1 00:00:00,000 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,820 Your support will help MIT OpenCourseWare continue to 4 00:00:06,820 --> 00:00:10,530 offer high-quality educational resources for free. 5 00:00:10,530 --> 00:00:13,550 To make a donation, or view additional materials from 6 00:00:13,550 --> 00:00:17,280 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,280 --> 00:00:18,530 ocw.mit.edu. 8 00:00:21,590 --> 00:00:22,050 Hi. 9 00:00:22,050 --> 00:00:26,560 I'm Jocelyn, and we're going to go over fall 2009, exam 1, 10 00:00:26,560 --> 00:00:29,480 problem number 3. 11 00:00:29,480 --> 00:00:32,060 With every question, we want to make sure we read the full 12 00:00:32,060 --> 00:00:35,070 problem first. So answer the following questions about the 13 00:00:35,070 --> 00:00:36,610 difluoro-iodate ion. 14 00:00:39,170 --> 00:00:41,290 Draw a three-dimensional representation of the 15 00:00:41,290 --> 00:00:44,480 molecular geometry around the central atom, not simply the 16 00:00:44,480 --> 00:00:45,770 Lewis structure. 17 00:00:45,770 --> 00:00:48,530 Show all atoms and bonds between them. 18 00:00:48,530 --> 00:00:50,620 Because this part seems pretty separate from the rest of 19 00:00:50,620 --> 00:00:52,880 them, we're going to start with that. 20 00:00:52,880 --> 00:00:59,250 So we have a difluoro-iodate ion. 21 00:00:59,250 --> 00:01:01,740 It has a positive charge. 22 00:01:01,740 --> 00:01:05,530 And it asks us to draw the molecular geometry in a 3D 23 00:01:05,530 --> 00:01:09,800 representation, and not just Lewis structure. 24 00:01:09,800 --> 00:01:13,200 However, to do the molecular geometry in 3D, we need to 25 00:01:13,200 --> 00:01:14,250 know the Lewis structure. 26 00:01:14,250 --> 00:01:17,550 So we're going to start with that. 27 00:01:17,550 --> 00:01:20,190 To draw the Lewis structure, we first need to figure out 28 00:01:20,190 --> 00:01:23,140 which atom is the central atom. 29 00:01:23,140 --> 00:01:27,700 And it's always a good idea to do that by looking at the 30 00:01:27,700 --> 00:01:30,140 relative electronegativities. 31 00:01:30,140 --> 00:01:35,680 As we know, fluorine is the most electronegative element 32 00:01:35,680 --> 00:01:38,940 in the periodic table, and so that's probably not going to 33 00:01:38,940 --> 00:01:40,500 be in the middle. 34 00:01:40,500 --> 00:01:43,385 Therefore, iodine, being the less electronegative, is going 35 00:01:43,385 --> 00:01:46,410 to be in the middle, and it will have 36 00:01:46,410 --> 00:01:49,170 fluorines on either side. 37 00:01:49,170 --> 00:01:53,620 This is our rough sketch of the molecular geometry. 38 00:01:53,620 --> 00:01:56,180 Not too important right now. 39 00:01:56,180 --> 00:02:00,470 Next we need to know how many electrons we have. Iodine has 40 00:02:00,470 --> 00:02:04,470 seven valence electrons, fluoride has seven valence 41 00:02:04,470 --> 00:02:11,270 electrons, and we have a positive charge, so we know 42 00:02:11,270 --> 00:02:13,160 that we're missing one. 43 00:02:13,160 --> 00:02:18,720 And that gives us 20 electrons. 44 00:02:18,720 --> 00:02:20,660 Now that we know how many electrons we have to work 45 00:02:20,660 --> 00:02:23,530 with, we need start filling them in. 46 00:02:23,530 --> 00:02:25,810 I always start with the outer electrons. 47 00:02:25,810 --> 00:02:29,770 Those are usually, hopefully, the most electronegative, and 48 00:02:29,770 --> 00:02:33,480 therefore, the most likely to have the octet rule satisfied. 49 00:02:33,480 --> 00:02:41,330 Also, if you remember, fluorine is in the second row, 50 00:02:41,330 --> 00:02:45,840 and therefore needs to have the octet rule satisfied. 51 00:02:45,840 --> 00:02:48,830 Iodine is a little bit lower in the periodic table, so the 52 00:02:48,830 --> 00:02:50,460 octet rule isn't as important. 53 00:02:50,460 --> 00:02:53,550 It can have an extended octet if we need it to. 54 00:02:53,550 --> 00:02:59,360 So we will work with iodine second. 55 00:02:59,360 --> 00:03:04,550 Let's start out with satisfying the octet rule for 56 00:03:04,550 --> 00:03:10,800 the fluorine, putting first the seven valence electrons 57 00:03:10,800 --> 00:03:14,320 that the fluorines brought themselves, and then noticing 58 00:03:14,320 --> 00:03:19,750 that each needs to share one from iodine, so that we have a 59 00:03:19,750 --> 00:03:21,790 full octet. 60 00:03:21,790 --> 00:03:23,620 So each has one bond with iodine. 61 00:03:26,740 --> 00:03:30,550 We've used 8 electrons, or 8 per fluorine. 62 00:03:30,550 --> 00:03:33,800 So we have a total of 16. 63 00:03:33,800 --> 00:03:35,820 We have 4 left, and we're going to put 64 00:03:35,820 --> 00:03:39,370 those on the iodine. 65 00:03:39,370 --> 00:03:43,340 Counting up how many electronics are around iodine, 66 00:03:43,340 --> 00:03:48,290 1, 2, 3, 4, 5, 6, 7, 8, we see that the octet rule is also 67 00:03:48,290 --> 00:03:49,820 satisfied for iodine. 68 00:03:49,820 --> 00:03:52,980 This is looking like a pretty good Lewis structure. 69 00:03:52,980 --> 00:03:55,640 Now, just to make sure it's really good, we're going to 70 00:03:55,640 --> 00:03:58,030 look at the formal charge. 71 00:03:58,030 --> 00:04:05,480 So if you recall, the formal charge is the number of 72 00:04:05,480 --> 00:04:16,490 valence electrons that the element usually has, and minus 73 00:04:16,490 --> 00:04:27,380 the number of unshared electrons, and 74 00:04:27,380 --> 00:04:28,630 the number of dots. 75 00:04:32,420 --> 00:04:38,940 So this sum subtracted from the valence electrons will 76 00:04:38,940 --> 00:04:41,980 give us the formal charge. 77 00:04:41,980 --> 00:04:47,010 Looking at the fluorine, we have seven valence electrons. 78 00:04:47,010 --> 00:04:52,260 We have 1, 2, 3, 4, 5, 6 unpaired, and 79 00:04:52,260 --> 00:04:56,300 one bond to the fluorine. 80 00:04:56,300 --> 00:05:00,220 So that gives us a formal charge of 0. 81 00:05:03,400 --> 00:05:05,645 The same goes for the other fluorine. 82 00:05:08,950 --> 00:05:12,210 So I guess its formal charge is 0. 83 00:05:12,210 --> 00:05:15,910 Now for the iodine, we have-- 84 00:05:15,910 --> 00:05:19,410 I'm going to do the formal charge down here-- 85 00:05:19,410 --> 00:05:23,930 it again has 7 valence electrons, usually. 86 00:05:23,930 --> 00:05:29,930 We have 4 unshared electrons, and 2 bonds. 87 00:05:29,930 --> 00:05:32,600 That gives us a formal charge of plus 1. 88 00:05:38,720 --> 00:05:42,550 Because we have a charged species here, the full 89 00:05:42,550 --> 00:05:45,110 molecule has a plus 1 charge. 90 00:05:45,110 --> 00:05:49,270 There's no way we cannot have a net formal charge. 91 00:05:49,270 --> 00:05:49,490 Right? 92 00:05:49,490 --> 00:05:51,850 We need to have a net formal charge of plus 1. 93 00:05:51,850 --> 00:05:55,490 And it makes sense that that formal charge is on the least 94 00:05:55,490 --> 00:05:58,280 electronegative atom. 95 00:05:58,280 --> 00:06:01,300 So this looks like the right Lewis 96 00:06:01,300 --> 00:06:04,970 structure for our purposes. 97 00:06:04,970 --> 00:06:08,050 One more thing that you might want to write down. 98 00:06:08,050 --> 00:06:09,870 I'm not sure if points were taken off from this. 99 00:06:09,870 --> 00:06:15,680 But technically, if you have a charged species, put brackets 100 00:06:15,680 --> 00:06:17,620 around your Lewis structure, and put the net 101 00:06:17,620 --> 00:06:21,230 charge on the outside. 102 00:06:21,230 --> 00:06:24,290 This is not, however, our final answer, right? 103 00:06:24,290 --> 00:06:26,080 This is the Lewis structure. 104 00:06:26,080 --> 00:06:32,230 But the question asks for the molecular geometry to be 105 00:06:32,230 --> 00:06:34,320 represented. 106 00:06:34,320 --> 00:06:37,480 So if we remember how we determine molecular geometry, 107 00:06:37,480 --> 00:06:42,880 we need to look at the number of electron domains around the 108 00:06:42,880 --> 00:06:44,450 essential atom. 109 00:06:44,450 --> 00:06:49,130 So we're going to move over here, and I'll redraw our 110 00:06:49,130 --> 00:06:50,380 Lewis structure. 111 00:06:58,750 --> 00:07:02,450 And remember, an electron domain can be either an 112 00:07:02,450 --> 00:07:04,530 electron pair, or a bond. 113 00:07:04,530 --> 00:07:13,790 So we see that iodine has 4 electron domains. 114 00:07:16,800 --> 00:07:23,180 And if we remember our skeletal geometry, we know 115 00:07:23,180 --> 00:07:24,430 that that's tetrahedral. 116 00:07:34,520 --> 00:07:40,930 Thus, to draw this in a more 3D fashion, we might want to 117 00:07:40,930 --> 00:07:49,190 say that one of the electron pairs is here, one of the 118 00:07:49,190 --> 00:07:54,750 electron pairs is over here, and then we have a fluorine 119 00:07:54,750 --> 00:07:59,825 coming out of the board, and a fluorine going into the board. 120 00:08:05,570 --> 00:08:07,120 We didn't really go over how to do 121 00:08:07,120 --> 00:08:09,050 three-dimensional drawings. 122 00:08:09,050 --> 00:08:15,310 So as long as you show something like, you know the 123 00:08:15,310 --> 00:08:20,820 fluorines will be an angle, not straight from the iodine, 124 00:08:20,820 --> 00:08:24,920 and that the electrons be on the other side, something to 125 00:08:24,920 --> 00:08:26,230 that effect. 126 00:08:26,230 --> 00:08:30,960 A little bit more detailed than just a Lewis structure, 127 00:08:30,960 --> 00:08:36,260 was what Professor Sadoway was looking for in this problem. 128 00:08:36,260 --> 00:08:37,650 And that would be your answer. 129 00:08:37,650 --> 00:08:41,440 And for this one especially, I'd say it's important to box 130 00:08:41,440 --> 00:08:42,390 your final answer. 131 00:08:42,390 --> 00:08:46,240 Because you hopefully would have figured out the Lewis 132 00:08:46,240 --> 00:08:49,120 structure, but this is not the correct answer. 133 00:08:49,120 --> 00:08:51,650 So you always want to signify what answer 134 00:08:51,650 --> 00:08:54,760 you want to be graded. 135 00:08:54,760 --> 00:08:58,890 Part B says, name the type of hybrid orbitals that the 136 00:08:58,890 --> 00:09:04,540 central atom forms. So we are almost all the way to 137 00:09:04,540 --> 00:09:09,150 answering this question, so I'll put it down here. 138 00:09:09,150 --> 00:09:13,380 Remember, hybrid orbitals are-- 139 00:09:13,380 --> 00:09:20,590 we have hybrid orbitals, because we can't really say 140 00:09:20,590 --> 00:09:24,990 what orbitals exactly are bonding to which atoms. We 141 00:09:24,990 --> 00:09:27,910 want to have 4 equivalent orbitals for 142 00:09:27,910 --> 00:09:29,170 this system, right? 143 00:09:29,170 --> 00:09:31,320 We have 4 electron domains, so we want 4 144 00:09:31,320 --> 00:09:33,410 equivalent bonding orbitals. 145 00:09:33,410 --> 00:09:37,670 And from our talks earlier in the lectures about 146 00:09:37,670 --> 00:09:45,130 hybridization, we know that if we want 4, we take 1S and 3P, 147 00:09:45,130 --> 00:09:50,240 and that gives us the hybridization of the SP3. 148 00:09:50,240 --> 00:09:53,550 if we only had 3 electron domains, it would be SP2, 149 00:09:53,550 --> 00:09:56,340 because we only use 3 orbitals. 150 00:09:56,340 --> 00:09:59,970 And the number of orbitals that go into the hybrid is the 151 00:09:59,970 --> 00:10:02,910 number of orbitals you will get out. 152 00:10:02,910 --> 00:10:06,210 So we know that our hybridization is SP3. 153 00:10:06,210 --> 00:10:11,290 And that goes along with the fact that our shape is 154 00:10:11,290 --> 00:10:12,540 tetrahedral. 155 00:10:14,520 --> 00:10:16,290 Now we need to name the molecular 156 00:10:16,290 --> 00:10:19,510 geometry of this molecule. 157 00:10:19,510 --> 00:10:24,630 so we already have the molecular shape and the kind 158 00:10:24,630 --> 00:10:26,430 of spacial arrangement is tetrahedral. 159 00:10:26,430 --> 00:10:31,600 However, when we're naming the molecular geometry, electron 160 00:10:31,600 --> 00:10:34,320 pairs no longer matter. 161 00:10:34,320 --> 00:10:44,820 So for part C, tetrahedral is our skeletal geometry, right? 162 00:10:44,820 --> 00:10:48,500 It shows us our special arrangement. 163 00:10:48,500 --> 00:10:52,360 The molecular geometry is ignoring those electron pairs. 164 00:10:52,360 --> 00:11:01,000 So it is called bent when we only have 2 bonds. 165 00:11:01,000 --> 00:11:08,080 And I think about that because in spectroscopy and ways that 166 00:11:08,080 --> 00:11:11,080 are used to study molecules, you can't see the electron 167 00:11:11,080 --> 00:11:12,700 clouds very well. 168 00:11:12,700 --> 00:11:17,550 So they could only see the iodine and the two fluorines, 169 00:11:17,550 --> 00:11:21,750 thus looking like a bent structure. 170 00:11:21,750 --> 00:11:25,460 So hopefully you're sticking with me here, because this 171 00:11:25,460 --> 00:11:27,050 problem has a lot of parts. 172 00:11:27,050 --> 00:11:28,810 And we're going to move on to part D now. 173 00:11:31,370 --> 00:11:37,420 Part D asks, is difluoro-iodate polar or 174 00:11:37,420 --> 00:11:41,250 nonpolar, and explain. 175 00:11:41,250 --> 00:11:44,030 If you put down polar or nonpolar and did not explain, 176 00:11:44,030 --> 00:11:48,020 we assume you guessed, and you didn't get any points. 177 00:11:48,020 --> 00:11:51,670 Explaining shows that you know why, and you understand the 178 00:11:51,670 --> 00:11:54,930 concepts behind this, and thus deserved 179 00:11:54,930 --> 00:11:57,230 points on this problem. 180 00:11:57,230 --> 00:11:59,530 So there's two different parts of this problem. 181 00:11:59,530 --> 00:12:04,400 First, to have a polar molecule, you need to have 182 00:12:04,400 --> 00:12:05,830 polar bonds. 183 00:12:05,830 --> 00:12:09,931 So I would first ask myself, are the bonds polar? 184 00:12:16,230 --> 00:12:19,460 We have a difference of electronegativity that you can 185 00:12:19,460 --> 00:12:24,380 look up on your periodic table and know is not negligible. 186 00:12:24,380 --> 00:12:25,630 And so yes. 187 00:12:27,500 --> 00:12:31,360 A difference in electronegativity signifies 188 00:12:31,360 --> 00:12:34,280 that our bonds are polar. 189 00:12:34,280 --> 00:12:37,240 However, not all molecules that have polar bonds are 190 00:12:37,240 --> 00:12:39,530 themselves polar, right? 191 00:12:39,530 --> 00:12:43,930 if we have spacial symmetry, those polar bonds 192 00:12:43,930 --> 00:12:45,200 could cancel out. 193 00:12:45,200 --> 00:12:49,393 So now we need to make sure, is the molecule polar? 194 00:12:55,620 --> 00:12:58,060 And for that, we need to have answered the 195 00:12:58,060 --> 00:12:59,770 first questions correctly. 196 00:12:59,770 --> 00:12:59,990 Right? 197 00:12:59,990 --> 00:13:02,520 We need to know that this is not a linear structure. 198 00:13:02,520 --> 00:13:05,540 If this was a linear structure, those polar bonds 199 00:13:05,540 --> 00:13:08,400 would be geometrically symmetric, and 200 00:13:08,400 --> 00:13:10,860 cancel each other out. 201 00:13:10,860 --> 00:13:16,340 But we all are very smart, and we got this part C right. 202 00:13:16,340 --> 00:13:18,620 We know that we have a bent structure. 203 00:13:18,620 --> 00:13:23,050 Thus, I'll rewrite this over here. 204 00:13:23,050 --> 00:13:32,450 We know that if we draw in our polarity vectors, we can 205 00:13:32,450 --> 00:13:38,070 realize that we'll have a net dipole pointing down. 206 00:13:38,070 --> 00:13:45,650 So the answer to this is yes, because of the difference in 207 00:13:45,650 --> 00:13:48,930 the electronegativity, and because 208 00:13:48,930 --> 00:13:55,160 of the spacial asymmetry. 209 00:14:01,320 --> 00:14:02,730 If you said both of those things, you 210 00:14:02,730 --> 00:14:04,340 would get full credit. 211 00:14:04,340 --> 00:14:05,610 If you said one or the other, you'd 212 00:14:05,610 --> 00:14:07,290 probably get part credit. 213 00:14:07,290 --> 00:14:12,080 But knowing to look at these two things shows that you 214 00:14:12,080 --> 00:14:17,520 fully understand what it means to have molecular polarity. 215 00:14:17,520 --> 00:14:17,840 All right. 216 00:14:17,840 --> 00:14:19,105 So now we're going to move to part E. 217 00:14:24,760 --> 00:14:28,660 This part is changing gears a little bit, so if you didn't 218 00:14:28,660 --> 00:14:30,960 quite understand the first part, you can 219 00:14:30,960 --> 00:14:33,460 still try this one. 220 00:14:33,460 --> 00:14:37,570 Part E, again, what is the question asking us? 221 00:14:37,570 --> 00:14:41,070 Determine the maximum wavelength of electromagnetic 222 00:14:41,070 --> 00:14:46,260 radiation capable of breaking the IS bond. 223 00:14:46,260 --> 00:14:55,740 So we need to find the maximum wavelengths which corresponds 224 00:14:55,740 --> 00:14:58,560 to the minimum energy. 225 00:14:58,560 --> 00:14:59,050 Right? 226 00:14:59,050 --> 00:15:04,290 Because we have e equals hc over over lambda for 227 00:15:04,290 --> 00:15:06,220 electromagnetic radiation. 228 00:15:06,220 --> 00:15:10,440 We're trying to figure out the lowest energy of photon that 229 00:15:10,440 --> 00:15:15,890 will break the IF bond. 230 00:15:15,890 --> 00:15:18,310 Now, what are we given to answer this question? 231 00:15:18,310 --> 00:15:24,920 We're given the energy of the homogeneous fluorine-fluorine 232 00:15:24,920 --> 00:15:30,850 bond, is 160 kilojoules per mole. 233 00:15:30,850 --> 00:15:37,290 And the energy of the homogeneous iodine-iodine bond 234 00:15:37,290 --> 00:15:43,970 is 150 kilojoules per mole. 235 00:15:43,970 --> 00:15:50,280 And in order to find the wavelength that will break the 236 00:15:50,280 --> 00:15:56,030 IS bond, we need to know the energy of the IS bond. 237 00:15:56,030 --> 00:15:58,100 Right? 238 00:15:58,100 --> 00:16:02,560 That's what we're working towards in this problem. 239 00:16:02,560 --> 00:16:08,200 So given the homogeneous energies, do we have a 240 00:16:08,200 --> 00:16:11,360 relationship that will help us determine the 241 00:16:11,360 --> 00:16:14,440 energy of the IS bond? 242 00:16:14,440 --> 00:16:17,870 If you recall, in lecture 9, Professor Sadoway went over 243 00:16:17,870 --> 00:16:21,630 the Pauling formula for determining a heterogeneous 244 00:16:21,630 --> 00:16:24,990 bond energy from the homogeneous bond energies, and 245 00:16:24,990 --> 00:16:26,240 the electronegativity. 246 00:16:28,890 --> 00:16:31,250 So I'll just write that up here. 247 00:16:31,250 --> 00:16:44,900 The Pauling formula is, in a generic form, the energy of an 248 00:16:44,900 --> 00:16:57,100 AB bond equals the square root of the product of the 249 00:16:57,100 --> 00:17:04,950 homogeneous bond energies plus a constant times the 250 00:17:04,950 --> 00:17:13,750 difference between the electronegativities squared. 251 00:17:13,750 --> 00:17:16,310 And I just said what all of these terms mean. 252 00:17:16,310 --> 00:17:20,340 Because even if you had this on your equation sheet in the 253 00:17:20,340 --> 00:17:23,480 test or something, a lot of people wrote this down, and 254 00:17:23,480 --> 00:17:25,150 then used it incorrectly. 255 00:17:25,150 --> 00:17:29,720 So it's really important to know not only what equations 256 00:17:29,720 --> 00:17:33,150 you have, but especially what each of the terms in the 257 00:17:33,150 --> 00:17:35,030 equations mean. 258 00:17:35,030 --> 00:17:39,640 So we're given the homogeneous bond energies, and we have a 259 00:17:39,640 --> 00:17:42,550 periodic table, or some other resource, that has the 260 00:17:42,550 --> 00:17:44,910 elecronegativities. 261 00:17:44,910 --> 00:17:48,150 Next step is to plug in the numbers and find the energy of 262 00:17:48,150 --> 00:17:50,780 the IF bond. 263 00:17:55,660 --> 00:18:21,430 So we have the square root of 150 kilojoules plus. 264 00:18:21,430 --> 00:18:26,770 And the exact values you get for the electronegativity may 265 00:18:26,770 --> 00:18:30,820 vary, depending on the source. 266 00:18:30,820 --> 00:18:32,720 But they'll be close enough, and 267 00:18:32,720 --> 00:18:34,980 relatively, probably very-- 268 00:18:34,980 --> 00:18:36,860 the difference between them will be very similar. 269 00:18:36,860 --> 00:18:38,580 So don't worry if your numbers were a little 270 00:18:38,580 --> 00:18:41,100 different than mine. 271 00:18:41,100 --> 00:18:43,500 So we wrote all that out, and plugging it into the 272 00:18:43,500 --> 00:18:53,160 calculator, you get that the IF bond has an energy of 323 273 00:18:53,160 --> 00:18:54,410 kilojoules per mole. 274 00:18:57,270 --> 00:19:04,020 Now that we've found the energy of the IS bond, are we 275 00:19:04,020 --> 00:19:05,960 done with the problem? 276 00:19:05,960 --> 00:19:07,170 You have an answer. 277 00:19:07,170 --> 00:19:09,500 You might be ready to move on to the next question. 278 00:19:09,500 --> 00:19:12,420 However, this isn't what the question is asking us, right? 279 00:19:12,420 --> 00:19:17,870 We are asked for the maximum wavelength that can break a 280 00:19:17,870 --> 00:19:19,500 bond of that energy. 281 00:19:19,500 --> 00:19:25,250 So we need to go back to our energy relation to our 282 00:19:25,250 --> 00:19:26,020 wavelengths. 283 00:19:26,020 --> 00:19:29,720 And we know that the wavelength 284 00:19:29,720 --> 00:19:35,100 equals hc over the energy. 285 00:19:35,100 --> 00:19:38,710 Now here's where some people tripped up a little bit. 286 00:19:38,710 --> 00:19:41,520 You have a value for the energy. 287 00:19:41,520 --> 00:19:44,700 h and c are constants. 288 00:19:44,700 --> 00:19:47,200 And you find the lambda. 289 00:19:47,200 --> 00:19:55,960 However, if you plugged that all in, your answer would be 290 00:19:55,960 --> 00:20:02,160 6.15 times 10 to the negative 31 meters. 291 00:20:02,160 --> 00:20:05,060 Some people, that was the answer that they found. 292 00:20:05,060 --> 00:20:05,850 And it makes sense. 293 00:20:05,850 --> 00:20:09,230 You have an energy, you have some constants, and you can 294 00:20:09,230 --> 00:20:10,530 get your lambda. 295 00:20:10,530 --> 00:20:14,080 However, if you think about it, does this wavelength 296 00:20:14,080 --> 00:20:16,610 actually make sense? 297 00:20:16,610 --> 00:20:20,160 The gamma ray radiation has a wavelength of about 10 to the 298 00:20:20,160 --> 00:20:22,130 negative 17. 299 00:20:22,130 --> 00:20:25,510 That's very, very, very energetic 300 00:20:25,510 --> 00:20:27,410 electromagnetic radiation. 301 00:20:27,410 --> 00:20:33,490 So this is 14 magnitudes smaller than gamma ray 302 00:20:33,490 --> 00:20:37,830 radiation, and it just doesn't make sense for a bond energy 303 00:20:37,830 --> 00:20:38,630 of this much. 304 00:20:38,630 --> 00:20:40,980 Plus it doesn't-- 305 00:20:40,980 --> 00:20:46,640 it just-- you should look at the answer you get, and try to 306 00:20:46,640 --> 00:20:50,290 be comfortable with if it makes sense or not. 307 00:20:50,290 --> 00:20:54,740 So where people got tripped up, is in the units. 308 00:20:54,740 --> 00:20:58,210 We found Pauling's formula gives you the bond energy in 309 00:20:58,210 --> 00:21:00,050 kilojoules per mole. 310 00:21:00,050 --> 00:21:03,850 However, if you look at Planck's constant, it has 311 00:21:03,850 --> 00:21:08,480 units of joules time seconds. 312 00:21:08,480 --> 00:21:10,810 The speed of light is in meters per second. 313 00:21:13,540 --> 00:21:17,180 And you want lambda in meters. 314 00:21:20,600 --> 00:21:24,540 To make everything or cancel out, that means we want the 315 00:21:24,540 --> 00:21:28,240 energy to be in joules. 316 00:21:28,240 --> 00:21:30,600 Not joules per mole, because we're looking at the 317 00:21:30,600 --> 00:21:33,330 wavelength of one photon, right? 318 00:21:33,330 --> 00:21:38,410 We're looking at the energy of one photon to break this bond. 319 00:21:38,410 --> 00:21:45,070 So we want the energy of the bond, not per mole of bonds, 320 00:21:45,070 --> 00:21:48,060 but just of one bond. 321 00:21:48,060 --> 00:21:53,500 So instead of just having this equation, we want to have put 322 00:21:53,500 --> 00:22:22,190 in h times c over 323, change it to joules, and then 323 00:22:22,190 --> 00:22:23,440 multiply by-- 324 00:22:25,866 --> 00:22:30,700 sorry, moles are on the bottom, 1 mole 325 00:22:30,700 --> 00:22:31,950 and Avogadro's number. 326 00:22:37,180 --> 00:22:39,860 And Planck's constant of the speed of light, you should 327 00:22:39,860 --> 00:22:43,710 have on your equation sheet, or somewhere else, some other 328 00:22:43,710 --> 00:22:47,640 resource that you have. You don't need to memorize those 329 00:22:47,640 --> 00:22:50,010 in most cases. 330 00:22:50,010 --> 00:23:00,570 If you remembered to divide by Avogadro's number, you got 3.7 331 00:23:00,570 --> 00:23:06,790 times 10 to the negative 7 meters, which equals 370 332 00:23:06,790 --> 00:23:08,830 nanometers. 333 00:23:08,830 --> 00:23:15,070 And if we recall, the light we can see is in the 400 to 700 334 00:23:15,070 --> 00:23:15,990 nanometer range. 335 00:23:15,990 --> 00:23:22,190 So this puts us in ultraviolet radiation, which makes sense 336 00:23:22,190 --> 00:23:26,720 that it would take that much energy to break a bond. 337 00:23:26,720 --> 00:23:30,510 The other problem some people had was forgetting to convert 338 00:23:30,510 --> 00:23:32,700 from kilojoules to joules. 339 00:23:32,700 --> 00:23:37,830 That would still give you a physical answer. 340 00:23:37,830 --> 00:23:43,810 But again, I can't stress unit cancel out enough. 341 00:23:43,810 --> 00:23:46,200 So make sure you know that since your Planck's constant 342 00:23:46,200 --> 00:23:48,450 is in joules, you want to make sure you have 343 00:23:48,450 --> 00:23:50,320 joules on the bottom.