solut13a - CHAPTER 13 BONDING: GENERAL CONCEPTS 387 H H C C...

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CHAPTER 13 BONDING: GENERAL CONCEPTS 387 C C C C C C H H H H H H C C C C C C H H H H H H Cl 54. We will use a hexagon to represent the six-membered carbon ring and we will omit the 4 hydrogen atoms and the three lone pairs of electrons on each chlorine. If no resonance exists, we could draw 4 different molecules: If the double bonds in the benzene ring exhibit resonance, then we can draw only three different dichlorobenzenes. The circle in the hexagon represents the delocalization of the three double bonds in the benzene ring (see Exercise 13.53). With resonance, all carbon-carbon bonds are equivalent. We can&t distinguish between a single and double bond between adjacent carbons that have a chlorine attached. That only 3 isomers are observed provides evidence for the existence of resonance. 55. Borazine (B 3 N 3 H 6 ) has 3(3) + 3(5) + 6(1) = 30 valence electrons. The possible resonance structures are similar to those of benzene in Exercise 13.53.
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388 CHAPTER 13 BONDING: GENERAL CONCEPTS F PF F F F H B H H 56. There are four different dimethylborazines. The circle in the above structures represents the ability of borazine to form resonance structures (see Exercise 13.55) and CH 3 is shorthand for three hydrogen atoms singly bonded to a carbon atom. There would be 5 structures if there were no resonance. All of the structures drawn above plus an additional one related to the first Lewis structure above. 57. In each case in this problem, the octet rule cannot be satisfied for the central atom. BeH 2 and BH 3 have too few electrons around the central atom and all the others have to many electrons around the central atom. Always try to satisfy the octet rule for every atom, but when it is impossible, the central atom is the species assumed to disobey the octet rule. PF 5 , 5 + 5(7) = 40 e - BeH 2 , 2 + 2(1) = 4 e - H - Be - H BH 3 , 3 + 3(1) = 6 e - Br 3 - , 3(7) + 1 = 22 e - N B N B N B H H H H H H N B N B N B H H H H H H N B N B N B H H H H H H N B N B N B H H H H H H
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CHAPTER 13 BONDING: GENERAL CONCEPTS 389 F S F F F F F FC l F F FB r F F SF 4 , 6 + 4(7) = 34 e - XeF 4 , 8 + 4(7) = 36 e - ClF 5 , 7 + 5(7) = 42 e - SF 6 , 6 + 6(7) = 48 e - 58. ClF 3 has 7 + 3(7) = 28 valence BrF 3 also has 28 valence electrons. electrons. We expand the octet of the central Cl atom in ClF 3 and the central Br atom in BrF 3 . 59. CO 3 2- has 4 + 3(6) + 2 = 24 valence electrons. Br Xe F F F F F S F F F Cl F F F F F C OO O C O O O C O O O 2- 2- 2-
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390 CHAPTER 13 BONDING: GENERAL CONCEPTS C O H O H O HCO 3 - has 1 + 4 + 3(6) + 1 = 24 valence electrons. H 2 CO 3 has 2(1) + 4 + 3(6) = 24 valence electrons. The Lewis structures for the reactants and products are: Bonds broken: Bonds formed: 2 C - O (358 kJ/mol) 1 C _ O (799 kJ/mol) 1 O - H (467 kJ/mol) 1 O - H (467 kJ/mol) H = 2(358) + 467 - [799 + 467] = -83 kJ; The carbon-oxygen double bond is stronger than two carbon-oxygen single bonds, hence CO 2 and H 2 O are more stable than H 2 CO 3 .
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solut13a - CHAPTER 13 BONDING: GENERAL CONCEPTS 387 H H C C...

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