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Unformatted text preview: goldstein (ddg625) – H08: MO Theory – mccord – (50970) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A sigma bond 1. stems from sp hybridization of orbitals. 2. is composed of non-bonding orbitals. 3. is always polar. 4. always exists in conjunction with a pi bond. 5. may exist alone or in conjunction with a pi bond. correct Explanation: A sigma, or single bond can exist by itself or with a pi bond, forming a double bond. 002 10.0 points In a new compound, it is found that the cen- tral carbon atom is sp 2 hybridized. This implies that 1. carbon has four lone pairs of electrons. 2. carbon has a tetrahedral electronic geom- etry. 3. carbon has four regions of high electron density. 4. carbon is also involved in a pi bond. cor- rect 5. carbon has four sigma bonds. 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. 003 10.0 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: 1 s and C: sp 2. H: 1 s and C: 2 p 3. H: 1 s and C: sp 2 correct 4. H: sp 2 and C: sp 2 5. H: 2 p and C: sp 3 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. 004 (part 1 of 4) 10.0 points Draw the Lewis structure for the following hydrocarbon molecule. The carbons are num- bered one to four starting with the far left carbon as one. CH 2 CCHCH 3 What is the molecular shape of carbon 3? 1. trigonal pyramidal 2. tetrahedral 3. square planar 4. angular 5. linear 6. bipyramidal 7. trigonal planar correct Explanation: The molecule has the structure C C C C Carbon 3 has bonds to two carbons and one hydrogen or RHED = 3. This is trigonal planar molecular geometry....
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