Exam 2 - Version 153 Exam 2 Mccord (52385) 1 This print-out...

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Unformatted text preview: Version 153 Exam 2 Mccord (52385) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. McCord CH301 This exam is only for McCords CH301 class. Bond Energies (kJ/mol) H C N O Br 436 413 391 463 366 H 346 305 358 276 C Single 163 201 243 N Bonds 146 201 O 193 Br 001 10.0 points Consider CN + , CN and CN . Based on a molecular orbital diagram, which of these will have the shortest bond length? 1. CN and CN are both the same and shorter than CN + . 2. CN correct 3. CN + 4. The bond lengths are all the same. 5. CN Explanation: The bond order is #bonding e - #antibonding e 2 The bond orders of the species are CN BO = 10- 4 2 = 3 CN BO = 9- 4 2 = 2 . 5 CN + BO = 8- 4 2 = 2 The higher the bond order, the shorter the bond length. 002 10.0 points How many valence electrons must be shown in the dot formula for the C 3 H 8 N 2 O molecule? 1. 36 correct 2. 30 3. 64 4. 44 5. 28 Explanation: To find the total number of valence elec- trons available for the dot formula we sum the number of valence electrons in each atom of the molecule. Carbon has 4 valence electrons and there are 3 C atoms. Hydrogen has 1 valence elec- trons and there are 8 H atoms. Nitrogen has 5 valence electrons and there are 2 N atoms. Oxygen has 6 valence electrons and there is 1 O atom. Total valence e = 4 3 (C atom) + 1 8 (H atom) + 5 2 (N atom) + 6 1 (O atom) = 36 003 10.0 points Give the bond angle surrounding each central atom: nitrogen, middle carbon, right carbon. H N C HO C H H H b b 1. 120,109.5,109.5 2. 109.5,109.5,109.5 3. 120,120,120 4. 120,90,109.5 Version 153 Exam 2 Mccord (52385) 2 5. 180,109.5,109.5 6. 120,180,120 7. 180,120,109.5 8. 120,120,109.5 correct Explanation: 004 10.0 points You can assume that these orbitals were the result of a homonuclear diatomic molecule. Consider the following molecular orbital di- agram: a b c What are the names of the labeled orbitals, a , b , and c , respectively? 1. 2 p , 2 p , 2 s 2. 2 p , 2 p , 2 s 3. 2 p , 2 p , 1 s 4. 2 p , 2 p , 2 s correct 5. 2 p , 2 p , 2 s Explanation: 005 10.0 points Carbon has a valence of four in nearly all of its compounds and can form chains and rings of C atoms. Consider the propyne structure. C C H H H C H a b c What hybridizations would you expect for the carbon atoms identified by a, b, and c, respectively? 1. sp 2 , sp, sp 2. sp 3 , sp, sp correct 3. sp 2 , sp 2 , sp 3 4. sp 3 , sp, sp 3 5. sp 3 , sp 2 , sp 3 6. sp 2 , sp, sp 2 7. sp 3 , sp 3 , sp 2 Explanation: C C H H H C H sp 3 sp sp 006 10.0 points If the neutral atom of an element has only 7 valence electrons it must be in which group?...
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This note was uploaded on 10/25/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Exam 2 - Version 153 Exam 2 Mccord (52385) 1 This print-out...

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