° D) 109.5°, 120°, 109.5° E) 109.5°, 90°, 120° Answer: D Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2

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9) The bond angles marked a, b, and c in the molecule below are about ________, ________, and ________, respectively.

A) 109.5°, 109.5°, 109.5° B) 120°, 109.5°, 120° C) 109.5°, 109.5°, 120° D) 90°, 180°, 90° E) 109.5°, 109.5°, 90° Answer: C Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 10) The central Xe atom in the XeF4 molecule has ________ unbonded electron pair(s) and ________ bonded electron pair(s) in its valence shell. A) 1, 4 B) 2, 4 C) 4, 0 D) 4, 1 E) 4, 2 Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 11) An electron domain consists of ________. a) a nonbonding pair of electrons b) a single bond c) a multiple bond A) a only B) b only C) c only D) a, b, and c E) b and c Answer: D Diff: 1 Var: 1 LO: 9.1 GO: G2

Page Ref: Sec. 9.2

12) The molecular geometry consists of ________. 4 Copyright © 2018 Pearson Education, Inc.

a) a nonbonding pair of electrons b) a single bond c) a multiple bond A) a only B) b only C) c only D) a, b, and c E) b and c Answer: E Diff: 1 Var: 1 LO: 9.1 GO: G2

Page Ref: Sec. 9.2

13) The electron-domain geometry and the molecular geometry of a molecule of the general formula ABn are ________. A) never the same B) always the same C) sometimes the same D) not related E) mirror images of one another Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 14) The electron-domain geometry and the molecular geometry of a molecule of the general formula ABn will always be the same if ________. A) there are no lone pairs on the central atom B) there is more than one central atom C) n is greater than four D) n is less than four E) the octet rule is obeyed Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2

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15) What is the molecular shape of H2O? A) T-shaped B) tetrahedral C) linear D) trigonal pyramidal E) bent Answer: E Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 16) PCl5 has ________ electron domains and a ________ molecular arrangement. A) 6, trigonal bipyramidal B) 6, tetrahedral C) 5, square pyramidal D) 5, trigonal bipyramidal E) 6, seesaw Answer: D Diff: 3 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 17) For molecules of the general formula ABn, n can be greater than four ________. A) for any element A B) only when A is an element from the third period or below the third period C) only when A is boron or beryllium D) only when A is carbon E) only when A is Xe Answer: B Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2

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Consider the following species when answering the following questions: (i) BCl3

(ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SF6

18) For which of the molecules is the molecular geometry (shape) the same as the VSEPR electron domain arrangement (electron domain geometry)? A) (i) and (ii) B) (i) and (iii) C) (ii) and (v) D) (iv) and (v) E) (v) only Answer: C Diff: 2 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 19) Of the molecules below, only ________ is polar. A) CCl4 B) CH4 C) SeF4 D) SiCl4 Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.3 LO: 9.2 GO: G2 20) Of the molecules below, only ________ is nonpolar. A) BF3 B) NF3 C) IF3 D) PBr3 E) BrCl3 Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.3 LO: 9.2 GO: G2

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21) Three monosulfur fluorides are observed: SF2, SF4, and SF6. Of these, ________ is/are polar. A) SF2 only B) SF2 and SF4 only C) SF4 only D) SF6 only E) SF2, SF4, and SF6 Answer: B Diff: 2 Var: 1 Page Ref: Sec. 9.3 LO: 9.2 GO: G2 22) The molecular geometry of the PF3 molecule is ________, and this molecule is ________. A) trigonal planar, polar B) trigonal planar, nonpolar C) trigonal pyramidal, polar D) trigonal pyramidal, nonpolar E) tetrahedral, unipolar Answer: C Diff: 2 Var: 1 Page Ref: Sec. 9.2, 9.3 LO: 9.1, 9.2 GO: G2 23) Of the following molecules, only ________ is polar. A) BeCl2 B) BF3 C) C D) SiH2Cl2 E) Cl2 Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.3 LO: 9.2 GO: G2 24) Of the following molecules, only ________ is polar. A) CCl4 B) BCl3 C) NCl3 D) BeCl2 E) Cl2 Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.3 LO: 9.2 GO: G2 8 Copyright © 2018 Pearson Education, Inc.

25) For molecules with only one central atom, how many lone pairs on the central atom guarantees molecular polarity? A) 1 B) 2 C) 1 or 2 D) 3 E) 1 or 3 Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.3 LO: 9.2 GO: G2 26) The molecular geometry of the CHF3 molecule is ________, and the molecule is ________. A) trigonal pyramidal, polar B) tetrahedral, nonpolar C) seesaw, nonpolar D) tetrahedral, polar E) seesaw, polar Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.2, 9.3 LO: 9.1, 9.2 GO: G2 27) The molecular geometry of the BCl3 molecule is ________, and this molecule is ________. A) trigonal pyramidal, polar B) trigonal pyramidal, nonpolar C) trigonal planar, polar D) trigonal planar, nonpolar E) trigonal bipyramidal, polar Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.2, 9.3 LO: 9.1, 9.2 GO: G2

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Consider the following species when answering the following questions: (i) BCl3

(ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SF6

28) Which of the molecules has a see-saw shape? A) (i) B) (ii) C) (iii) D) (iv) E) (v) Answer: C Diff: 3 Var: 1 Page Ref: Sec. 9.2 LO: 9.1 GO: G2 29) Which of the molecules has a square planar shape? A) (i) and (ii) B) (i) and (v) C) (iii) only D) (ii) and (iv) E) (iv) only Answer: E Diff: 2 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 30) The combination of two atomic orbitals results in the formation of ________ molecular orbitals. A) 1 B) 2 C) 3 D) 4 E) 0 Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2

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31) Of the following, the central atom is sp3d2 hybridized only in ________. A) PCl5 B) XeF4 C) PH3 D) Br3E) BeF2 Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 32) The sp3d2 atomic hybrid orbital set accommodates ________ electron domains. A) 2 B) 3 C) 4 D) 5 E) 6 Answer: E Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 33) The sp2 atomic hybrid orbital set accommodates ________ electron domains. A) 2 B) 3 C) 4 D) 5 E) 6 Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 34) The hybridizations of nitrogen in NF3 and NH3 are ________ and ________, respectively. A) sp2, sp2 B) sp, sp3 C) sp3, sp D) sp3, sp3 E) sp2, sp3 Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2

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35) The hybridizations of iodine in IF3 and IF5 are ________ and ________, respectively. A) sp3, sp3d B) sp3d, sp3d2 C) sp3d, sp3 D) sp3d2, sp3d E) sp3d2, sp3d2 Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 36) The hybridizations of bromine in BrF5 and of arsenic in AsF5 are ________ and ________, respectively. A) sp3, sp3d B) sp3d, sp3d2 C) sp3d, sp3 D) sp3d2, sp3d E) sp3d2, sp3d2 Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 37) The hybrid orbitals used for bonding by the sulfur atom in the SF4 molecule are ________ orbitals. A) sp B) sp2 C) sp3 D) sp3d E) sp3d2 Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2

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38) What are the hybrid orbitals used for bonding by Xe in a XeCl4 molecule? A) sp2 B) sp3d2 C) sp3 D) sp3d E) sp Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 39) The hybridization scheme for BeF2 is ________. A) sp B) sp2 C) sp3 D) sp3d E) sp3d2 Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 40) What is the hybridization of the carbon atom attached to the two oxygen atoms in the structure below?

A) sp3 B) sp C) sp3d2 D) sp2 E) sp3d Answer: D Diff: 2 Var: 1 LO: 9.4 GO: G2

Page Ref: Sec. 9.5

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41) What is the hybridization of the I atom in a IF5 molecule? A) sp2d2 B) sp3d C) sp3 D) sp3d2 E) sp2 Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 42) ________ hybrid orbitals are used for bonding by Xe in the XeF4 molecule. A) sp2 B) sp3 C) sp3d D) sp3d2 E) sp Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 Consider the following species when answering the following questions: (i) BCl3

(ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SF6

43) In which of the molecules is the central atom sp2 hybridized? A) (i) only B) (i) and (ii) C) (iii) and (iv) D) (iii) only E) (iv) and (v) Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2

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44) When four atomic orbitals are mixed to form hybrid orbitals, how many hybrid orbitals are formed? A) one B) six C) three D) four E) five Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 45) A triatomic molecule cannot be linear if the hybridization of the central atoms is ________. A) sp B) sp2 C) sp3 D) sp2 or sp3 E) sp2d or sp3d2 Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 46) Valence bond theory addresses all of the following except ________. A) molecular shape B) covalent bonding C) excited states of molecules D) hybridization E) multiple bonds Answer: C Diff: 2 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 47) A typical double bond ________. A) is stronger and shorter than a single bond B) consists of one σ bond and one π bond C) imparts rigidity to a molecule D) consists of two shared electron pairs E) All of the above answers are correct. Answer: E Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.5 GO: G2

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48) A typical triple bond ________. A) consists of one σ bond and two π bonds B) consists of three shared electrons C) consists of two σ bonds and one π bond D) consists of six shared electron pairs E) is longer than a single bond Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.5 GO: G2 49) In a SO42- ion, "localized" bonding electrons are associated with ________ particular atoms. A) zero B) one C) two D) three E) four Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.5 GO: G2 50) There are ________ σ bonds and ________ π bonds in H3C-CH2-CH CH-CH2-C CH. A) 14, 2 B) 10, 3 C) 12, 2 D) 13, 2 E) 16, 3 Answer: E Diff: 4 Var: 1 Page Ref: Sec. 9.6 LO: 9.5 GO: G2 51) Which of the following molecules or ions have various resonance structures? CO2 O3 CO32A) CO2, O3, and CO32B) CO2 and O3 C) O3 and CO32D) CO32- only E) None of the above will exhibit delocalized bonding. Answer: C Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2 16 Copyright © 2018 Pearson Education, Inc.

52) Which of the following molecules or ions will exhibit delocalized bonding? NO2- NH4+ N3A) NH4+ and N3B) NO2- only C) NO2-, NH4+, and N3D) N3- only E) NO2- and N3Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2 53) A molecule must have at least two resonance structures in order to display ________. A) delocalized σ bonding B) trigonal planar electron group geometry C) localized π bonding D) delocalized π bonding E) localized σ bonding Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2 54) In a C=C bond, the σ bond results from overlap of ________ orbitals and the π bond(s) result from overlap of ________ orbitals. A) sp2-hybrid, p-atomic B) sp2-atomic, p-hybrid C) sp3-hybrid, p-atomic D) sp-hybrid, p-atomic E) σ-atomic, π-hybrid Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2

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55) The carbon-hydrogen σ bond in ethylene, H2C CH2, results from the overlap of ________. A) sp2-hybrid and s-atomic orbitals B) sp hybrid orbitals C) sp3 hybrid orbitals D) s-hybrid and sp2-atomic orbitals E) s atomic orbitals Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2 56) The π bond in ethylene, H2C CH2, results from the overlap of ________. A) sp3 hybrid orbitals B) s atomic orbitals C) sp hybrid orbitals D) sp2 hybrid orbitals E) p atomic orbitals Answer: E Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2 57) A typical double bond consists of ________. A) three sigma bonds B) three pi bonds C) one sigma and two pi bonds D) one sigma and one pi bond E) three ionic bonds Answer: D Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.7 GO: G2 58) The N-H bond in ammonia consists of ________. A) one σ bond and one π bond B) one σ bond and no π bonds C) one σ bond and two π bonds D) two σ bonds and one π bond E) two σ bonds and two π bonds Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.6 LO: 9.7 GO: G2

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59) The hybridization of the terminal carbons in the H2C C CH2 molecule is ________. A) sp B) sp2 C) sp3 D) sp3d E) sp3d2 Answer: B Diff: 2 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 60) The hybridization of nitrogen in the H—C N: molecule is ________. A) sp B) s2p C) s3p D) sp2 E) sp3 Answer: A Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 61) The hybridization of carbon in the H-C C-H molecule is ________. A) sp2 B) sp C) s2p D) s3p E) sp3 Answer: B Diff: 1 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2

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62) The hybridization of the carbon atom labeled x in the molecule below is ________.

A) sp B) sp2 C) sp3 D) sp3d E) sp3d2 Answer: B Diff: 1 Var: 1 LO: 9.4 GO: G2

Page Ref: Sec. 9.5

63) The hybridization of the oxygen atom labeled x in the structure below is ________.

A) sp B) sp2 C) sp3 D) sp3d E) sp3d2 Answer: B Diff: 1 Var: 1 LO: 9.4 GO: G2

Page Ref: Sec. 9.5

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64) The hybridization and molecular shape of the carbon atom in carbon dioxide is ________. A) sp and linear B) sp3 and bent C) sp2 and linear D) sp2d and bent E) sp2d2 and linear Answer: A Diff: 2 Var: 1 Page Ref: Sec. 9.5 LO: 9.4 GO: G2 65) Electrons in ________ bonds remain localized between two atoms. Electrons in ________ bonds can become delocalized between more than two atoms. A) pi, sigma B) sigma, pi C) pi, pi D) sigma, sigma E) ionic, sigma Answer: B Diff: 2 Var: 1 Page Ref: Sec. 9.6 LO: 9.7 GO: G2 66) Structural changes around a ________ bond in the retinal portion of the rhodopsin molecule trigger the chemical reactions that result in vision. A) C=C B) C-C C) C C D) C-H E) C=N Answer: A Diff: 3 Var: 1 Page Ref: Sec. 9.6 LO: 9.6 GO: G2 67) The bond order of any molecule containing equal numbers of bonding and antibonding electrons is ________. A) 0 B) 1 C) 2 D) 3 E) 1/2 Answer: A Diff: 3 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 21 Copyright © 2018 Pearson Education, Inc.

68) In comparing the same two atoms bonded together, the ________ the bond order, the ________ the bond length, and the ________ the bond energy. A) greater, shorter, greater B) greater, greater, greater C) greater, longer, greater D) smaller, longer, smaller E) smaller, greater, greater Answer: D Diff: 4 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 69) In comparing the same two atoms bonded together, the ________ the bond order, the ________ the bond length, and the ________ the bond energy. A) greater, shorter, greater B) greater, greater, greater C) greater, longer, greater D) greater, greater, smaller E) smaller, greater, greater Answer: A Diff: 4 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 70) Based on molecular orbital theory, the bond orders of the H—H bonds in H2, H2+, and H2are ________, respectively A) 1, 0, and 0 B) 1, 1/2, and 0 C) 1, 0, and 1/2 D) 1, 1/2, and 1/2 E) 1, 2, and 0 Answer: D Diff: 4 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 71) Based on molecular orbital theory, the bond order of the H—H bond in the H2+ ion is ________. A) 0 B) 1/2 C) 1 D) 3/2 E) 2 Answer: B Diff: 4 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 22 Copyright © 2018 Pearson Education, Inc.

72) An antibonding π orbital contains a maximum of ________ electrons. A) 1 B) 2 C) 4 D) 6 E) 8 Answer: B Diff: 3 Var: 1 Page Ref: Sec. 9.8 LO: 9.8 GO: G2 73) According to MO theory, overlap of two s atomic orbitals produces ________. A) one bonding molecular orbital and one hybrid orbital B) two bonding molecular orbitals C) two bonding molecular orbitals and two antibonding molecular orbitals D) two bonding molecular orbitals and one antibonding molecular orbital E) one bonding molecular orbital and one antibonding molecular orbital Answer: E Diff: 3 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 74) A molecular orbital can accommodate a maximum of ________ electron(s). A) 1 B) 2 C) 4 D) 6 E) 12 Answer: B Diff: 3 Var: 1 Page Ref: Sec. 9.7, 9.8 LO: 9.8 GO: G2 75) Molecular Orbital theory correctly predicts paramagnetism of oxygen gas, O2. This is because ________. A) the bond order in O2 can be shown to be equal to 2 B) there are more electrons in the bonding orbitals than in the antibonding orbitals C) the energy of the π2p MOs is higher than that of the σ2p MO D) there are two unpaired electrons in the MO electron configuration of O2 E) the O—O bond distance is relatively short Answer: D Diff: 4 Var: 1 Page Ref: Sec. 9.8 LO: 9.8 GO: G2

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76) Molecular Orbital theory correctly predicts diamagnetism of fluorine gas, F2. This is because ________. A) the bond order in F2 can be shown to be equal to 1 B) there are more electrons in the bonding orbitals than in the antibonding orbitals C) all electrons in the MO electron configuration of F2 are paired D) the energy of the π2pMOs is higher than that of the σ2p MO E) the F—F bond enthalpy is very low Answer: C Diff: 4 Var: 1 Page Ref: S...