solut19 - CHAPTER NINETEEN THE REPRESENTATIVE ELEMENTS:...

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544 CHAPTER NINETEEN THE REPRESENTATIVE ELEMENTS: GROUPS 5A THROUGH 8A Group 5A Elements 1. NO 4 3- Both NO 4 3- and PO 4 3- have 32 valence electrons so both have similar Lewis structures. From the Lewis structure for NO 4 3- , the central N atom has a tetrahedral arrangement of electron pairs. N is small. There is probably not enough room for all 4 oxygen atoms around N. P is larger, thus, PO 4 3- is stable. PO 3 - PO 3 - and NO 3 - both have 24 valence electrons so both have similar Lewis structures. From the Lewis structure, PO 3 - has a trigonal arrangement of electron pairs about the central P atom (two single bonds and one double bond). P _ O bonds are not particularly stable, while N _ O bonds are stable. Thus, NO 3 - is stable. 2. H ° = 2(90. kJ) - [0 + 0] = 180. kJ; S ° = 2(211 J/K) - [192 + 205] = 25 J/K G ° = 2(87 kJ) - [0] = 174 kJ At the high temperatures in automobile engines, the reaction N 2 + O 2 2 NO becomes spontaneous since the favorable S ° term will become dominate. In the atmosphere, even though 2 NO N 2 + O 2 is spontaneous at the cooler temperatures of the atmosphere, it doesn’t occur because the rate is slow. Therefore, higher concentrations of NO are present in the atmosphere as compared to what is predicted by thermodynamics. 3- O N O O O P O O O
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CHAPTER 19 THE REPRESENTATIVE ELEMENTS: GROUPS 5A THROUGH 8A 545 N O O NN O O O O B F F F BNH F F F H H NNO O O NO 2 + NO 3. 2 NaN 3 (s) 2 Na(s) + 3 N 2 (g) K 273 _ K mol atm L 0.08206 L 70.0 _ atm 1.00 = RT PV = n N 2 = 3.12 mol N 2 needed to fill air bag. mass NaN 3 reacted = 3.12 mol N 2 × NaN mol NaN g 65.02 _ N mol 3 NaN mol 2 3 3 2 3 = 135 g NaN 3 4. As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around the small nitrogen atom, and the NX 3 molecule becomes less stable. 5. a. NO 2 , 5 + 2(6) = 17 e - N 2 O 4 , 2(5) + 4(6) = 34 e - plus other resonance structures plus other resonance structures b. BF 3 , 3 + 3(7) = 24 e - NH 3 , 5 + 3(1) = 8 e - BF 3 NH 3 , 24 + 8 = 32 e - In reaction a, NO 2 has an odd number of electrons so it is impossible to satisfy the octet rule. By dimerizing to form N 2 O 4 , the odd electron on two NO 2 molecules can pair up giving a species whose Lewis structure can satisfy the octet rule. In general odd electron species are very reactive. In reaction b, BF 3 can be considered electron deficient; boron has only six electrons around it. By forming BF 3 NH 3 , the boron atom satisfies the octet rule by accepting a lone pair of electrons from NH 3 to form a fourth bond. 6. For the reaction: N H H H
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CHAPTER 19 THE REPRESENTATIVE ELEMENTS: GROUPS 5A THROUGH 8A 546 NNO O O O 2 N 2 O + NON O OO O O O O O O O the activation energy must in some way involve the breaking of a nitrogen-nitrogen single bond. For the reaction: at some point nitrogen-oxygen bonds must be broken. N - N single bonds (160. kJ/mol) are weaker than N - O single bonds (201 kJ/mol). In addition, resonance structures indicate that there is more double bond character in the N - O bonds than in the N - N bond. Thus, NO 2 and NO are preferred by kinetics because of the lower activation energy.
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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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solut19 - CHAPTER NINETEEN THE REPRESENTATIVE ELEMENTS:...

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