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Version PREVIEW – Exam 1 – JOHNSON – (53140) 1 This Print VersionPREVIEW–Exam1–JOHNSON–(53140) 1 This print-out should have 30 questions. ·· Multiple-choice questions may continue on ·Cl· the next column or page – find all choices ··· ··· before answering. ·Cl·· C Cl·· · · · LDE Identifying Bonds 005 ·Cl·· · 001 10.0 points The molecule has tetrahedral electronic and Rank the labeled bonds in the molecule molecular geometry. The C−Cl bond is polar, below from least to most polar. but becuase of the symmetry of the molecule, b b b H O b the individual dipole moments cancel. The a b molecule is therefore nonpolar. b c b H C N e ChemPrin3e T03 39 d b b b b S S H 003 10.0 points b b b b Which of the following is NOT polar? 1. c < e < b < a < c 1. COCl2 2. c < d < e < a < b 2. ClF3 3. d < c < e < b < a correct − 3. BrO3 4. c < d < e < b < a + 4. CH3 correct 5. d < c < b < e < a 5. O3 Explanation: Bonds a, b, c, d, and e have a ∆EN of 1.4, 1.0, 0.5, 0.0, and 0.9, respectively. Explanation: COCl2 (the central atom is C) is polar be- Brodbelt 013 322 cause there are two different bonds: C=O 002 10.0 points and C-Cl which have different sized dipoles Which of the following statements about po- so their effects do not cancel. The other in- larity is false? correct choices have either 3, 4 or 5 RHED: one or more of these are lone pairs on the 1. Linear molecules can be polar. central atom which causes the polar bonds to be placed in positions where the dipole of at least one bond is not opposing another, caus- 2. Polar molecules must have a net dipole + ing the species to be polar. In CH3 there are moment. ◦ no lone pairs; the C-H bonds are at 120 so 3. Lone (unshared) pairs of electrons on the their effects cancel and the ion is nonpolar. central atom play an important role in influ- encing polarity. ChemPrin3e T03 22 004 10.0 points Which of the following has bond angles 4. CCl4 is a polar molecule. correct ◦ slightly less than 109 ? 5. Dipole moments can “cancel”, giving a + net non-polar molecule. 1. NH4 − Explanation: 2. BrO3 correct The Lewis Dot structure for CCl4 is VersionPREVIEW–Exam1–JOHNSON–(53140) 2 − 3. ClO4 5. 2; 6 − 4. BH4 6. 4; 2 3− 5. PO4 7. 3; 4 Explanation: 8. 3; 6 While all structures have four regions of electron density around the central atom, only − 9. 3; 2 BrO3 also has one of these regions as a lone pair. Since the lone pair requires more room, Explanation: the other bonds are repelled away from it slightly, folding them up rather like an um- N =8 × 3=25 brella and resulting in angles slightly less than ◦ × 109.5 . A =5+(7 2)+1=20 S = 24 − 20=4 Brodbelt 013 320 First, draw the molecule. The number of 005 10.0 points 3 regions of HED equals the number of bonds The sp hybridization has what percent s and the number of lone pairs. There are 4 character and what percent p character? shared, or bonded electrons, giving 2 regions of HED. The 2 F atoms have an octet. This 1. 50%; 50% means that 16 of the 20 available electrons are accounted for, leaving 4 electrons to be placed 2. 33%; 67% on the N atom as 2 lone pairs. Since there are 2 lone pairs, 2 bonds, that gives a total of 4 3. 67%; 33% regions of HED. 4. 75%; 25% Brodbelt 08 02 007 10.0 points 5. 25%; 75% correct Which molecular geometries can stem from Explanation: tetrahedral electronic geometry? 1 s = = 25% 1. tetrahedral, trigonal pyramidal, angular 4 correct 3 p = = 75% 4 2. trigonal pyramidal, seesaw Brodbelt 013 308 3. tetrahedral, T-shaped, linear 006 10.0 points − NF has (2, 3, 4) regions of high electron 2 4. only tetrahedral density and (2, 4, 6) bonded electrons. 5. tetrahedral, trigonal bipyramidal, linear, 1. 2; 2 angular, seesaw 2. 4; 4 correct Explanation: Tetrahedral, trigonal pyramidal and angu- 3. 2; 4 lar are three molecular geometries that can stem from tetrahedral electronic geometry. 4. 4; 6 Benzene Sigma Pi Bonds VersionPREVIEW–Exam1–JOHNSON–(53140) 3 008 10.0 points correct How many σ and π bonds, respectively, are found in benzene (C6H6)? 5. Antibonding orbitals are made from the overlap of sigma and pi orbitals. 1. 6; 6 Explanation: Antibonding orbitals are higher in energy 2. 12; 6 than their corresponding bonding orbitals. Molecules can exist with electrons in anti- 3. 12; 12 bonding orbitals, as long as the molecule does not place as many electrons in antibonding 4. correct 12; 3 orbitals as there are in bonding orbitals. 5. 3; 3 Msci 02 1186 Explanation: 011 10.0 points C2 is calculated to have a bond order of two Bonds in C2H2 according to MO theory. Pick the true state- 009 10.0 points ment. Which of the following bonds is not found in 1. acetylene (C2H2)? The molecule has exactly two pairs of bonding electrons and no more. 1. All of these bonds are found in C2H2. correct 2. The molecule is held together by two pi bonds. correct 2. π2p−2p 3. The molecule is held together by two sigma bonds. 3. σsp−sp 4. The molecule is paramagnetic. 4. σ1s−sp Explanation: 5. The molecule contains one sigma and one pi bond. Msci 02 1239 Explanation: 010 10.0 points The MO energy diagram will fill with 8 Which of the following is TRUE about anti- total electrons (4 from each C). 4 electrons fill bonding orbitals of the molecular orbital the- the σ2s and σ2s⋆ MO’s. The order of the MO’s ory? made with the overlapping 2p’s of the two C’s is 1. Unshared electrons are placed in anti- bonding orbitals. π2p,π2p <σ2p <π2p⋆ ,π2p⋆ <σ2p⋆ 2. The strongest bonds between atoms have So the 2 π2p’s fill first meaning C2 is held no electrons in antibonding orbitals. together with 2 π bonds and no σ’s. 3. Although antibonding orbitals may ac- LDE Bond Order 007 cept electrons, the electrons never remain for 012 10.0 points long. All of the species below have the same bond order except for one of them. Which is it? 4. Antibonding orbitals are higher in energy − than their corresponding bonding orbitals. 1. H2 VersionPREVIEW–Exam1–JOHNSON–(53140) 4 Msci 09 0604 − 2. B2 correct 015 10.0 points Which of the following substances has a delo- + 3. Ne2 calized bond? − 4. F2 1. CO + 2− 5. H2 2. CO3 correct Explanation: 3. CO2 All of the species have a bond order of 0.5 − except for B2 , which has a bond order of 1.5. − 4. ClO3 LDE Paramagnetism 005 5. NH3 013 10.0 points Which of the following species is paramag- Explanation: netic? Delocalized bonds occur whenever reso- nance occurs. In a molecule that exhibits 1. He2 resonance, the bond has partial double and partial single bond character. This means 2. N2 that electrons are delocalized around the res- 2− 2− onance bond. CO3 is the only compound 3. Be2 correct that exhibits resonance and therefore delocal- 2− ization. 4. C2 Explanation: Mlib 04 1137 2− 016 The species Be2 will have two unpaired 10.0 points electrons in degenerate π bonding orbitals. The volume of a gas varies directly with its Kelvin temperature, at constant pressure. LDE Bond Order 008 This is a statement of 014 10.0 points Rank the following species in terms of increas- 1. Boyle’s Law. − + 2− ing bond length: N2, O2 , Ne2 ,H2,B2 . 2. Dalton’s Law of partial pressure. + − 2− 1. Ne2 < H2 < O2 < N2 < B2 3. Henry’s Law. 2− − + 2. N2 < B2 < O2 < H2 < Ne2 correct 4. Charles’ Law. correct 3. − + 2− N2 < O2 < H2 < Ne2 < B2 Explanation: + 2− − Charles’ Law relates the volume and tem- 4. Ne < H2 < B < O < N2 2 2 2 perature of a fixed (definite) mass of gas at − 2− + constant pressure. The direct proportionality 5. N2 < O < B < Ne < H2 2 2 2 applies only if the temperature is expressed Explanation: on the absolute (Kelvin) scale. − + 2− The species N2, O2 , Ne2 ,H2 and B2 have bond orders of 3, 1.5, 0.5, 1 and 2, respec- LDE Ideal Gas Calculation 006 tively. Bond length is inversely proportional 017 10.0 points to bond strength. If a gaseous system initially at 789 torr and ◦ 215 C occupies a volume of 22.80 cm3, what VersionPREVIEW–Exam1–JOHNSON–(53140) 5 volume will it occupy if the pressure is reduced The ideal gas law is to 456 torr and the temperature is increased ◦ to 430 C? P V = n R T n P = 1. 9.14 cm3 V R T with unit of measure mol/L on each side. 2. Not enough information. Multiplying each by molar mass (MM) gives 3 n P 3. 5.86 cm · MM = · MM = ρ, V R T 3 4. 88.73 cm with units of g/L. 5. 56.83 cm3 correct ρ R T MM = Explanation: P It is not necessary to use L and atm for (8 g/L)(0.08206 L · atm/mol/K) = this problem because the relationship can be 1 atm ◦ written so that the units cancel ( C must be × (273.15 K) converted to K, though, because it is not an = 179.318 g/mol absolute measure): P1V1 P2V2 = Brad C12 006 T1 T2 019 10.0 points P1V1T2 If sufficient acid is used to react completely = V2 T1P2 with 21.0 grams of Mg 3 ◦ (789 torr)(22.80 cm )(703 C) 3 = 56.83 cm Mg(s) + 2HCl(aq) → MgCl2(aq)+ H2(g) (488 ◦C)(456 torr) what volume of hydrogen at STP would be ChemPrin3e T04 25 produced? 018 10.0 points If 250 mL of a gas at STP weighs 2 g, what is 1.
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