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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 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 . 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 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 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. (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 . 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 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 , the bond has partial double and partial 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 : 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. 22.40 liters the molar mass of the gas?

− 2. 9.68 liters 1. 44.8g · mol 1

− 3. 10.60 liters 2. 28.0g · mol 1

− 4. 4.84 liters 3. 56.0g · mol 1

− 5. 19.37 liters correct 4. 179 g · mol 1 correct Explanation: −1 5. 8.00 g · mol massini =21gMg Four quantities are required to describe the Explanation: behavior of gases: P (pressure), V (volume), V = 250 mL P = 1 atm T (temperature in Kelvin), and n (quantity ◦ T =0 C = 273.15 K m =2g in moles). Thus to know the volume of H2 The density of the sample is produced, we need to know P , T , and n. Fortunately, we know that the hydrogen is m 2g produced at STP. STP implies standard tem- ρ = = =8g/L V 0.25 L perature (1 atm or 760 torr) and temperature VersionPREVIEW–Exam1–JOHNSON–(53140) 6 ◦ (0 C or 273.15 K). Thus, to know the vol- ume of gas produced, we need to find n, the LDE Ranking Gases 003 number of moles of gas produced. 021 10.0 points Mg is the limiting reactant in the equation Which of the following molecules would described, and so we can determine the num- have the smallest a and b term, respec- ber of moles of H2 produced in the reaction: tively, in the van der Waals’ equation: 1 mol Mg O3, CHF3, SF5Cl, SiHCl3, Xe. mol H2 = 21.0 gMg 24.305 g Mg 1. Xe and O3, respectively correct × 1 mol H2 1 mol Mg 2. CHF3 and CHF3, respectively =0.804 mol H2 3. Xe and Xe, respectively We can then use the ideal gas law P V = n R T 4. Xe and SF5Cl, respectively to determine the volume of H2 gas produced: 5. SiHCl3 and O3, respectively n R T V = P Explanation: 0.08206 L · atm (0.864 mol) Xenon is the only non-polar species and  K · mol  = thus would have the smallest a term. 1 atm is the smallest in terms of molecular weight · (273.15 K) and would thus have the smallest b term. = 19.367 L Msci 12 1406 022 10.0 points Mlib 04 1049 In an improved version of the gas law, P is 020 10.0 points n2 a At constant temperature, the rate of effusion replaced by P + . In this expression,  V 2  of H2 is n2 a the second term, 2 , accounts for 1. one-half that of helium gas. V 1. the forces of attraction between 2. twice that of helium gas. molecules. correct

3. one-fourth that of gas. 2. the excluded volume of the molecules.

4. None of these 3. the size of the container.

5. four times that of oxygen gas. correct 4. the forces of repulsion between Explanation: molecules.

5. the size of the molecules. Rate of effusion of O2 MWH2 = Rate of effusion of H2 pMWO 2 Explanation: p EffO2 2 1 At high pressures, gas molecules are closer = =0.25 = EffH2 r32 4 together than they would be at lower pres- 1 sures. The attractive forces between gas EffO = EffH 2 4 2 molecules then become important. VersionPREVIEW–Exam1–JOHNSON–(53140) 7 When more molecules are present (greater n) 3. II only and when the molecules are close together (smaller V 2 in the denominator), the correc- 4. II and III only tion term becomes larger. In this correction term, large values of a indi- 5. I, II, and III cate strong attractive forces, and the correc- tion term as a whole is added to compensate 6. I only for attractive forces between gas molecules. Explanation: Chloromethane is held in solid form by both CIC T05 19 London forces and dipole-dipole forces. 023 10.0 points Which best describes the hydrogen bonding Dispersion Forces between two water molecules? 025 10.0 points hydrogen oxygen In which of these compounds would you find ONLY dispersion forces existing between the molecules? 1. I. HBr; II. NH3; III. CO2; 2. IV. CH2Cl2.

1. I and IV only

3. 2. III and IV only 3. II and IV only

4. I only 4. correct 5. III only correct Explanation: − 6. II only Note the water molecules line up so the δ + on the O of one molecule is aligned with a δ 7. II and III only on a H of another molecule. 8. I and II only ChemPrin3e 05 52 024 10.0 points 9. I and III only Chloromethane (CH3Cl) forms a molecular solid. What type of forces hold it in a solid 10. IV only configuration? I) London forces Explanation: II) dipole-dipole forces A nonpolar covalent molecule would have III) hydrogen bonding only dispersion forces with another nonpolar covalent molecule. 1. I and II only correct Msci 13 0302 2. III only 026 10.0 points The term used to describe resistance to flow VersionPREVIEW–Exam1–JOHNSON–(53140) 8 of a liquid is CH2OHCH2OH < CH2OHCHOHCH2OH

1. vaporization. Evaporation and IMF Ranking 028 10.0 points 2. capillary action. Rank the following compounds H2O NH3 HF CH3F 3. viscosity. correct in terms of increasing vapor pressure.

4. surface tension. 1. CH3F < HF < H2O < NH3

5. vapor pressure. 2. H2O < HF < NH3 < CH3F correct Explanation: 3. CH3F < NH3 < HF < H2O Viscosity is used to describe the resistance to flow of a liquid. 4. NH3 < HF < H2O < CH3F

ChemPrin3e 05 26 5. CH3F < H2O < HF < NH3 027 10.0 points Rank the liquids Explanation: I) C6H6 IV) CH2OHCH2OH II) CH3CH2OH V)H2O LDE Intermolecular Forces 007 III) CH2OHCHOHCH2OH 029 10.0 points ◦ in order of increasing viscosity at 25 C. Rank the following molecules in order of in- creasing intermolecular forces: C2H2, H2, 1. I < V < II < IV < III correct C(CH3)4,N2.

2. II < IV < I < V < III 1. N2 < H2 < C2H2 < C(CH3)4

3. I < V < III < IV < II 2. H2 < N2 < C(CH3)4 < C2H2

4. I < IV < II < V < III 3. N2 < N2 < C(CH3)4 < C2H2

5. IV < V < II < I < III 4. H2 < C2H2 < N2 < C(CH3)4

Explanation: 5. H2 < N2 < C2H2 < C(CH3)4 correct We need to consider both the strength of intermolecular forces and the tendency of Explanation: molecules to get tangled like spaghetti. All of the molecules are nonpolar and have Since glycerol (CH2OHCHOHCH2OH) can only instantaneous dipole forces. Instanta- form several hydrogen bonds per molecule neous dipole forces increase with molecular and it has a long chain structure, it has weight. both strong intermolecular forces and the tendency to get tangled. Ethylene glycol ChemPrin3e 05 20 (CH2OHCH2OH) is just one −CH−OH− unit 030 10.0 points smaller than glycerol, so we would expect it to Classify the solid dry ice (CO2). be less viscous than glycerol but more viscous than water. Ethanol (CH3CH2OH) is slightly 1. network more viscous than water, which is more vis- cous than benzene (C6H6). 2. None of these C6H6 < H2O < CH3CH2OH VersionPREVIEW–Exam1–JOHNSON–(53140) 9 3. ionic

4. molecular correct Explanation: Molecular solids consist of molecules held together by weak intermolecular forces. Ionic solids are held together by electro- static attraction between metal cations and non-metal anions. Covalent (or network) solids are like huge molecules held together by covalent bonds. in diamond is the most well-known example. Group IV B elements can form tetrahedral electronic geometries.