HW9 MO-solutions - nanni (arn437) HW9 MO vanden bout...

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nanni (arn437) – HW9 MO – vanden bout – (50985) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A sigma bond 1. always exists in conjunction with a pi bond. 2. is always polar. 3. is composed oF non-bonding orbitals. 4. stems From sp hybridization oF orbitals. 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 sigma bonds. 2. carbon has a tetrahedral electronic geom- etry. 3. carbon has Four lone pairs oF electrons. 4. carbon is also involved in a pi bond. cor- rect 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. 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: sp 2 and C: sp 2 2. H: 1 s and C: sp 2 correct 3. H: 2 p and C: sp 3 4. H: 1 s and C: 2 p 5. H: 1 s and C: sp 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 2) 10.0 points What is the expected bond order For the diatomic species B 2 ? 1. 4 2. 3 2 3. 1 correct 4. 2 5. 3 6. 1 2 Explanation: BO = 2 - 0 2 = 1 or BO = 2 + 2 - 2 2 = 1 2 p σ * 2 p π * 2 p σ 2 p π 2 p 2 p
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nanni (arn437) – HW9 MO – vanden bout – (50985) 2 ↑↓ 2 s ↑↓ σ * 2 s ↑↓ σ 2 s ↑↓ 2 s 005 (part 2 of 2) 10.0 points What is the magnetism and number of un- paired electrons of B 2 ? 1. paramagnetic; 2 correct 2. paramagnetic; 4 3. paramagnetic; 1 4. paramagnetic; 3 5. diamagnetic Explanation: 006 10.0 points According to molecular orbital theory, which of the following is NOT predicted to exist? 1. All are predicted to exist 2. He 3. He 2 - 4. He 2+ 5. He 2 correct Explanation: When the bond order is equal to 0, a molecule is predicted to not exist. B . O . = bonding e - - antibonding e - 2 For He 2 , the number of bonding electrons (2) equals the number of antibonding elec- trons, therefore the bond order is 0 and the molecule is predicted to not exist. 007
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HW9 MO-solutions - nanni (arn437) HW9 MO vanden bout...

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