1 00:00:00,030 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,820 Commons license. 3 00:00:03,820 --> 00:00:06,850 Your support will help MIT OpenCourseWare continue to 4 00:00:06,850 --> 00:00:10,520 offer high-quality educational resources for free. 5 00:00:10,520 --> 00:00:13,390 To make a donation or view additional materials from 6 00:00:13,390 --> 00:00:17,490 hundreds of MIT courses visit MIT OpenCourseWare at 7 00:00:17,490 --> 00:00:18,740 ocw.mit.edu. 8 00:00:21,620 --> 00:00:22,390 Hi I'm Sal. 9 00:00:22,390 --> 00:00:24,740 Today we're going to be doing problem number 5 of 10 00:00:24,740 --> 00:00:27,530 fall 2009, test 1. 11 00:00:27,530 --> 00:00:30,240 Make sure that you read the problem in full detail before 12 00:00:30,240 --> 00:00:31,720 attempting the problem. 13 00:00:31,720 --> 00:00:35,150 And there's also some things that you should know, pretty 14 00:00:35,150 --> 00:00:36,040 much background material. 15 00:00:36,040 --> 00:00:37,650 Before attempting the problem. 16 00:00:37,650 --> 00:00:41,030 And that's what I need, which is W I N. 17 00:00:41,030 --> 00:00:44,260 You will win if you know how to manipulate these equations. 18 00:00:44,260 --> 00:00:49,750 The first equation is the energy of two charged species 19 00:00:49,750 --> 00:00:54,370 as a function of their internuclear separation. 20 00:00:54,370 --> 00:00:59,350 And the other equation is a function of just the 21 00:00:59,350 --> 00:01:03,710 separation between two ion pairs at equilibrium, which is 22 00:01:03,710 --> 00:01:05,330 given by r naught. 23 00:01:05,330 --> 00:01:12,180 And r naught is just simply the sum of the radii of the 24 00:01:12,180 --> 00:01:15,070 cation and the radii of the anion. 25 00:01:15,070 --> 00:01:18,940 So if you look at this equation there's two 26 00:01:18,940 --> 00:01:21,940 components to it. 27 00:01:21,940 --> 00:01:25,960 The first component is actually the component that is 28 00:01:25,960 --> 00:01:31,100 responsible for the attraction of your charged species. 29 00:01:31,100 --> 00:01:34,480 And the second component is the repulsion of it. 30 00:01:34,480 --> 00:01:39,170 Because essentially ionic bonding happens because of 31 00:01:39,170 --> 00:01:41,170 your electrostatic interaction between an 32 00:01:41,170 --> 00:01:43,590 cation and an anion. 33 00:01:43,590 --> 00:01:45,990 So with this in mind we can go ahead and read the 34 00:01:45,990 --> 00:01:48,490 problem and solve it. 35 00:01:48,490 --> 00:01:50,010 So the problem reads as follows. 36 00:01:50,010 --> 00:01:51,380 On the same graph below-- 37 00:01:51,380 --> 00:01:54,860 and this is the actual graph that we're going to be doing 38 00:01:54,860 --> 00:01:57,760 the problem on-- 39 00:01:57,760 --> 00:02:01,510 it says on the same graph below, for one, BeF2, so 40 00:02:01,510 --> 00:02:06,460 beryllium fluoride, and two, BeO, which is beryllium oxide, 41 00:02:06,460 --> 00:02:09,890 sketch the variation of potential energy with 42 00:02:09,890 --> 00:02:13,360 internuclear separation arc between a cation and an anion 43 00:02:13,360 --> 00:02:14,690 pair in each compound. 44 00:02:14,690 --> 00:02:17,830 The diagram need not to be drawn to scale however you 45 00:02:17,830 --> 00:02:21,260 must convey the relative magnitudes of key features. 46 00:02:21,260 --> 00:02:27,430 So the first thing that I would do is write down my 47 00:02:27,430 --> 00:02:28,740 compounds that I'm using. 48 00:02:28,740 --> 00:02:35,650 So I'm focusing on beryllium fluoride, is the first one. 49 00:02:35,650 --> 00:02:39,210 So if I look at beryllium fluoride, I'm going to look at 50 00:02:39,210 --> 00:02:41,360 the relative magnitudes and charges when they're in their 51 00:02:41,360 --> 00:02:43,410 cation and anion state. 52 00:02:43,410 --> 00:02:49,930 So I know that for beryllium, beryllium forms a 53 00:02:49,930 --> 00:02:51,690 cation that is 2 plus. 54 00:02:51,690 --> 00:02:54,630 So it has a positive charge of 2, which means that it lost 55 00:02:54,630 --> 00:02:56,250 two electrons. 56 00:02:56,250 --> 00:03:06,490 So I can say that z plus for beryllium is plus 2 and for 57 00:03:06,490 --> 00:03:13,140 fluorine, when it's in the anion state, my charge is 58 00:03:13,140 --> 00:03:15,670 equal to minus 1. 59 00:03:15,670 --> 00:03:20,190 So 2 plus minus 1. 60 00:03:20,190 --> 00:03:24,610 So with this in mind, you can take this and it will help you 61 00:03:24,610 --> 00:03:25,760 solve the problem. 62 00:03:25,760 --> 00:03:30,440 Because like I said, the first part of the equation is pretty 63 00:03:30,440 --> 00:03:33,540 much the attraction part, which involves the product of 64 00:03:33,540 --> 00:03:36,370 your charges. 65 00:03:36,370 --> 00:03:38,960 And I want to go ahead into the same thing for-- 66 00:03:38,960 --> 00:03:40,440 so this is for BeF2. 67 00:03:40,440 --> 00:03:46,370 And for BeO we have the oxygen 2 minus. 68 00:03:46,370 --> 00:03:57,730 So my anion has a z minus of minus 2 and my cation just has 69 00:03:57,730 --> 00:04:02,800 z plus of plus 2. 70 00:04:02,800 --> 00:04:07,740 Now if you look at your equation over here. 71 00:04:07,740 --> 00:04:09,300 There's two things that will dominate at certain 72 00:04:09,300 --> 00:04:09,980 parameters. 73 00:04:09,980 --> 00:04:13,130 Now this value of n is known as the Born Exponent, and that 74 00:04:13,130 --> 00:04:15,370 has a value between 6 and 12. 75 00:04:15,370 --> 00:04:17,890 So it's always greater than 1. 76 00:04:17,890 --> 00:04:24,240 And this is actually what's nominating your repulsion, so 77 00:04:24,240 --> 00:04:27,250 I'll do some colored chalk. 78 00:04:27,250 --> 00:04:31,110 So for the attraction part I'll circle it. 79 00:04:31,110 --> 00:04:35,580 And the fact that it's being attractive means that one of 80 00:04:35,580 --> 00:04:37,940 these charges has to be negative that will make the 81 00:04:37,940 --> 00:04:39,390 overall component negative. 82 00:04:39,390 --> 00:04:42,400 Right here because you'll have q1 times q2. 83 00:04:42,400 --> 00:04:45,560 And in our case if we look up beryllium fluoride, you have 84 00:04:45,560 --> 00:04:48,220 beryllium 2 plus you have fluorine minus, so if you make 85 00:04:48,220 --> 00:04:51,720 the product of those two you get minus 2. 86 00:04:51,720 --> 00:04:55,100 And you have this other component, that is the 87 00:04:55,100 --> 00:04:56,360 repulsive part. 88 00:04:56,360 --> 00:05:02,650 So if I was to draw these independently, not as a sum, I 89 00:05:02,650 --> 00:05:05,900 know that for my attractive part, which is the negative, 90 00:05:05,900 --> 00:05:09,710 it should look something like so, as a function of r. 91 00:05:09,710 --> 00:05:16,470 And my repulsive should look something like so. 92 00:05:16,470 --> 00:05:20,510 And this makes sense because as your radius becomes smaller 93 00:05:20,510 --> 00:05:28,150 and smaller then essentially if you get below 1, your 94 00:05:28,150 --> 00:05:30,650 potential energy curve for your-- 95 00:05:30,650 --> 00:05:32,790 the component for your repulsion-- 96 00:05:32,790 --> 00:05:34,060 spikes up. 97 00:05:34,060 --> 00:05:39,040 So if I was to superimpose these two, then I should 98 00:05:39,040 --> 00:05:47,470 expect that the curve to look something like so. 99 00:05:47,470 --> 00:05:51,460 Now this is good because this is exactly how the curve 100 00:05:51,460 --> 00:05:52,410 should look. 101 00:05:52,410 --> 00:05:55,240 And there's some characteristic features here 102 00:05:55,240 --> 00:05:59,840 that we want to go ahead and use BeF2 BeO2, draw on the 103 00:05:59,840 --> 00:06:02,720 same diagram to compare together. 104 00:06:02,720 --> 00:06:07,970 And this point right here on your r-axis is simply just 105 00:06:07,970 --> 00:06:08,940 your r naught. 106 00:06:08,940 --> 00:06:11,870 So it's just the sum of your cation radius 107 00:06:11,870 --> 00:06:13,390 and your anion radius. 108 00:06:13,390 --> 00:06:22,010 And this valley here is your energy at r naught. 109 00:06:22,010 --> 00:06:23,590 Which is the second equation that you 110 00:06:23,590 --> 00:06:25,650 should know how to apply. 111 00:06:25,650 --> 00:06:29,500 So if I want to analyze my system, how do I know which 112 00:06:29,500 --> 00:06:32,130 one has a bigger radii than the other? 113 00:06:32,130 --> 00:06:35,250 Well I know that my cation is common. 114 00:06:35,250 --> 00:06:38,670 So in both cases I'm dealing with beryllium cations. 115 00:06:38,670 --> 00:06:42,150 But I'm dealing with a different anion. 116 00:06:42,150 --> 00:06:46,290 Now I know that both fluorine and oxygen anions have the 117 00:06:46,290 --> 00:06:48,852 same number of electrons, from looking at the Periodic Table 118 00:06:48,852 --> 00:06:53,580 of Elements, or their charges that they have. But fluorine 119 00:06:53,580 --> 00:06:55,340 has one more proton. 120 00:06:55,340 --> 00:07:00,400 So what that does is that it allows the fluorine to pull 121 00:07:00,400 --> 00:07:04,340 the electron cloud tighter together and that decreases 122 00:07:04,340 --> 00:07:04,920 your radius. 123 00:07:04,920 --> 00:07:10,460 So just by looking at the Periodic Table of Elements, I 124 00:07:10,460 --> 00:07:16,370 can already assume that my radius for my fluorine should 125 00:07:16,370 --> 00:07:22,080 be less than the radius of my O2 minus anion. 126 00:07:22,080 --> 00:07:27,170 So if I look at my system than I know that if I want to look 127 00:07:27,170 --> 00:07:29,820 at the internuclear separation, the r naughts of 128 00:07:29,820 --> 00:07:33,990 the two compounds, then I should assume-- 129 00:07:33,990 --> 00:07:37,120 or have an educated guess based on my 130 00:07:37,120 --> 00:07:38,630 data that I have here-- 131 00:07:38,630 --> 00:07:46,790 that r naught for beryllium fluoride should be less than 132 00:07:46,790 --> 00:07:51,570 the r naught for beryllium oxide. 133 00:07:51,570 --> 00:07:55,550 So this allows you to put the position of your radius on 134 00:07:55,550 --> 00:07:57,850 your curve. 135 00:07:57,850 --> 00:08:04,140 So I'm going to go ahead and just redraw the a new axis 136 00:08:04,140 --> 00:08:07,220 over here just to compare both of them so you 137 00:08:07,220 --> 00:08:09,275 won't confuse yourself. 138 00:08:12,860 --> 00:08:18,060 So we know this is the energy of our system. 139 00:08:18,060 --> 00:08:23,550 And we concluded that our, that the radius of your 140 00:08:23,550 --> 00:08:27,170 beryllium fluoride is less than the radius 141 00:08:27,170 --> 00:08:29,690 of beryllium oxide. 142 00:08:29,690 --> 00:08:38,690 So I can go ahead and mark this as r naught of BeF2 and 143 00:08:38,690 --> 00:08:44,400 I'll mark that as r naught of BeO. 144 00:08:44,400 --> 00:08:48,880 So this tells me exactly where the low point of my potential 145 00:08:48,880 --> 00:08:50,290 curve should lie. 146 00:08:50,290 --> 00:08:53,120 But how do I know how deep it sits in the well? 147 00:08:53,120 --> 00:08:59,150 Well that's going to be dictated by the equation that 148 00:08:59,150 --> 00:09:01,100 governs the energy at r naught. 149 00:09:01,100 --> 00:09:04,720 And that's our simple equation of when it's the 150 00:09:04,720 --> 00:09:06,090 function of r naught. 151 00:09:06,090 --> 00:09:08,540 And if you look at the equation, the equation is just 152 00:09:08,540 --> 00:09:10,840 the product of your two charges. 153 00:09:10,840 --> 00:09:14,460 So if we come back and we look at our analysis here, we know 154 00:09:14,460 --> 00:09:18,950 that if I multiply these two charges together, you know the 155 00:09:18,950 --> 00:09:26,080 combination of these two, should be, q1 or q-- 156 00:09:26,080 --> 00:09:28,425 beryllium 2 plus, I'll call that q1 and I'll 157 00:09:28,425 --> 00:09:30,070 call this one q2. 158 00:09:30,070 --> 00:09:32,920 The product of these two gives you minus 2. 159 00:09:32,920 --> 00:09:42,080 But for this one, q1 q2 gives us minus 4. 160 00:09:42,080 --> 00:09:47,400 So this tells me that the electrostatic interaction for 161 00:09:47,400 --> 00:09:50,450 beryllium oxide is a lot stronger 162 00:09:50,450 --> 00:09:51,670 than beryllium fluoride. 163 00:09:51,670 --> 00:09:55,680 And it makes sense because you have plus 2 and minus 2 being 164 00:09:55,680 --> 00:09:59,050 attracted to each other instead of plus 2, minus 1. 165 00:09:59,050 --> 00:10:06,460 So on my graph I can assume, I can guess, that my value of my 166 00:10:06,460 --> 00:10:11,630 beryllium oxide should sit lower, more negative, on the 167 00:10:11,630 --> 00:10:14,160 potential curve than beryllium fluoride. 168 00:10:14,160 --> 00:10:17,710 So if I designate this as the low point for my beryllium 169 00:10:17,710 --> 00:10:22,090 fluoride then my curve for this system, for that 170 00:10:22,090 --> 00:10:26,410 particular compound should take a certain type of shape. 171 00:10:26,410 --> 00:10:30,620 And so again, this is your energy, the 172 00:10:30,620 --> 00:10:34,170 bottom point of r naught. 173 00:10:34,170 --> 00:10:40,450 And if this is the energy given to the fact that we have 174 00:10:40,450 --> 00:10:45,530 a higher magnitude when you multiply the two charges 175 00:10:45,530 --> 00:10:49,370 together, then it should sit lower on the potential well. 176 00:10:49,370 --> 00:10:50,780 And this will be my low point. 177 00:10:50,780 --> 00:10:57,710 So my curve should also look something like so. 178 00:10:57,710 --> 00:11:01,326 So if you were able to do this on the exam you would have 179 00:11:01,326 --> 00:11:03,420 gotten full points, which is good. 180 00:11:03,420 --> 00:11:06,470 And, again the key points about this problem is just 181 00:11:06,470 --> 00:11:09,050 knowing what equations to use by reading the problem. 182 00:11:09,050 --> 00:11:13,390 The problem talked about internuclear separation, well 183 00:11:13,390 --> 00:11:16,350 we've learned in class how to treat those problems and how 184 00:11:16,350 --> 00:11:19,400 to use equations to be able to describe the way that the 185 00:11:19,400 --> 00:11:22,210 energy should look, which is a good visual. 186 00:11:22,210 --> 00:11:24,440 And by looking at the Periodic Table of Elements, just 187 00:11:24,440 --> 00:11:27,360 thinking about how many protons, how many electrons 188 00:11:27,360 --> 00:11:29,860 certain cations have, you should have an estimated guess 189 00:11:29,860 --> 00:11:31,960 of what your ionic radius should be. 190 00:11:31,960 --> 00:11:34,120 And with that you can go ahead and look at the math. 191 00:11:34,120 --> 00:11:37,010 Look where that parameter plays a role in your equation. 192 00:11:37,010 --> 00:11:40,360 And then decide whether or not it'll be greater or less or 193 00:11:40,360 --> 00:11:41,110 what have you. 194 00:11:41,110 --> 00:11:44,870 So with that in mind, I dare you to try the problem now and 195 00:11:44,870 --> 00:11:46,440 give it a good shot.