1 00:00:00,000 --> 00:00:02,400 The following content is provided under a creative 2 00:00:02,400 --> 00:00:03,860 commons license. 3 00:00:03,860 --> 00:00:06,840 Your support will help MIT OpenCourseWare continue to 4 00:00:06,840 --> 00:00:10,570 offer high quality educational resources for free. 5 00:00:10,570 --> 00:00:13,410 To make a donation or view additional materials from 6 00:00:13,410 --> 00:00:17,440 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,440 --> 00:00:18,690 ocw.mit.edu. 8 00:00:22,200 --> 00:00:24,230 PROFESSOR: Hi, I'm Jocelyn, and today we're going to go 9 00:00:24,230 --> 00:00:29,940 over to the Exam 2 from Fall 2009, and Problem 4, 10 00:00:29,940 --> 00:00:33,300 Parts (a) and (b). 11 00:00:33,300 --> 00:00:35,930 As always, we're going to start by reading the question. 12 00:00:35,930 --> 00:00:40,560 Boron exists in the gas state as the dimer of boron 2. 13 00:00:40,560 --> 00:00:44,550 Explain how the fact that B(2) is paramagnetic, two unpaired 14 00:00:44,550 --> 00:00:48,070 electrons, implies that, in this molecule, the pi 2p 15 00:00:48,070 --> 00:00:53,590 orbitals must lie at a lower energy than do the sigma 2p. 16 00:00:53,590 --> 00:00:58,240 OK, so what is this question actually asking you to do? 17 00:00:58,240 --> 00:01:10,450 Well, it's asking you why the energy of the pi 2p is less 18 00:01:10,450 --> 00:01:19,530 than the energy of the sigma 2p, given that boron 2 is 19 00:01:19,530 --> 00:01:20,780 paramagnetic. 20 00:01:26,270 --> 00:01:29,620 So the first thing we need to know is what paramagnetic 21 00:01:29,620 --> 00:01:32,580 means, and he actually gives that to you in the problem, so 22 00:01:32,580 --> 00:01:35,530 it means that you have two unpaired electrons. 23 00:01:35,530 --> 00:01:43,760 So this is unpaired electrons. 24 00:01:43,760 --> 00:01:50,770 We are asked here then to give a reason why the fact that 25 00:01:50,770 --> 00:01:54,490 there are unpaired electrons implies a certain ordering of 26 00:01:54,490 --> 00:01:58,490 the sigma and pi 2p orbitals. 27 00:01:58,490 --> 00:02:00,940 So the first thing we want to do, probably, is recognize 28 00:02:00,940 --> 00:02:06,160 that these are molecular orbitals, not atomic orbitals. 29 00:02:06,160 --> 00:02:10,200 Therefore, we probably want to start by drawing the molecular 30 00:02:10,200 --> 00:02:11,530 orbital diagram for boron. 31 00:02:14,460 --> 00:02:17,520 Now, how do we draw a molecular orbital diagram. 32 00:02:17,520 --> 00:02:19,890 Well, the first thing to do is to start 33 00:02:19,890 --> 00:02:22,170 with the atomic orbitals. 34 00:02:22,170 --> 00:02:30,280 So boron has five electrons, as we can find out from our 35 00:02:30,280 --> 00:02:41,850 periodic table, and they occupy the 1s, the 2s, and one 36 00:02:41,850 --> 00:02:45,090 electron in the 2p. 37 00:02:45,090 --> 00:02:47,840 Again, if this seems unfamiliar to you, you might 38 00:02:47,840 --> 00:02:51,160 want to go back to earlier in the course, where we went over 39 00:02:51,160 --> 00:02:53,940 electron configuration. 40 00:02:53,940 --> 00:02:56,550 However, when we're looking at molecular orbitals, we 41 00:02:56,550 --> 00:02:59,240 actually only care about the valence electrons, because 42 00:02:59,240 --> 00:03:02,400 those are what are involved in bonding. 43 00:03:02,400 --> 00:03:07,060 The core electrons are much, much lower in energy, so they 44 00:03:07,060 --> 00:03:09,180 aren't really involved in the bonding. 45 00:03:09,180 --> 00:03:13,290 So we're actually only going to look at the electrons in 46 00:03:13,290 --> 00:03:15,260 the second energy level. 47 00:03:15,260 --> 00:03:26,090 So we have three valence electrons per boron, and they 48 00:03:26,090 --> 00:03:33,940 are in the 2s and 2p, which we remember, there are three p 49 00:03:33,940 --> 00:03:36,470 orbitals in each energy level. 50 00:03:36,470 --> 00:03:39,530 So now let's draw those in. 51 00:03:39,530 --> 00:03:49,450 We have 2s and 2p. 52 00:03:49,450 --> 00:03:54,930 Remember to always have a scale here, and we're in an 53 00:03:54,930 --> 00:04:03,780 energy space, and this is not to scale, because we're just 54 00:04:03,780 --> 00:04:06,590 looking at relative energies, we don't care about the actual 55 00:04:06,590 --> 00:04:08,540 difference in energy. 56 00:04:08,540 --> 00:04:11,650 And because we have two borons-- 57 00:04:11,650 --> 00:04:15,580 so this is boron one-- 58 00:04:15,580 --> 00:04:17,700 I'm going to draw the same thing over here. 59 00:04:27,310 --> 00:04:31,020 Make sure they're at the same energy level because they are 60 00:04:31,020 --> 00:04:33,250 the exact same atom, and therefore, 61 00:04:33,250 --> 00:04:35,630 the same atomic orbitals. 62 00:04:35,630 --> 00:04:41,260 So this is, again, boron, and we have three 63 00:04:41,260 --> 00:04:44,190 electrons for each. 64 00:04:44,190 --> 00:04:48,360 We don't have to fill them in, but I'm going to do so, just 65 00:04:48,360 --> 00:04:49,610 to keep track. 66 00:04:52,110 --> 00:04:54,910 And again, we're filling according to the orbital 67 00:04:54,910 --> 00:04:59,620 filling rules gone over earlier in the class. 68 00:04:59,620 --> 00:05:02,380 So now that we have our atomic orbitals, we are going to 69 00:05:02,380 --> 00:05:05,326 combine them to, like, molecular orbitals. 70 00:05:05,326 --> 00:05:10,010 And we want to remember that s orbitals combine to make a 71 00:05:10,010 --> 00:05:18,420 sigma and a sigma bonding and a sigma anti-bonding. 72 00:05:24,220 --> 00:05:26,570 Remember that whenever you combine atomic 73 00:05:26,570 --> 00:05:29,990 orbitals, you create-- 74 00:05:29,990 --> 00:05:32,940 if you're combining two atomic orbitals, you create one of a 75 00:05:32,940 --> 00:05:36,070 higher energy and one of a lower energy, because the net 76 00:05:36,070 --> 00:05:39,120 energy stays the same. 77 00:05:39,120 --> 00:05:40,470 Oh sorry, that's crooked. 78 00:05:43,560 --> 00:05:48,280 All right, now we'll move on to the 2p, and because the 79 00:05:48,280 --> 00:05:54,710 problem told us that we need to explain why the pi bonding 80 00:05:54,710 --> 00:05:58,060 orbitals are lower in energy than the sigma bonding 81 00:05:58,060 --> 00:06:02,340 orbitals, we're going to start with that configuration. 82 00:06:02,340 --> 00:06:07,640 The actual order depends on the molecule, and so without 83 00:06:07,640 --> 00:06:10,410 any extra information, we wouldn't really know if the 84 00:06:10,410 --> 00:06:15,400 sigma was above or below the pi. 85 00:06:15,400 --> 00:06:22,630 So as the question, kind of, hints at: p orbitals make two 86 00:06:22,630 --> 00:06:28,675 pi bonds and one sigma bond. 87 00:06:32,600 --> 00:06:39,700 This is due to the orientation of the pi bonds, or the p 88 00:06:39,700 --> 00:06:42,990 orbitals in space. 89 00:06:42,990 --> 00:06:53,010 Then, as always, we need to put in the anti-bonding just 90 00:06:53,010 --> 00:06:56,190 to make sure we keep track of all of our orbitals. 91 00:07:03,070 --> 00:07:09,680 OK, so we combined one, two, three, four, five, six, seven, 92 00:07:09,680 --> 00:07:14,120 eight atomic orbitals, and we made one, two, three, four, 93 00:07:14,120 --> 00:07:18,440 five, six, seven, eight molecular orbitals. 94 00:07:18,440 --> 00:07:20,800 We want to make sure we always have the same number of 95 00:07:20,800 --> 00:07:23,760 orbitals before and after. 96 00:07:23,760 --> 00:07:27,920 So now we can fill in the electrons into the molecular 97 00:07:27,920 --> 00:07:30,400 orbitals: that's these in the middle here. 98 00:07:30,400 --> 00:07:33,950 And we use the same rules as we do when we're filling 99 00:07:33,950 --> 00:07:35,370 atomic orbitals. 100 00:07:35,370 --> 00:07:42,450 So we start at the lowest energy, pairing spins, if we 101 00:07:42,450 --> 00:07:47,810 only have one orbital in the energy level; but as here, we 102 00:07:47,810 --> 00:07:50,620 have degenerate orbitals, so we can actually 103 00:07:50,620 --> 00:07:53,090 have unpaired spins. 104 00:07:53,090 --> 00:07:57,070 So we have one, two, three, four, five, six electrons, and 105 00:07:57,070 --> 00:08:02,260 we have three from each boron atom, and so, we have the same 106 00:08:02,260 --> 00:08:06,210 number of electrons in our molecule. 107 00:08:06,210 --> 00:08:09,500 Now that we have our molecular orbital diagram, we can go 108 00:08:09,500 --> 00:08:11,500 back to the actual question. 109 00:08:11,500 --> 00:08:13,800 Some students stopped here and were like, I made the 110 00:08:13,800 --> 00:08:15,420 molecular orbital diagram. 111 00:08:15,420 --> 00:08:16,260 I'm done. 112 00:08:16,260 --> 00:08:22,140 But we need to remember that the question is asking us why 113 00:08:22,140 --> 00:08:26,610 the configuration we drew makes sense because of the 114 00:08:26,610 --> 00:08:28,790 unpaired electrons. 115 00:08:28,790 --> 00:08:31,940 So if we go back to our molecular orbital diagram, we 116 00:08:31,940 --> 00:08:38,050 see that because the pi 2p have two degenerate orbitals, 117 00:08:38,050 --> 00:08:40,850 we can have unpaired electrons. 118 00:08:40,850 --> 00:08:46,240 If we think about the sigma being below, we would have 119 00:08:46,240 --> 00:08:51,620 paired up those electrons, and therefore, gotten 0 unpaired 120 00:08:51,620 --> 00:08:54,870 electrons, and it would not be paramagnetic. 121 00:08:54,870 --> 00:08:56,130 So let's just try that out. 122 00:08:59,510 --> 00:09:02,370 Move over here, and we're going to draw just the 123 00:09:02,370 --> 00:09:10,180 molecular orbitals; since we already did the full process 124 00:09:10,180 --> 00:09:12,660 of making them from the atomic. 125 00:09:12,660 --> 00:09:19,330 So this is the sigma 2s, sigma star 2s. 126 00:09:19,330 --> 00:09:24,860 And now we're just looking at what if the sigma 2p was lower 127 00:09:24,860 --> 00:09:29,390 in energy than the pi 2p. 128 00:09:38,250 --> 00:09:43,330 And again, we would fill up our orbitals, and we see that 129 00:09:43,330 --> 00:09:48,250 because this lowest occupied molecular orbital is 130 00:09:48,250 --> 00:09:53,210 singularly degenerate, we pair up our electrons, and so if 131 00:09:53,210 --> 00:09:56,880 this were the case, boron would not be paramagnetic. 132 00:09:56,880 --> 00:10:01,950 However, because it is paramagnetic, we know that the 133 00:10:01,950 --> 00:10:06,660 pi 2p is lower in energy than the sigma 2p. 134 00:10:06,660 --> 00:10:09,610 So that's what this question was asking, and if you said 135 00:10:09,610 --> 00:10:12,740 all those things, great job, and you answered 136 00:10:12,740 --> 00:10:15,810 the question correctly. 137 00:10:15,810 --> 00:10:19,780 Now we're going to move on to Part (b), and it's very 138 00:10:19,780 --> 00:10:24,980 related to Part (a), so we're going to use the 139 00:10:24,980 --> 00:10:27,590 same diagrams here. 140 00:10:27,590 --> 00:10:32,540 So it asks, is the gas molecule B(2) 2 minus more or 141 00:10:32,540 --> 00:10:35,990 less stable than the gas molecule to B(2). 142 00:10:35,990 --> 00:10:37,610 Explain. 143 00:10:37,610 --> 00:10:41,560 So here he's asking about the boron dimer that has a 144 00:10:41,560 --> 00:10:45,620 negative 2 charge-- that is, it has two extra electrons. 145 00:10:45,620 --> 00:10:49,610 So we're still talking about boron; and thus the atomic 146 00:10:49,610 --> 00:10:56,200 orbitals that we're combining are exactly the same. 147 00:10:56,200 --> 00:10:57,720 I'll write that down. 148 00:10:57,720 --> 00:11:04,280 So we have we still have the 2s and the 2p 149 00:11:04,280 --> 00:11:05,530 from each of the boron. 150 00:11:09,030 --> 00:11:18,550 That's not the same energy level, and they still make the 151 00:11:18,550 --> 00:11:46,350 same molecular orbitals, but we have two extra electrons. 152 00:11:46,350 --> 00:11:53,710 So when we go to fill up our molecular orbitals, we now 153 00:11:53,710 --> 00:11:55,135 have eight instead of six. 154 00:12:03,870 --> 00:12:08,340 So not only is the boron 2 minus no longer paramagnetic, 155 00:12:08,340 --> 00:12:13,050 because we don't have unpaired electrons, but we also have a 156 00:12:13,050 --> 00:12:15,600 difference in the bond order. 157 00:12:15,600 --> 00:12:20,850 So if you don't remember what bonding order is, it is when-- 158 00:12:20,850 --> 00:12:22,100 I'll just put it here-- 159 00:12:28,820 --> 00:12:33,020 it's how we determine the strength of the interaction 160 00:12:33,020 --> 00:12:36,600 between the two atoms. And because he's asking is the 161 00:12:36,600 --> 00:12:40,310 boron 2 minus more or less stable, we want to figure out 162 00:12:40,310 --> 00:12:42,550 the strength of the interaction, and that will 163 00:12:42,550 --> 00:12:44,920 tell us which is more stable. 164 00:12:44,920 --> 00:12:55,040 So the bond order is the bonding electrons minus the 165 00:12:55,040 --> 00:13:05,290 anti-bonding divided by 2. 166 00:13:05,290 --> 00:13:09,470 Sorry that's a little messy, but I hope you get the idea. 167 00:13:09,470 --> 00:13:15,640 So for boron 2 minus, which we're talking about over here, 168 00:13:15,640 --> 00:13:26,550 the bonding order: we have two, four, six bonding 169 00:13:26,550 --> 00:13:30,780 electrons, electrons in bonding orbitals. 170 00:13:30,780 --> 00:13:34,430 And we have two in anti-bonding orbitals, divided 171 00:13:34,430 --> 00:13:39,780 by 2 and that gives us a bonding order of 2. 172 00:13:39,780 --> 00:13:42,970 So that's equivalent to saying there's a double bond between 173 00:13:42,970 --> 00:13:47,132 the boron 2 minus, the two borons in that molecule. 174 00:13:47,132 --> 00:13:54,030 So let's move back to our diagram for the neutral boron 175 00:13:54,030 --> 00:14:13,210 dimer, and we have two, three, four bonding, and two 176 00:14:13,210 --> 00:14:21,690 anti-bonding, and so that equals 1. 177 00:14:21,690 --> 00:14:28,470 So in the neutral dimer, we have a bonding order of 1, and 178 00:14:28,470 --> 00:14:34,330 in the charged 2 minus dimer, we have a bonding order of 2. 179 00:14:34,330 --> 00:14:39,010 From this, we can see that the 2 minus has more bonding 180 00:14:39,010 --> 00:14:42,250 interaction, and so, it will be more stable. 181 00:14:42,250 --> 00:14:45,600 So again, you can't just put that it's more stable, we need 182 00:14:45,600 --> 00:14:50,040 to have an explanation for why, and this is a way to 183 00:14:50,040 --> 00:14:51,700 explain that.