Exam 2 - McCord
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Exam 2 - McCord

Course Number: CHEM CH301, Fall 2010

College/University: University of Texas

Word Count: 1973

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Version 207 Exam 2 mccord (50970) This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. McCord CH301 2. 2p , 2p , 1s correct 1 3. 2p , 2p , 2s 4. 2p , 2p , 2s 5. 2p , 2p , 1s 001 10.0 points Which of the following elements would be expected to have the highest electronegativity? 1. As 2. Li 3. Al 4. P correct 5. Ge...

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207 Version Exam 2 mccord (50970) This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. McCord CH301 2. 2p , 2p , 1s correct 1 3. 2p , 2p , 2s 4. 2p , 2p , 2s 5. 2p , 2p , 1s 001 10.0 points Which of the following elements would be expected to have the highest electronegativity? 1. As 2. Li 3. Al 4. P correct 5. Ge Explanation: Electronegativity generally increases from left to right and from bottom to top of the Periodic Table. 002 10.0 points The energy diagram in this question is the molecular orbital diagram for a homonuclear diatomic molecule. The lines represent MOs but the labels have been removed. Consider the following molecular orbital diagram: Explanation: 003 10.0 points Which of the following symbols is used to depict a partial positive charge? 1. + 2. + 3. + 4. + correct 5. + 6. + Explanation: 004 10.0 points White phosphorus is composed of molecules of P4 in which each P atom is connected to three other P atoms in the shape of a tetrahedron. Does it obey the octet rule? 1. Unable to determine a b 2. No 3. Yes correct Explanation: From the structure P c What are the names of the labeled orbitals, a through c, respectively? 1. 2p , 2p , 1s P P it does obey the octet rule. P Version 207 Exam 2 mccord (50970) 3 005 10.0 points O Which is the most polar molecule? B1 ) O P O 1. BF3 O 2. Ar of Lewis structures. 3. CO2 4. CH4 5. CF4 6. H2 O correct 2 O P O 3 B2 ) O O Select the one from each pair that is likely to make the dominant contribution to a resonance hybrid. 1. A2 and B1 correct 2. A1 and B2 3. A2 and B2 Explanation: H2 O has angular molecular geometry. It is assymetrical and polar. 006 10.0 points How many and bonds are in the molecule below? 4. A1 and B1 Explanation: A1 ) N 1 N +1 O 0 A2 ) N 0 N +1 O 1 H C C N H H 1. 6 , 1 2. 6 , 2 3. 7 , 1 4. 5 , 2 correct 5. 5 , 1 The rst structure is preferred for the same reason. The second structure is preferred because it places the negative formal charge on the most electronegative atom. 3 3 1 1 O O 0 0 1 0 O P O O P O B1 ) B2 ) 1 0 1 O 1 O 008 10.0 points A molecule has two lone non-bonded pairs of electrons on the central atom and four atoms bonded to the central atom. What is its molecular shape and its hybridization? 1. square planar; sp3 d2 correct 2. square planar; sp3 3. tetrahedral; sp3 4. square pyramidal; sp3 d2 O 5. tetrahedral; sp3 d2 Explanation: Sigma bonds are the rst bond formed between any two bonded atoms. Any subsequent bond is a pi bond. 007 10.0 points Consider the following pairs A and B A1 ) N N O A2 ) N N Version 207 Exam 2 mccord (50970) 6. octahedral; sp3 d2 7. pyramidal; sp3 Explanation: One way to solve is to draw a sample and determine HED. Since HED = 6, you know electronic geometry is octahedral. The lone pairs are placed 180 degrees from each other leaving the molecular geometry square planar and hybridization is sp3 d2 . 009 10.0 points Which of the species NO, BrO, CH+ , BF 3 4 are radicals? 1. NO and BF only 4 2. BrO, CH+ and BF only 3 4 3. NO, CH+ 3 and BF 4 only central atom. correct 3. the atoms are joined by sigma bonds. 3 4. the number of bonding orbitals equals the number of anti-bonding orbitals. 5. there are more shared electrons than nonshared electrons. Explanation: The only time that the molecular shape and electronic arrangement are the same is when there are no lone pairs of electrons on the central atom. 011 10.0 points Which of the following has bond angles of exactly 120 ? 1. S2 3 2. O3 3. NO 2 4. HO 2 5. CS2 correct 3 Explanation: While CS2 and NO both have three re3 2 gions of electron density around the central atom, in NO one of these regions is a lone 2 pair which will repel the other regions more strongly, making the bond angle less than 120 . Only in CS2 would the bond angle be 3 exactly 120 . 012 10.0 points The length of an O H bond in water is approximately 1. 1.0 104 cm. 2. 1.0 108 cm. correct 3. 1.0 m. 4. 1.0 1012 cm. 4. NO and CH+ only 3 5. BrO and CH+ 3 only 6. NO and BrO only correct 7. BrO and BF+ 4 only Explanation: The Lewis structures are N O F B F Br O + F Radicals are species with an unpaired electron, so only NO and BrO are radicals. 010 10.0 points The electronic arrangment is the same as the molecular shape when 1. the molecule is not polar. 2. there are no lone pairs of electrons on the F H C H H Version 207 Exam 2 mccord (50970) Explanation: In water the O 9.6 1011 m. H bond has a length of 4 013 10.0 points The C+ cation has how many total electrons and how many valence electrons? 1. 6; 3 2. 7; 6 3. 7; 5 4. 6; 6 5. 7; 8 6. 6; 5 7. 6; 4 8. 5; 4 9. 5; 3 correct Explanation: C+ has lost one electron, giving it 5 total electrons and 3 valence electrons. 014 10.0 points A molecule has one lone pair of electrons on the central atom and three atoms bonded to the central atom. The central atom follows the octet rule. What is its electronic arrangement and its hybridization? 1. tetrahedral; sp2 2. pyramidal; sp3 3. angular; sp3 4. trigonal planar; sp3 5. trigonal planar; sp2 6. pyramidal; sp2 7. tetrahedral; sp3 correct Explanation: One lone pair plus 3 bonded pairs equals 4 electronic regions. This means that the central atom is tetrahedral and hybridization is sp3 . 015 10.0 points Consider the following covalent bond radii: Single Double Triple 77 pm 67 pm 60 pm 75 pm 60 pm 55 pm 74 pm 60 pm 102 pm - C N O S What is the approximate length of the NN bond in nitrogen hydride (HNNH) using the table values? 1. 120 pm correct 2. 55 pm 3. 135 pm 4. 60 pm 5. 150 pm 6. 110 pm 7. 75 pm Explanation: The bond is a double bond so you add 60 pm to 60 pm and get 120 pm. 016 0.0 points EXTRA CREDIT QUESTION: A reminder here to PLEASE bubble in your name, uteid, and your version number correctly. Do it now - dont wait till the end of the exam. Also, a reminder that there is NO CLASS tomorrow (Friday, October 8th). Thats right, you get to SKIP chemistry class. Ill see you on Monday, the 11th, and well start up on Chapter 5. This question starts out at zero points but could very well increase after the grading. Now, if more points are awarded (the Version 207 Exam 2 mccord (50970) curve) on assignment, this would you like them added to your score? 1. YES, I would like the points and the higher score. correct 2. NO, leave my score alone, I prefer the lower score Explanation: This should be a no-brainer. Most students want higher scores. If you picked yes, you got credit for the question and you got the extra points you asked for (if they were granted by your instructor). If you answered NO, you also got what you wanted... no points awarded. 017 10.0 points The lattice enthalpy of calcium oxide is the energy change for the reaction 1. CaO(s) Ca(g) + O(g) 1 2. Ca(s) + O2 (g) CaO(s) 2 1 3. CaO(s) Ca(g) + O2 (g) 2 4. Ca(g) + O(g) CaO(g) 5. CaO(s) Ca2+ (g) + O2 (g) correct Explanation: Lattice enthalpy is the energy change at constant pressure when a 1 mole of an ionic solid is broken into ions of its component elements in gaseous form at 298 K, 1 atm. 018 10.0 points Draw the Lewis structure for CO. 1. 2. 3. 4. O O O O C C C C 5. 6. 7. 8. 9. 10. O O O O O O C C C C C C correct 5 Explanation: C has 4 valence electrons; O has 6: C C O O Total number of valence electrons: 1 4 e = 4 e 1 6 e = 6 e 10 e O C or O C This conguration gives each atom a complete octet. 019 10.0 points Which bond is not found in ethylene (C2 H4 )? 1. sp2 sp2 2. 2p2p 3. 1ssp2 4. All of these bonds are found in C2 H4 . 5. sp2 2p correct Explanation: 020 10.0 points Which of the following molecules is nonpolar? 1. PH3 2. XeF4 correct Version 207 Exam 2 mccord (50970) 3. H2 O 4. O3 5. H2 O2 Explanation: XeF4 is symmetrical and therefore nonpolar. Ozone (O3 ) is the only polar molecule composed of non-polar bonds. The other molecules are all asymmetrical and therefore polar. 021 10.0 points Which of the following would have the highest melting point? 1. KF correct 2. KBr 3. KI 4. RbF 5. KCl Explanation: KF has the highest charge density. 022 10.0 points 6 Explanation: The best drawing will show the valence electrons, the charges, and the appropriate ratio of ions for MgO. 023 10.0 points In the Lewis dot structure of the molecule ClF3 , how many unbonded electron pairs are found around the central atom? 1. 0 2. 1 3. 3 4. 4 5. 2 correct Explanation: Draw the Lewis dot structure. ClF3 has 3 Cl F single bonds and 2 lone pairs on Cl. 024 10.0 points In the molecular orbital theory of HCl, the expected bonding molecular orbital (MO) would be 1. MO = H(1s) Cl(3p) 2. MO = H(1s) + Cl(3p) correct 3. MO = H(3p) + Cl(3p) 4. MO = H(1s) + Cl(3s) 5. MO = H(1s) Cl(3s) Explanation: H has 1s valence electron and Cl has 3p valence electrons. Bonding orbitals result from positive (in-phase) overlap of the atomic orbitals. 025 10.0 points ICl3 is sp3 d hybridized. What is the electronic and molecular geometry? 1. trigonal bipyramidal; T-shaped correct Which of the following is the best representation of the compound magnesium oxide? 1. None is appropriate because magnesium oxide is a covalent compound. 2. Mg 2+ ,2 O O 3. Mg , 4. Mg2+ , 5. 2 Mg+ , 6. 3 Mg 2+ + 2 O 2 correct O 3 ,2 O Version 207 Exam 2 mccord (50970) 2. tetrahedral; pyramidal 3. octahedral; T-shaped 4. trigonal planar; trigonal planar 5. trigonal bipyramidal, seesaw Explanation: The hybridization tells us that there are 5 regions of high electron density. Three of those regions are the bonded Cl atoms. The other two regions must be lone pairs of electrons on the central I atom. This corresponds to trigonal bipyramidal electronic geometry and T-shaped molecular geometry. 026 10.0 points Which of the following is diamagnetic? 1. O2 correct 2 2. O 2 3. O+ 2 4. O2 Explanation: All but O2 are paramagentic (possess un2 paired electrons). 027 10.0 points Which of the following compounds is nonpolar? 1. BH3 correct 2. PCl3 3. PH3 4. O3 5. NH3 Explanation: NH3 , PH3 and PCl3 are tetrahedral with one lone pair and are therefore not symmetric; their polar bonds do not cancel, and the 7 molecule is polar. O3 is bent, with a lone pair on the central oxygen and is thus polar. BH3 is trigonal planar. The polar B H bonds cancel and the molecule is not polar. 028 10.0 points What hybridization would you expect for the carbon atoms in the cyclic compound benzene (C6 H6 )? 1. sp3 2. sp 3. sp3 d 4. sp3 d2 5. sp2 correct Explanation: In each carbon one s orbital and two p orbitals hybridize to produce three sp2 orbitals which form sigma bonds between the carbons and between carbon and hydrogen. The remaining p orbitals form pi bonds between the carbons. 029 10.0 points What is the electronic geometry around nitrogen in the molecule CH3 CH2 NH2 ? 1. trigonal pyramidal 2. trigonal planar 3. bent 4. tetrahedral correct 5. square planar 6. linear Explanation: 030 10.0 points What is the molecular geometry of SbCl2 ? 5 Antimony (Sb) can be an exception to the Version 207 Exam 2 mccord (50970) octet rule, accepting up to twelve electrons. 1. trigonal bipyramidal 2. seesaw 3. square planar 4. square pyramidal correct 5. octahedral The N As N P Sb H H H H EN 2. 2 2. 2 = 0. 0 3. 0 2. 2 = 0. 8 2. 2 2. 2 = 0. 0 2. 2 2. 1 = 0. 1 8 H bond is the most polar of these. 033 10.0 points Which of the following has the largest radius? 1. S2 correct Explanation: 031 10.0 points Which of the following species has bonds with the most ionic character? 1. PCl3 2. NO2 3. P4 O10 4. CO2 5. SnO2 correct 2. K+ 3. Cl 4. Cl 5. S Explanation: Although size decreases from left to right across the periodic table due to increasing eective nuclear charge, the negative ions will be the largest ions, and the negative ion with the largest charge will have the largest radius since the electrons will repel each other and the eective nuclear charge will be insucient to overcome this repulsion. 034 10.0 points More than one Lewis formula resonance structure can be drawn for which of the following for minimum formal charge? 1. O3 correct 2. H2 O 3. SbH3 4. CO2 5. SF4 Explanation: Oxygen is double bonded to one of the oxygen atoms and singly bonded to the third oxygen atom. The double bond can shift from Explanation: EN is max for SnO2 . 032 10.0 points Identify the compound with the most polar bond. 1. NH3 correct 2. SbH3 3. PH3 4. AsH3 Explanation: Calculate the dierence in the electronegativities (EN): Version 207 Exam 2 mccord (50970) one of the positions to the other to give rise to resonance structures. 9

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Chapter36.1 TransportofVascularPlants Withincell o Redalgae Celltocell Longdistance25/01/200711:12:00QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture.o Activetransport Themovementofasubstanceacrossabiologicalmembrane aga
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Chapter35.1 Leaves VeinPatterns o Monocots o Eudicots18/01/200710:55:00oQuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture.o angiosperms parts o tip veinormidribends o midrib o lateralveins o margins o petiole o blade o r
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WestmorelandClass NotesChapters 9 and 10Chapter 9 Things you need to know about Cellular Respiration: For Glycolysis (See figure 9.8) 1. Glucose 2. Pyruvate (and how many of them produced at the end per glucose) 3. ATP (and how many produced) 4. NADH (
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16/06/200915:43:0016/06/200915:43:0016/06/200915:43:00Chapter416/06/200915:43:00BusinessWritingBasics Purposeful Persuasive Economical Readeroriented AdaptingandAlteringtheWritingProcess Schedulingtheprocess o Startwith25%research,25%writing,25% edit
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Stomach16/03/200711:35:0016/03/200711:35:00Adaptationsofvertebrates SomeDentalAdaptations 16/03/200711:35:00Dentition o ananimalsassortmentofteeth o structuralvariationreflectingdiet o specializeddentitioninvertebrates o nonmammalianvertebratesgener
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Experiment: taxic behavior in isopodsIntroduction Different animal species have specific habitat requirements. Sensory receptors allow them to detect and show behavioral responses to environmental variables such as humidity, light, mates, predators, and
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Title > main object & taxonomic info main biological idea main method used Abstract (100 - 200 words; 1 or 2 sentences per section) summary of purpose To determine if the assortment of Armadillidium vulgare is random or not in response to phototaxis and h
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Classification of Armadillidium vulgare Domain Domain Kingdom Eukarya Eukarya Animalia AnimaliaPhylum Arthropoda Arthropoda Segmentation, hard exoskeleton, and jointed appendages; includes spiders, scorpions, millipedes, centipedes, insects, crabs, lobst
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LEGALITYAcontractisnotvalidifitisnotformedforalegalpurpose. Whatisnotalegalpurpose? Usury Gambling Noinsurableinterest Kstocommitcriminalacts WhatKsareagainstpublicpolicy? Covenantsnottocompete,unlessreasonable Adhesioncontracts Exculpatoryclauses KstoDi