1 CHEM 1180 Fall 2013 Exam 2B 25 OCT 13 Name KEY Show your work for each problem to receive credit Guard your test and conduct yourself properly The last sheet of the exam is a periodic table for your use Please be sure to write your name on the back of the exam Partial credit will be given for correctly drawn Lewis structures when appropriate 1 6 points Give the approximate ideal bond angles for the following molecules a AlCl4Cl Cl Al Cl Tetrahedral electron domain geometry Bond angle 109 47 Cl b RnF4 F F Rn F Octahedral electron domain geometry Bond angle 90 F c PF3 F P F Tetrahedral electron domain geometry Bond angle 109 47 F 2 4 points Arrange the following compounds in order of increasing lattice energy LiF KBr NaCl KBr NaCl LiF Since all the ions have a single charge the smaller the ions the greater the lattice energy 3 2 points Use electronegativities to predict the type of bonding between palladium and platinum Pd 2 20 Pt 2 28 bonding is covalent 4 2 points Use electronegativities to predict the type of bonding between gold and zinc Au 2 54 Zn 1 65 bonding is ionic 2 5 4 points Briefly but completely describe the difference between the origin of a coordinate covalent bond with the origin of a conventional covalent bond A coordinate covalent bond is a covalent bond where one of the atoms has contributed both electrons to be shared 6 10 points Draw a Lewis structure for the nitrite ion Include all resonance structures nitrite NO2 e 5 2 6 1 18 e O N O O N O 7 6 points Draw a Lewis structure for the boron trichloride molecule e 3 3 7 24 e Cl B Cl Cl 8 8 points Give the electron domain and the molecular geometry for the sulfur dibromide molecule e 6 2 7 20 eBr S Br electron domain geometry tetrahedral molecular geometry bent 3 9 4 points Describe a pi bond in terms of the overlap of atomic orbitals A pi bond is created when p orbitals on adjacent atoms that are perpendicular to the axis connecting the atoms overlap with each other 10 8 points Give the hybridization of the atomic orbitals of the carbon atom and of the chlorine atoms in the phosgene COCl2 molecule e 4 6 2 7 24 eO Cl C Cl Carbon atom 3 electron domains sp2 hybridized orbitals Chlorine atoms 4 electron domains sp3 hybridized orbitals 11 6 points Using the table of bond enthalpies provided estimate the enthalpy of reaction for the following reaction H2 g Br2 g 2 HBr g Bond Bond H kJ mol H H 436 Bonds broken H H and Br Br H Br 364 Br Br 193 Bonds formed H Br and H Br Hrxn H H H H Br Br 2 H H Br 436kJ mol 193kJ mol 2 364kJ mol 99kJ mol 4 12 9 points Draw proper Lewis structures include all hydrogens and lone pairs for the following organic molecules a 2 hexanone H H H H H O H C C C C C C H H H H H H b propylmethylamine H H H H H C N C C C H H H H H H c heptanoic acid H H H H H H H O C C C C C C C H H H H H H O 13 4 points Name the functional group in the following molecule H H H H O C C C C H H H H H N C C H H H H What is circled is an amide group H 5 14 9 points Give correct names for the following organic compounds a H H H H O C C C C H H H H butanal H H b H H C C H H C C C H H cyclopentene c H H H H C H H C C C Cl H H H 1 chloro 2 methylpropane 6 15 8 points Which of the following molecules has a positive formal charge on its central atom Give the molecular geometries for the molecules Note that only bonding electrons are shown in the figures I I I As I F Cl I Cl I Cl Se Cl Cl Cl F F Br F F F The molecular geometry of all the molecules is octahedral 16 8 points Explain why metals conduct electricity well Your explanation should include a knowledge of the nature of metallic bonding as well as a knowledge of the energy levels of the valence electrons in a metal To conduct electricity a material must have electrons in an excited electronic state Since the valence electrons are spread out throughout the metal the energy levels are very close together There s almost no gap between the highest occupied energy level and the first excited state Therefore very little energy supplied by a very low voltage is needed to get the electrons to conduct through the material 17 6 points The As Cl bonds in the AsCl5 molecule do not have identical lengths Using the appropriate terminology state which of the bonds is longer Explain why The axial position in the trigonal bipyramidal electron domain geometry is longer than the bond in the equatorial position experiences more repulsion the equatorial bond The axial position in the trigonal bipyramidal electron domain geometry experiences 3 90 repulsions whereas the equatorial position experiences only 2 90 repulsions Thus an axial As Cl bond will be longer than an equatorial As Cl bond Cl Cl As Cl Cl Cl