HW07-VSEPR-POLARITY-VB-solutions
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HW07-VSEPR-POLARITY-VB-solutions

Course Number: CH 301, Fall 2012

College/University: University of Texas

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parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points 1 4. C8 H5 NO2 Explanation: Remember that the shorthand notation Consider the structural formula means C6 H5 OH 003 of phenol, the active ingredient in some oral anesthetics used in...

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(jrp3484) parga HW07-VSEPR-POLARITY-VB vandenbout (51335) This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points 1 4. C8 H5 NO2 Explanation: Remember that the shorthand notation Consider the structural formula means C6 H5 OH 003 of phenol, the active ingredient in some oral anesthetics used in sore-throat sprays. What is the molar mass of phenol? 10.0 points This structure C C C C C C 1. 94 g/mol correct CC is the carbon skeleton for a structural isomer of octane with the hydrogen atoms omitted. What is the molecular formula of this isomer? 2. 50 g/mol 3. 89 g/mol 1. C8 H18 correct 4. 17 g/mol Explanation: Remember that the shorthand notation means a phenyl group C6 H5 . The 10.0 points This is the condensed structural formula for acetaminophen, the active ingredient in the over-the-counter medication Tylenol. Explanation: The structure is H H H N O 3. C8 H8 4. C8 H17 chemical formula is C6 H5 OH 002 2. C8 H24 H C H H H H C C C H H OH H C C C CH3 What is the molecular formula of acetaminophen? 1. C8 H9 NO2 correct 2. C8 H11 NO2 3. C8 H8 NO H H C H H H C H H 004 10.0 points Consider the structure H parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) C C CH2 CH2 CH2 CH3 O C O CH3 of a COX-2 inhibitor. How many single bonds, double bonds, and triple bonds are represented by this condensed formula? 1. 31 single, 4 double, 1 triple correct 2. 31 single, 3 double, 1 triple 3. 13 single, 3 double, 1 triple C F correct 5. C CH2 3. C 4. C S 2 N Explanation: Dipole moments result because of the dierence in electronegativity (EN) between bonding atoms. The bonds with the largest dipole will have the largest dierence in EN. EN increases from bottom to top and left to right on the periodic table. C C and N N have the same EN, so there is no dipole moment. Between N, O, and F, F is the most EN, and has the largest dierence in EN with C. N bond has the largest Therefore, the C dipole. 4. 13 single, 4 double, 1 triple Explanation: The benzene ring can be shown as H H H C C C C C C H H H 007 10.0 points Consider a 3-atom molecule A B A for which B has a total of only four valence electrons enough to make two bonds. Predict the ABA bond angle. 1. 180 correct 2. 90 3. 60 005 10.0 points The electronegativity of H is 1. a lot less than that of C. 2. a lot more than that of C. 3. about equal to that of C. correct 4. 109.5 5. 120 Explanation: 008 10.0 points What is the shape of COCl2 ? Explanation: 1. T-shaped 006 10.0 points Which pair of bonded atoms has the largest dipole moment? 2. trigonal pyramidal 3. tetrahedral 1. N N 2. C O 4. seesaw 5. trigonal planar correct parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) Explanation: There are three regions of electron density (with no lone pairs) around the central atom: 8. 4; 0; 0; 0 O 3 9. 2; 0; 0; 2 10. 0; 0; 1; 1 C Cl Cl 009 10.0 points Which of the following has bond angles slightly LESS than 120 ? Explanation: N has 5 valence electrons and O has 6 valence electron. The total number of valence electrons in the ion (plus one to account for the ions negative charge) is 5 + (2 6) + 1 = 18 1. I 3 The Lewis structure for NO can be found 2 by connecting the oxygen atoms to nitrogen and placing eight electrons around each atom: 2. SO3 3. SF2 ON O 4. NO 3 5. O3 correct Explanation: While SO3 and O3 have three regions of electron density around the central atom, only O3 has a lone pair in one of these regions. The lone pair repels the other bonds more, which makes their angle slightly less than 120 . 010 (part 1 of 2) 10.0 points NO . 2 Draw the Lewis structure for How many single bonds, double bonds, triple bonds, and unshared pairs of electrons are on the central atom, in that order? and the central atom has one single bond, one double bond, and one unshared pair of electrons. 011 (part 2 of 2) 10.0 points Determine the molecular geometry of the ion NO . 2 1. trigonal-planar 2. tetrahedral 3. octahedral 4. None of these 1. 1; 0; 1; 0 5. linear 2. 3; 0; 0; 1 6. bent or angular correct 3. 1; 1; 0; 1 correct 7. trigonal-pyramidal 4. None of these 8. trigonal-bipyramidal 5. 2; 1; 0; 0 6. 0; 2; 0; 0 7. 0; 1; 1; 0 Explanation: The central atom N has 2 bonding regions and 1 lone pair of electrons. The three areas of high electron density make the electronic geometry trigonal planar and the molecular geometry bent or angular: parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) 4 1. square planar N O O 2. tetrahedral 3. linear 012 (part 1 of 3) 10.0 points What is the electronic geometry of IF ? 4 1. trigonal planar 2. T-shaped 3. trigonal bipyramidal 4. octahedral correct 5. tetrahedral 4. trigonal planar correct 5. trigonal pyramidal Explanation: 016 10.0 points Which of the following has bond angles of 90 , 120 , and 180 ? 1. SF4 correct Explanation: 2. XeF4 013 (part 2 of 3) 10.0 points What is the molecular geometry of IF ? 4 3. PF 6 4. ICl 4 1. linear 5. IF5 2. spherical 3. tetrahedral 4. square planar correct 5. pointy looking Explanation: 014 (part 3 of 3) 10.0 points Is IF non-polar ? 4 1. No, its polar. 2. Yes, its non-polar. correct 3. Cannot be determined from the structure. Explanation: Only SF4 has ve regions of electron density around the central atom; the rest have six. 017 10.0 points A central atom is surrounded by four chlorine atoms. Which of the following combinations is possible? 1. a trigonal bipyramidal electronic geometry and trigonal bipyramidal molecular geometry 2. an octahedral electronic geometry and square pyramidal molecular geometry 3. a trigonal bipyramidal electronic geometry and t-shaped molecular geometry Explanation: 015 10.0 points What is the geometry around the leftmost carbon in the molecule CH2 CHCH3? 4. an octahedral electronic geometry and tetrahedral molecular geometry 5. a trigonal bipyramidal electronic geome- parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) try and seesaw molecular geometry correct Explanation: 018 10.0 points Consider the compound peroxyacetylnitrate, an eye irritant in smog. O H3 C N O O Predict the indicated bond angle. 1. 109.5 2. 90 3. 30 atom except for SF4 . SF4 has a lone pair and square pyramidal geometry which gives it an asymmetric distribution of electrons around the central atom S, so SF4 is a polar molecule. 020 10.0 points Which of the following about statements polarity is false? O C O 5 1. Lone (unshared) pairs of electrons on the central atom play an important role in inuencing polarity. 2. Linear molecules can be polar. 3. Dipole moments can cancel, giving a net non-polar molecule. 4. CCl4 is a polar molecule. correct 4. 120 correct 5. 60 Explanation: There are 46 valence electrons; the structure does not show lone pairs on the O atoms nor the C H bonds. There will be three lone pairs on the terminal O singly bonded to N, and two lone pairs on all other O atoms. The indicated atom has RHED = 3, so the bond angle will be 120 . 019 10.0 points Which of the following is a polar molecule? 1. XeF2 2. SF4 correct 5. Polar molecules must have a net dipole moment. Explanation: The Lewis Dot structure for CCl4 is Cl Cl Cl Cl The molecule has tetrahedral electronic and molecular geometry. The CCl bond is polar, but becuase of the symmetry of the molecule, the individual dipole moments cancel. The molecule is therefore nonpolar. 021 10.0 points Which of the following molecules is nonpolar? 3. SO3 1. SO2 4. CO2 2. BF3 correct 5. CCl4 3. CH3 Br Explanation: Each molecule listed has polar bonds arranged symmetrically around the central C 4. H2 O 5. NF3 parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) Explanation: The structure for BF3 is F F B F There are no lone pairs on B. This is a symmetrical molecule and non-polar. 022 10.0 points CHF3 is (less,more) polar than CHI3 because 6 2. I and II only 3. I and III only 4. II only 5. I only 6. II and III only correct 7. None t the criteria. 8. III only 1. less; the C-H bond in CHF3 is a non-polar bond. 2. more; the C-F bonds are more polar than the C-I bonds. correct 3. more; the C-H bond in CHF3 is a nonpolar bond. 4. less; the tetrahedral geometry decreases the polarity of C-F bonds. 5. less; the three polar C-F bonds are symmetrical and cancel the dipole moments. Explanation: Since F is more electronegative than I, a C-F bond will have a greater dierence in electronegativities than a C-I bond. Explanation: I) Water has polar covalent bonds and is itself polar. II) Carbon dioxide contains polar covalent bonds but is not polar overall due to symmetry. III) Boron trichloride contains polar covalent bonds but is not polar overall due to symmetry. 024 10.0 points Which of the following molecules has the largest dipole moment? 1. H2 2. HCl correct 3. HBr 023 10.0 points Which of the molecules I) II) O H O 5. HI H C O Cl I II) B Cl 4. F2 Cl contains polar covalent bonds but is NOT itself a polar molecule? Explanation: The molecule with the largest dipole moment is the molecule with the most polar bond and the molecule with the largest dierence in electronegativities. Out of the choices listed, HCl is the molecule with the largest dipole moment. 025 10.0 points Classify the molecule PBr3 . 1. All t the criteria. 1. polar molecule with nonpolar bonds parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) 7 4. 33%; 67% 2. nonpolar molecule with nonpolar bonds 5. 67%; 33% 3. polar molecule with polar bonds correct Explanation: 4. nonpolar molecule with polar bonds 1 = 25% 4 3 p = = 75% 4 Explanation: 026 10.0 points Which of the following combinations of hybridization and molecular geometry is possible? 1. sp3 : octahedral s= 028 10.0 points What hybridization would you expect for Se when it is found in SeO2 ? 4 2. sp2 : linear 1. sp3 d 3. sp3 : trapezoidal 2. p 4. sp2 : tetrahedral 3. s2 p2 5. sp3 : trigonal pyramidal correct 4. sp3 correct 6. sp3 d : tetrahedral 5. sp3 d2 7. sp3 d : octahedral 6. sp 8. sp3 : trigonal planar 7. sp2 Explanation: Use the hybridization to conrm the electronic geometry; e.g., sp2 means three hybrid orbitals, three RHED and a trigonal planar electronic geometry. Verify if the molecular geometry is possible for the given electronic geometry: for example, for trigonal planar electronic geometry, only trigonal planar and bent are possible. 027 10.0 points The sp hybridization has what percent s character and what percent p character? 3 1. 50%; 50% 8. spd Explanation: N = 5 8 = 40 A = (6 5) + 2 = 32 S = 40 32 = 8 There are 8 shared or bonded electrons. The remaining electrons are placed around the O atoms. There are no lone pairs on the central Se atoms. This molecule has tetrahedral electronic and molecular geometry so there are 4 regions of high electron density. This corresponds to sp3 hybridization. 2. 75%; 25% 3. 25%; 75% correct 029 10.0 points Give the hybridization of each central atom: nitrogen, middle carbon, right carbon. parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) H H N C C HO H H 8 031 10.0 points sp2 hybrid orbitals have 1. trigonal bipyramidal symmetry. 1. sp2 , sp2 , sp2 2. tetrahedral symmetry. 2. sp3 , sp2 , sp3 3. trigonal planar symmetry. correct 3. sp, sp3 , sp3 4. trigonal pyramidal symmetry. 4. sp, sp, sp 5. linear symmetry. 5. sp2 , sp3 , sp3 6. sp2 , sp2 , sp3 correct Explanation: 7. sp3 , sp3 , sp3 032 10.0 points What is the hybridization at the atom that is indicated by the arrow in the following molecule? O 8. sp2 , sp, sp2 Explanation: Hybridization is chosen based on the electronic geometry around the central atom, which is based on the number of RHED around the central atom. The RHED are three, three, and four. 030 10.0 points What hybridization would you expect for C in ethyne (C2 H2 )? H 2. sp 3. sp correct 4. sp3 d 4 5. sp d Explanation: Ethyne is a linear molecule with two central atoms each with 2 areas of electron density. It has sp hybridization and this orbital forms the sigma bonds. Thus it has 2 unhybridized p orbitals which form the 2 pi bonds in the triple bond between the 2 carbon atoms. See page 160 of the course supplment. O H H O H H C C O H Acetylsalicylic acid (Aspirin) 1. 2sp3 d 2 C H 1. sp3 2. sp2 correct 3. sp 4. sp3 d2 5. sp3 d Explanation: 033 10.0 points Consider the following molecule: H parga (jrp3484) HW07-VSEPR-POLARITY-VB vandenbout (51335) What are the bond angles A and B? 1. A = 120 , B = 180 2. A = 120 , B = 120 3. A = 120 , B = 240 4. A = 109.5 , B = 109.5 5. A = 120 , B = 109.5 correct 6. A = 109.5 , B = 120 Explanation: The angle for A is centered on a carbon with sp2 hybridization and is therefore 120 . The angle for B is centered on an sp3 hybridized carbon (remember, the Hs arent shown but are implied in this type of structure) and are therefore 109.5 . 9

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parga (jrp3484) Homework 6 homann (57870)This print-out should have 15 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 10.0 pointsYou do a certain amount of work on anobject initially at
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parga (jrp3484) Homework 7 homann (57870)1cb.2. The force is zero at the two values of rthat satisfy a r 2 b r + c = 0 .3. The force goes to zero only at r = 0 .4. The force is zero only at r =bcorrect2a.5. The force goes to zero as r .6. Ther
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University of Texas - PHY - 303K
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 C
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
parga (jrp3484) HW10 radin (55095)This print-out should have 19 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.00115. A only6. B and C only7. A and B only10.0 pointsRecently, Stewart dr
University of Texas - MATH - 408 D
parga (jrp3484) HW11 radin (55095)This print-out should have 12 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001Consequently,temp = T (30) 160 F .00210.0 pointsRichard removes a turkey
University of Texas - MATH - 408 D
parga (jrp3484) HW12 radin (55095)This print-out should have 23 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.0011y4.410.0 points2If f is a continuous function on [0, 6] havingx(1)
University of Texas - MATH - 408 D
parga (jrp3484) HW13 radin (55095)This print-out should have 19 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.00110.0 points1Consequently,A. not have: ( f (x) > 0, x = 0);B. has: ( f (x
University of Texas - MATH - 408 D
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University of Texas - MATH - 408 D
parga (jrp3484) HW15 radin (55095)This print-out should have 13 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.0011Thus the critical points of A(x) occur at5x = ,25,2only one of whic
University of Texas - MATH - 408 D
parga (jrp3484) HW17 radin (55095)1yThis print-out should have 14 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001R210.0 pointsaUse properties of integrals to determine thevalue ofc
University of Texas - MATH - 408 D
parga (jrp3484) HW18 radin (55095)14This print-out should have 26 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.0012.3221010.0 points-12-2A function h has graph22-3-433.2
University of Texas - MATH - 408 D
parga (jrp3484) HW19 radin (55095)This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.0011keywords: IntSubst, IntSubstExam,00210.0 pointsEvaluate the integral10
University of Texas - MATH - 408 D
parga (jrp3484) HW20 radin (55095)This print-out should have 19 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.00164.4210.0 points2For which one of the following shaded regions is its
University of Texas - MATH - 408 D
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parga (jrp3484) HW22 radin (55095)This print-out should have 16 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001In this case,1(1 + 2x2 ) dx =I=0I=02x + x3310.Consequently10.0
University of Texas - MATH - 408 D
parga (jrp3484) HW23 radin (55095)This print-out should have 15 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001keywords:0021dx .+ 1)3/21. I = ln x 1. I = 13. I =13x2 1 x2 1 + C
University of Texas - MATH - 408 D
parga (jrp3484) HW24 radin (55095)This print-out should have 16 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.Note that this last HW is due Monday 12/3,9pm. Midterm 3 is Tuesday 12/4 in cla
University of Texas - CH - 302
parga (jrp3484) HW 9 Kinetics I vandenbout (51540)This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 10.0 pointsIf the average rate of formation of H2 (g) is3.
University of Texas - CH - 302
parga (jrp3484) HW 10 Kinetics II vandenbout (51540)This print-out should have 23 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 10.0 pointsA and B react to form C according to thesingl
University of Texas - CH - 302
parga (jrp3484) HW 11 Electrochemistry I vandenbout (51540)This print-out should have 27 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 (part 1 of 2) 10.0 pointsUsing oxidation and reduc
University of Texas - CH - 302
parga (jrp3484) HW 12 Electrochemistry II vandenbout (51540)This print-out should have 25 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 10.0 pointsHow many moles of Cl2 (g) are produced
University of Texas - CH - 302
parga (jrp3484) HW1 - Physical Equilibria vandenbout (51540)This print-out should have 29 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 10.0 pointsGiven that you have 14.5 moles of N2 ,
University of Texas - CH - 302
parga (jrp3484) HW2 Solutions vandenbout (51540)This print-out should have 35 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.001 10.0 pointsBoth ammonia and phosphine (PH3 ) are soluble in w