Chapter 14 Even Problem Solutions

Chapter 14 Even Problem Solutions - CHAPTER FOURTEEN...

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Unformatted text preview: CHAPTER FOURTEEN COVALENT BONDING: ORBITALS The Localized Electron Model and Hybrid Orbitals 10. H 2 CO has 2(1) + 4 + 6 = 12 valence electrons. The central carbon atom has a trigonal planar arrangement of the electron pairs which requires sp 2 hybridization. The two C – H sigma bonds are formed from overlap of the sp 2 hybrid orbitals from carbon with the hydrogen 1s atomic orbitals. The double bond between carbon and oxygen consists of one σ and one π bond. The oxygen atom, like the carbon atom, also has a trigonal planar arrangement of the electrons which requires sp hybridization. The σ bond in the double bond is formed from overlap of a carbon sp 2 hybrid orbital with an oxygen sp 2 hybrid orbital. The π bond in the double bond is formed from overlap of the unhybridized p atomic orbitals. Carbon and oxygen each have one unhybridized p atomic orbital which are parallel to each other. When two parallel p atomic orbitals overlap, a π bond results where the shared electron pair occupies the space above and below a line joining the atoms in the bond. C 2 H 2 has 2(4) + 2(1) = 10 valence electrons. Each carbon atom in C 2 H 2 is sp hybridized since each carbon atom is surrounded by two effective pairs of electrons, i.e., each carbon atom has a linear arrangement of the electrons. Since each carbon atom is sp hybridized, then each carbon atom has two unhybridized p atomic orbitals. The two C – H sigma bonds are formed from overlap of carbon sp hybrid orbitals with hydrogen 1s atomic orbitals. The triple bond is composed of one σ bond and two π bonds. The sigma bond between to the carbon atoms is formed from overlap of sp hybrid orbitals from each carbon atom. The two π bonds of the triple bond are formed from parallel overlap of the two unhybridized p atomic orbitals from each carbon. 12. For the molecules/ion in Exercise 13.75, all have central atoms with dsp 3 hybridization since all are based on the trigonal bipyramid arrangement of electron pairs. See Exercise 13.75 for the Lewis structures. For the molecules/ion in Exercise 13.76, all have central atoms with d 2 sp 3 hybridization since all are based on the octahedral arrangement of electron pairs. See Exercise 13.76 for the Lewis structures. 14. a. V-shaped 120° sp 2 Only one resonance form is shown. Resonance does not change the position of the atoms. We can predict the geometry and hybridization from any one of the resonance structures. a b. c. 1 2 3 4 5 d. Tetrahedral geometry about each S, 109.5°, sp 2 hybrids; V-shaped arrangement about peroxide O's, 109.5°, sp 3 hybrids plus two other resonance structures trigonal planar 120° tetrahedral sp 3 109.5° sp 2 e. trigonal pyramid < 109.5° sp 3 f. g....
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This note was uploaded on 04/07/2008 for the course CHEM 027 taught by Professor Mccallum during the Spring '08 term at Pacific.

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Chapter 14 Even Problem Solutions - CHAPTER FOURTEEN...

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