CH101 test 2 spring 17 Name CODE 555555 Please 1 Write your name on this white paper use it to work out the problems and hand it back in 2 Write bubble in on your scantron name ID number and the above code without that code I cannot grade your scantron results 3 All questions ask you to fill in a correct answer unless you are told otherwise 4 Only bubble in one answer for each question even though there may be another correct answer 5 Try to do easy questions first 6 Don t guess this is not Las Vegas work out the problem on paper eliminate answers etc Then copy your answer to the scantron 7 If and only if you really have no clue at the end do put something 8 Use PENCIL not pen on the scantron 9 If you are nervous remember to smile a little you are not the only one 10 If the question is not clear or you think I made a mistake please raise your hand 11 Before you hand in make sure everything is in order do not cause a traffic jam rather step out of the line to fix things 12 Make sure your paper scantron are on the right stack and that the order of that stack is not scrambled Question 1 2 Consider the NO diatomic molecule The left represents an energy diagram and on the right an image of one the molecular orbitals is shown Select a true statement in the following two questions Q1 a b c d e The MO marked a is a orbital originating from the 2p orbitals of the N and O atoms Energy level b represents a MO originating from the 2s orbitals of the N and O atoms The degenerate atomic orbital level c is mostly located on the nitrogen atom The energy level d represents a bonding orbital The triply degenerate level e represents the 2p atomic orbitals of oxygen I gave partial credit for e although both from the energies and the picture on the right it should be clear that it is actually the NITROGEN atom To answer the second question you have to fill the above diagram with the available valence electrons making sure you follow the principles of Aufbau Pauli and Hund Again select a true statement Q2 a The MO depicted in a is the partly filled antibonding HOMO b The bond order of the neutral molecule is 3 It is 2 c The anion NO in the ground state would have one electron in the level marked d no two in the unmarked HOMO d The molecule should be diamagnetic with a singlet spin state MS 0 only No it is paramagnetic doublet e In the isolated N atom the level marked e would have three unpaired spins True and I even gave you the clue that e in the previous question is wrong Question 3 Consider the C10NH13 molecule depicted above All of its atoms are visible The carbon atoms have been numbered and the nitrogen atom marked N The light balls are the hydrogen atoms Select a true statement a The atom C1 can best be described with sp3 hybridization No it should be sp2 b The oxidation number for atom C5 is zero Yes that is true because it only has carbon neighbors so all the shared pairs are a tie and must be split c The bond order between C5 C10 is one No it is 1 because it is part of a benzene ring d The bond angle C2 N C3 is approximately 120 degrees No the N atom should have a lone pair and the angles would be about 109 e The bond between C1 and C2 is shorter than between C7 and C8 Yes because C1 C2 is a real double bond and C7 C8 is not Question 4 Ionic and covalent bonding are two of the three extreme cases of chemical bonding But often bonding is intermediary in character between ionic and covalent Select a true statement to demonstrate your understanding of the subject Consult the appendix for electronegativity values A lot of you seem to have ignored this a Tungsten II telluride WTe is a strongly ionic compound as tungsten is a metal and tellurium is not No no no Look at the electronegativities W is even a little bit more electronegative than Te A lot of people did this wrong b Covalent bonding often leads to the formation of discrete molecules ionic bonding does not Please read page 79 ionic compounds are neither molecular nor directional c The oxidation number for C in CH4 is 4 but for B is B2H6 is 3 Yes that is true because C is a bit more electronegative than H and B a little less That makes C the winner and B the loser d Ionic bonding requires particular bond angles like 109 or 120 but covalent bonding is less directional Again consult page 79 e All of the above is true Question 5 Which of the following oxidation numbers for the underlined element is incorrect Apparently people ignore me when I italicize bold and underline things The only incorrect one here the strontium peroxide one Strontium is never 4 I had to reduce the point count for this question because few of you can spot an incorrect answer a b c d e Ag2SO4 KMnO3 CH2Br2 SrO2 H2C CH2 Ox Ag 1 Ox Mn 5 Ox C 0 Ox Sr 4 Ox C 2 Question 6 What are the correct names for these species SF6 1 2 3 4 5 sulfur VI fluoride sulfur hexafluoride sulfur hexafluoride sulfur VI fluoride sulfur hexafluoride Cu2O Ag2O CsH2AsO4 dicopper oxide copper oxide copper I oxide copper I oxide copper oxide disilver oxide silver I oxide silver oxide silver oxide silver I oxide cerium hydrogen arsenide cesium dihydrogen arsenite cesium dihydrogen arsenate cesium hydrogen arsenide cesium dihydrogen arsenate Question 7 Consider the azide anion N3 Perform a Lewis type calculation and consider formal charges possible resonance structures hybridization and oxidation states Select a true statement ER 3 8 24 VE 3 5 1 16 TP 8 Reconcile 8 SP 4 LP 4 There are three Lewis structures N N N N N N and N N N The first has formal charges 1 1 1 the other two 0 1 2 and 2 1 0 The first one is the best one The oxidation state of the middle N is 1 regardless of the Lewis structure a Lewis theory cannot deal with this molecular ion very well because it is electron deficient Of course not every atom has a full octet b There are four Lewis structures that fulfill the octet rule No there are three c The middle N atom should be described with a different hybridization in each Lewis structure No it has two electron regions regardless and should be described with sp in all cases d The structure with the best formal charges has two resonance structures N N N and N N N No that would be the N N N structure e Although the …