1 00:00:00,030 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,820 Your support will help MIT OpenCourseWare continue to 4 00:00:06,820 --> 00:00:10,570 offer high-quality educational resources for free. 5 00:00:10,570 --> 00:00:13,550 To make a donation or view additional materials from 6 00:00:13,550 --> 00:00:17,500 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,500 --> 00:00:18,750 ocw.mit.edu. 8 00:00:21,880 --> 00:00:25,820 Hi, I'm Jocelyn, and we're going to go over Fall 2009, 9 00:00:25,820 --> 00:00:29,590 Exam 1, problem number 2. 10 00:00:29,590 --> 00:00:32,640 So as with every problem, we want to first 11 00:00:32,640 --> 00:00:35,230 read the full question. 12 00:00:35,230 --> 00:00:38,590 In box notation, give the complete electron 13 00:00:38,590 --> 00:00:42,590 configuration of each of the following gas-phase species: 14 00:00:42,590 --> 00:00:46,780 calcium 2 minus and magnesium 4 plus. 15 00:00:46,780 --> 00:00:50,340 So the first thing to note here, is you need to know what 16 00:00:50,340 --> 00:00:53,520 electronic configuration is. 17 00:00:53,520 --> 00:00:56,140 If you don't, you might want to go back and review that 18 00:00:56,140 --> 00:00:59,520 material, but it is basically assigning 19 00:00:59,520 --> 00:01:02,840 electrons to orbitals. 20 00:01:02,840 --> 00:01:06,430 The next thing you need to know is what box notation is. 21 00:01:06,430 --> 00:01:10,070 And Professor Sadoway introduced that in class, and 22 00:01:10,070 --> 00:01:14,100 you'll see, as I go through this, a refresher of that, if 23 00:01:14,100 --> 00:01:15,350 you forgot. 24 00:01:18,210 --> 00:01:21,710 So to assign electrons to orbitals, we first need to 25 00:01:21,710 --> 00:01:26,820 know how many electrons we have. So let's start with the 26 00:01:26,820 --> 00:01:30,680 first part of this problem, which is calcium 2 minus. 27 00:01:33,580 --> 00:01:39,140 We will look at our periodic table, and see that neutral 28 00:01:39,140 --> 00:01:47,030 calcium has 20 electrons. 29 00:01:47,030 --> 00:01:50,260 So calcium with no charge has 20 electrons. 30 00:01:50,260 --> 00:01:54,970 The 2 minus tells us that it has 2 extra electrons. 31 00:01:54,970 --> 00:01:55,880 Right? 32 00:01:55,880 --> 00:02:00,150 Because electrons have a negative charge. 33 00:02:00,150 --> 00:02:03,770 So we have a total of 22 electrons that we need to 34 00:02:03,770 --> 00:02:07,120 assign into orbitals. 35 00:02:07,120 --> 00:02:11,160 In order to actually do the assigning, we need to follow 36 00:02:11,160 --> 00:02:15,030 the three rules that Professor Sadoway introduced in class. 37 00:02:15,030 --> 00:02:17,508 So I will put that over here. 38 00:02:30,110 --> 00:02:34,170 The first rule is that you need to fill them, at least 39 00:02:34,170 --> 00:02:40,650 for a ground state species, in order of increasing energy. 40 00:02:40,650 --> 00:02:43,310 That is, you want to fill up the lowest energy orbitals 41 00:02:43,310 --> 00:02:46,650 first, and then move to the higher energy orbitals. 42 00:02:46,650 --> 00:02:48,960 This makes sense because ground state species want to 43 00:02:48,960 --> 00:02:51,770 have the lowest total energy possible. 44 00:02:51,770 --> 00:02:53,020 So the first is-- 45 00:03:00,990 --> 00:03:03,020 Now, how do we figure that out? 46 00:03:03,020 --> 00:03:05,470 Well, there's a few ways. 47 00:03:05,470 --> 00:03:08,890 One way is a nice trick that you can do on the side of your 48 00:03:08,890 --> 00:03:11,780 page or something. 49 00:03:11,780 --> 00:03:15,310 And it goes like this. 50 00:03:15,310 --> 00:03:21,330 So we have our s, p, d, and f orbitals. 51 00:03:21,330 --> 00:03:23,500 Right? 52 00:03:23,500 --> 00:03:27,360 We know that s can be in the first energy level up. 53 00:03:34,280 --> 00:03:43,020 We know that p starts at the second energy level, d starts 54 00:03:43,020 --> 00:03:49,400 at the third, and f starts at the fourth. 55 00:03:52,250 --> 00:03:57,970 The tricky part is that you don't just fill up 1s, 2s, 2p, 56 00:03:57,970 --> 00:04:00,020 3s, 3p, 3d. 57 00:04:00,020 --> 00:04:03,070 There's a little bit of mixing up of the energy levels versus 58 00:04:03,070 --> 00:04:05,480 the actual energy of the orbital. 59 00:04:05,480 --> 00:04:10,836 So a nice side of the page trick is to draw diagonal 60 00:04:10,836 --> 00:04:18,104 lines down through these numbers. 61 00:04:24,660 --> 00:04:28,100 And if you follow these arrows, you'll fill up in 62 00:04:28,100 --> 00:04:30,040 order of increasing energy. 63 00:04:30,040 --> 00:04:33,710 For example, after you fill up the 2p, you're going to fill 64 00:04:33,710 --> 00:04:42,380 up the 3s then the 3p, the 4s, and then you'll go to the 3d. 65 00:04:42,380 --> 00:04:45,140 If that doesn't make sense, you might want to go and 66 00:04:45,140 --> 00:04:51,340 review a little bit about the orbitals in your textbook, or 67 00:04:51,340 --> 00:04:55,180 other resources. 68 00:04:55,180 --> 00:05:00,250 The other way you can think about the ordering of the 69 00:05:00,250 --> 00:05:03,610 energy of the orbitals, is to be familiar with 70 00:05:03,610 --> 00:05:05,330 the periodic table. 71 00:05:05,330 --> 00:05:09,610 So we're just going to have an aside over here. 72 00:05:09,610 --> 00:05:19,570 And a general, very rough sketch of the periodic table 73 00:05:19,570 --> 00:05:23,260 would look something like this. 74 00:05:23,260 --> 00:05:24,030 OK? 75 00:05:24,030 --> 00:05:27,640 You have your alkali, alkali earth metals over here. 76 00:05:27,640 --> 00:05:31,880 You have aluminum, oxygen. 77 00:05:31,880 --> 00:05:35,780 Your halides, your noble gases over there. 78 00:05:35,780 --> 00:05:39,170 But the way I drew it here is in what we call blocks. 79 00:05:39,170 --> 00:05:48,050 So this is your s block, and then we have our p block over 80 00:05:48,050 --> 00:05:57,470 here, d block, and the the f block is down here. 81 00:06:01,450 --> 00:06:04,320 So once you get more familiar with electron configurations, 82 00:06:04,320 --> 00:06:09,780 you can actually looks towards the periodic table, and know, 83 00:06:09,780 --> 00:06:12,160 depending upon where your element is in the periodic 84 00:06:12,160 --> 00:06:14,210 table, what its electron configuration is. 85 00:06:14,210 --> 00:06:15,710 Or at least the electron configuration 86 00:06:15,710 --> 00:06:18,350 of the valence electrons. 87 00:06:18,350 --> 00:06:18,710 OK. 88 00:06:18,710 --> 00:06:24,300 So moving back to our trick, which we'll use now, we know 89 00:06:24,300 --> 00:06:29,630 that we have 22 electrons, and we know the order that we need 90 00:06:29,630 --> 00:06:31,700 to fill the orbitals. 91 00:06:31,700 --> 00:06:35,810 So let's start writing out our boxes, which is how Professor 92 00:06:35,810 --> 00:06:38,320 Sadoway wanted us to answer this problem. 93 00:06:38,320 --> 00:06:49,070 So we have 1s, our 2s, 2p. 94 00:06:49,070 --> 00:06:52,870 If you're unsure why I'm only drawing 1 orbital for s and 3 95 00:06:52,870 --> 00:06:56,920 orbitals for p, again, you might want to go review that 96 00:06:56,920 --> 00:06:59,010 beginning orbital material. 97 00:06:59,010 --> 00:07:07,160 So we filled, we have 1s, 2s, 2p, 3s, then we're going to do 98 00:07:07,160 --> 00:07:11,110 the 3s here, we're going to do the 3p. 99 00:07:15,040 --> 00:07:23,430 And after the 3p, we do the 4s, and after the 100 00:07:23,430 --> 00:07:29,500 4s, we do the 3d. 101 00:07:29,500 --> 00:07:31,570 Which has 5 boxes, right? 102 00:07:40,120 --> 00:07:43,870 So now that we know which orbitals we're going to fill, 103 00:07:43,870 --> 00:07:48,850 we need to know some more rules about filling them. 104 00:07:48,850 --> 00:07:55,420 So the second rule, over here, is the Pauli exclusion 105 00:07:55,420 --> 00:08:07,450 principle, which says that no two electrons can have the 106 00:08:07,450 --> 00:08:09,780 same quantum numbers. 107 00:08:09,780 --> 00:08:12,790 For us, that means that we basically can only have two 108 00:08:12,790 --> 00:08:16,760 electrons per orbital, and of opposite spin. 109 00:08:16,760 --> 00:08:20,160 So that's something we need to remember for this. 110 00:08:20,160 --> 00:08:25,040 The third is Hund's Rule. 111 00:08:31,210 --> 00:08:34,870 And that is, we want to have as many unpaired 112 00:08:34,870 --> 00:08:37,090 electrons as possible. 113 00:08:37,090 --> 00:08:39,650 So that will come into play when we're filling up our 114 00:08:39,650 --> 00:08:41,700 boxes in just a second. 115 00:08:41,700 --> 00:08:46,700 If these are confusing, again, we want to look 116 00:08:46,700 --> 00:08:47,950 at the lecture material. 117 00:08:50,410 --> 00:08:55,630 So I forgot my boxes here. 118 00:08:55,630 --> 00:08:59,160 Now let's start the fun part of filling up the elections. 119 00:08:59,160 --> 00:09:03,440 So we fill up 1s, spin up and down, the 120 00:09:03,440 --> 00:09:05,660 2s, spin up and down. 121 00:09:05,660 --> 00:09:16,380 So that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 122 00:09:16,380 --> 00:09:22,880 16, 17, 18, 19, 20, and we have 2 more. 123 00:09:22,880 --> 00:09:28,000 So in the 3d, we're going to do 21, 22. 124 00:09:28,000 --> 00:09:31,050 And I separate them out because of Hund's rule. 125 00:09:31,050 --> 00:09:34,000 We want to maximize the number of unpaired spins. 126 00:09:34,000 --> 00:09:38,180 They both spin down, or both spin up, but they will be the 127 00:09:38,180 --> 00:09:42,030 same spin, and they will be in different orbitals. 128 00:09:42,030 --> 00:09:45,260 So this was the answer for this part of the problem. 129 00:09:45,260 --> 00:09:47,010 Now let's move to the second part. 130 00:09:47,010 --> 00:09:49,760 And maybe you can try by yourself, and then 131 00:09:49,760 --> 00:09:52,590 we can do it here. 132 00:09:52,590 --> 00:09:53,020 All right. 133 00:09:53,020 --> 00:09:56,690 Now we're going to do the second part of this first part 134 00:09:56,690 --> 00:09:58,870 of the problem number 2. 135 00:09:58,870 --> 00:10:02,540 So it asks us for the electron configuration in box notation 136 00:10:02,540 --> 00:10:05,050 of magnesium 4 plus. 137 00:10:05,050 --> 00:10:08,310 So again, we're going to determine how many electrons 138 00:10:08,310 --> 00:10:20,520 we have. Which, neutral magnesium has 12 electrons. 139 00:10:20,520 --> 00:10:25,160 But the 4 plus signifies that we have 4 less electrons than 140 00:10:25,160 --> 00:10:31,480 neutrality, and so we subtract 4, and we have 8. 141 00:10:31,480 --> 00:10:34,790 With 8 electrons to work with, we again want to write down 142 00:10:34,790 --> 00:10:51,740 our boxes using the energy level tricks that we have. And 143 00:10:51,740 --> 00:10:54,130 now we can fill them up. 144 00:10:54,130 --> 00:10:58,680 The Pauli Exclusion Principle tells us 1 spin up and 1 spin 145 00:10:58,680 --> 00:11:06,100 down per orbital, and Hund's rule tells us that when we 146 00:11:06,100 --> 00:11:10,400 fill up this 3p, we're going to try to maximize the number 147 00:11:10,400 --> 00:11:16,770 of unpaired electrons, and then, when we have to, we pair 148 00:11:16,770 --> 00:11:18,670 up the spins. 149 00:11:18,670 --> 00:11:23,300 So that's another main mistake that people make, is they pair 150 00:11:23,300 --> 00:11:27,930 up too many spins, violate the Pauli exclusion principle, 151 00:11:27,930 --> 00:11:31,870 putting 2 electrons of the same spin in one orbital, or 152 00:11:31,870 --> 00:11:33,400 something of that nature. 153 00:11:33,400 --> 00:11:37,790 So this again is the correct answer, and would-- oh! 154 00:11:37,790 --> 00:11:38,320 I'm sorry. 155 00:11:38,320 --> 00:11:41,310 This is 2p. 156 00:11:41,310 --> 00:11:44,040 I'm sure you guys caught that. 157 00:11:44,040 --> 00:11:47,910 And would have awarded you full points in this part. 158 00:11:47,910 --> 00:11:48,390 All right. 159 00:11:48,390 --> 00:11:50,430 Now we're going to start the second part of this 160 00:11:50,430 --> 00:11:51,840 problem, part B. 161 00:11:51,840 --> 00:11:55,280 So first we want to read what the question is asking. 162 00:11:55,280 --> 00:11:59,030 Give the chemical identities of the species with these 163 00:11:59,030 --> 00:12:02,410 ground state electron configurations. 164 00:12:02,410 --> 00:12:04,890 So now we're working in the opposite direction. 165 00:12:04,890 --> 00:12:09,420 In part A, we were given a chemical species and asked for 166 00:12:09,420 --> 00:12:11,160 the electron configuration. 167 00:12:11,160 --> 00:12:14,760 Now we are given the electron configuration, and asked for 168 00:12:14,760 --> 00:12:17,480 the chemical species. 169 00:12:17,480 --> 00:12:24,940 So the first one has the electron 170 00:12:24,940 --> 00:12:26,100 configuration of xenon. 171 00:12:26,100 --> 00:12:33,150 So this is a noble gas configuration, plus what turns 172 00:12:33,150 --> 00:12:37,530 out to be 27 electrons. 173 00:12:37,530 --> 00:12:41,490 One way you can do this, is to go to xenon, which is in the 174 00:12:41,490 --> 00:12:48,310 fifth row of the Periodic Table, and count through 27 175 00:12:48,310 --> 00:12:59,550 electrons, remembering that when you get to the d on the 176 00:12:59,550 --> 00:13:05,620 sixth and seventh row, you need to go to the f block, 177 00:13:05,620 --> 00:13:07,210 which is signified in this problem by 178 00:13:07,210 --> 00:13:10,690 having f block electrons. 179 00:13:10,690 --> 00:13:13,430 Count through the f block, the d block, 180 00:13:13,430 --> 00:13:19,090 and end up at thallium. 181 00:13:19,090 --> 00:13:23,350 Or if you are familiar with the Periodic Table enough that 182 00:13:23,350 --> 00:13:30,620 you can recognize the valence electrons are in 6p. 183 00:13:33,240 --> 00:13:36,090 And the 6p only has one electron in it. 184 00:13:36,090 --> 00:13:39,450 So that tells me, xenon is in row 5. 185 00:13:39,450 --> 00:13:43,710 This element is going to be in row 6. 186 00:13:43,710 --> 00:13:48,390 I go to the first column of the p block, which signifies 187 00:13:48,390 --> 00:13:52,680 that there's one electron in that 6p orbital, and I see 188 00:13:52,680 --> 00:13:53,980 that it's thallium. 189 00:13:53,980 --> 00:13:55,260 That's the quicker way to do it. 190 00:13:55,260 --> 00:13:59,040 On an exam, you might want to be that familiar with the 191 00:13:59,040 --> 00:14:01,810 periodic table to be able to do that, because you're 192 00:14:01,810 --> 00:14:05,090 certainly time-crunched, a lot of times. 193 00:14:05,090 --> 00:14:09,130 So because that's a neutral atom, that method will work 194 00:14:09,130 --> 00:14:11,780 very easily. 195 00:14:11,780 --> 00:14:17,890 The second part is a little trickier, because it has a 196 00:14:17,890 --> 00:14:20,780 charged species. 197 00:14:20,780 --> 00:14:25,250 So we have a net charge of 4 plus, an electron 198 00:14:25,250 --> 00:14:32,265 configuration of argon and 3d5. 199 00:14:36,540 --> 00:14:39,260 This may seem a little confusing at first glance, 200 00:14:39,260 --> 00:14:47,090 because we always fill the 4s before the 3d. 201 00:14:47,090 --> 00:14:51,970 But when you're ionizing an atom, it turns out you take 202 00:14:51,970 --> 00:14:55,450 electrons from the 4s first, even though 203 00:14:55,450 --> 00:14:56,730 it's of lower energy. 204 00:14:56,730 --> 00:14:59,300 So that's a little subtlety that's not too important to 205 00:14:59,300 --> 00:15:01,340 this problem, but it may have tripped you up when you were 206 00:15:01,340 --> 00:15:04,790 looking at this, and certainly caused some problems for the 207 00:15:04,790 --> 00:15:06,040 people in this class. 208 00:15:08,140 --> 00:15:10,390 Again, we go to the periodic table and we see where argon 209 00:15:10,390 --> 00:15:14,200 is, and then we go to the next row. 210 00:15:14,200 --> 00:15:17,520 The way I like to think about this, is this is the electron 211 00:15:17,520 --> 00:15:25,070 configuration, missing 4 electrons. 212 00:15:25,070 --> 00:15:30,380 To find the element that has this charge, I'm just going to 213 00:15:30,380 --> 00:15:33,590 add back in the electrons, go to that position in the 214 00:15:33,590 --> 00:15:38,070 periodic table, and that is my elemental species. 215 00:15:38,070 --> 00:15:42,670 So we want argon plus, instead of 5 electrons-- 216 00:15:42,670 --> 00:15:45,000 I'm sorry, the problem says this is a 3-- 217 00:15:48,040 --> 00:15:52,660 instead plus 3 electrons, we are going to go to argon plus 218 00:15:52,660 --> 00:15:54,480 7 electrons. 219 00:15:54,480 --> 00:15:58,450 And going through the periodic table, you'll see that that 220 00:15:58,450 --> 00:15:59,700 leaves you with manganese. 221 00:16:02,950 --> 00:16:07,400 Adding the 4 plus, because that is part of this chemical 222 00:16:07,400 --> 00:16:12,480 identity, we get the answer of the problem. 223 00:16:12,480 --> 00:16:15,410 So again, the charged species make this a little more 224 00:16:15,410 --> 00:16:16,520 complicated. 225 00:16:16,520 --> 00:16:20,650 The easiest way is to go back and add in the missing 226 00:16:20,650 --> 00:16:25,300 electrons, because that is your chemical species, and 227 00:16:25,300 --> 00:16:31,030 then remember to signify that there actually is that charge 228 00:16:31,030 --> 00:16:34,120 for your answer. 229 00:16:34,120 --> 00:16:45,240 So moving on to part C, it asks us, write the quantum 230 00:16:45,240 --> 00:16:48,240 numbers of one of the 3d and one of the 4s 231 00:16:48,240 --> 00:16:50,110 electrons in iron. 232 00:16:50,110 --> 00:16:53,660 So this requires a knowledge of what quantum numbers are, 233 00:16:53,660 --> 00:16:57,040 and what values they can have. 234 00:16:57,040 --> 00:17:01,280 So let's start with the 4s, because that's a little more 235 00:17:01,280 --> 00:17:03,850 straightforward. 236 00:17:03,850 --> 00:17:07,400 He tells us that we have an n quantum number, l quantum 237 00:17:07,400 --> 00:17:10,730 number, m and s. 238 00:17:10,730 --> 00:17:15,580 And, hopefully, we all know that n is the principal 239 00:17:15,580 --> 00:17:16,630 quantum number. 240 00:17:16,630 --> 00:17:18,090 It's basically your energy level. 241 00:17:21,800 --> 00:17:25,130 l gives you what type of orbital you're in. 242 00:17:25,130 --> 00:17:25,400 Right? 243 00:17:25,400 --> 00:17:27,520 If you're in the s, the p, the d. 244 00:17:27,520 --> 00:17:30,870 So it just tells you your orbital shape, basically. 245 00:17:35,440 --> 00:17:39,010 m gives you the angular identity of the 246 00:17:39,010 --> 00:17:40,250 orbital you're in. 247 00:17:40,250 --> 00:17:42,570 We didn't really talk about that in depth. 248 00:17:42,570 --> 00:17:47,330 We just need to know, it tells you which of the p orbitals 249 00:17:47,330 --> 00:17:50,680 you're in, or which of the d orbitals you're in. 250 00:17:50,680 --> 00:17:53,334 So I'll say, which orbital. 251 00:17:58,280 --> 00:18:01,910 And the s is your spin. 252 00:18:01,910 --> 00:18:02,510 Right? 253 00:18:02,510 --> 00:18:06,740 So for a 4s electron, it doesn't actually matter what 254 00:18:06,740 --> 00:18:09,210 the identity of our species is. 255 00:18:09,210 --> 00:18:13,160 So that he tells us is iron is kind of extra information. 256 00:18:13,160 --> 00:18:18,300 A 4s electron already has 3 of its quantum numbers set. 257 00:18:18,300 --> 00:18:27,680 So we know that n equals 4, for s, l equals 0, and m, 258 00:18:27,680 --> 00:18:31,220 which orbital, s only has 1 orbital. 259 00:18:31,220 --> 00:18:36,780 So m always equals 0 for s, and your spin. 260 00:18:36,780 --> 00:18:39,950 And I'm going to make that a script, just so it's less 261 00:18:39,950 --> 00:18:49,400 confusing, is plus or minus 1/2. 262 00:18:49,400 --> 00:18:56,860 So the answer to this problem could have been 4, 0, 0, 1/2, 263 00:18:56,860 --> 00:19:01,170 or 4, 0, 0, minus 1/2. 264 00:19:03,670 --> 00:19:05,950 Three of them are set by the fact that 265 00:19:05,950 --> 00:19:07,730 this is the 4s orbital. 266 00:19:07,730 --> 00:19:11,280 The spin can either be plus or minus 1/2. 267 00:19:11,280 --> 00:19:14,260 If you're unsure of where I got these numbers, and why 268 00:19:14,260 --> 00:19:16,380 they have to be such, you may want to review 269 00:19:16,380 --> 00:19:18,820 the lecture on that. 270 00:19:18,820 --> 00:19:30,220 So moving onto the 3d quantum numbers, we know that n is 3, 271 00:19:30,220 --> 00:19:45,610 l for d is 2, m is negative 2, negative 1, 0, 1, or 2, right? 272 00:19:45,610 --> 00:19:50,750 Because there's 5 different orbitals in a d set. 273 00:19:50,750 --> 00:19:57,010 And then again, s can be plus or minus 1/2. 274 00:19:57,010 --> 00:20:04,070 So picking one set of those, you could have said 3, 2, 275 00:20:04,070 --> 00:20:11,290 minus 2, plus 1/2, or any combination of 3, 2, and 276 00:20:11,290 --> 00:20:13,940 the m and the s. 277 00:20:13,940 --> 00:20:15,860 You don't have to get the exact same answer as your 278 00:20:15,860 --> 00:20:18,090 neighbor or someone else you're-- well, you shouldn't 279 00:20:18,090 --> 00:20:22,740 be looking your neighbor's paper, or a classmate, but 280 00:20:22,740 --> 00:20:27,640 these do signify that you're in a 3d electron.