Version PREVIEW – Quiz 3 – JOHNSON – (53755) 1This print-out should have 8 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.LDE Identifying Bonds 001001 10.0 pointsHow many polar ionic, po lar covalent andnon-po lar covalent bonds are present in theLewis structure shown below? Ba se your an-swer on simple ∆EN considerations.BBFbbbbbbFbbbbbbNbbHH1. 0; 3; 32. 2; 4; 03. 0; 5; 14. 2; 3; 1 correct5. 0; 6; 0Explanation:Based on the simple rule that ∆EN > 1.5is ionic, only the two B—F bonds are ionic,for a total of 2. B ased on the simple rule that1.5 > ∆EN > 0 is polar covalent, t he B—Nbond and two N—H bonds are polar covalent,for a total of 3. Lastly, any bond betweenan atom and another of the same a tom isnon-po lar covalent, so the one B—B bond isnon-po lar covalent, for a to tal of 1.VSEPR Electron Geometry 001002 10.0 pointsIn the Lewis structure for dimethylamine(CH3NHCH3), what are the bond angles, hy-bridization and electronic geometry of the ni-trogen atom?1. 109.5◦, sp3, tetrahedral correct2. 109.5◦, sp2, trigonal planar3. 180◦, sp, linear4. 90◦, sp3, tetrahedral5. 120◦, sp2, trigonal planarExplanation:VSEPR Electron Geometry 002003 10.0 pointsOf the following choices, which has a moleculepaired incorrectly with either its bond an-gle(s), electronic geometry, or hybridization?1. NH4+, 109.5◦2. XeF2, linear correct3. CO2, 180◦4. SiH4, tetra hedral5. BF3, sp26. SCl6, sp3d2Explanation:The xenon atom in XeF2, has two bondedfluorine atoms and 3 non-b onding pai rs ofelectrons, for a total of 5 regions of high elec-tron density. Its electronic geometry is there-fore trigo nal bipyramidal, not linear (whichwould be its molecular geometry).LDE VSEPR Molecular Geometry 001004 10.0 pointsWhat are the molecular geometries of thelabeled atoms in the Lewis structure below?Note: only bonding electrons are shown.ObCHCaHHPcOOO1. trigonal planar; bent; tetrahedral cor-rect2. trigonal planar; linear; trig onal bipyrami-dalVersion PREVIEW – Quiz 3 – JOHNSON – (53755) 23. trig onal pyramidal; linear; see-saw4. bent; trigonal pyramidal; t-shaped5. bent; tetrahedral; t-shapedExplanation:Atom a has three bonded atoms and nonon-bo nding pairs of electrons and is thereforetrigonal planar. Atom b has two bondedatoms and two non-bonding pairs of electronsand is therefore bent. Atom c has four bondedatoms and no non-bonding pairs o f electronsand is therefore tetrahedral.LDE VSEPR Molecular Geometry 002005 10.0 pointsWhich of the following molecules is/are po-lar?I)CFHCFHII)CFHCHFIII)CFFCHH1. II only2. I, III correct3. I, II4. II, III5. I only6. III only7. I, II, IIIExplanation:Molecule II is symmetrical and therefore itsindividual dipol e moments cancel, making itnon-po lar. Molecules I and III are asymmet-rical and t herefore polar.LDE VB Sigma Pi Bonds 001006 10.0 pointsThe Lewis structure for benzene (C6H6)has how many σ (sigma) and how many π (pi)bonds?C HCHCHCHCHCH1. 6; 32. 12; 3 correct3. 9; 64. 12; 05. 15; 0Explanation:LDE VB Hybridization 001007 10.0 pointsWhich of the foll owing statements concerninghybrid orbita ls is/are true?I) Hybrid orbitals are energetically degen-erate.II) Any element can form sp3d2hybrid or-bitals.III) Hybridizing a 2s and a 2p orbital wouldproduce one sp hybrid orbital.1. II, III2. I, II, III3. III only4. I only correctVersion PREVIEW – Quiz 3 – JOHNSON – (53755) 35. II only6. I, III7. I, IIExplanation:Statement I is true; hybridization was de-veloped as a theoretical framework to ex-plain the energetic degeneracy o f bonds inmolecules. Statement II is false; hybridiza-tion involving d orbitals requires access toempty d orbitals, a nd thus begins in period3. Statement III is false; the number of or-bitals used to hybridize is always equal to thenumber of hybridized orbitals, so using a 2sand a 2p orbit al would result in two sp hybridorbitals.LDE VB Hybridization 002008 10.0 pointsWhat orbitals were used to form the π (pi)bond in the thionoester molecule below?CHHHOCHS1. 2s, 3p2. 2p, 3p correct3. sp3, 3s4. sp2, 3s5. sp3,
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