solut14a - CHAPTER 14 COVALENT BONDING: ORBITALS 421 Note:...

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CHAPTER 14 COVALENT BONDING: ORBITALS 421 Note: The ring structures are all shorthand notation for rings of carbon atoms. In piperine, the first ring contains 6 carbon atoms and the second ring contains 5 carbon atoms (plus nitrogen). Also notice that CH 3 , CH 2 and CH are shorthand for a carbon atoms singly bonded to hydrogen atoms. b. piperine: 0 sp, 11 sp 2 and 6 sp 3 carbons; capsaicin: 0 sp, 9 sp 2 and 9 sp 3 carbons c. The nitrogens are sp 3 hybridized in each molecule. d. a. 120 ° e. 109.5 ° i. 120 °
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422 CHAPTER 14 COVALENT BONDING: ORBITALS b. 120 ° f. 109.5 ° j. 109.5 ° c. 120 ° g. 120 ° k. 120 ° d. 120 ° h. 109.5 ° l. 109.5 ° 22. a. To complete the Lewis structure, add two lone pairs to each sulfur atom. b. See the Lewis structure. The four carbon atoms in the ring are all sp 2 hybridized and the two sulfur atoms are sp 3 hybridized. c. 23 σ and 9 π bonds. Note: CH 3 , CH 2 and CH are shorthand for carbon atoms singly bonded to hydrogen atoms. 23. To complete the Lewis structure, just add lone pairs of electrons to satisfy the octet rule for the atoms with fewer than eight electrons. a. The two nitrogens in the ring with double bonds are sp 2 hybridized. The other three nitrogens are sp 3 hybridized. b. The five carbon atoms in the ring with one nitrogen are all sp 3 hybridized. The four carbon atoms in the other ring with double bonds are all sp 2 hybridized. c. Angles a and b: 109.5 ° ; angles c, d, and e: 120 ° d. 31 sigma bonds e. 3 pi bonds (Each double bond consists of one sigma and one pi bond.)
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CHAPTER 14 COVALENT BONDING: ORBITALS 423 The Molecular Orbital (MO) Model 24. Bonding molecular orbitals have maximum electron density between the bonded atoms and are lower in energy than the atomic orbitals from which they are formed. Antibonding molecular orbitals have minimal electron density between the bonded atoms and are higher in energy than the atomic orbitals from which they are formed. 25. Bond energy is directly proportional to bond order. Bond length is inversely proportional to bond order. Bond energy and bond length can be measured. 26. If we calculate a non-zero bond order for a molecule, we predict that it can exist (is stable). a. H 2 + :( σ 1s ) 1 B.O. = (1-0)/2 = 1/2, stable H 2 σ 1s ) 2 B.O. = (2-0)/2 = 1, stable H 2 - σ 1s ) 2 ( σ 1s *) 1 B.O. = (2-1)/2 = 1/2, stable H 2 2- σ 1s ) 2 ( σ 1s *) 2 B.O. = (2-2)/2 = 0, not stable b. He 2 2+ σ 1s ) 2 B.O. = (2-0)/2 = 1, stable He 2 + ( σ 1s ) 2 ( σ 1s *) 1 B.O. = (2-1)/2 = 1/2, stable He 2 σ 1s ) 2 ( σ 1s *) 2 B.O. = (2-2)/2 = 0, not stable c. Be 2 σ 2s ) 2 ( σ 2s *) 2 B.O. = (2-2)/2 = 0, not stable B 2 σ 2s ) 2 ( σ 2s *) 2 ( π 2p ) 2 B.O. = (4-2)/2 = 1, stable Li 2 σ 2s ) 2 B.O. = (2-0)/2 = 1, stable 27. Paramagnetic: Unpaired electrons are present. Measure the mass of a substance in the presence and absence of a magnetic field. A substance with unpaired electrons will be attracted by the magnetic field, giving an apparent increase in mass in the presence of the field. Greater number of unpaired electrons will give greater attraction and greater observed mass increase.
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This homework help was uploaded on 04/06/2008 for the course CHEM 142 taught by Professor Zoller,williamh during the Fall '07 term at University of Washington.

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solut14a - CHAPTER 14 COVALENT BONDING: ORBITALS 421 Note:...

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