Version038–Exam2–McCord–(53130) 1 This print-out should have 31 questions. 3. sp3 Multiple-choice questions may continue on the next column or page – find all choices 4. Pure pz orbitals are used in bonding. before answering. 5. sp3d McCord CH301tt Explanation: The structure is b b b This exam is only for b O McCord’s Tues/Thur CH301 class. C If Quest works properly, you should be able H3C CH3 to see your score for this exam by 11:30 PM tonight. Do realize however that Quest has 003 10.0 points been taking up to 8 hours to get through the Which of the molecules b b b grading, so your score might appear bit by bit. b O I will post an announcement on our web page I) and/or Quest when I feel like all bubblesheets H H have been graded. b b b b b b Cl PLEASE carefully bubble in your UTEID and Version Number! We av- II) b b B b b b erage about 20-30 students per exam that b b b b Cl Clb b misbubble this information. Get it right! b b b b b b b b III) b Cl Cl b b b b 001 10.0 points contains polar covalent bonds but is NOT Choose the species that is incorrectly matched itself a polar molecule? with electronic geometry about the central atom. 1. III only 1. CF4 : tetrahedral 2. II and III only 2. BeBr2 : linear 3. I and II only 3. H2O : tetrahedral 4. All fit the criteria. 4. PF3 : pyramidal correct 5. None fit the criteria. 5. NH3 : tetrahedral 6. II only correct Explanation: 7. I and III only 002 10.0 points 8. I only Identify the hybrid orbitals used by the underlined atom in acetone (CH3COCH3). Explanation: I) Water has polar covalent bonds and is 1. sp2 correct itself polar. II) Boron trichloride contains polar covalent 2. sp bonds but is not polar overall due to symmetry. Version038–Exam2–McCord–(53130) 2 III) Chlorine gas has nonpolar bonds only. b b b H C S b 004 10.0 points 10. b b b b Which of the following is the correct Lewis b H b formula for thioformaldehyde (CH2S)? b b b b b b Explanation: b b b H C S b The Lewis formula for thioformaldehyde 1. b b b b b b b b b b H b b b b S (CH2S) is b b b b b b H C b b b H C S b H 2. b b b b b b b H b b b 005 10.0 points The molecule AX3 is polar and obeys the b b octet rule; therefore, the central atom A H C S 3. b b b H b 1. has no lone pairs of electrons. b b 2. b b has two lone pairs of electrons. b b S C 4. b 3. correct b has one lone pair of electrons. H b H b b b 4. has four lone pairs of electrons. b b b S b 5. has three lone pairs of electrons. b b b Explanation: 5. b H C b b b The molecule AX3 could obey the octet b b H b rule in one of two ways. The molecule could b b contain two A−X single bonds and one A−X double bond, yielding the structure H C S X 6. b b A X b H b b b X b b This molecule would have a trigonal planar b S b geometry. It is symmetrical and therefore 7. correct H C non-polar, so this is not the correct structure for our molecule. H The second possible dot structure for AX3 is ·· X A X 8. HS C H X b b b b This molecule obeys the octet rule and has b H C S b a pyramidal molecular geometry. Assuming 9. b b b b there is an electronegativity difference be- b H b b b tween A and X, this molecule would be polar. Version038–Exam2–McCord–(53130) 3 This structure has one lone pair on A. what is the interaction energy between a zinc − ion (Zn2+) and a sulfide ion (S2 ) in a hy- 006 10.0 points pothetical structure in which the inter-ionic Which of the following compounds would be distances are the same as that of NaCl? expected to have the shortest bonds? 1. 1.140 kJ/mol − 1. NO3 2. 1140 kJ/mol 2. CCl4 3. 3040 kJ/mol correct − 3. NO2 correct 4. 760 kJ/mol 4. NCl3 5. 1520 kJ/mol Explanation: triple bonds < double bonds < single Explanation: − bonds. The more bonds, the shorter the bond qZn =+2C qS = 2C − length. Resonance structures have bonds qNa =+1C qCl = 1C somewhere between single and double bonds. VNaCl = 760 kJ/mol The more bonding atoms sharing the 2 elec- If r is the distance between the two ions (the trons, the weaker the bond and the longer the sum of the ionic radii), the energy interaction − − between ions is given by Coulomb’s Law: bond length. NO2 and NO3 are both reso- − nance structures, but NO2 has fewer bonding q1 q2 atoms. Therefore it has stronger bonds, and V = shorter bond lengths. 4 π ǫ0 r q1 q2 r = 007 10.0 points 4 π ǫ0 V A double bond consists of 1. two pi bonds. rNaCl = rZnS qNa qCl qZn q = S 2. one sigma bond containing four elec- 4 π ǫ0 VNaCl 4 π ǫ0 VZnS qZn qS VNaCl trons. VZnS = qNa qCl 3. one sigma and one pi bond. correct (+2 C)(−2 C)(−760 kJ/mol) = (+1 C)(−1 C) 4. two sigma bonds. = −3040 kJ/mol , 5. two beta bonds. the energy released by the interaction. 6. an alpha and a beta bond. 009 10.0 points Consider the following pairs A and B Explanation: b b b b b b b b b The sigma bonds overlap along a line drawn A1) N N O A2) b N N O b b b b b b between the two nuclei, and the pi bonds − b b − 3 b b 3 b b b b overlap perpendicular to this line. b b b b O O b b b b b b b b b B1) O P O b 2 O P O b b B ) 008 10.0 points b b b b b b If the interaction energy between a sodium ion b b b b b O b b O b and a chloride ion in table salt is 760 kJ/mol, b b b b Version038–Exam2–McCord–(53130) 4 of Lewis structures. onal bipyramidal. Select the one from each pair that is likely to make the dominant contribution to a reso- 011 10.0 points nance hybrid. Consider the species − a) I2, b) O3, c)I3 , d) CS2, e) CO. 1. A1 and B1 Which of the species is/are polar? 2. A2 and B1 correct 1. c) and e) only 3. A1 and B2 2. b) and e) only correct 4. A2 and B2 3. e) only Explanation: b b b b b b 4. b) and c) only b b b A1) N N O A2) b N N O b b b b b b −1 +1 0 0 +1 −1 Explanation: Of the species listed, only O3 and CO are The second structure is preferred because it polar. CO is polar due to the difference in places the negative formal charge on the most electronegativity between O and C; O3 is po- electronegative atom. − b b − 3 b b 3 lar because it has 3 RHED and one lone pair b b − b b − b O b 1 b O b 1 on the central atom. This lone pair is an 0 0 b b b b b b − b b 0 b 1 area where negative charge is concentrated, 1 O P O b 2 O P O B ) b b b b B ) b b b b so this results in the molecule having an over- −1 0 b b b b 2 b b all dipole moment. In the other species, I and −1 O −1 b O b b b b b CS2 are both linear and in the case of CS2, The first structure is preferred for the same the two opposing dipoles of the C-S bonds reason. − will cancel. Finally I3 has 5 RHED and three lone pairs on the central atom but they are ar- ◦ 010 10.0 points ranged at 120 so their effects cancel and the ion is nonpolar. On the basis of valence shell electron pair repulsion theory, the electronic geometry of 012 10.0 points XeF2 is Which of the following has bond angles ◦ slightly less than 109.5 ? 1. trigonal bipyramidal. correct 2− 1. CS3 2. linear. 2. COCl2 3. octahedral. 3. AsF3 correct 4. square planar. 4. COS 5. trigonal planar. Explanation: 5. SO2 b b b b b b b b b F Explanation: b b b b Xe b b b b b AsF3 has four regions of electron density, b b F b b one of which is a lone pair, which repels the For the central atom Xe, HED = 5, lone other regions, making those bond angles less ◦ pairs = 3, and the electronic geometry is trig- than 109.5 . Version038–Exam2–McCord–(53130) 5 013 10.0 points 3.