Version001–PracticeExam2–mccord–(49230) 1 b b This print-out should have 25 questions. b b b b 5. Ca + Ca + I b −→ Multiple-choice questions may continue on b − the next column or page – find all choices b b 2 + + b b before answering. Ca + Ca + b I b −→ Ca2I bb b b b b b b b b b b b b Important Message: Please please 6. Ca + b I b + b I b + b I b −→ b b b b b b − − − please remember to CORRECTLY bubble in b b b b b b 3+ b b b b b b your name, uteid, and version number for Ca + b I b + b I b + b I b −→ your exam. We are averaging about 20 stu- bb bb bb dents who fail to do this. Think of it as the Ca2I3 easiest question on the exam. If you get it b b b b wrong, you’ll get a zero when we grade it. Be- 7. Ca + b I b −→ b b − cause Quest is not that user friendly, it will b b + b b take at least 24 hours to get your avoidable Ca + b I b −→ CaI mistake fixed. It is a huge pain in the ass - bb b b know that. Please take the time to bubble b b b 8. Ca b + I b −→ in your information carefully. Lets have NO b − mistakes on this information please. Thank b b 2 2+ b b you so much. I’m only putting this plea on Ca + b I b −→ CaI bb this practice exam and not on the real exam. b b b b b b b b b So just remember to do it. 9. Ca b + Ca b + Ca b + b I + b I −→ b b - Dr. McCord − − b b 3 b b 3 2+ 2+ 2+ b b b b Ca +Ca +Ca + b I b + b I b −→ bb bb Ca3I2 Holt da 6 3 rev 2b 001 10.0 points b b b b b b 10. Ca + Ca + Ca + b I −→ Use electron-dot notation to demonstrate b − the formation of ionic compounds involving b b 3 + + + b b the elements Ca and I. Ca + Ca + Ca + b I b −→ Ca3I bb 1. None of these Explanation: b b b b b b b b b b b 2+ b b b b b b b b b Ca 2. Ca + Ca + I b + I b + I b −→ gives up two electrons to form the Ca b b b − − − b b b b 2 b b 2 b b 2 b b b 3+ 3+ b b b b b b cation, and I acquires an electron to form Ca +Ca + b I b + b I b + b I b −→ b b bb bb bb − b b b b Ca2I3 the b I b anion. bb b b b b Two iodine anions combine with one cal- b b 3. Ca + b I −→ cium cation to form an electrically neutral b − b b 3 compound: b b b b 3+ b b b b −→ Ca + I CaI b b b b b b b b bb Ca + I + I −→ b b b b b b b b − − b b b b b b b b b b b b −→ 2+ b b b b 4. Ca + I + I b b b b −→ b b b b Ca + I + I CaI2 − − bb bb b b b b 2+ b b b b Ca + b I b + b I b −→ CaI2 correct Mlib 01 2023 bb bb 002 10.0 points Choose the pair of names and formulas that Version001–PracticeExam2–mccord–(49230) 2 do NOT match. the latter are smaller than than their counter- parts in the former and consequently have a 1. As4O6 : tetraarsenic oxide correct greater lattice energy. 2. SO3 : sulfur trioxide Campion 05 Ex 01 07 004 10.0 points 3. NO : nitrogen monoxide If the interaction energy between a sodium ion and a chloride ion in table salt is 760 kJ/mol, 4. Cl2O7 : dichlorine heptoxide what is the interaction energy between a zinc − ion (Zn2+) and a sulfide ion (S2 ) in a hy- 5. N2O5 : dinitrogen pentoxide pothetical structure in which the inter-ionic distances are the same as that of NaCl? Explanation: Cl2O7,N2O5, SO3, and NO are all covalent 1. 1.140 kJ/mol molecules and are named correctly using the appropriate prefixes. 2. 3040 kJ/mol correct As4O6 is also a covalent compound; appro- priate prefixes are used in the name. It should 3. 1520 kJ/mol be correctly named tetraarsenic hexoxide to indicate the presence of six oxygen atoms in 4. 1140 kJ/mol each molecule. 5. 760 kJ/mol LDE Rank Lattice Energy 004 003 10.0 points Explanation: − Rank the crystal lattice energy of the salts qZn =+2C qS = 2C qNa =+1C qCl = −1C Al2O3, CaCl2, CaO, NaF, Mg3(PO4)2 from least to greatest: VNaCl = 760 kJ/mol If r is the distance between the two ions (the sum of the ionic radii), the energy interaction 1. NaF < CaO < CaCl2 < Mg3(PO4)2 < between ions is given by Coulomb’s Law: Al2O3 q1 q2 2. NaF < CaCl2 < CaO < Mg3(PO4)2 < V = Al2O3 correct 4 π ǫ0 r q1 q2 r = 3. CaO < NaF < CaCl2 < Al2O3 < 4 π ǫ0 V Mg3(PO4)2 rNaCl = rZnS 4. Mg3(PO4)2 < NaF < < CaO CaCl2 < qNa qCl qZn qS Al2O3 = 4 π ǫ0 VNaCl 4 π ǫ0 VZnS qZn qS VNaCl 5. Mg3(PO4)2 < NaF < CaCl2 < CaO < VZnS = qNa qCl Al2O3 (+2 C)(−2 C)(−760 kJ/mol) = Explanation: (+1 C)(−1 C) Sodium fluoride has the least lattice energy = −3040 kJ/mol , because of its small charges, then calcium chloride and calcium oxide with incremen- the energy released by the interaction. tally increasing charges. Both magnesium phosphate and aluminum oxide have identi- Internuclear Distance 1 cal magnitudes for their charges, but both of 005 10.0 points Version001–PracticeExam2–mccord–(49230) 3 Consider a potential energy diagram for the Cl interaction of a sodium ion (Na+) with a chlo- − ride ion (Cl ). Which of the following state- ments is/are true? I) Repulsive forces predominate at very Cl small internuclear distances. II) The minimum potential energy occurs 1. C6H6Cl2 when attractive forces are greatest. III) Attractive forces are linearly dependent 2. C4H12Cl2 on the internuclear distance. 3. C6H6 1. I and II 4. C6H4Cl2 correct 2. III only 5. C6H4Cl 3. I and III 6. C4H4Cl2 4. I only correct Explanation: 5. II and III LDE Molecular Polarity 004 6. I, II and III 007 10.0 points 7. II only Consider the labeled bonds in the molecule below and rank them from least to most polar Explanation: in terms of difference in electronegativity. b b It is true that repulsive forces predominate b b H b F b at very small internuclear distances; this is the e b result of the positively charged nuclei electro- d b statically repelling each other, which occurs H c B a C C N b despite that fact that sodium ion and chlo- b b b b b b b b b b b b b b b F F O b F ride ion have opposite charges overall. The b b b b b b attractive forces actually continue to increase 1. a < b < c < d < e as you force the ions closer together, but the repulsive forces increase more, thus offsetting 2. d < c < e < a < b correct the attractive forces and producing a net in- crease in potential energy. Thus, the mini- 3. e < b < d < c < a mum potential energy does not occur at max- imum attractive force, but rather at a ”sweet 4. c < d < e < b < a spot” where the sum of attractive and repul- sive forces is minimized. Attractive forces are 5. d < c < a < e < b a rectangular hyperbola that asymptotically approach zero as the internuclear distance ap- Explanation: proacues infinity. The polarity of a bond is proportional to the difference in electronegativity of the two Line Drawing to Formula bonded atoms. The electronegatavity differ- 006 10.0 points ences of the labeled bonds are 0.5, 1.0, 0.1, 0.0 and 0.4, respectively. Determine the molecular formula for the molecule: Lewis CO dash Version001–PracticeExam2–mccord–(49230) 4 008 10.0 points available from the atoms: Which of the following is the correct Lewis formula for carbon monoxide (CO)? A=4 × 2(C atom)+1 × 4(H atoms) − = 12 e b b 1. b O C b correct bb Hydrogen can form only one bond, so car- b b 2. b O C b bon atoms must serve as the central atoms. bb We place the most symmetrical arrangement b b b b possible. 3. O C b b b b The correct dot structure for the molecule b b b should show a complete octet (8 electrons) b b 4. O C b around the carbon atoms, two electrons b b b b around each hydrogen atom, and a total of bb b b 12 valence electrons for the entire structure: b b 5. b O C b bb b b H H C C 6. O C H H b b b b 7. b O C b As can be seen above, the carbon-carbon b b bond is a double bond. b b − b (If using the S = N A rule to determine 8. O C b b b the dot structure, × × − b b N = (8 2)+(2 4) = 24 e and b b − 9.