# solut14b - CHAPTER 14 COVALENT BONDING ORBITALS 431 49...

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CHAPTER 14 COVALENT BONDING: ORBITALS 431 49. Considering only the twelve valence electrons in O 2 , the MO models would be: O 2 ground state Arrangement of electrons consistent with the Lewis structure (double bond and no unpaired electrons). It takes energy to pair electrons in the same orbital. Thus, the structure with no unpaired electrons is at a higher energy; it is an excited state. 50. a. Yes, both have 4 sets of electrons about the P. We would predict a tetrahedral structure for both. See part d for the Lewis structures. b. The hybridization is sp 3 for P in each structure since both structures exhibit a tetrahedral arrangement of electron pairs. c. P has to use one of its d orbitals to form the π bond since the p orbitals are all used to form the hybrid orbitals. d. Formal charge = number of valence electrons of an atom - [(number of lone pair electrons) + 1/2 (number of shared electrons)]. The formal charges calculated for the O and P atoms are next to the atoms in the following Lewis structures. Cl P O P O -1 0 0 +1

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432 CHAPTER 14 COVALENT BONDING: ORBITALS C Cl Cl O NN F F Cl I Cl Cl a a In both structures, the formal charges of the Cl atoms are all zeros. The structure with the P _ O bond is favored on the basis of formal charge since it has a zero formal charge for all atoms. 51. a. COCl 2 has 4 + 6 + 2(7) = 24 valence electrons. trigonal planar polar 120 ° sp 2 b. N 2 F 2 has 2(5) + 2(7) = 24 valence electrons. Can also be: V-shaped about both Ns; 120 ° about both Ns; Both Ns: sp 2 polar nonpolar These are distinctly different molecules. c. COS has 4 + 6 + 6 = 16 valence electrons. linear, polar, 180 ° , sp d. ICl 3 has 7 + 3(7) = 28 valence electrons. T-shaped polar a. 90 ° dsp 3 52. For carbon, nitrogen and oxygen atoms to have formal charge values of zero, each C atom will form four bonds to other atoms and have no lone pairs of electrons, each N atom will form three bonds to other atoms and have one lone pair of electrons, and each O atom will form two bonds to other atoms and have two lone pairs of electrons. Following these bonding require-ments gives the following two resonance structures for vitamin B 6 : F F OCS
CHAPTER 14 COVALENT BONDING: ORBITALS 433 a. 21 σ bonds; 4 π bonds (The electrons in the 3 π bonds in the ring are delocalized.) b. angles a, c, and g: 109.5 ° ; angles b, d, e and f: 120 ° c. 6 sp 2 carbons; the 5 carbon atoms in the ring are sp 2 hybridized as is the carbon with the double bond to oxygen.

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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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solut14b - CHAPTER 14 COVALENT BONDING ORBITALS 431 49...

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