Exam2-711111-elab

Exam2-711111-elab - Exam 2 Friday Special code 711111 NAME...

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Exam 2 Friday Oct 14 2011. Special code 711111 NAME: ______________ Pleas both write in and bubble in only the following items: *Your name as LASTNAME_FIRSTNAME (please no middle initials of double names) *Your students ID number *Your special code (see above) Please write your name on the white paper. Now. Please solve your problems on the white paper first, then copy your answer to the bubble sheet Please use the empty reverse sides of the white paper as scrap paper. Please read through the questions first. Some are easier than others. Do the easy ones first and leave your brain to stew on the harder ones a bit. Please don’t leave questions unanswered. If you still don’t know the answer at the end, make an educated guess. You have 20% chance to guess right. (Unanswered: 0%. ..) Please finish filling everything out before handing in the paper. Do not hold up the line. It is rude. Please do not mess up the stack with exam papers. It is even ruder to your fellow students. Please put your white paper and bubble sheet on the right pile. Your teacher will put up an improvised sign with the special code on it to guide you. Please raise your hand if I screwed up with something. I try not to, but I am human too. Good luck!
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Question 1 Consider the molecule CH 2 N 2 . There are actually a number of different molecules with this overall composition. We will consider the one that has a linear C-N-N shape with two hydrogen atoms attached to the carbon atom called diazomethane. Use a Lewis calculation to determine its most likely electronic structure and consider the following statements pertaining to the most stable structure. A. One of the ‘resonance’ structures representing the most stable electronic structure has a triple bond between the nitrogen atoms B. The central nitrogen atom has formal charge (FC) and oxidation number (OX) both equal to zero C. The central nitrogen atom has a FC and an OX that are equal, but not zero D. The oxidation number of carbon is -4 E. None of the above The only feasible structure is H2C=N=N> . Any structure with a triple bond generates problems with the octet rule. There is a single structure therefore. The FC on the central N is zero the Ox is -2 so B and C are wrong. The Ox on carbon is zero (win from H, lose to N). None of the answers was correct. And yes that can very well be a valid answer. Please stop (second) guessing. Question 2 Select a false statement A. Sodium bromide is a compound with largely ionic bonding B. A phosphide is a compound with phosphorus in oxidation state OX=+5 C. The oxidation number of chlorine in chlorate is OX=+7 D. A phosphite is a compound with phosphorus in oxidation state OX=+3 E. A vanadate is a compound with vanadium in oxidation state OX=+5 B is false: phosphides have Ox=-3 C is false because OX=+7 is the per chlorate Question 3 Select that which contains the smallest number of atoms A. 1 mole of methane CH 4 B. 9 grams of water C. 1 kilogram of mercury D. A and B are equally small and smaller than C
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This note was uploaded on 11/08/2011 for the course CH 101 taught by Professor Bigham during the Spring '08 term at N.C. State.

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Exam2-711111-elab - Exam 2 Friday Special code 711111 NAME...

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