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. O C S = 24 − 12 = 12 e . b b b b This would indicate 6 bonds and corre-
bb b b spond to the structure shown above.) b 10. b O C b bb b Lewis BF3 dash Explanation: 010 10.0 points The Lewis formula for carbon monoxide Which of the following is the correct Lewis
b b formula for boron trifluoride (BF3)? (CO) is b O C b
b b
b Mlib 03 1055 F b 009 10.0 points b b b 1. b F B
What kind of carbon-carbon bond is in a b b b b molecule of ethylene (C2H4)? b F b b b
1. ionic b b b b bb
b b b F B F b 2. single 2. b b bb
b b b F b 3. double correct bb
b b bb
b b 4. triple b F B F b 3. b b bb correct
b b Explanation: b F b To draw the dot structure for C2H4 we must bb first calculate the number of valence electrons Version001–PracticeExam2–mccord–(49230) 5
b b b b FB 3. 1 4. b b
F b F b b b 4. 2
5. 5 F B F 5.
b b b F b
bb Explanation: b b
b b b b
b b b b bb b b
The Lewis structure is b F Xe F b
b b b b b b b F B F b 6. bb
b b LDE Resonant Species 002 b F b bb 012 10.0 points Which of the following species exhibit reso- bb
b nance/delocalization? F B F b 7. bb I) HCN
b b b b II) O3 F − bb 2 III) CO3
b b
b 1. I, III F b 8. FB 2. III only F 3. I only
4. II, III correct 9. FBF F 5. I, II, III b b F B F 10. 6. II only
b b b F b bb 7. I, II Explanation: The Lewis formula for boron trifluoride b b bb Explanation: b b b F B F b Both ozone and carbonate have a single (BF3) is b b bb
b b pair of resonant electrons and are famous ex- b F b
bb amples of resonant molecules. Cyanide can- not have resonance since hydrogen can only ChemPrin3e T02 23 form a single bond. 011 10.0 points ChemPrin3e 02 46 Draw the Lewis structure of xenon diflu- 013 10.0 points oride and give the number of lone pairs of Consider the following sets A, B, and C electrons around the central atom. 2− bb bb A1) N C N 1. bb bb 4 2− bb
b b A2) b N C N b 2. 3 correct bb Version001–PracticeExam2–mccord–(49230) 6
bb − − − 3 − 3 bb b b b b b b b b b b b b 0 − O b b O b O b 1 O −1 bb bb 0 bb bb bb bb 0 bb bb b b b b +1 b b b b O As O O As O B2) bb bb C2) O I O B1) bb bb C1) O I O bb bb bb bb −1 0 b b b b b b b b b b − b b b b b b O 1 O −1 O O bb bb bb bb − − − bb 3 b b − bb 3 b b b b b b b b b b 0 − b b 1 O b O b O O 0 −1 bb − bb bb bb 3 bb bb bb bb b b b +3 b b b O As O b b O As O B3) bb bb C3) O I O B2) bb bb C2) O I O bb bb bb bb 0 −1 b b b b b b b b b b b b b b b b 0 O −1 O O O bb bb bb − bb b b 3− b b b b O b O b ChemPrin3e T03 26
bb bb bb bb 014 10.0 points b b O As O b b B3) bb bb C3) O I O Which of the following has bond angles bb bb ◦ slightly less than 109.5 ? b b b b b O b b O b bb − of ions. 1. I3 Determine the formal charge on each atom and identify the structure of lowest energy for 2. O3 each set A, B, and C. + 3. CH3 1. A2,B2,C2 4. HOCl correct 2. A2,B3,C3 − 5. NO2 3. A1,B1,C1 Explanation: Only HOCl has four regions of electron den- 4. A1,B2,C3 sity around the central atom; the two lone pairs repel the bonds more, making the bond 5. A2,B3,C2 ◦ angle less than 109.5 .
6. A1,B2,C1 correct Mlib 03 2087 015 10.0 points Explanation: 2− Which of the following is planar? bb bb A1) N C N bb bb 1. NH3 −1 0 −1 2− 2. SO3 2− bb
b b A2) b N C N b − bb 3. NO3 correct 0 0 −2 + 4. H3O
bb 3− − b b 5. PF3 − b b b b 1 O b O b 0 −1 0 bb +1 bb bb bb Explanation: b b 0 b O As O b O I O B1) bb bb C1) bb bb All except the nitrate ion have trigonal −1 0 b b pyramidal molecular geometries; the nitrate b b b b −1 b b −1 O O bb bb ion has trigonal planer geometry. Version001–PracticeExam2–mccord–(49230) 7
ChemPrin3e T03 16 LDE VB Hybridization 001 016 10.0 points 018 10.0 points What is the shape (molecular geometry) of Which of the following statements concerning + IF4 ? hybrid orbitals is/are true? I) Hybrid orbitals are energetically degen- 1. square planar erate. II) Any element can form sp3d2 hybrid or- 2. tetrahedral bitals. III) Hybridizing a 2s and a 2p orbital would 3. seesaw correct produce one single sp hybrid orbital.
4. trigonal bipyramidal 1. III only
5. T-shaped 2. II, III Explanation: 3. I only correct There are five regions of electron density (including one lone pair) around the central 4. II only atom:
b b b b b b + 5. I, II
b b
b b F I F b b b b b
b 6. I, III
b b
b b
b b
b F F b b b 7. I, II, III ChemPrin3e T03 35 Explanation: 017 10.0 points Statement I is true; hybridization was de- Which of the following is polar? veloped as a theoretical framework to ex- plain the energetic degeneracy of bonds in 1. correct IF5 molecules. Statement II is false; hybridiza- tion involving d orbitals requires access to 2. SF6 empty d orbitals, and thus begins in period 3. Statement III is false; the number of or- 3. − ICl4 bitals used to hybridize is always equal to the number of hybridized orbitals, so using a 2s 4. XeF4 and a 2p orbital would result in two sp hybrid orbitals. 5. PCl5 Explanation: LDE VB Sigma Pi Bonds 005 − XeF4 and ICl4 have 6 RHED with two 019 10.0 points RHED being lone pairs situated opposite each How many different types of sigma (σ) bonds other so their effects cancel. Only IF5 has an are found in ethanoic acid, CH3COOH? In unbalanced number of lone pairs which places other words, how many different combina- one of the polar I F bonds opposite a lone tions of atomic orbitals are used when forming pair. the σ bonds in ethanoic acid. This unopposed dipole and the lone pair itself make it polar. The others have either 5 1. 5 or 6 RHED and no lone pairs on the central atom. 2. 4 Version001–PracticeExam2–mccord–(49230) 8
3. 7 5. sp3, sp, sp correct
4. 3 correct 6. sp3, sp3, sp2
5. none of the above 7. sp2, sp, sp2 Explanation: Explanation: All of the σ bonds found in ethanoic acid are sp3 − 1s,sp3 − sp2, or sp2 − sp2. 3 sp Hsp
LDE VB Sigma Pi Bonds 004 ← ← H C C C H 020 10.0 points ← How many sigma (σ) and pi (π) bonds are in H sp the Lewis structure for C(COOH)4?
1. 8 σ, 4 π LDE MO Diagram 001 022 10.0 points 2. 12 σ, 4 π Consider the following molecular orbital dia- gram: 3. 16 σ, 0 π a 4. 12 σ, 0 π
5. 16 σ, 4 π correct Explanation: b
ChemPrin3e 03 76 c 021 10.0 points
Carbon has a valence of four in nearly all of its compounds and can form chains and rings What are the names of the labeled orbitals, of C atoms. Consider the propyne structure. a, b, and c, respectively? H a c ∗ ∗ 1. π2p,σ2p,σ2s ← ← H C C C H ∗ ∗ ← 2. π2p,σ2p,σ2s
H b ∗ ∗ 3. σ2p,π2p,σ1s What hybridizations would you expect for the carbon atoms identified by a, b, and c, ∗ 4. σ2p,π2p,σ2s correct respectively? ∗ ∗ 5. σ2p,π2p,σ2s 1. sp3, sp2, sp3 Explanation: 2. sp2, sp2, sp3 LDE Bond Order 009 3. sp2, sp, sp 023 10.0 points All of the species below have the same bond 4. sp3, sp, sp3 order except for one of them. Which is it? Version001–PracticeExam2–mccord–(49230) 9
+ 1. Ne2 4. II only
− 2. B2 correct 5. III only
− 3. H2 6. I only
− 4. F2 7. I and II
+ Explanation: 5. H2 − + Li2 and H2 both have an odd number Explanation: of electrons and therefore must be paramag- netic. O2 has 16 total electrons, the last two All of the species have a bond order of 0.5 ∗ − of which must go into separate degenerate π except for B2 , which has a bond order of 1.5. anti-bonding orbitals. Msci 09 0116 024 10.0 points An antibonding orbital is formed when
1. a px-orbital overlaps a pz-orbital.
2. a free electron is present in the molecule.
3. an s-orbital overlaps a p-orbital.
4. None of these is correct.
5. the overlap of the corresponding atomic orbitals leads to destructive interference. cor- rect
Explanation: In-phase overlap of two atomic orbitals re- sults in molecular bonding orbitals; out-of- phase overlap of two atomic orbitals results in molecular antibonding orbitals.
LDE Paramagnetism 004 025 10.0 points Which of the following species is/are param- agnetic? − I) Li2 II) O2 + III) H2
1. II and III
2. I and III
3. I, II and III correct