HW7S - Gauthier Joseph – Homework 7 – Due midnight –...

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Unformatted text preview: Gauthier, Joseph – Homework 7 – Due: Oct 16 2007, midnight – Inst: Vandenbout 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Here are more questions finishing up Chap- ter 14. Remember, the HWS could be over- loaded when the actual due time arrives Tues- day night. Room Assignments for EXAM 2 A-K WEL 1.316 L-O WEL 2.122 P-Z WEL 1.308 001 (part 1 of 1) 10 points sp 2 hybrid orbitals have 1. trigonal bipyramidal symmetry. 2. trigonal pyramidal symmetry. 3. linear symmetry. 4. tetrahedral symmetry. 5. trigonal planar symmetry. correct Explanation: 002 (part 1 of 1) 10 points In the molecule, C 2 H 4 , what are the atomic orbitals that participate in forming the sigma bond between the C and H atoms? 1. H: 2 p and C: sp 3 2. H: 1 s and C: 2 p 3. H: 1 s and C: sp 4. H: sp 2 and C: sp 2 5. H: 1 s and C: sp 2 correct Explanation: The electrons in the H 1 s orbital and the electrons in the C sp 2 hybrid orbital partici- pate in forming the bond. 003 (part 1 of 1) 10 points A sigma bond 1. always exists in conjunction with a pi bond. 2. is always polar. 3. stems from sp hybridization of orbitals. 4. may exist alone or in conjunction with a pi bond. correct 5. is composed of non-bonding orbitals. Explanation: A sigma, or single bond can exist by itself or with a pi bond, forming a double bond. 004 (part 1 of 1) 10 points In a new compound, it is found that the cen- tral carbon atom is sp 2 hybridized. This implies that 1. carbon is also involved in a pi bond. cor- rect 2. carbon has four lone pairs of electrons. 3. carbon has four sigma bonds. 4. carbon has a tetrahedral electronic geom- etry. 5. carbon has four regions of high electron density. Explanation: Carbon forms four bonds. sp 2 hybridiza- tion enables a C atom to bond to three other atoms, leaving one electron in a p orbital, which will form a pi bond. 005 (part 1 of 4) 10 points Draw the Lewis structure for the following hydrocarbon molecule. The carbons are num- bered one to four starting with the far left Gauthier, Joseph – Homework 7 – Due: Oct 16 2007, midnight – Inst: Vandenbout 2 carbon as one....
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This homework help was uploaded on 03/19/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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HW7S - Gauthier Joseph – Homework 7 – Due midnight –...

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