[{"Name":"Lewis Theory of Chemical Bonding","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Lewis Theory and Ionic Bonds","Duration":"4m 53s","ChapterTopicVideoID":20271,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"In this video, we\u0027ll discuss one of"},{"Start":"00:02.550 ","End":"00:06.660","Text":"the simplest theories of chemical bonding, Lewis theory."},{"Start":"00:06.660 ","End":"00:09.705","Text":"Between 1916 and 1919,"},{"Start":"00:09.705 ","End":"00:12.030","Text":"3 people, Gilbert Lewis,"},{"Start":"00:12.030 ","End":"00:18.695","Text":"Irving Langmuir, and Walter Kossel developed a very simple theory of chemical bonding."},{"Start":"00:18.695 ","End":"00:22.865","Text":"Nowadays it\u0027s usually called the Lewis theory of chemical bonding."},{"Start":"00:22.865 ","End":"00:28.285","Text":"The first thing they said was that valence electrons participate in chemical bonding,"},{"Start":"00:28.285 ","End":"00:31.910","Text":"and then that electrons can be transferred from one atom to"},{"Start":"00:31.910 ","End":"00:35.465","Text":"another to form ions which then attract to each other."},{"Start":"00:35.465 ","End":"00:37.805","Text":"They call this ionic bonds,"},{"Start":"00:37.805 ","End":"00:40.370","Text":"and we\u0027ll talk about it in this video."},{"Start":"00:40.370 ","End":"00:43.955","Text":"Alternatively, electrons can be shared between atoms."},{"Start":"00:43.955 ","End":"00:45.905","Text":"This is called covalent bonds,"},{"Start":"00:45.905 ","End":"00:48.575","Text":"and we\u0027ll talk about it in the next video."},{"Start":"00:48.575 ","End":"00:50.900","Text":"Atoms, apart from hydrogen,"},{"Start":"00:50.900 ","End":"00:52.160","Text":"lithium, and beryllium,"},{"Start":"00:52.160 ","End":"00:54.995","Text":"acquire 8 electrons in the valence shell,"},{"Start":"00:54.995 ","End":"00:56.965","Text":"just like noble gases."},{"Start":"00:56.965 ","End":"00:58.985","Text":"We call this the octet rule."},{"Start":"00:58.985 ","End":"01:04.115","Text":"Hydrogen, lithium, and beryllium themselves acquire only 2 electrons."},{"Start":"01:04.115 ","End":"01:06.515","Text":"We call that a duplet."},{"Start":"01:06.515 ","End":"01:08.240","Text":"In order to start the C,"},{"Start":"01:08.240 ","End":"01:12.320","Text":"we have to write the Lewis symbols for the various elements."},{"Start":"01:12.320 ","End":"01:15.274","Text":"We write the name of the element in the center"},{"Start":"01:15.274 ","End":"01:18.710","Text":"and around it the number of valence electrons."},{"Start":"01:18.710 ","End":"01:20.345","Text":"Let\u0027s start with hydrogen."},{"Start":"01:20.345 ","End":"01:22.985","Text":"We know that hydrogen only has 1 electron."},{"Start":"01:22.985 ","End":"01:25.440","Text":"We know that helium has 2 electrons."},{"Start":"01:25.440 ","End":"01:27.800","Text":"We know now that they\u0027re paired."},{"Start":"01:27.800 ","End":"01:32.675","Text":"But even then, it was known that helium only has 2 electrons."},{"Start":"01:32.675 ","End":"01:35.420","Text":"Lithium, which belongs to the second period,"},{"Start":"01:35.420 ","End":"01:38.300","Text":"has 1 electron in the valence shell."},{"Start":"01:38.300 ","End":"01:40.550","Text":"Beryllium has 2,"},{"Start":"01:40.550 ","End":"01:43.640","Text":"but unlike helium, they are unpaired."},{"Start":"01:43.640 ","End":"01:46.000","Text":"Boron has 3."},{"Start":"01:46.000 ","End":"01:49.005","Text":"Carbon has 4."},{"Start":"01:49.005 ","End":"01:51.460","Text":"Nitrogen has 5."},{"Start":"01:51.460 ","End":"01:54.485","Text":"But now we start to pair the electrons."},{"Start":"01:54.485 ","End":"02:01.475","Text":"We will have 1 pair and 3 unpaired electrons."},{"Start":"02:01.475 ","End":"02:04.370","Text":"Oxygen has 6 electrons,"},{"Start":"02:04.370 ","End":"02:08.355","Text":"2 pairs and 2 unpaired electrons."},{"Start":"02:08.355 ","End":"02:10.920","Text":"Fluorine has 7 electrons,"},{"Start":"02:10.920 ","End":"02:14.790","Text":"3 pairs and 1 unpaired,"},{"Start":"02:14.790 ","End":"02:18.825","Text":"and neon has 4 pairs of electrons."},{"Start":"02:18.825 ","End":"02:22.085","Text":"Neon has an octet of electrons."},{"Start":"02:22.085 ","End":"02:25.760","Text":"Now we\u0027re going to write the Lewis structures for ionic compounds."},{"Start":"02:25.760 ","End":"02:29.960","Text":"Ionic compounds consists of a metal and a non-metal."},{"Start":"02:29.960 ","End":"02:32.135","Text":"We\u0027re going to take 2 examples,"},{"Start":"02:32.135 ","End":"02:35.540","Text":"sodium chloride and barium chloride."},{"Start":"02:35.540 ","End":"02:37.115","Text":"Before we do this,"},{"Start":"02:37.115 ","End":"02:39.845","Text":"let\u0027s remember that sodium chloride,"},{"Start":"02:39.845 ","End":"02:43.535","Text":"just NaCL and BaCl_2 are formula units."},{"Start":"02:43.535 ","End":"02:49.220","Text":"Sodium chloride is really an ionic crystal consisting of many sodium ions,"},{"Start":"02:49.220 ","End":"02:54.360","Text":"NA plus, and chloride ions Cl minus in a 1:1 ratio."},{"Start":"02:54.360 ","End":"02:58.430","Text":"Barium chloride is an ionic crystal consisting of barium 2 plus"},{"Start":"02:58.430 ","End":"03:03.150","Text":"ions and chlorine minus in a 1:2 ratio."},{"Start":"03:03.150 ","End":"03:06.800","Text":"They\u0027re really solids, ionic solids."},{"Start":"03:06.800 ","End":"03:11.590","Text":"But we\u0027re going to describe just the formula unit."},{"Start":"03:11.590 ","End":"03:14.119","Text":"Here\u0027s sodium and chlorine."},{"Start":"03:14.119 ","End":"03:18.905","Text":"Sodium just has 1 electron in the valence shell,"},{"Start":"03:18.905 ","End":"03:23.255","Text":"whereas chlorine has 7 electrons in the valence shell."},{"Start":"03:23.255 ","End":"03:27.960","Text":"It\u0027s below fluorine in the periodic table."},{"Start":"03:29.470 ","End":"03:36.925","Text":"Now, sodium donates its single electron to chlorine."},{"Start":"03:36.925 ","End":"03:41.690","Text":"Now chlorine has 8 electrons surrounding it."},{"Start":"03:41.690 ","End":"03:44.080","Text":"It has an octet of electrons."},{"Start":"03:44.080 ","End":"03:46.650","Text":"Now sodium has lost an electron."},{"Start":"03:46.650 ","End":"03:48.495","Text":"So sodium plus."},{"Start":"03:48.495 ","End":"03:51.360","Text":"Chlorine has an additional electron."},{"Start":"03:51.360 ","End":"03:53.145","Text":"So it\u0027s chlorine minus,"},{"Start":"03:53.145 ","End":"03:55.565","Text":"and it has 8 electrons around it."},{"Start":"03:55.565 ","End":"03:59.045","Text":"Now let\u0027s consider barium fluoride."},{"Start":"03:59.045 ","End":"04:06.240","Text":"Barium has 2 valence electrons and fluorine has 7."},{"Start":"04:06.240 ","End":"04:09.020","Text":"Barium can donate electron to one of"},{"Start":"04:09.020 ","End":"04:13.465","Text":"the fluorine atoms and an electron to the other fluorine atom."},{"Start":"04:13.465 ","End":"04:16.499","Text":"Now barium has lost 2 electrons."},{"Start":"04:16.499 ","End":"04:18.375","Text":"So it\u0027s barium 2 plus."},{"Start":"04:18.375 ","End":"04:21.230","Text":"Each fluorine has an additional electron."},{"Start":"04:21.230 ","End":"04:23.300","Text":"So it\u0027s fluorine minus,"},{"Start":"04:23.300 ","End":"04:28.405","Text":"and now each one has an octet of electrons around it."},{"Start":"04:28.405 ","End":"04:35.045","Text":"Now sodium plus and chlorine minus can form a crystal, stable crystal."},{"Start":"04:35.045 ","End":"04:43.835","Text":"Barium 2 plus and 2 fluorine minus ions can form also a stable crystal."},{"Start":"04:43.835 ","End":"04:47.990","Text":"Of course, the energetics of this have to be described as well."},{"Start":"04:47.990 ","End":"04:53.670","Text":"In this video we talked about ionic bonding according to the Lewis theory."}],"ID":21062},{"Watched":false,"Name":"Lewis Theory and Covalent Bonds","Duration":"7m 7s","ChapterTopicVideoID":20272,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.410","Text":"In the previous video,"},{"Start":"00:01.410 ","End":"00:04.365","Text":"we talked about Lewis theory and ionic bonds."},{"Start":"00:04.365 ","End":"00:07.305","Text":"In this video, we\u0027ll talk about covalent bonds."},{"Start":"00:07.305 ","End":"00:11.910","Text":"We\u0027re going to describe the Lewis structure of covalent compounds."},{"Start":"00:11.910 ","End":"00:15.735","Text":"Covalent compounds take place between nonmetals."},{"Start":"00:15.735 ","End":"00:18.150","Text":"We\u0027re going to describe 2 examples,"},{"Start":"00:18.150 ","End":"00:23.325","Text":"water H_2O and oxygen difluoride, which is OF_2."},{"Start":"00:23.325 ","End":"00:25.125","Text":"Let\u0027s begin with water."},{"Start":"00:25.125 ","End":"00:27.840","Text":"Hydrogen has 1 valence electron,"},{"Start":"00:27.840 ","End":"00:31.605","Text":"and oxygen has 6 valence electrons,"},{"Start":"00:31.605 ","End":"00:36.585","Text":"2 pairs and 2 single electrons that are unpaired."},{"Start":"00:36.585 ","End":"00:38.505","Text":"When the molecule is formed,"},{"Start":"00:38.505 ","End":"00:42.240","Text":"1 electron from hydrogen shares with"},{"Start":"00:42.240 ","End":"00:48.080","Text":"1 electron from the oxygen to form what we call a bond pair."},{"Start":"00:48.080 ","End":"00:51.470","Text":"Now we have a bond pair between hydrogen and oxygen,"},{"Start":"00:51.470 ","End":"00:54.275","Text":"and the other one between oxygen and hydrogen."},{"Start":"00:54.275 ","End":"00:57.305","Text":"I\u0027ve indicated 1 electron in red and 1 in blue,"},{"Start":"00:57.305 ","End":"01:01.700","Text":"but it doesn\u0027t really matter because electrons are indistinguishable."},{"Start":"01:01.700 ","End":"01:07.370","Text":"Another way of writing this is to indicate a bond pair by a single line,"},{"Start":"01:07.370 ","End":"01:10.670","Text":"that\u0027s a single covalent bond."},{"Start":"01:10.670 ","End":"01:14.179","Text":"A bond pair to form"},{"Start":"01:14.179 ","End":"01:22.230","Text":"is single covalent bond."},{"Start":"01:22.230 ","End":"01:24.025","Text":"We have 2 bond pairs."},{"Start":"01:24.025 ","End":"01:29.930","Text":"In addition, we can add the pairs of electrons on the oxygen."},{"Start":"01:31.590 ","End":"01:37.670","Text":"These pairs of electrons are called lone pairs."},{"Start":"01:40.500 ","End":"01:44.665","Text":"We have lone pairs and bond pairs."},{"Start":"01:44.665 ","End":"01:47.095","Text":"Now if we look at H_2O,"},{"Start":"01:47.095 ","End":"01:51.630","Text":"we can count the number of electrons around each atom."},{"Start":"01:51.630 ","End":"01:54.280","Text":"Around hydrogen, there are 2 electrons,"},{"Start":"01:54.280 ","End":"01:56.575","Text":"that we call it duplet."},{"Start":"01:56.575 ","End":"01:59.620","Text":"Around oxygen, there are 8 electrons,"},{"Start":"01:59.620 ","End":"02:02.135","Text":"we call that an octet,"},{"Start":"02:02.135 ","End":"02:07.510","Text":"another duplet, and we can do the same thing in this picture."},{"Start":"02:07.510 ","End":"02:09.129","Text":"If we look at the oxygen,"},{"Start":"02:09.129 ","End":"02:11.320","Text":"2 electrons for each bond pair,"},{"Start":"02:11.320 ","End":"02:15.715","Text":"and 2 electrons for each lone pair giving a total of 8."},{"Start":"02:15.715 ","End":"02:19.275","Text":"Now let\u0027s look at OF_2."},{"Start":"02:19.275 ","End":"02:23.340","Text":"Fluorine has 7 valence electrons,"},{"Start":"02:23.340 ","End":"02:27.505","Text":"3 pairs and 1 unpaired electron."},{"Start":"02:27.505 ","End":"02:34.930","Text":"Oxygen has 6 valence electrons as we saw before."},{"Start":"02:36.060 ","End":"02:40.430","Text":"You\u0027ll see electrons in the other fluorine."},{"Start":"02:41.720 ","End":"02:46.280","Text":"The unpaired electrons, fluorine can share with"},{"Start":"02:46.280 ","End":"02:52.000","Text":"the unpaired electron on the oxygen to form a bond pair,"},{"Start":"02:52.000 ","End":"02:55.560","Text":"and each bond pair we saw before,"},{"Start":"02:55.560 ","End":"02:58.850","Text":"can be written as a single straight line."},{"Start":"02:58.850 ","End":"03:04.135","Text":"Once again, we can add all the lone pairs."},{"Start":"03:04.135 ","End":"03:14.830","Text":"We have 3 lone pairs around fluorine and 2 lone pairs around the oxygen."},{"Start":"03:22.940 ","End":"03:28.430","Text":"Once again, we can count the number of electrons around each atom,"},{"Start":"03:28.430 ","End":"03:30.700","Text":"around fluorine there\u0027s an octet,"},{"Start":"03:30.700 ","End":"03:33.180","Text":"around oxygen another octet,"},{"Start":"03:33.180 ","End":"03:35.810","Text":"around fluorine another octet."},{"Start":"03:35.810 ","End":"03:42.095","Text":"One thing we should notice is that we\u0027ve counted some of the electrons more than once."},{"Start":"03:42.095 ","End":"03:45.664","Text":"For example, fluorine has 7 electrons,"},{"Start":"03:45.664 ","End":"03:47.835","Text":"oxygen has 6,"},{"Start":"03:47.835 ","End":"03:49.455","Text":"and fluorine has 7."},{"Start":"03:49.455 ","End":"03:51.285","Text":"That\u0027s a total of 20,"},{"Start":"03:51.285 ","End":"03:55.350","Text":"and if we counted an octet,"},{"Start":"03:55.350 ","End":"03:59.790","Text":"an octet and an octet that means we\u0027ve counted 24 electrons."},{"Start":"03:59.790 ","End":"04:03.935","Text":"Some electrons are being counted more than once."},{"Start":"04:03.935 ","End":"04:05.885","Text":"Now we\u0027re going to describe"},{"Start":"04:05.885 ","End":"04:12.245","Text":"a coordinate covalent bond which is a variant on the covalent bond."},{"Start":"04:12.245 ","End":"04:14.180","Text":"Now in our previous examples,"},{"Start":"04:14.180 ","End":"04:16.535","Text":"H_2O and OF_2,"},{"Start":"04:16.535 ","End":"04:19.895","Text":"each atom contributed 1 electron to the bond pair."},{"Start":"04:19.895 ","End":"04:24.860","Text":"However, sometimes 1 atom contributes both electrons to the bond pair."},{"Start":"04:24.860 ","End":"04:28.580","Text":"This is called a coordinate covalent bond."},{"Start":"04:28.580 ","End":"04:32.600","Text":"Once it\u0027s formed, it seemed distinguishable from a covalent bond."},{"Start":"04:32.600 ","End":"04:33.910","Text":"But when it\u0027s formed,"},{"Start":"04:33.910 ","End":"04:36.905","Text":"it\u0027s a coordinate covalent bond."},{"Start":"04:36.905 ","End":"04:38.775","Text":"Let\u0027s take an example."},{"Start":"04:38.775 ","End":"04:43.895","Text":"Going to discuss the reaction of ammonia with hydrogen chloride."},{"Start":"04:43.895 ","End":"04:47.390","Text":"Let\u0027s look at the Lewis structure of ammonia."},{"Start":"04:47.390 ","End":"04:53.710","Text":"Ammonia has 3 single bonds connecting the nitrogen with each of the hydrogens."},{"Start":"04:53.710 ","End":"04:56.660","Text":"In addition, it has a lone pair,"},{"Start":"04:56.660 ","End":"04:58.910","Text":"so that makes an octet,"},{"Start":"04:58.910 ","End":"05:02.555","Text":"2 electrons in each of the bond pairs,"},{"Start":"05:02.555 ","End":"05:04.790","Text":"and 2 electrons from the lone pair,"},{"Start":"05:04.790 ","End":"05:07.105","Text":"a total of 8 electrons."},{"Start":"05:07.105 ","End":"05:09.680","Text":"HCl has a single bond,"},{"Start":"05:09.680 ","End":"05:12.709","Text":"so hydrogen has a duplet around it,"},{"Start":"05:12.709 ","End":"05:15.875","Text":"and chlorine has a bond pair,"},{"Start":"05:15.875 ","End":"05:20.025","Text":"and another 3 lone pairs."},{"Start":"05:20.025 ","End":"05:24.785","Text":"Now what happens when they combine is that"},{"Start":"05:24.785 ","End":"05:29.405","Text":"all the electrons in this bond pair go to the chlorine."},{"Start":"05:29.405 ","End":"05:33.410","Text":"Now, chlorine has 1 electron more than it started with."},{"Start":"05:33.410 ","End":"05:38.360","Text":"It started off with 7 and now it has 8,"},{"Start":"05:38.360 ","End":"05:39.900","Text":"so it has an additional electrons,"},{"Start":"05:39.900 ","End":"05:44.920","Text":"so it\u0027s chlorine minus with an octet around it."},{"Start":"05:45.710 ","End":"05:50.675","Text":"Hydrogen is left without any electrons at all."},{"Start":"05:50.675 ","End":"05:53.240","Text":"Hydrogen is H plus,"},{"Start":"05:53.240 ","End":"05:56.960","Text":"so we\u0027ve gone to H plus plus Cl minus."},{"Start":"05:56.960 ","End":"06:03.345","Text":"Now the lone pair on the nitrogen combines with the hydrogen plus."},{"Start":"06:03.345 ","End":"06:07.985","Text":"The nitrogen donates its lone pair to hydrogen plus,"},{"Start":"06:07.985 ","End":"06:12.770","Text":"and we get NH_4 plus."},{"Start":"06:12.770 ","End":"06:15.880","Text":"Here are the single bonds."},{"Start":"06:15.880 ","End":"06:21.665","Text":"We have 4 single bonds between the nitrogen and 4 hydrogens,"},{"Start":"06:21.665 ","End":"06:27.270","Text":"and the whole species has a positive charge, it\u0027s called ammonium."},{"Start":"06:30.380 ","End":"06:36.350","Text":"Now we have ammonium plus and chlorine minus."},{"Start":"06:36.350 ","End":"06:40.080","Text":"We have ammonium chloride."},{"Start":"06:46.700 ","End":"06:52.760","Text":"Now NH_3 ammonia is called a Lewis base because it donates a pair of"},{"Start":"06:52.760 ","End":"06:59.420","Text":"electrons and HCL is called a Lewis acid because it accepts a pair of electrons."},{"Start":"06:59.420 ","End":"07:03.470","Text":"In this video, we discussed single covalent bonds,"},{"Start":"07:03.470 ","End":"07:07.350","Text":"and also coordinate covalent bonds."}],"ID":21063},{"Watched":false,"Name":"Multiple Covalent Bonds","Duration":"7m 30s","ChapterTopicVideoID":20273,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.643","Text":"In the previous videos,"},{"Start":"00:01.643 ","End":"00:06.450","Text":"we talked about ionic and single covalent bonds according to the Lewis theory."},{"Start":"00:06.450 ","End":"00:10.475","Text":"In this video we\u0027ll learn about multiple covalent bonds,"},{"Start":"00:10.475 ","End":"00:13.635","Text":"so we\u0027re going to talk about multiple covalent bonds."},{"Start":"00:13.635 ","End":"00:14.880","Text":"Why do we need them?"},{"Start":"00:14.880 ","End":"00:18.570","Text":"Sometimes, forming only single covalent bonds does"},{"Start":"00:18.570 ","End":"00:22.410","Text":"not allow the atoms to obtain an octet configuration,"},{"Start":"00:22.410 ","End":"00:26.295","Text":"and we can remedy this by forming multiple covalent bonds."},{"Start":"00:26.295 ","End":"00:27.570","Text":"There are 2 sorts,"},{"Start":"00:27.570 ","End":"00:30.165","Text":"double and triple bonds."},{"Start":"00:30.165 ","End":"00:32.495","Text":"Let\u0027s begin with double bonds."},{"Start":"00:32.495 ","End":"00:34.475","Text":"We\u0027re going to take it as example,"},{"Start":"00:34.475 ","End":"00:37.030","Text":"carbon dioxide, CO_2."},{"Start":"00:37.030 ","End":"00:41.080","Text":"We have oxygen, it has 6 valence electron;"},{"Start":"00:41.080 ","End":"00:51.510","Text":"2 pairs and 2 unpaired electrons and carbon has 4 unpaired electrons."},{"Start":"00:51.730 ","End":"00:56.210","Text":"If we form a single bond between the oxygen and the carbon,"},{"Start":"00:56.210 ","End":"00:59.040","Text":"and the carbon and the other oxygen,"},{"Start":"00:59.960 ","End":"01:03.245","Text":"we will still have electrons left."},{"Start":"01:03.245 ","End":"01:08.500","Text":"We have 2 pairs on the oxygen and a single unpaired electron,"},{"Start":"01:08.500 ","End":"01:11.030","Text":"and the same in the other one."},{"Start":"01:12.200 ","End":"01:16.790","Text":"If we count how many electrons we have around each atom,"},{"Start":"01:16.790 ","End":"01:20.356","Text":"we find that around the oxygen we have 2,"},{"Start":"01:20.356 ","End":"01:23.565","Text":"4, 6, 7 electrons."},{"Start":"01:23.565 ","End":"01:25.774","Text":"Around carbon, we have 1,"},{"Start":"01:25.774 ","End":"01:28.305","Text":"2, 3, 4 electrons."},{"Start":"01:28.305 ","End":"01:31.410","Text":"Around oxygen, again, we have 7."},{"Start":"01:31.410 ","End":"01:34.453","Text":"None of the atoms have achieved an octet,"},{"Start":"01:34.453 ","End":"01:41.180","Text":"we can draw the same structure here with losing a single bond as a straight line."},{"Start":"01:41.180 ","End":"01:43.345","Text":"We have 2 electrons,"},{"Start":"01:43.345 ","End":"01:46.455","Text":"and other 2, and 1 unpaired."},{"Start":"01:46.455 ","End":"01:49.630","Text":"We have 2 unpaired electrons on the carbon."},{"Start":"01:49.630 ","End":"01:54.455","Text":"How are we going to achieve octets on each of the atoms?"},{"Start":"01:54.455 ","End":"01:58.850","Text":"What we\u0027re going to do is to form a bond pair between"},{"Start":"01:58.850 ","End":"02:03.895","Text":"the red one here and this on oxygen and the blue one on the carbon."},{"Start":"02:03.895 ","End":"02:05.810","Text":"That will give us another bond,"},{"Start":"02:05.810 ","End":"02:07.610","Text":"that\u0027s a double bond."},{"Start":"02:07.610 ","End":"02:10.145","Text":"The same thing between carbon-oxygen,"},{"Start":"02:10.145 ","End":"02:15.770","Text":"we take this blue one and this red one and form another bond pair."},{"Start":"02:15.770 ","End":"02:19.850","Text":"We have 2 bond pairs between oxygen and carbon,"},{"Start":"02:19.850 ","End":"02:23.885","Text":"and another 2 bond pairs between carbon and oxygen."},{"Start":"02:23.885 ","End":"02:28.935","Text":"We indicate this as 2 parallel lines,"},{"Start":"02:28.935 ","End":"02:30.590","Text":"so that\u0027s a double bond."},{"Start":"02:30.590 ","End":"02:34.970","Text":"We have 2 double bonds and we still have electrons leftover."},{"Start":"02:34.970 ","End":"02:38.258","Text":"We have 2 lone pairs on each oxygen,"},{"Start":"02:38.258 ","End":"02:43.325","Text":"so now we can count up how many electrons we have around each atom."},{"Start":"02:43.325 ","End":"02:45.995","Text":"Around the oxygen we have 2, 4, 6, 8."},{"Start":"02:45.995 ","End":"02:48.360","Text":"That\u0027s 8."},{"Start":"02:48.360 ","End":"02:50.790","Text":"Around the carbon we have 4 bonds,"},{"Start":"02:50.790 ","End":"02:52.740","Text":"so that\u0027s 8 electrons."},{"Start":"02:52.740 ","End":"02:54.420","Text":"Around this oxygen,"},{"Start":"02:54.420 ","End":"02:56.795","Text":"again, we have 8 electrons."},{"Start":"02:56.795 ","End":"03:01.560","Text":"We have an octet on every atom."},{"Start":"03:01.560 ","End":"03:06.980","Text":"By forming double bonds we\u0027ve achieved octet status for all the atoms."},{"Start":"03:06.980 ","End":"03:09.380","Text":"Let\u0027s discuss triple bonds."},{"Start":"03:09.380 ","End":"03:13.480","Text":"I\u0027m going to take the example of nitrogen, N_2."},{"Start":"03:13.480 ","End":"03:17.820","Text":"Nitrogen has 5 valence electrons,"},{"Start":"03:17.820 ","End":"03:21.845","Text":"a pair and 3 unpaired electrons."},{"Start":"03:21.845 ","End":"03:23.855","Text":"Same on the other nitrogen,"},{"Start":"03:23.855 ","End":"03:28.295","Text":"a pair and 3 unpaired electrons."},{"Start":"03:28.295 ","End":"03:35.255","Text":"If we form single bond then we\u0027ll still have electrons leftover,"},{"Start":"03:35.255 ","End":"03:39.365","Text":"a pair and 2 unpaired electrons."},{"Start":"03:39.365 ","End":"03:45.360","Text":"A pair on this nitrogen and 2 unpaired electrons."},{"Start":"03:45.360 ","End":"03:51.485","Text":"What we can do is look first and see how many electrons we have around each atom."},{"Start":"03:51.485 ","End":"03:52.820","Text":"Around the first nitrogen,"},{"Start":"03:52.820 ","End":"03:54.827","Text":"we have 2, 3, 4, 5,"},{"Start":"03:54.827 ","End":"03:58.815","Text":"6, 6 electrons."},{"Start":"03:58.815 ","End":"04:00.494","Text":"Around the other nitrogen,"},{"Start":"04:00.494 ","End":"04:02.190","Text":"of course, we also have 6,"},{"Start":"04:02.190 ","End":"04:07.625","Text":"and we can draw this with a straight line for the electron pair,"},{"Start":"04:07.625 ","End":"04:09.920","Text":"for the bond pair."},{"Start":"04:09.920 ","End":"04:13.770","Text":"We have the same electrons as we had before."},{"Start":"04:13.840 ","End":"04:19.820","Text":"How can we proceed to make octets around each of the nitrogens?"},{"Start":"04:19.820 ","End":"04:27.245","Text":"What we can do is to take this electron and this electron to form another pair,"},{"Start":"04:27.245 ","End":"04:31.690","Text":"and this electron and this electron to form a third pair."},{"Start":"04:31.690 ","End":"04:33.665","Text":"Once we\u0027ve done that,"},{"Start":"04:33.665 ","End":"04:36.920","Text":"we will have a triple bond,"},{"Start":"04:36.920 ","End":"04:40.040","Text":"3 bond pairs between the 2 nitrogens,"},{"Start":"04:40.040 ","End":"04:42.380","Text":"and what we have left is"},{"Start":"04:42.380 ","End":"04:48.920","Text":"this lone pair on the nitrogen and this lone pair on the other nitrogen."},{"Start":"04:48.920 ","End":"04:54.494","Text":"We can look at each nitrogen and see that we have 3 bond pairs,"},{"Start":"04:54.494 ","End":"04:57.950","Text":"that\u0027s 6 electrons and a lone pair"},{"Start":"04:57.950 ","End":"05:02.935","Text":"making a total of 8 electrons around each of the nitrogens."},{"Start":"05:02.935 ","End":"05:08.150","Text":"By forming the triple bond, we\u0027ve achieved octets."},{"Start":"05:08.150 ","End":"05:12.545","Text":"Each of the nitrogens now has an octet around it."},{"Start":"05:12.545 ","End":"05:18.695","Text":"Lewis theory is a very simple quite primitive theory and it isn\u0027t always correct."},{"Start":"05:18.695 ","End":"05:21.230","Text":"Let\u0027s take the example of oxygen,"},{"Start":"05:21.230 ","End":"05:23.840","Text":"the ordinary O_2 molecule."},{"Start":"05:23.840 ","End":"05:30.167","Text":"We know that we have 6 electrons around the oxygen."},{"Start":"05:30.167 ","End":"05:37.360","Text":"2 pairs, and 2 unpaired electrons and the same around the other oxygen."},{"Start":"05:40.100 ","End":"05:46.405","Text":"If we form a pair between this red one and blue one,"},{"Start":"05:46.405 ","End":"05:48.970","Text":"we\u0027re still left with electrons."},{"Start":"05:48.970 ","End":"05:53.875","Text":"We have 2 lone pairs"},{"Start":"05:53.875 ","End":"06:00.080","Text":"and an unpaired electrons around this oxygen and the same around the other oxygen,"},{"Start":"06:03.260 ","End":"06:07.505","Text":"2 bond pairs and a single unpaired electron."},{"Start":"06:07.505 ","End":"06:13.080","Text":"If we take the red electrons on this oxygen and the blue electrons on this oxygen"},{"Start":"06:13.080 ","End":"06:18.685","Text":"and form another bond pair so that we now have a double bond as here,"},{"Start":"06:18.685 ","End":"06:22.760","Text":"then we can indicate the electrons that are left."},{"Start":"06:24.300 ","End":"06:28.225","Text":"We have 2 lone pairs on each oxygen,"},{"Start":"06:28.225 ","End":"06:31.180","Text":"and that would seem to be the end of things."},{"Start":"06:31.180 ","End":"06:38.375","Text":"2 lone pairs on each oxygen and 2 bond pairs making a total of 8 electrons."},{"Start":"06:38.375 ","End":"06:42.684","Text":"We have an octet around each of the atoms,"},{"Start":"06:42.684 ","End":"06:46.040","Text":"however, there\u0027s one thing wrong with this."},{"Start":"06:46.040 ","End":"06:53.915","Text":"What it doesn\u0027t explain is the experimental fact that oxygen is paramagnetic."},{"Start":"06:53.915 ","End":"06:56.165","Text":"If you remember from a previous video,"},{"Start":"06:56.165 ","End":"07:01.320","Text":"paramagnetic means that it has unpaired electrons."},{"Start":"07:06.230 ","End":"07:09.470","Text":"In this model, this Lewis model,"},{"Start":"07:09.470 ","End":"07:12.202","Text":"it looks as if all the electrons are paired."},{"Start":"07:12.202 ","End":"07:15.178","Text":"They\u0027re either bond pairs or lone pairs,"},{"Start":"07:15.178 ","End":"07:20.225","Text":"so it doesn\u0027t explain the fact that oxygen is paramagnetic."},{"Start":"07:20.225 ","End":"07:26.479","Text":"We need a more sophisticated theory based on quantum mechanics to explain this fact,"},{"Start":"07:26.479 ","End":"07:29.940","Text":"and we\u0027ll see that in a future video."}],"ID":21064},{"Watched":false,"Name":"Polar Covalent Bonds and Dipole Moments","Duration":"7m 7s","ChapterTopicVideoID":20274,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"In previous videos, we learned about ionic and covalent bonds."},{"Start":"00:04.530 ","End":"00:08.010","Text":"In this video we\u0027ll learn about polar covalent bonds that"},{"Start":"00:08.010 ","End":"00:12.015","Text":"are intermediate between ionic and covalent bonds."},{"Start":"00:12.015 ","End":"00:14.565","Text":"Let\u0027s go over what we\u0027ve learned already."},{"Start":"00:14.565 ","End":"00:16.785","Text":"First of all, ionic bonds."},{"Start":"00:16.785 ","End":"00:18.225","Text":"In an ionic bond,"},{"Start":"00:18.225 ","End":"00:22.755","Text":"the electrons are completely transferred from 1 atom to another."},{"Start":"00:22.755 ","End":"00:25.275","Text":"For example, in NaCl,"},{"Start":"00:25.275 ","End":"00:29.820","Text":"we have Na^plus, and Cl^minus ions."},{"Start":"00:29.820 ","End":"00:34.845","Text":"The electron has been completely transferred from Na to Cl."},{"Start":"00:34.845 ","End":"00:36.705","Text":"What about covalent bonds?"},{"Start":"00:36.705 ","End":"00:37.860","Text":"In a covalent bond,"},{"Start":"00:37.860 ","End":"00:40.910","Text":"the electrons are shared equally between the atoms."},{"Start":"00:40.910 ","End":"00:46.895","Text":"Non-polar or pure covalent bonds only occur between identical elements."},{"Start":"00:46.895 ","End":"00:49.555","Text":"For example, H_2 or Cl_2."},{"Start":"00:49.555 ","End":"00:53.360","Text":"Here the electrons are equally shared between the 2 hydrogens,"},{"Start":"00:53.360 ","End":"00:54.860","Text":"or the 2 chlorines."},{"Start":"00:54.860 ","End":"00:58.190","Text":"Now we want to talk about polar covalent bonds."},{"Start":"00:58.190 ","End":"00:59.960","Text":"In a polar covalent bond,"},{"Start":"00:59.960 ","End":"01:03.650","Text":"the electrons are not shared equally between the atoms."},{"Start":"01:03.650 ","End":"01:06.935","Text":"One atom has more of them than the other."},{"Start":"01:06.935 ","End":"01:12.290","Text":"All covalent bonds between non-identical atoms are polar to some extent."},{"Start":"01:12.290 ","End":"01:17.855","Text":"One atom attracts the bond pair electrons more strongly than the other."},{"Start":"01:17.855 ","End":"01:20.795","Text":"Now, which 1 attracts more strongly?"},{"Start":"01:20.795 ","End":"01:26.855","Text":"We find out that the more metallic atom has a partial positive charge,"},{"Start":"01:26.855 ","End":"01:34.675","Text":"Delta plus, and the more non-metallic atom has a partial negative charge, Delta minus."},{"Start":"01:34.675 ","End":"01:36.705","Text":"For example, in HCl,"},{"Start":"01:36.705 ","End":"01:40.535","Text":"H is positively charged and Cl is negatively charged."},{"Start":"01:40.535 ","End":"01:44.734","Text":"But it\u0027s not as if a complete electron charge is transferred,"},{"Start":"01:44.734 ","End":"01:47.735","Text":"only partial electron charge is transferred."},{"Start":"01:47.735 ","End":"01:54.260","Text":"We\u0027d write H^Delta plus, Cl^Delta minus."},{"Start":"01:54.260 ","End":"01:57.845","Text":"Now, we can learn about how much of"},{"Start":"01:57.845 ","End":"02:02.299","Text":"the electron charge is transferred from the dipole moment,"},{"Start":"02:02.299 ","End":"02:05.455","Text":"which is something can be measured experimentally."},{"Start":"02:05.455 ","End":"02:10.310","Text":"Now, 2 atoms that are polar covalent bond form a dipole moment."},{"Start":"02:10.310 ","End":"02:12.380","Text":"Mu is a dipole moment."},{"Start":"02:12.380 ","End":"02:14.630","Text":"That\u0027s the magnitude of the dipole moment,"},{"Start":"02:14.630 ","End":"02:21.635","Text":"and is equal to Delta times d. Delta is a partial electron charge that\u0027s transferred,"},{"Start":"02:21.635 ","End":"02:24.815","Text":"and d is the distance between the atoms."},{"Start":"02:24.815 ","End":"02:26.554","Text":"Let\u0027s take an example."},{"Start":"02:26.554 ","End":"02:29.510","Text":"If an electron charge is transferred,"},{"Start":"02:29.510 ","End":"02:34.415","Text":"that\u0027s Delta equal to 1.602 times 10^minus 19 coulomb,"},{"Start":"02:34.415 ","End":"02:37.514","Text":"and that\u0027s a complete electron charge,"},{"Start":"02:37.514 ","End":"02:41.180","Text":"and the distance between the atoms is a 100 picometers,"},{"Start":"02:41.180 ","End":"02:44.795","Text":"what is the value of Mu, the dipole moment?"},{"Start":"02:44.795 ","End":"02:47.915","Text":"Mu is equal to Delta,"},{"Start":"02:47.915 ","End":"02:54.860","Text":"which is 1.602 times 10^minus 19 coulombs, times 100 picometers."},{"Start":"02:54.860 ","End":"02:58.745","Text":"That\u0027s 100 times 10^minus 12 meters."},{"Start":"02:58.745 ","End":"03:00.080","Text":"Multiply that out."},{"Start":"03:00.080 ","End":"03:08.075","Text":"We get 1.602 times 10^minus 29 coulombs times meters."},{"Start":"03:08.075 ","End":"03:14.135","Text":"Now, there\u0027s a unit of dipole moment named after a famous scientist called Peter Debye,"},{"Start":"03:14.135 ","End":"03:18.650","Text":"who lived from 1884-1966."},{"Start":"03:18.650 ","End":"03:21.950","Text":"The Debye, written as D,"},{"Start":"03:21.950 ","End":"03:23.890","Text":"is a non-SI unit,"},{"Start":"03:23.890 ","End":"03:26.915","Text":"but it\u0027s still often used for dipole moments."},{"Start":"03:26.915 ","End":"03:35.785","Text":"One Debye is equal to 3.334 times 10^minus 30 coulomb meters."},{"Start":"03:35.785 ","End":"03:40.760","Text":"The dipole moment that we calculated before is"},{"Start":"03:40.760 ","End":"03:46.445","Text":"1.602 times 10^minus 29 coulombs times meters,"},{"Start":"03:46.445 ","End":"03:49.555","Text":"and we work out how many Debyes that is."},{"Start":"03:49.555 ","End":"03:55.545","Text":"That\u0027s 1.602 times 10^minus 29 coulombs times meters."},{"Start":"03:55.545 ","End":"04:05.835","Text":"We have here 1 Debye is equivalent to 3.334 times 10^minus 29 coulomb times meter."},{"Start":"04:05.835 ","End":"04:08.420","Text":"If we multiply out,"},{"Start":"04:08.420 ","End":"04:10.760","Text":"then the coulomb times meter cancels,"},{"Start":"04:10.760 ","End":"04:12.245","Text":"and we\u0027re left with Debye,"},{"Start":"04:12.245 ","End":"04:14.900","Text":"and if we divide the numbers,"},{"Start":"04:14.900 ","End":"04:17.950","Text":"we get 4.80 Debye."},{"Start":"04:17.950 ","End":"04:22.100","Text":"That\u0027s very useful because it\u0027s a much smaller number."},{"Start":"04:22.100 ","End":"04:24.395","Text":"It\u0027s a more realizable number,"},{"Start":"04:24.395 ","End":"04:26.420","Text":"more understandable number,"},{"Start":"04:26.420 ","End":"04:31.220","Text":"than using 3.334 times 10^minus 30."},{"Start":"04:31.220 ","End":"04:33.335","Text":"Let\u0027s take an example."},{"Start":"04:33.335 ","End":"04:38.825","Text":"The dipole moment of the HCl bond is 1.03 Debye."},{"Start":"04:38.825 ","End":"04:43.750","Text":"Which percentage of electron charge is transferred from H to Cl?"},{"Start":"04:43.750 ","End":"04:48.335","Text":"We\u0027re given that the length of the bond is 127 picometers."},{"Start":"04:48.335 ","End":"04:50.180","Text":"Now, we know the dipole moment,"},{"Start":"04:50.180 ","End":"04:54.493","Text":"we know the value of the distance between H and Cl,"},{"Start":"04:54.493 ","End":"04:56.940","Text":"so we can calculate the Delta,"},{"Start":"04:56.940 ","End":"05:06.553","Text":"the charge transferred is equal to Mu over d. Mu is 1.03 Debye,"},{"Start":"05:06.553 ","End":"05:14.735","Text":"and each Debye is 3.34 times 10^minus 30 coulombs meter divided by Debye,"},{"Start":"05:14.735 ","End":"05:21.355","Text":"and the length of the bond is 127 times 10^minus 12 meters."},{"Start":"05:21.355 ","End":"05:24.330","Text":"Debye cancels with Debye,"},{"Start":"05:24.330 ","End":"05:27.665","Text":"meter cancels with meter,"},{"Start":"05:27.665 ","End":"05:34.045","Text":"and we\u0027re left with 2.70 times 10^minus 20 coulomb."},{"Start":"05:34.045 ","End":"05:37.335","Text":"Now, we know the charge of an electron."},{"Start":"05:37.335 ","End":"05:42.755","Text":"The percentage charge is 2.70 times 10^minus 20"},{"Start":"05:42.755 ","End":"05:48.934","Text":"coulombs divided by 1.60 times 10^minus 19 coulomb,"},{"Start":"05:48.934 ","End":"05:51.760","Text":"and that is the charge of the electron."},{"Start":"05:51.760 ","End":"05:53.210","Text":"If we want percentage,"},{"Start":"05:53.210 ","End":"05:54.785","Text":"we multiply it by 100,"},{"Start":"05:54.785 ","End":"05:56.330","Text":"and we do the calculation,"},{"Start":"05:56.330 ","End":"05:59.215","Text":"we get 17 percent."},{"Start":"05:59.215 ","End":"06:06.360","Text":"The percentage charge is 17 percent of a full electron charge."},{"Start":"06:06.360 ","End":"06:10.979","Text":"Now, there\u0027s a concept called percent ionic character,"},{"Start":"06:10.979 ","End":"06:16.400","Text":"and the percentage or percent ionic character is the percent of the charge of"},{"Start":"06:16.400 ","End":"06:22.535","Text":"the electron transferred from atom A to B in an A-B bond."},{"Start":"06:22.535 ","End":"06:28.390","Text":"For HCl, the percentage ionic character is 17 percent,"},{"Start":"06:28.390 ","End":"06:30.770","Text":"exactly what we calculated before."},{"Start":"06:30.770 ","End":"06:33.665","Text":"Now, what about the sign of the dipole moment?"},{"Start":"06:33.665 ","End":"06:36.500","Text":"Now, the modern convention according to"},{"Start":"06:36.500 ","End":"06:44.225","Text":"IUPAC is that the dipole moment points towards the positive charge."},{"Start":"06:44.225 ","End":"06:47.540","Text":"But most textbooks still use the opposite convention,"},{"Start":"06:47.540 ","End":"06:49.775","Text":"and that\u0027s the convention that we\u0027re going to use."},{"Start":"06:49.775 ","End":"06:54.210","Text":"If we have HCl, this is Delta plus,"},{"Start":"06:54.210 ","End":"07:01.955","Text":"Cl^Delta minus, we\u0027re going to say that the dipole moment points towards Cl."},{"Start":"07:01.955 ","End":"07:08.070","Text":"In this video, we learned about polar covalent bonds and dipole moments."}],"ID":21065},{"Watched":false,"Name":"Electronegativity 1","Duration":"7m 28s","ChapterTopicVideoID":23063,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.575","Text":"In the previous video,"},{"Start":"00:01.575 ","End":"00:04.305","Text":"we learned about polar covalent bonds."},{"Start":"00:04.305 ","End":"00:07.755","Text":"In this video, we\u0027ll learn about electronegativity,"},{"Start":"00:07.755 ","End":"00:10.710","Text":"how polar is a polar covalent bond?"},{"Start":"00:10.710 ","End":"00:13.515","Text":"Not all polar covalent bonds are equally polar."},{"Start":"00:13.515 ","End":"00:18.194","Text":"Some atoms attract the bond pair electrons more strongly than others."},{"Start":"00:18.194 ","End":"00:21.839","Text":"In order to know which atom attracts electrons more strongly,"},{"Start":"00:21.839 ","End":"00:24.735","Text":"we use the electronegativity."},{"Start":"00:24.735 ","End":"00:29.745","Text":"By comparing the electronegativity chi of 2 atoms A and B,"},{"Start":"00:29.745 ","End":"00:32.775","Text":"we can estimate the polarity of the A-B bond."},{"Start":"00:32.775 ","End":"00:38.580","Text":"The atom with the higher electronegativity attracts electrons more strongly."},{"Start":"00:38.580 ","End":"00:40.970","Text":"There are many scales of electronegativity."},{"Start":"00:40.970 ","End":"00:42.800","Text":"Here, we\u0027ll talk about 2 of them,"},{"Start":"00:42.800 ","End":"00:47.564","Text":"1 due to Mulliken and the other due to Pauling."},{"Start":"00:47.564 ","End":"00:51.590","Text":"Mulliken won the Nobel Prize in 1966."},{"Start":"00:51.590 ","End":"00:55.490","Text":"He defined electronegativity chi as"},{"Start":"00:55.490 ","End":"01:01.740","Text":"half the sum of the ionization energy and the electron affinity."},{"Start":"01:01.740 ","End":"01:06.350","Text":"He used non-SI units called electron volts."},{"Start":"01:06.350 ","End":"01:12.670","Text":"1 electron volt per mole is the same as 96.485 kilojoules per mole."},{"Start":"01:12.670 ","End":"01:17.795","Text":"Now, if the ionization energy and electron affinity are slow,"},{"Start":"01:17.795 ","End":"01:20.520","Text":"that means it\u0027s easy to remove electron from the atom"},{"Start":"01:20.520 ","End":"01:23.635","Text":"and an atom does not attract electrons strongly,"},{"Start":"01:23.635 ","End":"01:26.525","Text":"then we\u0027ll get a low value of chi,"},{"Start":"01:26.525 ","End":"01:28.820","Text":"and that\u0027s indicative of metals."},{"Start":"01:28.820 ","End":"01:32.060","Text":"Metals have low values of electronegativity."},{"Start":"01:32.060 ","End":"01:33.560","Text":"If, on the other hand,"},{"Start":"01:33.560 ","End":"01:37.700","Text":"the ionization energy is high, I is high,"},{"Start":"01:37.700 ","End":"01:40.940","Text":"and the Electron affinity is also high,"},{"Start":"01:40.940 ","End":"01:43.400","Text":"that means it\u0027s difficult to remove the electron from"},{"Start":"01:43.400 ","End":"01:46.835","Text":"the atom and the atom attracts electrons strongly,"},{"Start":"01:46.835 ","End":"01:49.910","Text":"then we\u0027ll get a high value of chi."},{"Start":"01:49.910 ","End":"01:55.140","Text":"Substances with high values of chi are non-metals."},{"Start":"01:56.380 ","End":"01:59.600","Text":"Let\u0027s consider 4 examples."},{"Start":"01:59.600 ","End":"02:03.165","Text":"Lithium, which is a metal, fluorine,"},{"Start":"02:03.165 ","End":"02:07.665","Text":"which is 1 of the most extreme non-metals,"},{"Start":"02:07.665 ","End":"02:08.975","Text":"carbon and oxygen,"},{"Start":"02:08.975 ","End":"02:10.600","Text":"which are both non-metals."},{"Start":"02:10.600 ","End":"02:12.580","Text":"Let\u0027s start off with lithium."},{"Start":"02:12.580 ","End":"02:16.285","Text":"The ionization energy is 519 kilojoules per mole."},{"Start":"02:16.285 ","End":"02:19.990","Text":"The electron affinity is 60 kilojoules per mole."},{"Start":"02:19.990 ","End":"02:23.260","Text":"That gives chi of 290 kilojoules per mole,"},{"Start":"02:23.260 ","End":"02:27.010","Text":"which translates into 3 eVs per mole."},{"Start":"02:27.010 ","End":"02:31.480","Text":"That\u0027s considered a low electronegativity,"},{"Start":"02:31.480 ","End":"02:33.650","Text":"so that\u0027s low EN."},{"Start":"02:35.130 ","End":"02:37.614","Text":"Now let\u0027s take fluorine."},{"Start":"02:37.614 ","End":"02:39.445","Text":"If we do the same thing for fluorine,"},{"Start":"02:39.445 ","End":"02:43.115","Text":"we discover that chi is 10.4 eVs per mole"},{"Start":"02:43.115 ","End":"02:47.785","Text":"to considered a high value of electronegativity."},{"Start":"02:47.785 ","End":"02:51.620","Text":"A low value for lithium,"},{"Start":"02:51.620 ","End":"02:53.225","Text":"which is a metal,"},{"Start":"02:53.225 ","End":"02:55.610","Text":"and a high volume for fluorine,"},{"Start":"02:55.610 ","End":"02:56.900","Text":"which is a non-metal."},{"Start":"02:56.900 ","End":"02:59.270","Text":"If we do the same for carbon and oxygen,"},{"Start":"02:59.270 ","End":"03:02.690","Text":"we get 6.3 eVs per mole for carbon,"},{"Start":"03:02.690 ","End":"03:07.935","Text":"intermediate between lithium and fluorine and 7.5"},{"Start":"03:07.935 ","End":"03:13.880","Text":"eVs per mole for oxygen also intermediate between lithium and fluorine."},{"Start":"03:13.880 ","End":"03:17.000","Text":"Now, in order to find the polarity of the various bonds,"},{"Start":"03:17.000 ","End":"03:20.585","Text":"we take the difference in electronegativities."},{"Start":"03:20.585 ","End":"03:23.870","Text":"For example, the difference in electronegativity"},{"Start":"03:23.870 ","End":"03:28.325","Text":"between lithium and fluorine gives us 7.4."},{"Start":"03:28.325 ","End":"03:31.835","Text":"Between carbon and fluorine, 3.1."},{"Start":"03:31.835 ","End":"03:35.345","Text":"Between oxygen and fluorine, 2.9."},{"Start":"03:35.345 ","End":"03:38.390","Text":"Between carbon and oxygen, 1.2,"},{"Start":"03:38.390 ","End":"03:42.305","Text":"which suggests that this is the order of the polarity."},{"Start":"03:42.305 ","End":"03:44.959","Text":"LiF more polar than CF,"},{"Start":"03:44.959 ","End":"03:46.370","Text":"more polar than OF,"},{"Start":"03:46.370 ","End":"03:48.620","Text":"more polar than CO. Now,"},{"Start":"03:48.620 ","End":"03:51.830","Text":"another scale is based on Pauling."},{"Start":"03:51.830 ","End":"03:55.685","Text":"This is little famous Linus Pauling who got 2 Nobel Prizes,"},{"Start":"03:55.685 ","End":"04:00.400","Text":"1 in chemistry in 1954 and 1 for peace and 1962."},{"Start":"04:00.400 ","End":"04:06.170","Text":"The values I\u0027m going to use here are based on his 1960 book."},{"Start":"04:06.170 ","End":"04:12.500","Text":"Later on, other people corrected it and some textbooks use the corrected values,"},{"Start":"04:12.500 ","End":"04:16.010","Text":"but I\u0027m using the values of 1960 book."},{"Start":"04:16.010 ","End":"04:20.060","Text":"Pauling based his values on the dissociation energies,"},{"Start":"04:20.060 ","End":"04:24.565","Text":"which we\u0027ll call D in units of electron volts per mole."},{"Start":"04:24.565 ","End":"04:30.980","Text":"He said the difference in electronegativity between atom A and atom B,"},{"Start":"04:30.980 ","End":"04:33.425","Text":"which he always took his positive number,"},{"Start":"04:33.425 ","End":"04:35.330","Text":"he took the absolute value of that,"},{"Start":"04:35.330 ","End":"04:39.755","Text":"is equal to the difference in the dissociation energy of"},{"Start":"04:39.755 ","End":"04:47.450","Text":"A-B bond and the average of the dissociation energy of an A-A bond and the B-B bond."},{"Start":"04:47.450 ","End":"04:53.400","Text":"If DAB is greater than the average of DAA and DBB,"},{"Start":"04:53.400 ","End":"04:56.330","Text":"then there will be a large difference in"},{"Start":"04:56.330 ","End":"05:00.200","Text":"the electronegativities and the bond will be polar."},{"Start":"05:00.200 ","End":"05:05.120","Text":"Now, let\u0027s look at the trends in electronegativity according to Pauling."},{"Start":"05:05.120 ","End":"05:08.810","Text":"We\u0027re only considering the main group elements."},{"Start":"05:08.810 ","End":"05:12.350","Text":"The electronegativity, just like the ionization energy,"},{"Start":"05:12.350 ","End":"05:17.945","Text":"increases from left to right across a period and decreases down a group."},{"Start":"05:17.945 ","End":"05:24.050","Text":"The blue here is electronegativity and the red is ionization energy,"},{"Start":"05:24.050 ","End":"05:30.515","Text":"and they increase across the periodic table and decrease down the periodic table."},{"Start":"05:30.515 ","End":"05:32.480","Text":"Let\u0027s take some examples."},{"Start":"05:32.480 ","End":"05:37.185","Text":"Let\u0027s consider first group 1, which is metals."},{"Start":"05:37.185 ","End":"05:41.070","Text":"Lithium is 1, sodium 0.93,"},{"Start":"05:41.070 ","End":"05:46.595","Text":"potassium 0.82, rubidium 0.82, cesium 0.79."},{"Start":"05:46.595 ","End":"05:51.540","Text":"So we see that the decrease down group 1."},{"Start":"05:51.540 ","End":"05:53.600","Text":"Let\u0027s consider group 17,"},{"Start":"05:53.600 ","End":"05:56.045","Text":"which are halogens, non-metals."},{"Start":"05:56.045 ","End":"05:59.150","Text":"The highest value is for fluorine, which is 4,"},{"Start":"05:59.150 ","End":"06:01.790","Text":"and then it decreases chlorine 3,"},{"Start":"06:01.790 ","End":"06:05.140","Text":"bromine 2.8, iodine 2.5."},{"Start":"06:05.140 ","End":"06:08.900","Text":"We can see that the metals in group 1 have"},{"Start":"06:08.900 ","End":"06:13.040","Text":"much lower values than the non-metals in group 17."},{"Start":"06:13.040 ","End":"06:15.980","Text":"A few other values that will be useful, just now,"},{"Start":"06:15.980 ","End":"06:18.095","Text":"are hydrogen 2.1,"},{"Start":"06:18.095 ","End":"06:22.220","Text":"carbon 2.5, and oxygen 3.5."},{"Start":"06:22.220 ","End":"06:24.110","Text":"Now there are some rules of thumb."},{"Start":"06:24.110 ","End":"06:27.020","Text":"If the electronegativity difference is greater than 2,"},{"Start":"06:27.020 ","End":"06:29.015","Text":"we say the bond is ionic."},{"Start":"06:29.015 ","End":"06:31.280","Text":"If it\u0027s between 0.4 and 2,"},{"Start":"06:31.280 ","End":"06:33.980","Text":"we say the bond is polar covalent."},{"Start":"06:33.980 ","End":"06:36.425","Text":"If it\u0027s less than 0.4,"},{"Start":"06:36.425 ","End":"06:38.120","Text":"we say the bond is covalent."},{"Start":"06:38.120 ","End":"06:40.805","Text":"Obviously, this is all very approximate."},{"Start":"06:40.805 ","End":"06:45.485","Text":"Let\u0027s take some examples of differences in electronegativity."},{"Start":"06:45.485 ","End":"06:48.739","Text":"For LiF, the difference is 3,"},{"Start":"06:48.739 ","End":"06:51.245","Text":"so it\u0027s definitely ionic, it\u0027s greater than 2."},{"Start":"06:51.245 ","End":"06:53.495","Text":"For CF, it\u0027s 1.4,"},{"Start":"06:53.495 ","End":"06:55.040","Text":"so it\u0027s polar covalent."},{"Start":"06:55.040 ","End":"06:57.420","Text":"It lies between 0.4 and 2."},{"Start":"06:57.420 ","End":"07:01.350","Text":"For OF, it\u0027s 0.5 also polar covalent and CO,"},{"Start":"07:01.350 ","End":"07:03.470","Text":"1, also polar covalent."},{"Start":"07:03.470 ","End":"07:06.620","Text":"Notice, this particular, the Pauling scale,"},{"Start":"07:06.620 ","End":"07:13.925","Text":"seems to suggest that CO is more polar than OF."},{"Start":"07:13.925 ","End":"07:15.440","Text":"Whereas according to Mulliken,"},{"Start":"07:15.440 ","End":"07:17.045","Text":"it was the opposite way around,"},{"Start":"07:17.045 ","End":"07:20.255","Text":"but they\u0027re both certainly polar covalent."},{"Start":"07:20.255 ","End":"07:24.980","Text":"In this video, we introduced the concept of electronegativity."},{"Start":"07:24.980 ","End":"07:26.450","Text":"Then the next video,"},{"Start":"07:26.450 ","End":"07:29.040","Text":"we\u0027ll continue the discussion."}],"ID":23905},{"Watched":false,"Name":"Electronegativity - Part 2","Duration":"2m 42s","ChapterTopicVideoID":20276,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.585","Text":"In the previous video, we introduced electronegativity."},{"Start":"00:03.585 ","End":"00:06.300","Text":"In this video, we\u0027ll discuss the correlation between"},{"Start":"00:06.300 ","End":"00:10.335","Text":"the electronegativity difference and ionic character."},{"Start":"00:10.335 ","End":"00:16.530","Text":"We\u0027re going to plot the electronegativity difference versus percentage ionic character."},{"Start":"00:16.530 ","End":"00:21.765","Text":"It turns out that many values of the percentage ionic character,"},{"Start":"00:21.765 ","End":"00:28.515","Text":"like close to the curve given by percentage ionic character is equal to 100 times"},{"Start":"00:28.515 ","End":"00:35.570","Text":"1 minus the exponential of minus 0.25 times Delta EN^2,"},{"Start":"00:35.570 ","End":"00:41.195","Text":"where Delta EN is equal to the EN,"},{"Start":"00:41.195 ","End":"00:47.645","Text":"electronegativity of atom A minus electronegativity of atom B."},{"Start":"00:47.645 ","End":"00:50.240","Text":"Here\u0027s the plot, here\u0027s the curve."},{"Start":"00:50.240 ","End":"00:54.930","Text":"On the y-axis we have percentage ionic character,"},{"Start":"00:55.880 ","End":"01:00.730","Text":"and on the x-axis we have Delta EN."},{"Start":"01:00.830 ","End":"01:03.846","Text":"The scale is 0, 20, 40,"},{"Start":"01:03.846 ","End":"01:09.600","Text":"60, 80, and 100 percent."},{"Start":"01:09.600 ","End":"01:14.090","Text":"Let\u0027s take 2 examples; HCl and LiF."},{"Start":"01:14.090 ","End":"01:15.829","Text":"Let\u0027s start with HCl."},{"Start":"01:15.829 ","End":"01:20.705","Text":"The difference in electronegativities between hydrogen and chlorine is 0.9."},{"Start":"01:20.705 ","End":"01:24.500","Text":"We can calculate the percentage ionic character from"},{"Start":"01:24.500 ","End":"01:29.450","Text":"the equation we had before when we put Delta EN equal to 0.9,"},{"Start":"01:29.450 ","End":"01:32.515","Text":"we can calculate that to be 18 percent."},{"Start":"01:32.515 ","End":"01:33.890","Text":"In a previous video,"},{"Start":"01:33.890 ","End":"01:39.680","Text":"we calculated the volume from a dipole moment and we got 17 percent."},{"Start":"01:39.680 ","End":"01:42.620","Text":"17 and 18 are rather close,"},{"Start":"01:42.620 ","End":"01:47.390","Text":"and so we can write the percentage ionic character of HCl."},{"Start":"01:47.390 ","End":"01:51.970","Text":"We can draw it as slightly lower than this curve."},{"Start":"01:51.970 ","End":"01:54.065","Text":"What about LiF?"},{"Start":"01:54.065 ","End":"01:56.420","Text":"The Delta EN here is 3."},{"Start":"01:56.420 ","End":"01:59.869","Text":"We calculate the percentage ionic character from the equation,"},{"Start":"01:59.869 ","End":"02:05.170","Text":"putting Delta EN equal to 3 and we get 89.5 percent."},{"Start":"02:05.170 ","End":"02:09.460","Text":"The value we calculated from the dipole moment is 85 percent."},{"Start":"02:09.460 ","End":"02:12.080","Text":"That\u0027s a bit lower,"},{"Start":"02:12.080 ","End":"02:14.630","Text":"but still quite close to the curve."},{"Start":"02:14.630 ","End":"02:20.660","Text":"Something we should note is that the ionic character never reaches 100 percent."},{"Start":"02:20.660 ","End":"02:24.020","Text":"Even an ionic bond always has"},{"Start":"02:24.020 ","End":"02:33.940","Text":"a small component of covalent character."},{"Start":"02:36.680 ","End":"02:42.870","Text":"On this video, we talked about the ionic character of bonds."}],"ID":21067},{"Watched":false,"Name":"Electrostatic Potential Maps","Duration":"4m 3s","ChapterTopicVideoID":20277,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.920","Text":"In previous videos, we talked about electronegativity, and polar bonds."},{"Start":"00:04.920 ","End":"00:09.255","Text":"In this video, we\u0027ll learn about electrostatic potential maps."},{"Start":"00:09.255 ","End":"00:12.585","Text":"Let\u0027s recall what we learned about electronegativity."},{"Start":"00:12.585 ","End":"00:15.570","Text":"We learned about 2 scales; the Mulliken scale,"},{"Start":"00:15.570 ","End":"00:19.635","Text":"which is based on ionization energy and electron affinity,"},{"Start":"00:19.635 ","End":"00:21.105","Text":"and the Pauling scale,"},{"Start":"00:21.105 ","End":"00:24.165","Text":"which is based on bond dissociation energies."},{"Start":"00:24.165 ","End":"00:28.020","Text":"Now both these scales are more than 50 years old."},{"Start":"00:28.020 ","End":"00:30.080","Text":"Nowadays, we can also use"},{"Start":"00:30.080 ","End":"00:33.500","Text":"approximate quantum mechanical calculations to get"},{"Start":"00:33.500 ","End":"00:37.310","Text":"a more accurate picture of the electron distribution in the molecule."},{"Start":"00:37.310 ","End":"00:40.925","Text":"Let\u0027s recall what we learned about boundary surface diagrams."},{"Start":"00:40.925 ","End":"00:43.220","Text":"When we studied atomic orbitals,"},{"Start":"00:43.220 ","End":"00:46.460","Text":"we saw that 1 way of representing an orbital is to draw"},{"Start":"00:46.460 ","End":"00:50.870","Text":"a surface within which the probability of finding the electron is,"},{"Start":"00:50.870 ","End":"00:52.115","Text":"say, 90 percent."},{"Start":"00:52.115 ","End":"00:55.255","Text":"We call the boundary surface diagrams."},{"Start":"00:55.255 ","End":"00:58.355","Text":"Supposing we want to represent a p orbital,"},{"Start":"00:58.355 ","End":"01:00.620","Text":"for example, px orbital."},{"Start":"01:00.620 ","End":"01:05.140","Text":"This is the x-axis and this is the z-axis."},{"Start":"01:05.140 ","End":"01:10.170","Text":"We can draw the boundary surface diagram like this."},{"Start":"01:11.090 ","End":"01:19.900","Text":"You have 2 distorted ellipsoids on either side of the z-axis."},{"Start":"01:19.900 ","End":"01:21.240","Text":"That tells us,"},{"Start":"01:21.240 ","End":"01:27.615","Text":"the probability of finding the electron is 90 percent within this structure,"},{"Start":"01:27.615 ","End":"01:29.115","Text":"which looks somewhat,"},{"Start":"01:29.115 ","End":"01:32.565","Text":"like a bean on between this surface,"},{"Start":"01:32.565 ","End":"01:34.360","Text":"which also looks like a bean."},{"Start":"01:34.360 ","End":"01:38.750","Text":"Now we\u0027re going to learn about electrostatic potential maps,"},{"Start":"01:38.750 ","End":"01:44.579","Text":"which can be written as elpots."},{"Start":"01:44.579 ","End":"01:50.960","Text":"For molecules, we can draw a surface within which the total electron density is,"},{"Start":"01:50.960 ","End":"01:52.510","Text":"say, 90 percent."},{"Start":"01:52.510 ","End":"01:56.570","Text":"Instead of just electron density within an orbital,"},{"Start":"01:56.570 ","End":"02:00.955","Text":"we\u0027re talking about the total electron density of all the orbitals."},{"Start":"02:00.955 ","End":"02:03.950","Text":"We\u0027re going to take as examples, HCl,"},{"Start":"02:03.950 ","End":"02:05.930","Text":"which we know is polar covalent,"},{"Start":"02:05.930 ","End":"02:09.095","Text":"and NaCl, which we know is ionic."},{"Start":"02:09.095 ","End":"02:13.775","Text":"The left-hand side we have HCl and the right-hand side we have NaCl."},{"Start":"02:13.775 ","End":"02:23.000","Text":"Within this surface, there\u0027s a 90 percent probability of finding the electrons."},{"Start":"02:23.000 ","End":"02:26.600","Text":"Electron density is 90 percent within this shape,"},{"Start":"02:26.600 ","End":"02:29.285","Text":"and here it\u0027s 90 percent within this shape."},{"Start":"02:29.285 ","End":"02:31.730","Text":"But, also we see there are colors."},{"Start":"02:31.730 ","End":"02:35.760","Text":"These colors indicate the electrostatic potential."},{"Start":"02:35.760 ","End":"02:40.085","Text":"This gives us information about the distribution of electron charge in a molecule."},{"Start":"02:40.085 ","End":"02:43.285","Text":"Supposing we take a positive charge,"},{"Start":"02:43.285 ","End":"02:44.935","Text":"and bring it from a far."},{"Start":"02:44.935 ","End":"02:47.840","Text":"Now this is the negative side of the molecule,"},{"Start":"02:47.840 ","End":"02:49.940","Text":"this is Cl minus,"},{"Start":"02:49.940 ","End":"02:51.320","Text":"so this is negative."},{"Start":"02:51.320 ","End":"02:56.650","Text":"The positive charge reaches a negative charge,"},{"Start":"02:56.650 ","End":"02:58.815","Text":"and there is attraction between them."},{"Start":"02:58.815 ","End":"03:01.400","Text":"The energy decreases."},{"Start":"03:01.400 ","End":"03:04.430","Text":"I\u0027m going to indicate this by the color red."},{"Start":"03:04.430 ","End":"03:07.490","Text":"However, if we go to the other side of the molecule,"},{"Start":"03:07.490 ","End":"03:10.775","Text":"where we have any plus this positive charge here,"},{"Start":"03:10.775 ","End":"03:15.030","Text":"the positive charge will experience repulsion."},{"Start":"03:15.030 ","End":"03:18.530","Text":"The energy of interaction will increase."},{"Start":"03:18.530 ","End":"03:21.605","Text":"We\u0027re going to indicate that by blue."},{"Start":"03:21.605 ","End":"03:26.015","Text":"Now these pictures are called electrostatic potential maps."},{"Start":"03:26.015 ","End":"03:29.030","Text":"The other side we have 1 for HCl."},{"Start":"03:29.030 ","End":"03:36.440","Text":"Here the colors are not so extreme because it\u0027s polar covalent rather than being ionic."},{"Start":"03:36.440 ","End":"03:41.330","Text":"The colors we use vary from red through yellow to blue."},{"Start":"03:41.330 ","End":"03:45.445","Text":"The red indicates negative charge,"},{"Start":"03:45.445 ","End":"03:48.545","Text":"and the blue indicates positive charge."},{"Start":"03:48.545 ","End":"03:51.649","Text":"In NaCl, Na is very positive,"},{"Start":"03:51.649 ","End":"03:53.240","Text":"Cl is very negative,"},{"Start":"03:53.240 ","End":"03:55.850","Text":"and in HCl, H is slightly positive,"},{"Start":"03:55.850 ","End":"03:58.120","Text":"and Cl slightly negative."},{"Start":"03:58.120 ","End":"04:03.779","Text":"In this video, we learned about electrostatic potential maps."}],"ID":21068},{"Watched":false,"Name":"Strategy for Drawing Lewis Structures","Duration":"5m 40s","ChapterTopicVideoID":20278,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.150","Text":"In this video, we\u0027ll learn how to draw plausible Lewis structures for molecules and ions."},{"Start":"00:06.150 ","End":"00:10.905","Text":"We\u0027re going to show the steps that a strategy for writing Lewis structures."},{"Start":"00:10.905 ","End":"00:14.040","Text":"Step 1 is to count the number of valence electrons"},{"Start":"00:14.040 ","End":"00:17.385","Text":"and adjust for charge if we\u0027re talking about an ion."},{"Start":"00:17.385 ","End":"00:21.345","Text":"Step 2 arrange the atoms to form a skeletal structure."},{"Start":"00:21.345 ","End":"00:24.570","Text":"We have a few things that can help us here."},{"Start":"00:24.570 ","End":"00:27.620","Text":"For example, hydrogen is always a terminal atom."},{"Start":"00:27.620 ","End":"00:29.630","Text":"Carbon is always a central atom."},{"Start":"00:29.630 ","End":"00:32.810","Text":"Central atoms have low electronegativity."},{"Start":"00:32.810 ","End":"00:35.750","Text":"Structures tend to be compact and symmetrical."},{"Start":"00:35.750 ","End":"00:37.805","Text":"Of course, there are exceptions."},{"Start":"00:37.805 ","End":"00:41.690","Text":"Step 3, connect the atoms with single covalent bonds."},{"Start":"00:41.690 ","End":"00:46.025","Text":"Step 4 count the electrons not yet used."},{"Start":"00:46.025 ","End":"00:50.554","Text":"Step 5, complete the octet of terminal atoms."},{"Start":"00:50.554 ","End":"00:53.000","Text":"Step 6, if electrons remain,"},{"Start":"00:53.000 ","End":"00:55.190","Text":"place them on central atoms,"},{"Start":"00:55.190 ","End":"01:00.815","Text":"step 7, check where all the atoms have octets or duplets for hydrogen."},{"Start":"01:00.815 ","End":"01:02.570","Text":"If they don\u0027t have octets,"},{"Start":"01:02.570 ","End":"01:06.760","Text":"form multiple bonds until the octet rule is satisfied."},{"Start":"01:06.760 ","End":"01:10.775","Text":"Let\u0027s start with cases where there\u0027s 1 central atom."},{"Start":"01:10.775 ","End":"01:14.140","Text":"The example we\u0027re going to discuss is H_2CO,"},{"Start":"01:14.140 ","End":"01:17.000","Text":"who\u0027s popular name is formaldehyde,"},{"Start":"01:17.000 ","End":"01:19.790","Text":"and it\u0027s more proper name is methanol."},{"Start":"01:19.790 ","End":"01:23.540","Text":"The first step is to count the number of valence electrons."},{"Start":"01:23.540 ","End":"01:25.160","Text":"We have 2 hydrogens,"},{"Start":"01:25.160 ","End":"01:26.525","Text":"so 1 for each."},{"Start":"01:26.525 ","End":"01:30.020","Text":"We have 4 for the carbon and 6 for the oxygen."},{"Start":"01:30.020 ","End":"01:33.410","Text":"That gives us a total of 12 valence electrons."},{"Start":"01:33.410 ","End":"01:35.945","Text":"The hydrogens have to be terminal."},{"Start":"01:35.945 ","End":"01:39.935","Text":"Carbon has a lower electronegativity than oxygen,"},{"Start":"01:39.935 ","End":"01:41.825","Text":"and carbon is always central."},{"Start":"01:41.825 ","End":"01:44.575","Text":"Oxygen is a terminal atom."},{"Start":"01:44.575 ","End":"01:47.415","Text":"Now we can draw single bonds."},{"Start":"01:47.415 ","End":"01:50.865","Text":"Step 4, count how many electrons remain."},{"Start":"01:50.865 ","End":"01:55.170","Text":"We had 12 and we\u0027ve used 6 electrons,"},{"Start":"01:55.170 ","End":"01:56.700","Text":"2 for each bond,"},{"Start":"01:56.700 ","End":"01:57.960","Text":"we have 3 bonds."},{"Start":"01:57.960 ","End":"01:59.550","Text":"So it\u0027s 6 electrons,"},{"Start":"01:59.550 ","End":"02:01.190","Text":"we have 6 leftover."},{"Start":"02:01.190 ","End":"02:03.440","Text":"Now, each hydrogen already has a duplet,"},{"Start":"02:03.440 ","End":"02:05.210","Text":"so we don\u0027t have to do anything about them."},{"Start":"02:05.210 ","End":"02:10.789","Text":"We can put the 6 electrons on the terminal atom, oxygen."},{"Start":"02:10.789 ","End":"02:14.840","Text":"We\u0027ve used all the electrons,"},{"Start":"02:14.840 ","End":"02:19.054","Text":"but we\u0027ve notice that carbon does not have an octet,"},{"Start":"02:19.054 ","End":"02:20.200","Text":"it has 2, 4,"},{"Start":"02:20.200 ","End":"02:21.920","Text":"6 electrons around it."},{"Start":"02:21.920 ","End":"02:24.695","Text":"In order to complete the structure,"},{"Start":"02:24.695 ","End":"02:31.365","Text":"we can take 2 electrons from the oxygen and form a double bond."},{"Start":"02:31.365 ","End":"02:34.375","Text":"Now we have H-C-H,"},{"Start":"02:34.375 ","End":"02:36.710","Text":"and a double bond to the oxygen."},{"Start":"02:36.710 ","End":"02:39.650","Text":"Now that the octet rule is satisfied,"},{"Start":"02:39.650 ","End":"02:41.840","Text":"this is a final structure."},{"Start":"02:41.840 ","End":"02:44.285","Text":"Let\u0027s consider polyatomic ion."},{"Start":"02:44.285 ","End":"02:50.030","Text":"An example I\u0027ve chosen is CO_3 2 minus, that\u0027s carbonate."},{"Start":"02:51.520 ","End":"02:54.050","Text":"How many electrons does it have?"},{"Start":"02:54.050 ","End":"02:55.430","Text":"Carbon has 4,"},{"Start":"02:55.430 ","End":"03:02.945","Text":"each of the oxygens has 6 and we have 2 for the charge on the carbonate anion."},{"Start":"03:02.945 ","End":"03:06.005","Text":"Altogether we have 24 valence electrons."},{"Start":"03:06.005 ","End":"03:10.475","Text":"Carbon must be a central atom and all the oxygens are equivalent."},{"Start":"03:10.475 ","End":"03:12.700","Text":"We can draw them round about."},{"Start":"03:12.700 ","End":"03:17.705","Text":"We have a single bond between carbon and each of the oxygens."},{"Start":"03:17.705 ","End":"03:19.340","Text":"How many electrons have we used,"},{"Start":"03:19.340 ","End":"03:21.905","Text":"we\u0027ve use 2, 4, 6 electrons."},{"Start":"03:21.905 ","End":"03:23.525","Text":"We had 24,"},{"Start":"03:23.525 ","End":"03:25.805","Text":"so 18 electrons remain."},{"Start":"03:25.805 ","End":"03:30.745","Text":"We can complete the octets on the terminal atoms that step 5,"},{"Start":"03:30.745 ","End":"03:32.130","Text":"we have 18 electrons,"},{"Start":"03:32.130 ","End":"03:33.330","Text":"3 terminal atoms,"},{"Start":"03:33.330 ","End":"03:35.620","Text":"that\u0027s 6 on each."},{"Start":"03:47.360 ","End":"03:50.230","Text":"Now we\u0027ve used all the electrons,"},{"Start":"03:50.230 ","End":"03:54.045","Text":"but we notice that carbon doesn\u0027t have an octet,"},{"Start":"03:54.045 ","End":"03:57.995","Text":"it has 2, 4, 6 electrons only."},{"Start":"03:57.995 ","End":"04:04.870","Text":"We can take a pair of electrons on the oxygen to form a double bond."},{"Start":"04:04.870 ","End":"04:07.075","Text":"Now, if we check,"},{"Start":"04:07.075 ","End":"04:09.760","Text":"we see that carbon has an octet."},{"Start":"04:09.760 ","End":"04:11.620","Text":"It has 2, 4, 6,"},{"Start":"04:11.620 ","End":"04:13.450","Text":"8 electrons around it,"},{"Start":"04:13.450 ","End":"04:15.580","Text":"and oxygen still has its octet."},{"Start":"04:15.580 ","End":"04:17.195","Text":"It has 2, 4,"},{"Start":"04:17.195 ","End":"04:19.825","Text":"6, 8 electrons around it."},{"Start":"04:19.825 ","End":"04:22.240","Text":"Now all the atoms are satisfied."},{"Start":"04:22.240 ","End":"04:29.840","Text":"We\u0027ll see that this structure is only 1 of 3 possible structures in a later video."},{"Start":"04:29.840 ","End":"04:33.395","Text":"Now what happens when we have more than 1 central atom?"},{"Start":"04:33.395 ","End":"04:38.840","Text":"We\u0027re going to take the example H_2NNH_2, that\u0027s hydrazine."},{"Start":"04:38.840 ","End":"04:42.755","Text":"We have 4 hydrogens, that\u0027s 4 electrons."},{"Start":"04:42.755 ","End":"04:44.805","Text":"We have 2 nitrogens,"},{"Start":"04:44.805 ","End":"04:49.340","Text":"each with 5 electrons making a total of 14 valence electrons."},{"Start":"04:49.340 ","End":"04:51.770","Text":"Hydrogen has to be terminal."},{"Start":"04:51.770 ","End":"04:53.810","Text":"We draw this structure."},{"Start":"04:53.810 ","End":"04:56.674","Text":"We can draw the single bonds."},{"Start":"04:56.674 ","End":"04:59.240","Text":"Now how many electrons have we use?"},{"Start":"04:59.240 ","End":"05:00.590","Text":"We have 1, 2, 3,"},{"Start":"05:00.590 ","End":"05:01.940","Text":"4, 5 bonds."},{"Start":"05:01.940 ","End":"05:03.650","Text":"We\u0027ve used 10 electrons,"},{"Start":"05:03.650 ","End":"05:04.940","Text":"we had 14,"},{"Start":"05:04.940 ","End":"05:07.375","Text":"so 4 electrons remain."},{"Start":"05:07.375 ","End":"05:12.515","Text":"All this terminal hydrogens have duplicates from the single bonds."},{"Start":"05:12.515 ","End":"05:15.170","Text":"We have to adjust the nitrogens."},{"Start":"05:15.170 ","End":"05:18.965","Text":"We have 4 electrons and we have 2 nitrogens."},{"Start":"05:18.965 ","End":"05:22.925","Text":"We can put 2 electrons on each nitrogen."},{"Start":"05:22.925 ","End":"05:25.820","Text":"Now, each nitrogen has an octet,"},{"Start":"05:25.820 ","End":"05:27.065","Text":"2, 4,"},{"Start":"05:27.065 ","End":"05:28.385","Text":"6, 8,"},{"Start":"05:28.385 ","End":"05:32.810","Text":"and so the nitrogens have octets and the hydrogens have duplets,"},{"Start":"05:32.810 ","End":"05:35.045","Text":"and that\u0027s the final structure."},{"Start":"05:35.045 ","End":"05:40.920","Text":"In this video, we learned how to construct plausible Lewis structures."}],"ID":21069},{"Watched":false,"Name":"Exercise 1","Duration":"4m 10s","ChapterTopicVideoID":23596,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.970","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.970 ","End":"00:05.475","Text":"Write the Lewis symbols for the following atoms."},{"Start":"00:05.475 ","End":"00:08.445","Text":"We\u0027re going to start with a, we have neon."},{"Start":"00:08.445 ","End":"00:11.625","Text":"In order to write a Lewis symbol for an atom,"},{"Start":"00:11.625 ","End":"00:15.435","Text":"we begin by writing the chemical symbol of the atom."},{"Start":"00:15.435 ","End":"00:18.150","Text":"In our case, neon it\u0027s Ne."},{"Start":"00:18.150 ","End":"00:24.570","Text":"Now the chemical symbol represents the nucleus and the inner shell electrons of the atom."},{"Start":"00:24.570 ","End":"00:27.510","Text":"Now what we have to do is add dots."},{"Start":"00:27.510 ","End":"00:32.790","Text":"The dots that we\u0027re going to place around our chemical symbol are the valence electrons,"},{"Start":"00:32.790 ","End":"00:35.775","Text":"meaning the outer shell electrons of the atom."},{"Start":"00:35.775 ","End":"00:37.770","Text":"Now for s-block elements,"},{"Start":"00:37.770 ","End":"00:41.775","Text":"the number of valence electrons equals the group number."},{"Start":"00:41.775 ","End":"00:43.430","Text":"For the p-block elements,"},{"Start":"00:43.430 ","End":"00:47.790","Text":"the number of valence electrons equals the group number minus 10."},{"Start":"00:47.790 ","End":"00:49.940","Text":"Meaning if we look at neon,"},{"Start":"00:49.940 ","End":"00:52.175","Text":"neon is in group 18."},{"Start":"00:52.175 ","End":"00:54.800","Text":"In order to know the number of valence electrons,"},{"Start":"00:54.800 ","End":"00:56.879","Text":"we\u0027d have to put around the neon,"},{"Start":"00:56.879 ","End":"01:01.485","Text":"we\u0027re going to take our Group number 18 and we\u0027re going to subtract 10 from it,"},{"Start":"01:01.485 ","End":"01:05.400","Text":"meaning we need to draw 8 valence electrons around neon."},{"Start":"01:05.400 ","End":"01:08.445","Text":"You remember neon is a noble gas."},{"Start":"01:08.445 ","End":"01:11.019","Text":"We\u0027re going to write 1, 2, 3,"},{"Start":"01:11.019 ","End":"01:15.280","Text":"4 and add another 4,"},{"Start":"01:15.470 ","End":"01:19.560","Text":"1, 2, 3, 4."},{"Start":"01:19.560 ","End":"01:21.915","Text":"This is our Lewis symbol for neon."},{"Start":"01:21.915 ","End":"01:24.495","Text":"Now we\u0027re going to go on to b,"},{"Start":"01:24.495 ","End":"01:27.330","Text":"in b we have phosphorus."},{"Start":"01:27.330 ","End":"01:31.505","Text":"Again we\u0027re going to write the chemical symbol for phosphorus,"},{"Start":"01:31.505 ","End":"01:35.780","Text":"which is P. If you look in your periodic table,"},{"Start":"01:35.780 ","End":"01:40.535","Text":"you will see that phosphorus is in group 15."},{"Start":"01:40.535 ","End":"01:44.375","Text":"Meaning that again, we have to take the group number 15 minus"},{"Start":"01:44.375 ","End":"01:48.785","Text":"10 and we get 5 valence electrons."},{"Start":"01:48.785 ","End":"01:50.884","Text":"We\u0027re going to write them 1, 2,"},{"Start":"01:50.884 ","End":"01:53.400","Text":"3, 4, 5."},{"Start":"01:55.390 ","End":"01:58.710","Text":"That\u0027s a Lewis symbol for b."},{"Start":"01:58.760 ","End":"02:01.690","Text":"In c we have cesium."},{"Start":"02:01.690 ","End":"02:06.180","Text":"Again writing the chemical symbol is Cs."},{"Start":"02:06.400 ","End":"02:10.580","Text":"Cesium is in the first group,"},{"Start":"02:10.580 ","End":"02:15.295","Text":"meaning we only have to put 1 electron for cesium."},{"Start":"02:15.295 ","End":"02:18.550","Text":"In d, we have Iodine."},{"Start":"02:19.780 ","End":"02:22.760","Text":"Iodine is in the 17th group,"},{"Start":"02:22.760 ","End":"02:27.190","Text":"meaning the number of valence electrons equals 17 minus 10."},{"Start":"02:27.190 ","End":"02:29.235","Text":"We get 7 electrons."},{"Start":"02:29.235 ","End":"02:35.860","Text":"That\u0027s 1, 2, 3, 4, 5, 6, 7."},{"Start":"02:36.350 ","End":"02:40.510","Text":"There is the Lewis symbol for iodine."},{"Start":"02:40.510 ","End":"02:43.529","Text":"In e, we have strontium."},{"Start":"02:44.770 ","End":"02:47.210","Text":"Strontium is in the second group,"},{"Start":"02:47.210 ","End":"02:49.775","Text":"meaning it will have 2 valence electrons."},{"Start":"02:49.775 ","End":"02:52.585","Text":"It\u0027s an s-block element."},{"Start":"02:52.585 ","End":"02:55.480","Text":"In f, we have aluminum."},{"Start":"02:55.640 ","End":"02:58.275","Text":"Aluminum is in group 13."},{"Start":"02:58.275 ","End":"03:00.350","Text":"Again, since it\u0027s a p-block element,"},{"Start":"03:00.350 ","End":"03:05.210","Text":"we take 13 minus 10 we get 3 electrons so you"},{"Start":"03:05.210 ","End":"03:11.310","Text":"write 3 dots around the aluminum."},{"Start":"03:11.310 ","End":"03:15.220","Text":"That\u0027s our Lewis symbol for aluminum."},{"Start":"03:15.220 ","End":"03:18.340","Text":"Now in g, we have tin."},{"Start":"03:19.580 ","End":"03:22.950","Text":"Tin is in group 14."},{"Start":"03:22.950 ","End":"03:26.505","Text":"Again, 14 minus 10 is 4 electrons."},{"Start":"03:26.505 ","End":"03:29.940","Text":"We\u0027re going to draw 4 dots around the tin."},{"Start":"03:29.940 ","End":"03:33.135","Text":"That\u0027s our Lewis symbol for tin."},{"Start":"03:33.135 ","End":"03:40.760","Text":"Again, we have the Lewis symbol for neon with 8 valence electrons."},{"Start":"03:40.760 ","End":"03:45.250","Text":"We have Lewis symbol for phosphorus with 5 valence electrons."},{"Start":"03:45.250 ","End":"03:48.440","Text":"The Lewis symbol for cesium with 1 valence electron,"},{"Start":"03:48.440 ","End":"03:52.460","Text":"the Lewis symbol for iodine with 7 valence electrons."},{"Start":"03:52.460 ","End":"03:56.150","Text":"Lewis symbol for strontium with 2 valence electrons."},{"Start":"03:56.150 ","End":"03:58.085","Text":"The Lewis symbol for aluminum with"},{"Start":"03:58.085 ","End":"04:05.410","Text":"3 valence electrons and the Lewis symbol for tin with 4 valence electrons."},{"Start":"04:05.590 ","End":"04:07.850","Text":"Those are our final answers."},{"Start":"04:07.850 ","End":"04:10.410","Text":"Thank you very much for watching."}],"ID":24507},{"Watched":false,"Name":"Exercise 2","Duration":"3m 49s","ChapterTopicVideoID":23597,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.760","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.760 ","End":"00:05.400","Text":"Write the Lewis symbols for the following ions."},{"Start":"00:05.400 ","End":"00:08.530","Text":"So in A, we have the chloride anion."},{"Start":"00:09.530 ","End":"00:13.140","Text":"The first thing we\u0027re going to do in order to write the Lewis symbol for"},{"Start":"00:13.140 ","End":"00:17.520","Text":"the chloride anion is to check how many valence electrons we have in chlorine."},{"Start":"00:17.520 ","End":"00:19.470","Text":"Now, if you look at your periodic table,"},{"Start":"00:19.470 ","End":"00:22.260","Text":"you will see that chlorine is in the 17th period."},{"Start":"00:22.260 ","End":"00:24.750","Text":"I want to remind you that for main group elements,"},{"Start":"00:24.750 ","End":"00:28.560","Text":"the number of valence electrons equals the number of the group for"},{"Start":"00:28.560 ","End":"00:33.600","Text":"the s-block elements and then group number minus 10 for the p-block elements."},{"Start":"00:33.600 ","End":"00:36.165","Text":"So chlorine is a p-block element,"},{"Start":"00:36.165 ","End":"00:39.990","Text":"meaning the number of valence electrons equals the group number,"},{"Start":"00:39.990 ","End":"00:43.965","Text":"which is 17 minus 10."},{"Start":"00:43.965 ","End":"00:47.305","Text":"So we get 7 electrons as valence electrons."},{"Start":"00:47.305 ","End":"00:51.310","Text":"First of all, we have 7 valence electrons in chlorine,"},{"Start":"00:51.310 ","End":"00:54.535","Text":"but remember we\u0027re talking about the chloride anion."},{"Start":"00:54.535 ","End":"00:57.630","Text":"So we have a charge of minus 1,"},{"Start":"00:57.630 ","End":"00:59.775","Text":"meaning we have another electron."},{"Start":"00:59.775 ","End":"01:02.530","Text":"So in all, we\u0027re going to take our 7 electrons and add"},{"Start":"01:02.530 ","End":"01:06.665","Text":"another electron and this comes to 8 electrons,"},{"Start":"01:06.665 ","End":"01:09.805","Text":"meaning the chloride anion has 8 electrons."},{"Start":"01:09.805 ","End":"01:15.673","Text":"Again, draw the 8 electrons around the chlorine,"},{"Start":"01:15.673 ","End":"01:19.225","Text":"and since it\u0027s an ion, we\u0027re going to put it in square brackets,"},{"Start":"01:19.225 ","End":"01:21.715","Text":"and we\u0027re going to add the charge."},{"Start":"01:21.715 ","End":"01:26.423","Text":"That\u0027s the Lewis symbol for the chloride anion."},{"Start":"01:26.423 ","End":"01:28.250","Text":"Now we\u0027re going on to b,"},{"Start":"01:28.250 ","End":"01:31.290","Text":"in b we have the calcium cation."},{"Start":"01:32.900 ","End":"01:36.810","Text":"Again, we\u0027re going to check how many valence electrons calcium has,"},{"Start":"01:36.810 ","End":"01:38.180","Text":"and if you look at your periodic table,"},{"Start":"01:38.180 ","End":"01:42.470","Text":"you will see the calcium is in the second period and it\u0027s an S block element,"},{"Start":"01:42.470 ","End":"01:47.275","Text":"meaning that the period number is the number of valence electrons."},{"Start":"01:47.275 ","End":"01:49.125","Text":"So in calcium,"},{"Start":"01:49.125 ","End":"01:52.270","Text":"we have 2 valence electrons,"},{"Start":"01:52.270 ","End":"01:54.075","Text":"and now we have to look at our charge."},{"Start":"01:54.075 ","End":"01:55.410","Text":"In a calcium cation,"},{"Start":"01:55.410 ","End":"01:56.750","Text":"the charge is plus 2,"},{"Start":"01:56.750 ","End":"01:59.300","Text":"meaning we have to take off 2 electrons."},{"Start":"01:59.300 ","End":"02:00.320","Text":"We\u0027re missing 2 electrons,"},{"Start":"02:00.320 ","End":"02:04.930","Text":"so it\u0027s minus 2 electrons and this comes to 0 electrons."},{"Start":"02:04.930 ","End":"02:11.055","Text":"Again, we take the calcium put it in square brackets, and we add the charge."},{"Start":"02:11.055 ","End":"02:17.435","Text":"I didn\u0027t add any dots because our electrons came out to 0."},{"Start":"02:17.435 ","End":"02:21.305","Text":"So that\u0027s a calcium plus 2 cation."},{"Start":"02:21.305 ","End":"02:26.015","Text":"Now we\u0027re going to take a look at C. In C we have the cesium cation."},{"Start":"02:26.015 ","End":"02:28.490","Text":"Again, if you look at the periodic table,"},{"Start":"02:28.490 ","End":"02:31.830","Text":"you\u0027ll see that cesium is in the first period,"},{"Start":"02:32.990 ","End":"02:36.210","Text":"meaning it has 1 valence electron."},{"Start":"02:36.210 ","End":"02:38.760","Text":"In our case, the charge is plus 1,"},{"Start":"02:38.760 ","End":"02:41.070","Text":"meaning we\u0027re missing 1 electron."},{"Start":"02:41.070 ","End":"02:42.860","Text":"Again, for cesium,"},{"Start":"02:42.860 ","End":"02:44.435","Text":"we have 0 electrons,"},{"Start":"02:44.435 ","End":"02:48.440","Text":"meaning we don\u0027t write any dots around the cesium."},{"Start":"02:48.440 ","End":"02:53.215","Text":"We just add the square brackets and our plus 1 charge."},{"Start":"02:53.215 ","End":"02:57.755","Text":"So that\u0027s the Lewis symbol for the cesium cation."},{"Start":"02:57.755 ","End":"02:59.360","Text":"Now we\u0027re going to go onto D,"},{"Start":"02:59.360 ","End":"03:01.740","Text":"we have the lead cation."},{"Start":"03:03.350 ","End":"03:05.690","Text":"If you look at your periodic table,"},{"Start":"03:05.690 ","End":"03:09.140","Text":"you\u0027ll see that lead is in the 14th period and it is also a p-block element,"},{"Start":"03:09.140 ","End":"03:12.050","Text":"meaning the number of valence electrons equals 14,"},{"Start":"03:12.050 ","End":"03:13.790","Text":"the group number minus 10."},{"Start":"03:13.790 ","End":"03:16.160","Text":"So that comes to 4 electrons,"},{"Start":"03:16.160 ","End":"03:19.220","Text":"which is the number of valence electrons for lead."},{"Start":"03:19.220 ","End":"03:20.630","Text":"But in our case again,"},{"Start":"03:20.630 ","End":"03:21.650","Text":"we have to look at the charge,"},{"Start":"03:21.650 ","End":"03:22.775","Text":"we have plus 2,"},{"Start":"03:22.775 ","End":"03:25.780","Text":"meaning we\u0027re missing 2 electrons."},{"Start":"03:25.780 ","End":"03:28.965","Text":"So this comes to 2 electrons."},{"Start":"03:28.965 ","End":"03:33.942","Text":"So around the lead, we have to draw 2 dots for the electrons,"},{"Start":"03:33.942 ","End":"03:38.440","Text":"and then we add the square brackets and the charge."},{"Start":"03:38.440 ","End":"03:47.060","Text":"That\u0027s the Lewis symbol for the lead cation in D. These are our final answers."},{"Start":"03:47.060 ","End":"03:49.800","Text":"Thank you very much for watching."}],"ID":24508},{"Watched":false,"Name":"Exercise 3 - part a","Duration":"3m 46s","ChapterTopicVideoID":23601,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.865","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.865 ","End":"00:08.850","Text":"Write Lewis structures for the following structures. We\u0027re going to start with A."},{"Start":"00:08.850 ","End":"00:13.839","Text":"A, we have chloroform and we have to write the Lewis structure for chloroform."},{"Start":"00:13.839 ","End":"00:15.614","Text":"In order to write Lewis structures,"},{"Start":"00:15.614 ","End":"00:17.355","Text":"we\u0027re going to have to follow a number of steps."},{"Start":"00:17.355 ","End":"00:20.540","Text":"The first step, we\u0027re going to call it A,"},{"Start":"00:20.540 ","End":"00:24.575","Text":"is to determine the total number of valence electrons in this structure."},{"Start":"00:24.575 ","End":"00:26.900","Text":"Here, we have carbon, hydrogen,"},{"Start":"00:26.900 ","End":"00:29.090","Text":"and 3 chlorines, so we\u0027re going to look at our carbon."},{"Start":"00:29.090 ","End":"00:31.445","Text":"We see that it\u0027s in Group 14."},{"Start":"00:31.445 ","End":"00:35.855","Text":"Therefore, we\u0027re going to write 4 valence electrons, 4 carbon."},{"Start":"00:35.855 ","End":"00:37.100","Text":"Next, we have hydrogen,"},{"Start":"00:37.100 ","End":"00:39.320","Text":"which is in the first group."},{"Start":"00:39.320 ","End":"00:41.875","Text":"We\u0027re going to add 1 valence electron,"},{"Start":"00:41.875 ","End":"00:43.580","Text":"and we have now 3 chlorines,"},{"Start":"00:43.580 ","End":"00:45.860","Text":"so we\u0027re going to add 3 times 7."},{"Start":"00:45.860 ","End":"00:47.150","Text":"Since if we look at chlorine,"},{"Start":"00:47.150 ","End":"00:50.060","Text":"we can see that it\u0027s in Group 17."},{"Start":"00:50.060 ","End":"00:54.680","Text":"This comes to a total of 26 valence electrons,"},{"Start":"00:54.680 ","End":"00:56.690","Text":"and that is the first step."},{"Start":"00:56.690 ","End":"00:59.105","Text":"The next step, we\u0027ll call it B,"},{"Start":"00:59.105 ","End":"01:03.270","Text":"is to identify the central and terminal atoms."},{"Start":"01:04.010 ","End":"01:08.660","Text":"Central atoms are generally those with the lowest electronegativity."},{"Start":"01:08.660 ","End":"01:16.505","Text":"Carbon is the central atom and chlorine is a terminal atom."},{"Start":"01:16.505 ","End":"01:19.400","Text":"It has high electronegativity."},{"Start":"01:19.400 ","End":"01:23.120","Text":"Hydrogen is also a terminal atom because it can only"},{"Start":"01:23.120 ","End":"01:26.630","Text":"accommodate 2 electrons in its valence shell."},{"Start":"01:26.630 ","End":"01:29.555","Text":"Therefore, hydrogen is terminal atom."},{"Start":"01:29.555 ","End":"01:36.360","Text":"So we identified the central and its terminal atoms and the next step is to connect them."},{"Start":"01:39.950 ","End":"01:43.420","Text":"Now, I drew 4 bonds here."},{"Start":"01:44.840 ","End":"01:48.990","Text":"Now sometimes you will see these bonds as 2 dots,"},{"Start":"01:49.730 ","End":"01:53.010","Text":"but here, I\u0027m writing bonds."},{"Start":"01:53.010 ","End":"01:56.345","Text":"The next step, which we\u0027ll call it C,"},{"Start":"01:56.345 ","End":"01:59.690","Text":"is to subtract the number of bonds from our valence electrons,"},{"Start":"01:59.690 ","End":"02:02.890","Text":"meaning we have 26 valence electrons, remember?"},{"Start":"02:02.890 ","End":"02:06.635","Text":"We\u0027re going to subtract 8 valence electrons."},{"Start":"02:06.635 ","End":"02:08.690","Text":"Why? Because we have 4 bonds,"},{"Start":"02:08.690 ","End":"02:10.475","Text":"meaning 2, 4,"},{"Start":"02:10.475 ","End":"02:13.025","Text":"6, 8 since we have 2 electrons in each bond."},{"Start":"02:13.025 ","End":"02:19.179","Text":"So 26 minus 8 is going to give us 18 more electrons."},{"Start":"02:20.950 ","End":"02:27.055","Text":"We have 18 more electrons to write in our Lewis structure."},{"Start":"02:27.055 ","End":"02:29.525","Text":"Now, with the electrons that are left,"},{"Start":"02:29.525 ","End":"02:32.450","Text":"we first need to complete the octets of the terminal atoms,"},{"Start":"02:32.450 ","End":"02:34.220","Text":"and then we go to the central atom."},{"Start":"02:34.220 ","End":"02:37.280","Text":"Let\u0027s complete the octets for our chlorines."},{"Start":"02:37.280 ","End":"02:39.580","Text":"So each chlorine, we have to add 1,"},{"Start":"02:39.580 ","End":"02:41.735","Text":"2, 3, 4, 5, 6."},{"Start":"02:41.735 ","End":"02:44.970","Text":"That way, the chlorine will have an octet, 1, 2, 3,"},{"Start":"02:44.970 ","End":"02:47.010","Text":"4, 5, 6, 1,"},{"Start":"02:47.010 ","End":"02:49.380","Text":"2, 3, 4, 5, 6."},{"Start":"02:49.380 ","End":"02:52.610","Text":"Now we added 6 electrons to 3 chlorine atoms,"},{"Start":"02:52.610 ","End":"02:54.470","Text":"meaning we added 18 electrons,"},{"Start":"02:54.470 ","End":"02:56.395","Text":"so we added all of the electrons."},{"Start":"02:56.395 ","End":"02:59.000","Text":"We have no electrons left for the central atom."},{"Start":"02:59.000 ","End":"03:00.230","Text":"Now we have to look at"},{"Start":"03:00.230 ","End":"03:05.150","Text":"our Lewis structure and we have to see if it\u0027s a satisfactory structure or not."},{"Start":"03:05.150 ","End":"03:10.230","Text":"For this, we have to see if our atoms have octets."},{"Start":"03:11.960 ","End":"03:14.570","Text":"The hydrogen, again,"},{"Start":"03:14.570 ","End":"03:17.465","Text":"can form only 1 bond to another atom."},{"Start":"03:17.465 ","End":"03:19.460","Text":"So our hydrogen is good."},{"Start":"03:19.460 ","End":"03:23.450","Text":"We can also see that our chlorines have 8 electrons,"},{"Start":"03:23.450 ","End":"03:25.910","Text":"meaning they have an octet, and if we look at our carbon,"},{"Start":"03:25.910 ","End":"03:27.560","Text":"we also see 1, 2, 3,"},{"Start":"03:27.560 ","End":"03:30.590","Text":"4 bonds times 2 is 8 electrons,"},{"Start":"03:30.590 ","End":"03:32.930","Text":"so our carbon also has an octet."},{"Start":"03:32.930 ","End":"03:37.580","Text":"Therefore, we have a satisfactory Lewis structure."},{"Start":"03:37.580 ","End":"03:48.880","Text":"The Lewis structure for chloroform is what we did. Now we\u0027re going to go on to B."}],"ID":24512},{"Watched":false,"Name":"Exercise 3 - part b","Duration":"10m 4s","ChapterTopicVideoID":23602,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:02.490","Text":"Again, we\u0027re doing b,"},{"Start":"00:02.490 ","End":"00:04.590","Text":"which is hydrogen sulfide."},{"Start":"00:04.590 ","End":"00:09.510","Text":"We want to write the Lewis structure so we have to start with the first step."},{"Start":"00:09.510 ","End":"00:12.075","Text":"Now, again, in the first step, we\u0027re going to find"},{"Start":"00:12.075 ","End":"00:15.930","Text":"the number of valence electrons for the structure."},{"Start":"00:15.930 ","End":"00:17.850","Text":"Again, we have hydrogen,"},{"Start":"00:17.850 ","End":"00:19.935","Text":"which we know is in the first group,"},{"Start":"00:19.935 ","End":"00:24.060","Text":"and therefore we have 1 valence electron from each hydrogen,"},{"Start":"00:24.060 ","End":"00:25.230","Text":"but we have 2 hydrogen,"},{"Start":"00:25.230 ","End":"00:27.555","Text":"so that\u0027s 2 times 1,"},{"Start":"00:27.555 ","End":"00:30.420","Text":"plus we have sulfur."},{"Start":"00:30.420 ","End":"00:34.520","Text":"If you look at sulfur, you will see that it is in the 16th group,"},{"Start":"00:34.520 ","End":"00:37.960","Text":"meaning that it has 6 valence electrons."},{"Start":"00:37.960 ","End":"00:40.970","Text":"All of this gives us 8 electrons."},{"Start":"00:40.970 ","End":"00:46.800","Text":"Now, again, in B, we have to identify the central and the terminal atoms."},{"Start":"00:46.800 ","End":"00:49.580","Text":"We said already that hydrogen is a terminal atom."},{"Start":"00:49.580 ","End":"00:55.414","Text":"Therefore, we\u0027re going to write the hydrogen as terminal and the sulfur as central."},{"Start":"00:55.414 ","End":"00:59.670","Text":"Next step, we\u0027re going to connect them by bonds."},{"Start":"00:59.690 ","End":"01:02.460","Text":"Next step, we\u0027re going to take"},{"Start":"01:02.460 ","End":"01:07.030","Text":"our 8 electrons and subtract the number of electrons in bonds."},{"Start":"01:07.030 ","End":"01:10.960","Text":"We have 2 electrons in 1 bond and 2 electrons in the second one,"},{"Start":"01:10.960 ","End":"01:12.905","Text":"meaning we have 4 electrons."},{"Start":"01:12.905 ","End":"01:16.805","Text":"We have 4 electrons left for our structure."},{"Start":"01:16.805 ","End":"01:19.430","Text":"Now, again, we usually begin with a terminal atoms,"},{"Start":"01:19.430 ","End":"01:23.020","Text":"but since we have hydrogen and we know that it only forms 1 bond,"},{"Start":"01:23.020 ","End":"01:25.860","Text":"so hydrogen can\u0027t accept any other electrons,"},{"Start":"01:25.860 ","End":"01:27.275","Text":"we\u0027re going to go to the sulfur."},{"Start":"01:27.275 ","End":"01:29.030","Text":"We have to add in 4 electrons,"},{"Start":"01:29.030 ","End":"01:33.440","Text":"so we\u0027re going to add in two here and two here."},{"Start":"01:33.440 ","End":"01:36.050","Text":"Now, if we look at our Lewis structure,"},{"Start":"01:36.050 ","End":"01:41.135","Text":"we can see that the hydrogen again has 1 bond which is exactly as it\u0027s supposed to have."},{"Start":"01:41.135 ","End":"01:44.589","Text":"If we look at the sulfur, we can see that it has an octet,"},{"Start":"01:44.589 ","End":"01:49.300","Text":"1,2,3,4,5,6,7,8 electrons around it."},{"Start":"01:49.300 ","End":"01:52.460","Text":"That is a satisfactory Lewis structure,"},{"Start":"01:52.460 ","End":"01:55.210","Text":"and that is our answer for B."},{"Start":"01:55.210 ","End":"01:57.500","Text":"Now, we\u0027re going to go onto C,"},{"Start":"01:57.500 ","End":"01:59.670","Text":"which we have chlorine."},{"Start":"01:59.950 ","End":"02:04.190","Text":"In C, we have to write Lewis structure for chlorine."},{"Start":"02:04.190 ","End":"02:09.710","Text":"So, again, the first step is to determine the number of valence electrons."},{"Start":"02:09.710 ","End":"02:12.140","Text":"Since we have 2 chlorines,"},{"Start":"02:12.140 ","End":"02:14.555","Text":"2 times 7,"},{"Start":"02:14.555 ","End":"02:18.655","Text":"because remember, chlorine is in the Group 17."},{"Start":"02:18.655 ","End":"02:23.620","Text":"That equals 14 electrons for chlorine."},{"Start":"02:23.950 ","End":"02:28.430","Text":"Next step is to identify the central and terminal atoms held over."},{"Start":"02:28.430 ","End":"02:29.540","Text":"We only have 2 chlorines,"},{"Start":"02:29.540 ","End":"02:31.994","Text":"therefore we have 1 chlorine at one side and 1 chlorine at"},{"Start":"02:31.994 ","End":"02:35.940","Text":"the other side and we\u0027re going to put a bond in between."},{"Start":"02:36.470 ","End":"02:40.700","Text":"The third step is to subtract the number"},{"Start":"02:40.700 ","End":"02:44.000","Text":"of electrons in bonds from the total number of valence electrons."},{"Start":"02:44.000 ","End":"02:46.415","Text":"We have a total number 14 and we\u0027re"},{"Start":"02:46.415 ","End":"02:51.125","Text":"subtracting 2 since we have 1 bond which contains 2 electrons."},{"Start":"02:51.125 ","End":"02:54.250","Text":"This gives us another 12 electrons."},{"Start":"02:54.250 ","End":"02:57.885","Text":"Now, we have to take our atoms, our chlorines,"},{"Start":"02:57.885 ","End":"03:00.360","Text":"and add the 12 electrons,"},{"Start":"03:00.360 ","End":"03:03.010","Text":"we\u0027re going to add 6 to each chlorine."},{"Start":"03:05.990 ","End":"03:09.474","Text":"Now, we\u0027re going to look at our Lewis structure."},{"Start":"03:09.474 ","End":"03:12.140","Text":"We see that around each chlorine,"},{"Start":"03:12.140 ","End":"03:13.160","Text":"we have 8 electrons,"},{"Start":"03:13.160 ","End":"03:17.480","Text":"1,2,3,4,5,6 and another bond adding another 2,"},{"Start":"03:17.480 ","End":"03:20.270","Text":"meaning we have 8 electrons on each chlorine."},{"Start":"03:20.270 ","End":"03:25.530","Text":"This is a satisfactory Lewis structure and this is our answer"},{"Start":"03:25.530 ","End":"03:31.410","Text":"for C. Now,"},{"Start":"03:31.410 ","End":"03:33.490","Text":"we\u0027re going to go onto D."}],"ID":24513},{"Watched":false,"Name":"Exercise 4","Duration":"5m 26s","ChapterTopicVideoID":23603,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.100 ","End":"00:04.754","Text":"We\u0027re going to solve the following exercise."},{"Start":"00:04.754 ","End":"00:08.475","Text":"Write Lewis structures for the following ionic compounds."},{"Start":"00:08.475 ","End":"00:11.650","Text":"In a, we have calcium iodide."},{"Start":"00:13.100 ","End":"00:15.750","Text":"We\u0027re going to take a look at calcium,"},{"Start":"00:15.750 ","End":"00:18.899","Text":"and also at iodine."},{"Start":"00:18.899 ","End":"00:21.540","Text":"Now if you take a look at your periodic table,"},{"Start":"00:21.540 ","End":"00:25.275","Text":"you\u0027ll see that calcium is in the second period and it is also an s-block element,"},{"Start":"00:25.275 ","End":"00:28.680","Text":"meaning that it has 2 valence electrons."},{"Start":"00:28.680 ","End":"00:31.215","Text":"Iodine is a p-block element,"},{"Start":"00:31.215 ","End":"00:32.850","Text":"and is in the 17th period,"},{"Start":"00:32.850 ","End":"00:35.760","Text":"meaning it has 7 valence electrons."},{"Start":"00:35.760 ","End":"00:37.320","Text":"Because again, in the p-block elements,"},{"Start":"00:37.320 ","End":"00:43.940","Text":"we take the period number which is 17 minus 10 = 7 electrons,"},{"Start":"00:43.940 ","End":"00:48.710","Text":"7 valence electrons for the iodine."},{"Start":"00:48.710 ","End":"00:53.555","Text":"Now, in order to reach a noble gas configuration,"},{"Start":"00:53.555 ","End":"00:55.625","Text":"the calcium has to lose 2 electrons."},{"Start":"00:55.625 ","End":"00:57.440","Text":"In order to reach an octet,"},{"Start":"00:57.440 ","End":"00:59.735","Text":"iodine has to gain 1 electron."},{"Start":"00:59.735 ","End":"01:04.250","Text":"We\u0027re going to take 1 electron for the calcium and pass it on to the iodine."},{"Start":"01:04.250 ","End":"01:06.395","Text":"The iodine has an octet,"},{"Start":"01:06.395 ","End":"01:10.580","Text":"but the calcium, still has 1 electron which we have to get rid of."},{"Start":"01:10.580 ","End":"01:15.090","Text":"For this we\u0027re going to add another iodine again with"},{"Start":"01:15.090 ","End":"01:18.650","Text":"7 valence electrons and the calcium is going to"},{"Start":"01:18.650 ","End":"01:23.710","Text":"pass another electron to the second iodine."},{"Start":"01:23.730 ","End":"01:29.328","Text":"Now, if we take a look, we can see that the calcium has noble gas configuration,"},{"Start":"01:29.328 ","End":"01:34.270","Text":"and both of the iodine\u0027s have an octet."},{"Start":"01:34.410 ","End":"01:39.500","Text":"The way we\u0027re going to write this down is we have the calcium."},{"Start":"01:39.900 ","End":"01:43.195","Text":"Now, since the calcium passed on 2 electrons,"},{"Start":"01:43.195 ","End":"01:50.410","Text":"its charge is +2 and we have the iodine,"},{"Start":"01:51.810 ","End":"01:54.070","Text":"which received the electron,"},{"Start":"01:54.070 ","End":"01:57.370","Text":"but we have 2 so, we have to multiply the iodine by 2."},{"Start":"01:57.370 ","End":"02:01.075","Text":"Now, the number of electrons iodine has is 8."},{"Start":"02:01.075 ","End":"02:06.240","Text":"Remember it received an octet and it has a charge of -1."},{"Start":"02:06.240 ","End":"02:09.130","Text":"This is our answer for a."},{"Start":"02:11.510 ","End":"02:13.760","Text":"We\u0027re going to go on to b, b,"},{"Start":"02:13.760 ","End":"02:16.290","Text":"we have potassium sulfide."},{"Start":"02:18.530 ","End":"02:25.040","Text":"Again, we\u0027re just going to take a look at potassium and sulfur in the beginning."},{"Start":"02:25.040 ","End":"02:27.050","Text":"If you take a look at your periodic table,"},{"Start":"02:27.050 ","End":"02:28.940","Text":"we see that potassium is in the first period,"},{"Start":"02:28.940 ","End":"02:32.090","Text":"meaning it has 1 valence electron."},{"Start":"02:32.090 ","End":"02:36.785","Text":"You will see that sulfur is in the 16th period and again, it\u0027s a p-block elements,"},{"Start":"02:36.785 ","End":"02:43.020","Text":"meaning it has 16 minus 10 electrons so it has 6 valence electrons."},{"Start":"02:48.590 ","End":"02:52.150","Text":"Now the sulfur needs to gain 2 electrons in order"},{"Start":"02:52.150 ","End":"02:55.195","Text":"to reach an octet and the potassium has to lose"},{"Start":"02:55.195 ","End":"03:01.730","Text":"1 electron in order to gain a noble gas configuration, electron configurations."},{"Start":"03:01.730 ","End":"03:03.130","Text":"We\u0027re going to take this electron,"},{"Start":"03:03.130 ","End":"03:04.720","Text":"we\u0027re going to pass it on,"},{"Start":"03:04.720 ","End":"03:07.960","Text":"from the potassium to the sulfur."},{"Start":"03:07.960 ","End":"03:11.613","Text":"The sulfur has 7 valence electrons,"},{"Start":"03:11.613 ","End":"03:14.065","Text":"and the potassium has noble gas configuration,"},{"Start":"03:14.065 ","End":"03:18.220","Text":"so in order to reach an octet for the sulfur,"},{"Start":"03:18.220 ","End":"03:21.504","Text":"what we\u0027re going to do is we\u0027re going to add in another potassium."},{"Start":"03:21.504 ","End":"03:27.670","Text":"Then this electron is going to go from the potassium to the sulfur."},{"Start":"03:27.670 ","End":"03:34.530","Text":"Now, the potassium has noble gas configuration and the sulfur has an octet."},{"Start":"03:35.050 ","End":"03:44.745","Text":"This comes out to 2 potassium cations"},{"Start":"03:44.745 ","End":"03:46.380","Text":"with a +1 charge,"},{"Start":"03:46.380 ","End":"03:49.924","Text":"since each potassium lost 1 electron,"},{"Start":"03:49.924 ","End":"03:52.630","Text":"we have 1 sulfur,"},{"Start":"03:52.630 ","End":"04:00.700","Text":"which gained 2 electrons so the charge is -2 and it has 8 electrons."},{"Start":"04:02.540 ","End":"04:05.890","Text":"This is our answer for b."},{"Start":"04:08.390 ","End":"04:12.370","Text":"In c, we have strontium oxide."},{"Start":"04:13.790 ","End":"04:18.190","Text":"We\u0027re going to take a look at strontium and oxygen."},{"Start":"04:18.190 ","End":"04:22.300","Text":"Oxygen is in the 16th period, just like sulfur."},{"Start":"04:22.300 ","End":"04:25.820","Text":"Therefore, it has 6 valence electrons."},{"Start":"04:30.090 ","End":"04:33.520","Text":"Strontium is in the second period."},{"Start":"04:33.520 ","End":"04:37.915","Text":"Therefore, it has 2 valence electrons."},{"Start":"04:37.915 ","End":"04:42.250","Text":"In this case in order for the oxygen to reach a complete octet and"},{"Start":"04:42.250 ","End":"04:46.105","Text":"the strontium to acquire a noble gas electron configuration,"},{"Start":"04:46.105 ","End":"04:52.550","Text":"we\u0027re just going to pass 2 electrons from the strontium to the oxygen."},{"Start":"04:53.520 ","End":"04:58.516","Text":"Now the strontium has a charge of +2,"},{"Start":"04:58.516 ","End":"05:02.519","Text":"since 2 electrons were passed to the oxygen."},{"Start":"05:02.810 ","End":"05:09.230","Text":"The oxygen has an octet around it,"},{"Start":"05:09.230 ","End":"05:15.900","Text":"8 electrons and a -2 charge."},{"Start":"05:17.260 ","End":"05:24.860","Text":"This here is our answer for c. These are our final answers."},{"Start":"05:24.860 ","End":"05:27.390","Text":"Thank you very much for watching."}],"ID":24514},{"Watched":false,"Name":"Exercise 5 - part a","Duration":"3m 49s","ChapterTopicVideoID":23598,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:03.435","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.435 ","End":"00:05.970","Text":"Write Lewis structures for the following structures."},{"Start":"00:05.970 ","End":"00:07.800","Text":"We\u0027re going to start with a and we have"},{"Start":"00:07.800 ","End":"00:11.765","Text":"chloroform and we have to write the Lewis structure for chloroform."},{"Start":"00:11.765 ","End":"00:13.994","Text":"In order to write Lewis structures,"},{"Start":"00:13.994 ","End":"00:15.735","Text":"we\u0027re going to have to follow a number of steps."},{"Start":"00:15.735 ","End":"00:18.930","Text":"The first step, we\u0027re going to call it a,"},{"Start":"00:18.930 ","End":"00:22.550","Text":"is to determine the total number of valence electrons in this structure."},{"Start":"00:22.550 ","End":"00:23.930","Text":"Here we have carbon,"},{"Start":"00:23.930 ","End":"00:25.280","Text":"hydrogen and 3 chlorines,"},{"Start":"00:25.280 ","End":"00:26.660","Text":"so we\u0027re going to look at our carbon."},{"Start":"00:26.660 ","End":"00:29.015","Text":"We see that it\u0027s in group 14."},{"Start":"00:29.015 ","End":"00:33.410","Text":"Therefore, we\u0027re going to write 4 valence electrons for carbon."},{"Start":"00:33.410 ","End":"00:34.640","Text":"Next we have hydrogen,"},{"Start":"00:34.640 ","End":"00:36.840","Text":"which is in the first group,"},{"Start":"00:36.840 ","End":"00:39.260","Text":"so we\u0027re going to add 1 valence electron,"},{"Start":"00:39.260 ","End":"00:41.120","Text":"and we have now 3 chlorine,"},{"Start":"00:41.120 ","End":"00:43.410","Text":"so we\u0027re going to add 3 times 7."},{"Start":"00:43.410 ","End":"00:44.720","Text":"Since if we look at chlorine,"},{"Start":"00:44.720 ","End":"00:47.615","Text":"we can see that it\u0027s in group 17."},{"Start":"00:47.615 ","End":"00:54.830","Text":"This comes to a total of 26 valence electrons and that is the first step."},{"Start":"00:54.830 ","End":"00:57.245","Text":"The next step, we\u0027ll call it b,"},{"Start":"00:57.245 ","End":"01:01.000","Text":"is to identify the central and terminal atoms."},{"Start":"01:01.000 ","End":"01:05.390","Text":"Now, central atoms are generally those with the lowest electronegativity."},{"Start":"01:05.390 ","End":"01:08.015","Text":"Carbon is the central atom,"},{"Start":"01:08.015 ","End":"01:11.030","Text":"and chlorine is a terminal atom,"},{"Start":"01:11.030 ","End":"01:13.640","Text":"it has high electronegativity."},{"Start":"01:13.640 ","End":"01:16.940","Text":"Now hydrogen is also a terminal atom because it can"},{"Start":"01:16.940 ","End":"01:20.120","Text":"only accommodate 2 electrons in its valence shell."},{"Start":"01:20.120 ","End":"01:22.730","Text":"Therefore, hydrogen is terminal atom."},{"Start":"01:22.730 ","End":"01:24.590","Text":"We identify the central,"},{"Start":"01:24.590 ","End":"01:26.210","Text":"and it\u0027s terminal atoms."},{"Start":"01:26.210 ","End":"01:29.220","Text":"The next step is to connect them."},{"Start":"01:32.150 ","End":"01:35.380","Text":"I drew 4 bonds here."},{"Start":"01:36.830 ","End":"01:40.950","Text":"Now sometimes you will see these bonds as 2 dots."},{"Start":"01:41.680 ","End":"01:44.870","Text":"But here I\u0027m writing bonds."},{"Start":"01:44.870 ","End":"01:48.110","Text":"The next step, which we\u0027ll call it c,"},{"Start":"01:48.110 ","End":"01:51.470","Text":"is to subtract the number of bonds from our valence electrons,"},{"Start":"01:51.470 ","End":"01:54.640","Text":"meaning we have 26 valence electrons remember,"},{"Start":"01:54.640 ","End":"01:58.710","Text":"we\u0027re going to subtract 8 valence electrons."},{"Start":"01:58.710 ","End":"02:01.050","Text":"Why? Because we have 4 bonds,"},{"Start":"02:01.050 ","End":"02:02.835","Text":"meaning 2, 4,"},{"Start":"02:02.835 ","End":"02:05.405","Text":"6, 8, since we have 2 electrons in each bond."},{"Start":"02:05.405 ","End":"02:11.159","Text":"So 26 minus 8 is going to give us 18 more electrons."},{"Start":"02:12.580 ","End":"02:18.200","Text":"We have 18 more electrons to write in our Lewis structure."},{"Start":"02:18.200 ","End":"02:20.780","Text":"Now, with the electrons that are left,"},{"Start":"02:20.780 ","End":"02:23.720","Text":"we first need to complete the octets on the terminal atoms,"},{"Start":"02:23.720 ","End":"02:25.250","Text":"and then we go to the central atom."},{"Start":"02:25.250 ","End":"02:28.550","Text":"Let\u0027s complete the octets for our chlorines."},{"Start":"02:28.550 ","End":"02:30.860","Text":"Each chlorine we have to add 1,"},{"Start":"02:30.860 ","End":"02:33.010","Text":"2, 3, 4, 5, 6."},{"Start":"02:33.010 ","End":"02:36.240","Text":"That way the chlorine will have an octet, 1, 2, 3,"},{"Start":"02:36.240 ","End":"02:38.280","Text":"4, 5, 6; 1,"},{"Start":"02:38.280 ","End":"02:40.650","Text":"2, 3, 4, 5, 6."},{"Start":"02:40.650 ","End":"02:43.880","Text":"Now we added 6 electrons to 3 chlorine atoms,"},{"Start":"02:43.880 ","End":"02:45.710","Text":"meaning we added 18 electrons,"},{"Start":"02:45.710 ","End":"02:48.110","Text":"so we added all of the electrons."},{"Start":"02:48.110 ","End":"02:50.960","Text":"We have no electrons left for the central atom."},{"Start":"02:50.960 ","End":"02:53.780","Text":"Now we have to look at our Lewis structure,"},{"Start":"02:53.780 ","End":"02:56.870","Text":"and we have to see if it\u0027s a satisfactory structure or not."},{"Start":"02:56.870 ","End":"03:01.680","Text":"For this, we have to see if our atoms have octets,"},{"Start":"03:03.230 ","End":"03:08.945","Text":"and hydrogen again can form only 1 bond to another atom,"},{"Start":"03:08.945 ","End":"03:10.940","Text":"so hydrogen is good."},{"Start":"03:10.940 ","End":"03:14.930","Text":"We can also see that our chlorines have 8 electrons,"},{"Start":"03:14.930 ","End":"03:16.190","Text":"meaning they have enough octets,"},{"Start":"03:16.190 ","End":"03:17.390","Text":"and if we look at our carbon,"},{"Start":"03:17.390 ","End":"03:19.040","Text":"we also see 1, 2, 3,"},{"Start":"03:19.040 ","End":"03:22.040","Text":"4 bonds times 2 is 8 electrons,"},{"Start":"03:22.040 ","End":"03:24.140","Text":"so our carbon also has an octet."},{"Start":"03:24.140 ","End":"03:28.190","Text":"Therefore, we have a satisfactory Lewis structure."},{"Start":"03:28.190 ","End":"03:35.310","Text":"The Lewis structure for chloroform is what we did. Now we\u0027re gonna go on to b."}],"ID":24509},{"Watched":false,"Name":"Exercise 5 - part b","Duration":"3m 33s","ChapterTopicVideoID":23599,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"Now we\u0027re going to go on to b."},{"Start":"00:02.790 ","End":"00:05.850","Text":"In b we have hydrogen sulfide."},{"Start":"00:05.850 ","End":"00:07.590","Text":"Again, we\u0027re doing b,"},{"Start":"00:07.590 ","End":"00:08.940","Text":"which is hydrogen sulfide,"},{"Start":"00:08.940 ","End":"00:10.980","Text":"and we want to write the Lewis structure."},{"Start":"00:10.980 ","End":"00:13.230","Text":"We have to start with the first step."},{"Start":"00:13.230 ","End":"00:15.465","Text":"Now again, in the first step, we\u0027re going to find"},{"Start":"00:15.465 ","End":"00:19.320","Text":"the number of valence electrons for the structure."},{"Start":"00:19.320 ","End":"00:21.240","Text":"Again, we have hydrogen,"},{"Start":"00:21.240 ","End":"00:23.400","Text":"which we know is in the first group."},{"Start":"00:23.400 ","End":"00:27.450","Text":"Therefore we have 1 valence electron from each hydrogen,"},{"Start":"00:27.450 ","End":"00:28.620","Text":"but we have 2 hydrogens,"},{"Start":"00:28.620 ","End":"00:30.714","Text":"so that\u0027s 2 times 1,"},{"Start":"00:30.714 ","End":"00:33.840","Text":"plus we have sulfur."},{"Start":"00:33.840 ","End":"00:38.315","Text":"If you look at sulfur, you will see that it is in the 16th group,"},{"Start":"00:38.315 ","End":"00:41.660","Text":"meaning that it has 6 valence electrons."},{"Start":"00:41.660 ","End":"00:44.695","Text":"In all of this gives us 8 electrons."},{"Start":"00:44.695 ","End":"00:50.525","Text":"Now again in b, we have to identify the central and the terminal atoms."},{"Start":"00:50.525 ","End":"00:53.330","Text":"We said already that hydrogen is a terminal atom."},{"Start":"00:53.330 ","End":"00:55.370","Text":"Therefore, we\u0027re going to write the hydrogen is terminal,"},{"Start":"00:55.370 ","End":"00:58.489","Text":"and the sulfur is central."},{"Start":"00:58.489 ","End":"01:01.945","Text":"Next step we\u0027re going to connect them by bonds,"},{"Start":"01:01.945 ","End":"01:06.340","Text":"and next step, we\u0027re going to take our 8 electrons,"},{"Start":"01:06.340 ","End":"01:09.380","Text":"and subtract the number of electrons in bonds."},{"Start":"01:09.380 ","End":"01:13.640","Text":"So we have 2 electrons in 1 bond and 2 electrons in the second one we need,"},{"Start":"01:13.640 ","End":"01:15.309","Text":"we have 4 electrons."},{"Start":"01:15.309 ","End":"01:20.195","Text":"We have 4 electrons left for our structure."},{"Start":"01:20.195 ","End":"01:22.850","Text":"Now again, we usually begin with a terminal atoms,"},{"Start":"01:22.850 ","End":"01:23.900","Text":"but since we have hydrogen,"},{"Start":"01:23.900 ","End":"01:26.450","Text":"and we know that it only forms 1 bond,"},{"Start":"01:26.450 ","End":"01:29.300","Text":"so hydrogen can\u0027t accept any other electrons."},{"Start":"01:29.300 ","End":"01:30.710","Text":"We\u0027re going to go to the sulfur."},{"Start":"01:30.710 ","End":"01:32.480","Text":"We have to add in 4 electrons."},{"Start":"01:32.480 ","End":"01:36.235","Text":"We\u0027re going to add in 2 here, and 2 here."},{"Start":"01:36.235 ","End":"01:38.180","Text":"Now if we look at our Lewis structure,"},{"Start":"01:38.180 ","End":"01:43.265","Text":"we can see that the hydrogen again has 1 bond which is exactly as it\u0027s supposed to have."},{"Start":"01:43.265 ","End":"01:46.719","Text":"If we look at the sulfur, we can see that it has an octet,"},{"Start":"01:46.719 ","End":"01:48.570","Text":"1, 2, 3, 4, 5, 6,"},{"Start":"01:48.570 ","End":"01:50.970","Text":"7, 8 electrons around it."},{"Start":"01:50.970 ","End":"01:53.540","Text":"That is a satisfactory Lewis structure,"},{"Start":"01:53.540 ","End":"01:55.925","Text":"and that is our answer for b."},{"Start":"01:55.925 ","End":"01:59.740","Text":"Now we\u0027re going to go on to c which we have chlorine."},{"Start":"01:59.740 ","End":"02:03.470","Text":"In c, we have to write Lewis structure for chlorine."},{"Start":"02:03.470 ","End":"02:08.705","Text":"Again, the first step is to determine the number of valence electrons."},{"Start":"02:08.705 ","End":"02:11.150","Text":"We have 2 times since we have 2 chlorines,"},{"Start":"02:11.150 ","End":"02:14.240","Text":"2 times 7 because remember,"},{"Start":"02:14.240 ","End":"02:17.640","Text":"chlorine is in the group 17."},{"Start":"02:17.640 ","End":"02:22.590","Text":"That equals 14 electrons for chlorine."},{"Start":"02:22.660 ","End":"02:26.630","Text":"Next step is to identify the central, and terminal atoms."},{"Start":"02:26.630 ","End":"02:28.280","Text":"However, we only have 2 chlorines."},{"Start":"02:28.280 ","End":"02:31.250","Text":"Therefore, we have 1 chlorine at one side and 1 chlorine on the other side,"},{"Start":"02:31.250 ","End":"02:34.500","Text":"and we\u0027re going to put a bond in-between."},{"Start":"02:35.060 ","End":"02:39.260","Text":"The third step is to subtract the number"},{"Start":"02:39.260 ","End":"02:42.560","Text":"of electrons in bonds from the total number of valence electrons."},{"Start":"02:42.560 ","End":"02:44.570","Text":"We have a total number 14,"},{"Start":"02:44.570 ","End":"02:50.054","Text":"and we\u0027re subtracting 2 since we have 1 bond which contains 2 electrons."},{"Start":"02:50.054 ","End":"02:52.695","Text":"This gives us another 12 electrons,"},{"Start":"02:52.695 ","End":"02:58.460","Text":"and now we have to take our atoms on chlorines and add the 12 electrons."},{"Start":"02:58.460 ","End":"03:01.050","Text":"We\u0027re going to add 6-8 chlorine."},{"Start":"03:03.160 ","End":"03:06.725","Text":"Now we\u0027re going to look at our Lewis structure."},{"Start":"03:06.725 ","End":"03:10.400","Text":"We see that around each chlorine we have 8 electrons,"},{"Start":"03:10.400 ","End":"03:11.810","Text":"1, 2, 3, 4, 5,"},{"Start":"03:11.810 ","End":"03:14.720","Text":"6, and another bond adding another 2,"},{"Start":"03:14.720 ","End":"03:17.510","Text":"meaning we have 8 electrons on each chlorine."},{"Start":"03:17.510 ","End":"03:21.365","Text":"This is a satisfactory Lewis structure,"},{"Start":"03:21.365 ","End":"03:27.360","Text":"and this is our answer for c. Now we\u0027re going to go on to d."}],"ID":24510},{"Watched":false,"Name":"Exercise 5 - part c","Duration":"7m ","ChapterTopicVideoID":23600,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.980 ","End":"00:06.020","Text":"Now we\u0027re going to continue on to d. In d we have CH_4O."},{"Start":"00:06.020 ","End":"00:09.210","Text":"Now again, the first step counting the valence electrons."},{"Start":"00:09.210 ","End":"00:16.730","Text":"In carbon we said we have 4 since carbon is in the 14th group, group number 14."},{"Start":"00:16.730 ","End":"00:20.500","Text":"We have 1 valence electron for each hydrogen,"},{"Start":"00:20.500 ","End":"00:24.555","Text":"so we have 4 times 1 since we have 4 hydrogens,"},{"Start":"00:24.555 ","End":"00:25.959","Text":"and for the oxygen,"},{"Start":"00:25.959 ","End":"00:30.067","Text":"as you can see, it is in the 16th group,"},{"Start":"00:30.067 ","End":"00:34.930","Text":"so we have 6 valence electrons for the oxygen."},{"Start":"00:34.930 ","End":"00:40.225","Text":"That gives us a total of 14 valence electrons."},{"Start":"00:40.225 ","End":"00:46.045","Text":"In b we have to decide which atoms are terminal and which atoms are central."},{"Start":"00:46.045 ","End":"00:48.490","Text":"Again, we know that our carbon is going to be"},{"Start":"00:48.490 ","End":"00:52.640","Text":"central and we know that the hydrogens are terminal atoms."},{"Start":"00:52.640 ","End":"00:56.120","Text":"Oxygen is also usually a terminal atom."},{"Start":"00:56.120 ","End":"00:58.360","Text":"However, in this case,"},{"Start":"00:58.360 ","End":"01:07.070","Text":"the only way we can connect the atoms is if we connect 3 hydrogens to the carbon,"},{"Start":"01:07.070 ","End":"01:13.584","Text":"and then connect the carbon to the oxygen and the oxygen to another hydrogen."},{"Start":"01:13.584 ","End":"01:16.690","Text":"Because the hydrogens can only form bond, therefore,"},{"Start":"01:16.690 ","End":"01:20.160","Text":"this is the only way our structure can be connected."},{"Start":"01:20.160 ","End":"01:22.480","Text":"Now we\u0027re going to count the number of bonds we have."},{"Start":"01:22.480 ","End":"01:23.620","Text":"We have 1, 2,"},{"Start":"01:23.620 ","End":"01:25.315","Text":"3, 4, 5 bonds,"},{"Start":"01:25.315 ","End":"01:28.540","Text":"meaning 5 times 2 since we have 2 electrons in each bond,"},{"Start":"01:28.540 ","End":"01:30.640","Text":"we have 10 electrons in our bond."},{"Start":"01:30.640 ","End":"01:33.980","Text":"We\u0027re going to take the total number of valence electrons,"},{"Start":"01:33.980 ","End":"01:36.845","Text":"which was 14 from a,"},{"Start":"01:36.845 ","End":"01:41.755","Text":"minus 10 electrons, which is the number of electrons in the bonds,"},{"Start":"01:41.755 ","End":"01:45.725","Text":"and this gives us 4 electrons to add to our structure."},{"Start":"01:45.725 ","End":"01:47.870","Text":"Now again, the hydrogens can only form"},{"Start":"01:47.870 ","End":"01:51.980","Text":"1 bond so we\u0027re not going to add any electrons to them."},{"Start":"01:51.980 ","End":"01:55.020","Text":"If we look at our carbon it already has an octet,"},{"Start":"01:55.020 ","End":"01:57.405","Text":"it already has 8 electrons around it."},{"Start":"01:57.405 ","End":"02:00.634","Text":"We\u0027re going to add these electrons to the oxygen."},{"Start":"02:00.634 ","End":"02:04.280","Text":"We\u0027re going to add 1, 2, 3, 4."},{"Start":"02:04.280 ","End":"02:07.100","Text":"As you can see now that we added these electrons,"},{"Start":"02:07.100 ","End":"02:11.420","Text":"the oxygen also has an octet."},{"Start":"02:11.420 ","End":"02:17.550","Text":"Therefore, this is a satisfactory structure and this is our Lewis structure for"},{"Start":"02:17.550 ","End":"02:25.085","Text":"d. Now we\u0027re going to go on to e. In e we can see that we have the ammonium ion."},{"Start":"02:25.085 ","End":"02:28.564","Text":"Again, we\u0027re going to start with counting the number of valence electrons."},{"Start":"02:28.564 ","End":"02:33.740","Text":"Nitrogen is in the 15th group giving us 5 electrons."},{"Start":"02:33.740 ","End":"02:37.440","Text":"We\u0027re going to add 4 more electrons for"},{"Start":"02:37.440 ","End":"02:40.960","Text":"the hydrogen since hydrogen is in the first group,"},{"Start":"02:40.960 ","End":"02:43.490","Text":"and it gives 1 valence electron."},{"Start":"02:43.490 ","End":"02:47.930","Text":"Now there\u0027s 1 more thing that we have to take into consideration."},{"Start":"02:47.930 ","End":"02:54.410","Text":"This time we\u0027re dealing with an ion and this ion has a charge of plus 1."},{"Start":"02:54.410 ","End":"03:02.310","Text":"We have to take off 1 electron from our valence electron since it has a charge of plus 1."},{"Start":"03:03.080 ","End":"03:06.180","Text":"This gives us 5 plus 4 minus 1,"},{"Start":"03:06.180 ","End":"03:09.049","Text":"which gives us a total of 8 electrons."},{"Start":"03:09.049 ","End":"03:11.690","Text":"Now the second step again is to decide"},{"Start":"03:11.690 ","End":"03:14.240","Text":"who are the terminal atoms and who are the central."},{"Start":"03:14.240 ","End":"03:16.880","Text":"Now again, we said that the hydrogens have to be terminal,"},{"Start":"03:16.880 ","End":"03:20.430","Text":"so the nitrogen is going to be our central atom."},{"Start":"03:21.130 ","End":"03:26.175","Text":"We\u0027re going to draw bonds between the nitrogen, and hydrogens."},{"Start":"03:26.175 ","End":"03:29.220","Text":"These bonds give us 1, 2, 3,"},{"Start":"03:29.220 ","End":"03:33.680","Text":"4, 4 bonds, meaning 8 electrons."},{"Start":"03:33.680 ","End":"03:36.860","Text":"The third step is to take the total number of valence electrons,"},{"Start":"03:36.860 ","End":"03:41.290","Text":"which is 8 minus the number of electrons inside the bonds."},{"Start":"03:41.290 ","End":"03:45.860","Text":"That\u0027s going to be minus 8 and it\u0027s going to give us 0 electrons to add to our structure."},{"Start":"03:45.860 ","End":"03:48.095","Text":"Now if we look at our structure,"},{"Start":"03:48.095 ","End":"03:50.450","Text":"we can see that it\u0027s a satisfactory structure."},{"Start":"03:50.450 ","End":"03:56.180","Text":"Again, the hydrogens have 1 bond and the nitrogen has 4 bonds around it,"},{"Start":"03:56.180 ","End":"04:00.385","Text":"meaning it has 8 electrons around it, it has an octet."},{"Start":"04:00.385 ","End":"04:03.270","Text":"The ions we add square brackets too,"},{"Start":"04:03.270 ","End":"04:09.745","Text":"and we write the charge of the ion on the top right of the square brackets."},{"Start":"04:09.745 ","End":"04:12.315","Text":"That\u0027s a plus 1 charge."},{"Start":"04:12.315 ","End":"04:18.550","Text":"That\u0027s our Lewis structure for e, the ammonium ion."},{"Start":"04:20.030 ","End":"04:25.135","Text":"In f we have phosphite ion, PO_3."},{"Start":"04:25.135 ","End":"04:27.620","Text":"Again, the first step to write"},{"Start":"04:27.620 ","End":"04:30.605","Text":"a Lewis structure is to count the number of valence electrons."},{"Start":"04:30.605 ","End":"04:32.540","Text":"The phosphorus is in group 15,"},{"Start":"04:32.540 ","End":"04:35.185","Text":"meaning it has 5 valence electrons."},{"Start":"04:35.185 ","End":"04:41.423","Text":"We\u0027re going to add 3 times since we have 3 oxygens,"},{"Start":"04:41.423 ","End":"04:43.670","Text":"and we\u0027re going to look at our oxygen and we know that it\u0027s in"},{"Start":"04:43.670 ","End":"04:47.675","Text":"the 16th group so we\u0027re going to add 3 times 6 valence electrons."},{"Start":"04:47.675 ","End":"04:51.925","Text":"Now, don\u0027t forget that we also have to add 3 more electrons because of the charge."},{"Start":"04:51.925 ","End":"04:56.445","Text":"Since we have a 3 minus,"},{"Start":"04:56.445 ","End":"04:59.795","Text":"then we have to add 3 electrons to our valence electrons."},{"Start":"04:59.795 ","End":"05:04.175","Text":"This is going to give us a total of 26 valence electrons."},{"Start":"05:04.175 ","End":"05:08.435","Text":"Now the next step is to decide the central atoms and the terminal atoms."},{"Start":"05:08.435 ","End":"05:12.890","Text":"This time the phosphorus is going to be a central atom and the oxygens are going to"},{"Start":"05:12.890 ","End":"05:18.245","Text":"be terminal atoms since they have a higher electronegativity."},{"Start":"05:18.245 ","End":"05:20.720","Text":"Again, we\u0027re going to put bonds between them."},{"Start":"05:20.720 ","End":"05:25.220","Text":"In this case, we have 3 bonds meaning 6 electrons are in these bonds."},{"Start":"05:25.220 ","End":"05:28.520","Text":"The third step is to take the total number of electrons,"},{"Start":"05:28.520 ","End":"05:31.770","Text":"which is 26 minus 6 electrons."},{"Start":"05:31.770 ","End":"05:36.740","Text":"It gives us 20 more electrons to place in our structure."},{"Start":"05:36.740 ","End":"05:39.110","Text":"We\u0027re going to start with our terminal atoms,"},{"Start":"05:39.110 ","End":"05:40.534","Text":"and we\u0027re going to give it an octet."},{"Start":"05:40.534 ","End":"05:45.030","Text":"We\u0027re going to add 6 electrons to each oxygen since it has"},{"Start":"05:45.030 ","End":"05:50.340","Text":"2 electrons already to give them an octet."},{"Start":"05:50.340 ","End":"05:52.110","Text":"Since we gave 3 oxygen,"},{"Start":"05:52.110 ","End":"05:54.195","Text":"6 electrons each,"},{"Start":"05:54.195 ","End":"05:56.765","Text":"we used 18 electrons."},{"Start":"05:56.765 ","End":"05:58.745","Text":"Remember we have 20, so we have 2 left."},{"Start":"05:58.745 ","End":"06:03.535","Text":"We\u0027re going to add these to the central phosphorus."},{"Start":"06:03.535 ","End":"06:05.315","Text":"If we check our structure,"},{"Start":"06:05.315 ","End":"06:07.775","Text":"every oxygen has an octet"},{"Start":"06:07.775 ","End":"06:11.000","Text":"and if we look at the phosphorus in the middle, it also has an octet."},{"Start":"06:11.000 ","End":"06:13.280","Text":"Therefore, this is a satisfactory structure."},{"Start":"06:13.280 ","End":"06:15.170","Text":"Now remember it\u0027s an ion, therefore,"},{"Start":"06:15.170 ","End":"06:17.540","Text":"we\u0027re going to add square brackets and we\u0027re"},{"Start":"06:17.540 ","End":"06:20.495","Text":"going to write the charge of the ion on the top right."},{"Start":"06:20.495 ","End":"06:26.675","Text":"This is our Lewis structure for f, the phosphite ion."},{"Start":"06:26.675 ","End":"06:28.685","Text":"That is our final answer."},{"Start":"06:28.685 ","End":"06:31.590","Text":"Thank you very much for watching."}],"ID":24511},{"Watched":false,"Name":"Exercise 6 - part a","Duration":"6m 1s","ChapterTopicVideoID":23607,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.925","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.925 ","End":"00:07.675","Text":"Calculate the formal charge of each atom in the following Lewis structures."},{"Start":"00:07.675 ","End":"00:09.975","Text":"We\u0027re going to begin with a."},{"Start":"00:09.975 ","End":"00:13.050","Text":"In a, we have carbon and we have oxygen,"},{"Start":"00:13.050 ","End":"00:16.920","Text":"and we need to calculate the formal charge in each case."},{"Start":"00:16.920 ","End":"00:19.695","Text":"I want to remind you that the formal charge"},{"Start":"00:19.695 ","End":"00:24.300","Text":"equals the number of valence electrons in the free atom,"},{"Start":"00:24.300 ","End":"00:30.600","Text":"that\u0027s the number of electrons in the free atom,"},{"Start":"00:30.600 ","End":"00:36.880","Text":"minus the number of lone pair electrons,"},{"Start":"00:40.190 ","End":"00:48.730","Text":"minus 1/2 of the number of the bond pair electrons."},{"Start":"00:52.100 ","End":"00:56.179","Text":"Again, let\u0027s take a look at a and we\u0027re going to start with carbon."},{"Start":"00:56.179 ","End":"01:00.080","Text":"We\u0027re going to calculate the formal charge of carbon."},{"Start":"01:00.080 ","End":"01:05.060","Text":"The number of valence electrons in the free atom of carbon are 4,"},{"Start":"01:05.060 ","End":"01:08.030","Text":"because remember, carbon is in the 14th group."},{"Start":"01:08.030 ","End":"01:12.810","Text":"Now we\u0027re going to have to subtract the number of lone pair electrons."},{"Start":"01:12.810 ","End":"01:14.130","Text":"Now, if we look at the carbon,"},{"Start":"01:14.130 ","End":"01:16.949","Text":"you can see that it has 4 bonds around it,"},{"Start":"01:16.949 ","End":"01:19.400","Text":"and there are no lone pair electrons."},{"Start":"01:19.400 ","End":"01:25.875","Text":"That\u0027s going to be 0 minus 1/2 of the number of the bond pair electrons."},{"Start":"01:25.875 ","End":"01:29.585","Text":"Since we have 4 bonds around the carbon,"},{"Start":"01:29.585 ","End":"01:34.320","Text":"that means we have 8 electrons because remember you have 2 electrons per bond,"},{"Start":"01:34.910 ","End":"01:41.940","Text":"so 1/2 times 8 is the number of bond pair."},{"Start":"01:41.940 ","End":"01:46.840","Text":"This equals 4 minus 0 minus 4."},{"Start":"01:47.600 ","End":"01:51.985","Text":"Now, our carbon has a formal charge of 0."},{"Start":"01:51.985 ","End":"01:58.040","Text":"We\u0027re going to write the formal charge of carbon equals 0."},{"Start":"01:58.040 ","End":"02:00.860","Text":"Now, we\u0027re going to calculate the formal charge of the oxygens."},{"Start":"02:00.860 ","End":"02:03.320","Text":"Now if we look at our Lewis structure,"},{"Start":"02:03.320 ","End":"02:06.780","Text":"we can see that we have 2 different oxygens: 1,"},{"Start":"02:06.780 ","End":"02:13.805","Text":"are the oxygens down here and the upper oxygen is actually different."},{"Start":"02:13.805 ","End":"02:15.410","Text":"Let\u0027s begin with the upper oxygen."},{"Start":"02:15.410 ","End":"02:18.065","Text":"We\u0027re just going to call it 1."},{"Start":"02:18.065 ","End":"02:21.810","Text":"Oxygen 1, the formal charge,"},{"Start":"02:21.810 ","End":"02:25.250","Text":"equals, again the number of electrons in the free atom."},{"Start":"02:25.250 ","End":"02:28.430","Text":"Now remember that the oxygen is in the 16th group,"},{"Start":"02:28.430 ","End":"02:33.575","Text":"meaning it has 6 valence electrons minus the number of lone pair electrons."},{"Start":"02:33.575 ","End":"02:37.340","Text":"Here we have 2 sets of lone pair electrons,"},{"Start":"02:37.340 ","End":"02:40.669","Text":"meaning we have 4 lone pair electrons."},{"Start":"02:40.669 ","End":"02:46.625","Text":"We\u0027re going to subtract 4 minus 1/2 of the number of the bond pair minus,"},{"Start":"02:46.625 ","End":"02:49.130","Text":"if we look at the upper oxygen again,"},{"Start":"02:49.130 ","End":"02:52.130","Text":"we have 2 bonds connected to the oxygen,"},{"Start":"02:52.130 ","End":"02:59.880","Text":"meaning, 1/2 times 4."},{"Start":"03:01.340 ","End":"03:04.670","Text":"If we have 2 bonds, we have 4 electrons because again,"},{"Start":"03:04.670 ","End":"03:07.824","Text":"in 1 bond we have 2 electrons."},{"Start":"03:07.824 ","End":"03:10.555","Text":"This is 1/2 times 4."},{"Start":"03:10.555 ","End":"03:16.915","Text":"If we calculate this, we get 6 minus 4 minus 2,"},{"Start":"03:16.915 ","End":"03:19.680","Text":"which gives us a 0."},{"Start":"03:19.930 ","End":"03:27.305","Text":"The formal charge of oxygen 1 also equals 0."},{"Start":"03:27.305 ","End":"03:31.260","Text":"Now, we\u0027re going onto the oxygen 2."},{"Start":"03:31.520 ","End":"03:36.770","Text":"The formal charge of oxygen 2 equals, again,"},{"Start":"03:36.770 ","End":"03:38.810","Text":"oxygen is in the 16th group,"},{"Start":"03:38.810 ","End":"03:44.175","Text":"meaning it has 6 valence electrons minus the number of lone pair electrons."},{"Start":"03:44.175 ","End":"03:45.840","Text":"Here we can see we have 1,"},{"Start":"03:45.840 ","End":"03:47.100","Text":"2, 3, 4, 5,"},{"Start":"03:47.100 ","End":"03:53.930","Text":"6 lone pair electrons minus 1/2 the number of the bond pair electrons."},{"Start":"03:53.930 ","End":"03:57.685","Text":"Since we have 1 bond which is 2 electrons,"},{"Start":"03:57.685 ","End":"04:02.770","Text":"we have to do 1/2 times 2 electrons."},{"Start":"04:03.170 ","End":"04:07.380","Text":"That\u0027s going to give us 6 minus 6 minus 1,"},{"Start":"04:07.380 ","End":"04:10.450","Text":"which is going to give us a negative 1."},{"Start":"04:10.940 ","End":"04:15.005","Text":"The formal charge of O_2,"},{"Start":"04:15.005 ","End":"04:18.320","Text":"the second oxygen, equals minus 1."},{"Start":"04:18.450 ","End":"04:21.740","Text":"Again, the carbon,"},{"Start":"04:21.930 ","End":"04:24.760","Text":"which we calculated its formal charge,"},{"Start":"04:24.760 ","End":"04:26.815","Text":"the formal charge came out to 0."},{"Start":"04:26.815 ","End":"04:31.165","Text":"The formal charge of the upper oxygen also equals 0."},{"Start":"04:31.165 ","End":"04:36.310","Text":"The formal charge of the 2 lower oxygens equals minus 1 each."},{"Start":"04:36.310 ","End":"04:42.774","Text":"Now the sum of the formal charges of the atoms in a Lewis structure equals"},{"Start":"04:42.774 ","End":"04:50.020","Text":"the charge on the ion or if we have a neutral Lewis structure, it equals 0."},{"Start":"04:50.020 ","End":"04:53.740","Text":"In our case we have an ion and the charge is minus 2."},{"Start":"04:53.740 ","End":"04:58.735","Text":"The sum of our formal charges has to equal minus 2."},{"Start":"04:58.735 ","End":"05:03.130","Text":"As we can see if we take the sum of carbon,"},{"Start":"05:03.130 ","End":"05:04.585","Text":"which is the formal charge, is 0,"},{"Start":"05:04.585 ","End":"05:10.720","Text":"so 0 plus the formal charge of the upper oxygen,"},{"Start":"05:10.720 ","End":"05:12.990","Text":"which is also 0."},{"Start":"05:12.990 ","End":"05:16.480","Text":"Add the formal charge of each lower oxygen,"},{"Start":"05:16.480 ","End":"05:19.350","Text":"meaning plus negative 1,"},{"Start":"05:19.350 ","End":"05:22.095","Text":"plus negative 1,"},{"Start":"05:22.095 ","End":"05:25.040","Text":"we get minus 2."},{"Start":"05:25.040 ","End":"05:29.360","Text":"As you can see, this equals the charge of the ion."},{"Start":"05:29.480 ","End":"05:34.475","Text":"Our answers for this question are the formal charge of carbon came out to 0,"},{"Start":"05:34.475 ","End":"05:37.130","Text":"the formal charge of the upper oxygen equals 0,"},{"Start":"05:37.130 ","End":"05:42.090","Text":"and the formal charge of the 2 lower oxygens equals minus 1."},{"Start":"05:42.190 ","End":"05:46.320","Text":"That\u0027s our answer for a. Now, we\u0027re going to go on to b."}],"ID":24518},{"Watched":false,"Name":"Exercise 6 - part b","Duration":"5m 37s","ChapterTopicVideoID":23592,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.135","Text":"Now we\u0027re going to go on to b."},{"Start":"00:03.135 ","End":"00:05.775","Text":"We\u0027re going to begin with the sulfur."},{"Start":"00:05.775 ","End":"00:08.444","Text":"The formal charge of sulfur,"},{"Start":"00:08.444 ","End":"00:12.945","Text":"again, equals the number of the valence electrons in the free atom."},{"Start":"00:12.945 ","End":"00:14.700","Text":"Sulfur is in the 16th group,"},{"Start":"00:14.700 ","End":"00:19.230","Text":"meaning it has 6 valence electrons minus the number of lone pair."},{"Start":"00:19.230 ","End":"00:20.850","Text":"As you can see around the sulfur,"},{"Start":"00:20.850 ","End":"00:23.130","Text":"there are no lone pair electrons,"},{"Start":"00:23.130 ","End":"00:28.235","Text":"so that\u0027s 0 minus 0.5 times how many bond pair electrons do we have."},{"Start":"00:28.235 ","End":"00:30.341","Text":"Since we have 1, 2, 3,"},{"Start":"00:30.341 ","End":"00:33.295","Text":"4, 4 bonds around the sulfur,"},{"Start":"00:33.295 ","End":"00:38.180","Text":"we have 8 bond pair electrons because again,"},{"Start":"00:38.180 ","End":"00:40.675","Text":"every bond contains 2 electrons."},{"Start":"00:40.675 ","End":"00:45.750","Text":"This is going to give us 6 minus 4 and we\u0027re going to get 2."},{"Start":"00:45.750 ","End":"00:49.810","Text":"The formal charge of sulfur is 2."},{"Start":"00:55.090 ","End":"00:59.885","Text":"Now we\u0027re going to calculate the formal charges of the oxygens in b."},{"Start":"00:59.885 ","End":"01:01.370","Text":"Again, if you look,"},{"Start":"01:01.370 ","End":"01:04.235","Text":"the upper oxygen is the same as we had in a,"},{"Start":"01:04.235 ","End":"01:06.995","Text":"is the same as the upper oxygen."},{"Start":"01:06.995 ","End":"01:10.280","Text":"The formal charges of the lower oxygens are"},{"Start":"01:10.280 ","End":"01:12.920","Text":"going to be the same as we had in a, of lower oxygen."},{"Start":"01:12.920 ","End":"01:15.305","Text":"Let\u0027s begin with the upper oxygen."},{"Start":"01:15.305 ","End":"01:20.690","Text":"Again, the number of electrons in the free atom of valence electrons of oxygen"},{"Start":"01:20.690 ","End":"01:25.430","Text":"is 6 because it\u0027s in the 16th group minus the number of lone pair electrons."},{"Start":"01:25.430 ","End":"01:29.510","Text":"If we look at the upper oxygens and we have 4 lone pair electrons, 1, 2, 3,"},{"Start":"01:29.510 ","End":"01:34.065","Text":"4 minus 0.5 of the number of bond pair electrons."},{"Start":"01:34.065 ","End":"01:36.440","Text":"How many bond pair electrons do we have?"},{"Start":"01:36.440 ","End":"01:39.784","Text":"We have 2 bonds connected to the oxygen,"},{"Start":"01:39.784 ","End":"01:44.550","Text":"meaning that we have 4 electrons in there."},{"Start":"01:44.550 ","End":"01:50.415","Text":"That\u0027s going to give us 6 minus 4 minus 2."},{"Start":"01:50.415 ","End":"01:52.985","Text":"That\u0027s going to give us a formal charge of 0."},{"Start":"01:52.985 ","End":"01:55.790","Text":"The upper oxygen, we\u0027ll call it oxygen 1,"},{"Start":"01:55.790 ","End":"02:00.180","Text":"and again equals 0."},{"Start":"02:00.190 ","End":"02:02.660","Text":"Now we\u0027re going to look at the lower oxygen,"},{"Start":"02:02.660 ","End":"02:05.075","Text":"we\u0027re going to call it oxygen 2."},{"Start":"02:05.075 ","End":"02:10.305","Text":"Again, we begin with 6 electrons minus,"},{"Start":"02:10.305 ","End":"02:13.620","Text":"in this case, we have 6 lone pair electrons,"},{"Start":"02:13.620 ","End":"02:15.393","Text":"1, 2, 3, 4, 5,"},{"Start":"02:15.393 ","End":"02:18.860","Text":"6 minus 0.5 times the bond pair electrons,"},{"Start":"02:18.860 ","End":"02:20.720","Text":"which in our case, since we have 1 bond,"},{"Start":"02:20.720 ","End":"02:22.324","Text":"we only have 2 electrons."},{"Start":"02:22.324 ","End":"02:25.955","Text":"So this gives us 6 minus 6 minus 1,"},{"Start":"02:25.955 ","End":"02:28.990","Text":"which equals a negative 1."},{"Start":"02:28.990 ","End":"02:34.890","Text":"The formal charge on oxygen 2 equals minus 1."},{"Start":"02:34.890 ","End":"02:38.435","Text":"The formal charges that we found for the sulfur,"},{"Start":"02:38.435 ","End":"02:41.380","Text":"we found a formal charge of plus 2."},{"Start":"02:41.380 ","End":"02:43.120","Text":"For the upper oxygen,"},{"Start":"02:43.120 ","End":"02:45.870","Text":"which is a 1 oxygen 1,"},{"Start":"02:45.870 ","End":"02:47.720","Text":"the formal charge equals 0,"},{"Start":"02:47.720 ","End":"02:49.870","Text":"for the lower oxygens for each one,"},{"Start":"02:49.870 ","End":"02:52.920","Text":"the formal charge equals minus 1."},{"Start":"02:52.920 ","End":"02:56.675","Text":"Now, if we take these formal charges and sum them up,"},{"Start":"02:56.675 ","End":"03:02.110","Text":"we expect them to equals 0 since we\u0027re talking about a neutral molecule."},{"Start":"03:02.110 ","End":"03:04.690","Text":"Let\u0027s sum them up. We have 2,"},{"Start":"03:04.690 ","End":"03:08.210","Text":"which is the formal charge of sulfur plus 0,"},{"Start":"03:08.210 ","End":"03:12.710","Text":"which is a formal charge of the upper oxygen plus minus 1,"},{"Start":"03:12.710 ","End":"03:17.360","Text":"which is the first lower oxygen plus another minus 1 for the second."},{"Start":"03:17.360 ","End":"03:20.405","Text":"This is going to give us 0."},{"Start":"03:20.405 ","End":"03:22.520","Text":"The sum of the formal charges equals 0,"},{"Start":"03:22.520 ","End":"03:28.305","Text":"and this is a neutral molecule. That\u0027s good."},{"Start":"03:28.305 ","End":"03:32.110","Text":"What we found is the formal charge again of sulfur."},{"Start":"03:32.110 ","End":"03:34.330","Text":"The formal charge of sulfur equals 2,"},{"Start":"03:34.330 ","End":"03:36.875","Text":"formal charge of the upper oxygen,"},{"Start":"03:36.875 ","End":"03:38.225","Text":"01 equals zero,"},{"Start":"03:38.225 ","End":"03:43.620","Text":"and the formal charge of the lower oxygens each equal minus 1."},{"Start":"03:43.640 ","End":"03:47.630","Text":"Now we\u0027re going to go on to c. In c,"},{"Start":"03:47.630 ","End":"03:49.790","Text":"we need to calculate the formal charge"},{"Start":"03:49.790 ","End":"03:52.820","Text":"of the nitrogen and the formal charge of the hydrogen."},{"Start":"03:52.820 ","End":"03:56.255","Text":"Let\u0027s begin with the nitrogen formal charge"},{"Start":"03:56.255 ","End":"03:59.870","Text":"equals again the number of valence electrons in the free atom."},{"Start":"03:59.870 ","End":"04:03.605","Text":"Now, remember that nitrogen is in the 15th group,"},{"Start":"04:03.605 ","End":"04:10.880","Text":"meaning that the valence electrons equals 5 minus the number of lone pair electrons."},{"Start":"04:10.880 ","End":"04:16.710","Text":"In our case, this is 0 minus 0.5 times the bond pair electrons."},{"Start":"04:16.710 ","End":"04:18.500","Text":"We have 4 bonds around the nitrogen,"},{"Start":"04:18.500 ","End":"04:21.110","Text":"so that\u0027s 8 electrons."},{"Start":"04:21.110 ","End":"04:27.150","Text":"This gives us 5 minus 4 which equals 1."},{"Start":"04:28.190 ","End":"04:33.870","Text":"The formal charge for nitrogen equals 1."},{"Start":"04:33.870 ","End":"04:38.100","Text":"Now we\u0027re going to calculate the formal charge for the hydrogens."},{"Start":"04:39.070 ","End":"04:41.390","Text":"Now, the hydrogen is in the first group,"},{"Start":"04:41.390 ","End":"04:42.560","Text":"so this equals 1,"},{"Start":"04:42.560 ","End":"04:44.555","Text":"so we have 1 valence electron"},{"Start":"04:44.555 ","End":"04:47.720","Text":"minus the number of lone pair electrons, which we have is 0."},{"Start":"04:47.720 ","End":"04:52.400","Text":"That\u0027s minus 0 minus 0.5 times the number of the bond pair electrons."},{"Start":"04:52.400 ","End":"04:54.570","Text":"Hydrogen is connected to 1 bond,"},{"Start":"04:54.570 ","End":"04:56.430","Text":"so that gives us 2 electrons."},{"Start":"04:56.430 ","End":"05:01.665","Text":"We get 1 minus 1, which equals 0."},{"Start":"05:01.665 ","End":"05:06.905","Text":"The formal charge at the hydrogen equals 0."},{"Start":"05:06.905 ","End":"05:11.450","Text":"Now let\u0027s sum up the formal charges and see that we get the charge of the ion."},{"Start":"05:11.450 ","End":"05:13.960","Text":"The formal charge on the nitrogen equals 1"},{"Start":"05:13.960 ","End":"05:19.105","Text":"plus 4 times the formal charge of hydrogen, which equals 0."},{"Start":"05:19.105 ","End":"05:21.610","Text":"This gives us a total of 1,"},{"Start":"05:21.610 ","End":"05:25.465","Text":"and that equals the charge on the ion, which is plus 1."},{"Start":"05:25.465 ","End":"05:27.470","Text":"The formal charge of nitrogen,"},{"Start":"05:27.470 ","End":"05:29.870","Text":"which we calculated equals 1 or plus 1,"},{"Start":"05:29.870 ","End":"05:33.740","Text":"and the formal charge of the hydrogen that we calculated equals 0."},{"Start":"05:33.740 ","End":"05:36.960","Text":"Now we\u0027re going to go on to d."}],"ID":24503},{"Watched":false,"Name":"Exercise 6 - part c","Duration":"4m 38s","ChapterTopicVideoID":23593,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:05.250","Text":"Now we\u0027re going to continue with d. We\u0027re going to start with our carbon."},{"Start":"00:05.250 ","End":"00:07.485","Text":"We\u0027re going to calculate the formal charge of the carbons."},{"Start":"00:07.485 ","End":"00:12.405","Text":"Again, carbon has 4 valence electrons minus the number of lone pair electrons."},{"Start":"00:12.405 ","End":"00:13.530","Text":"In this case, we have"},{"Start":"00:13.530 ","End":"00:18.135","Text":"0 lone pair electrons minus half times the number of bond pair electrons."},{"Start":"00:18.135 ","End":"00:20.430","Text":"Since carbon has 4 bonds around it,"},{"Start":"00:20.430 ","End":"00:24.210","Text":"it means that it has 8 electrons in the bonds."},{"Start":"00:24.210 ","End":"00:26.790","Text":"This equals 4 minus 4,"},{"Start":"00:26.790 ","End":"00:28.935","Text":"and this gives us a 0,"},{"Start":"00:28.935 ","End":"00:33.525","Text":"meaning that the formal charge for carbon equals 0."},{"Start":"00:33.525 ","End":"00:36.935","Text":"Next, we\u0027re going to calculate the formal charge for the hydrogen."},{"Start":"00:36.935 ","End":"00:41.600","Text":"The hydrogen is in the first group giving us 1 valence electron minus, as we can see,"},{"Start":"00:41.600 ","End":"00:43.790","Text":"there are 0 lone pair electrons around the hydrogen,"},{"Start":"00:43.790 ","End":"00:49.350","Text":"so there\u0027s going to be minus 0 minus 1/2 times the bond pair electrons."},{"Start":"00:49.350 ","End":"00:52.400","Text":"We can see there\u0027s 1 bond connected to the hydrogen,"},{"Start":"00:52.400 ","End":"00:54.725","Text":"so meaning there are 2 electrons in that bond,"},{"Start":"00:54.725 ","End":"01:03.145","Text":"so this gives us 1 minus 1 equals 0 so the formal charge for hydrogen also equals 0."},{"Start":"01:03.145 ","End":"01:05.570","Text":"Now we\u0027re going to go on to the fluorine."},{"Start":"01:05.570 ","End":"01:07.430","Text":"The fluorine is in 17th group giving"},{"Start":"01:07.430 ","End":"01:11.490","Text":"its 7 valence electrons minus the number of lone pair electrons."},{"Start":"01:11.490 ","End":"01:12.650","Text":"As we can see for each one,"},{"Start":"01:12.650 ","End":"01:15.214","Text":"we have 1, 2, 3, 4, 5,"},{"Start":"01:15.214 ","End":"01:21.060","Text":"6 lone pair electrons minus 1/2 times the bond pair electrons."},{"Start":"01:21.060 ","End":"01:23.480","Text":"Again, the fluorine is connected to 1 bond,"},{"Start":"01:23.480 ","End":"01:25.945","Text":"meaning we have 2 electrons in there."},{"Start":"01:25.945 ","End":"01:29.640","Text":"This gives us 7 minus 6 minus 1,"},{"Start":"01:29.640 ","End":"01:31.305","Text":"which equals 0,"},{"Start":"01:31.305 ","End":"01:35.495","Text":"so the formal charge for fluorine equals 0."},{"Start":"01:35.495 ","End":"01:38.570","Text":"Now as you can see, all the formal charges in our case,"},{"Start":"01:38.570 ","End":"01:42.350","Text":"for all the atoms equals 0 and the sum also equals 0."},{"Start":"01:42.350 ","End":"01:45.185","Text":"This makes sense because this is a neutral molecule,"},{"Start":"01:45.185 ","End":"01:49.510","Text":"so the sum of the formal charges equals zero."},{"Start":"01:49.510 ","End":"01:52.610","Text":"We found that the formal charge for carbon equals 0,"},{"Start":"01:52.610 ","End":"01:54.590","Text":"the formal charge for hydrogen equals 0,"},{"Start":"01:54.590 ","End":"01:57.380","Text":"and the formal charge for fluorine equals 0."},{"Start":"01:57.380 ","End":"02:00.935","Text":"Now we\u0027re going to go on to e. We\u0027re going to begin with carbon."},{"Start":"02:00.935 ","End":"02:05.125","Text":"The number of valence electrons equals 4 minus, again,"},{"Start":"02:05.125 ","End":"02:06.500","Text":"we have no lone pair electrons,"},{"Start":"02:06.500 ","End":"02:10.325","Text":"so it\u0027s minus 0 minus 1/2 times the bond pair electrons."},{"Start":"02:10.325 ","End":"02:12.110","Text":"Carbon has 4 bonds around it,"},{"Start":"02:12.110 ","End":"02:14.425","Text":"meaning 8 electrons in its bonds,"},{"Start":"02:14.425 ","End":"02:16.470","Text":"which equals 4 minus 4,"},{"Start":"02:16.470 ","End":"02:17.940","Text":"which equals 0,"},{"Start":"02:17.940 ","End":"02:21.615","Text":"so the formal charge for carbon equals 0."},{"Start":"02:21.615 ","End":"02:24.680","Text":"Next, we\u0027re going to calculate the formal charge for sulfur,"},{"Start":"02:24.680 ","End":"02:27.625","Text":"so we\u0027re going to start with 6 valence electrons."},{"Start":"02:27.625 ","End":"02:29.705","Text":"Sulfur is in the 16th group,"},{"Start":"02:29.705 ","End":"02:32.030","Text":"minus the lone pair electrons,"},{"Start":"02:32.030 ","End":"02:34.380","Text":"we have around 8 sulfur atom."},{"Start":"02:34.380 ","End":"02:35.796","Text":"We have 1, 2, 3,"},{"Start":"02:35.796 ","End":"02:40.550","Text":"4 lone pair electrons minus 1/2 of the bond pair electrons."},{"Start":"02:40.550 ","End":"02:44.185","Text":"Since we have 2 bonds next to each sulfur."},{"Start":"02:44.185 ","End":"02:48.785","Text":"That\u0027s 4 electrons because in each bond again we have 2 electrons,"},{"Start":"02:48.785 ","End":"02:52.745","Text":"so this equals 6 minus 4 minus 2,"},{"Start":"02:52.745 ","End":"02:54.650","Text":"which gives us 0."},{"Start":"02:54.650 ","End":"02:58.990","Text":"Here the formal charge for sulfur equals 0."},{"Start":"02:59.030 ","End":"03:02.950","Text":"Again the sum of the formal charges equals 0,"},{"Start":"03:02.950 ","End":"03:06.240","Text":"and we have a neutral molecule,"},{"Start":"03:06.240 ","End":"03:07.920","Text":"so that makes a lot of sense."},{"Start":"03:07.920 ","End":"03:10.190","Text":"The formal charge that we found for carbon equals"},{"Start":"03:10.190 ","End":"03:13.760","Text":"0 and the formal charge that we found for sulfur equals 0."},{"Start":"03:13.760 ","End":"03:19.370","Text":"Now we\u0027re going to go on to f. Again,"},{"Start":"03:19.370 ","End":"03:20.450","Text":"we\u0027re going to start with carbon,"},{"Start":"03:20.450 ","End":"03:23.270","Text":"which begins with 4 valence electrons minus,"},{"Start":"03:23.270 ","End":"03:27.965","Text":"we see that we have 2 lone pair electrons minus 1/2 times."},{"Start":"03:27.965 ","End":"03:29.990","Text":"We have 3 bonds,"},{"Start":"03:29.990 ","End":"03:33.690","Text":"meaning 6 bond pair electrons,"},{"Start":"03:33.690 ","End":"03:36.480","Text":"and this equals 4 minus 2,"},{"Start":"03:36.480 ","End":"03:40.335","Text":"minus 3, and this is going to give us a minus 1,"},{"Start":"03:40.335 ","End":"03:45.810","Text":"so the formal charge for the carbon equals minus 1."},{"Start":"03:46.240 ","End":"03:49.770","Text":"Now we\u0027re going to look at our nitrogen."},{"Start":"03:50.140 ","End":"03:53.510","Text":"In this case where we\u0027re beginning with 5 valence electrons,"},{"Start":"03:53.510 ","End":"03:55.295","Text":"since the nitrogen is in the 15th group,"},{"Start":"03:55.295 ","End":"03:58.490","Text":"then we\u0027re going to take off the 2 lone pair electrons"},{"Start":"03:58.490 ","End":"04:04.220","Text":"minus 1/2 of the bond pair electrons,"},{"Start":"04:04.220 ","End":"04:05.990","Text":"and nitrogen is connected to 3 bonds,"},{"Start":"04:05.990 ","End":"04:08.480","Text":"so that\u0027s also 6 electrons,"},{"Start":"04:08.480 ","End":"04:14.030","Text":"so that gives us a total of 5 minus 2 minus 3, giving us 0."},{"Start":"04:14.030 ","End":"04:18.805","Text":"The formal charge for nitrogen equals 0."},{"Start":"04:18.805 ","End":"04:22.640","Text":"Now, if we sum up the formal charges we have minus 1 plus 0,"},{"Start":"04:22.640 ","End":"04:23.930","Text":"and that equals minus 1,"},{"Start":"04:23.930 ","End":"04:26.390","Text":"which is the charge of the ion."},{"Start":"04:26.390 ","End":"04:29.930","Text":"The formal charges that we found for carbon equals minus 1"},{"Start":"04:29.930 ","End":"04:33.770","Text":"and the formal charges that we found for nitrogen equals 0."},{"Start":"04:33.940 ","End":"04:36.140","Text":"Those are our final answers."},{"Start":"04:36.140 ","End":"04:38.670","Text":"Thank you very much for watching."}],"ID":24504},{"Watched":false,"Name":"Formal Charge","Duration":"6m 50s","ChapterTopicVideoID":20279,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.905","Text":"In the previous video,"},{"Start":"00:01.905 ","End":"00:05.385","Text":"we learned how to write plausible Lewis structures."},{"Start":"00:05.385 ","End":"00:08.714","Text":"Sometimes there\u0027s more than one possible structure."},{"Start":"00:08.714 ","End":"00:12.885","Text":"In this video, we\u0027ll learn how to decide which one is more probable."},{"Start":"00:12.885 ","End":"00:14.490","Text":"In order to do so,"},{"Start":"00:14.490 ","End":"00:18.015","Text":"we use a device called the formal charge."},{"Start":"00:18.015 ","End":"00:22.290","Text":"The formal charge is defined as the number of electrons in"},{"Start":"00:22.290 ","End":"00:24.690","Text":"a free atom minus the number of electrons"},{"Start":"00:24.690 ","End":"00:28.215","Text":"assigned to the atom where we draw the Lewis structure."},{"Start":"00:28.215 ","End":"00:30.420","Text":"It\u0027s the number of electrons in the free atom"},{"Start":"00:30.420 ","End":"00:34.500","Text":"minus the number of electrons in lone pair,"},{"Start":"00:34.500 ","End":"00:37.575","Text":"plus half the number of electrons in a bond pair."},{"Start":"00:37.575 ","End":"00:42.390","Text":"In other words, we calculate the number of electrons in the free atom."},{"Start":"00:42.390 ","End":"00:50.165","Text":"From that, we take off 2 electrons for each lone pair and 1 electron for each bond pair."},{"Start":"00:50.165 ","End":"00:52.310","Text":"Now, it should be emphasized,"},{"Start":"00:52.310 ","End":"00:54.320","Text":"this is not a real charge."},{"Start":"00:54.320 ","End":"00:57.110","Text":"But it indicates to what extent electrons in"},{"Start":"00:57.110 ","End":"01:01.555","Text":"each atom have been reassigned in forming the structure."},{"Start":"01:01.555 ","End":"01:06.815","Text":"Also, it doesn\u0027t take into account that if we have 2 atoms;"},{"Start":"01:06.815 ","End":"01:10.190","Text":"A and B and a bond pair between them,"},{"Start":"01:10.190 ","End":"01:17.795","Text":"it may be that one and a half electrons belongs to B and only half to A."},{"Start":"01:17.795 ","End":"01:23.090","Text":"Here, it\u0027s assumed that each atom has 1 electron,"},{"Start":"01:23.090 ","End":"01:27.455","Text":"so the atoms equally share the bond pair."},{"Start":"01:27.455 ","End":"01:29.389","Text":"Let\u0027s take an example,"},{"Start":"01:29.389 ","End":"01:37.175","Text":"calculate the formal charges of the atoms in H_2NOH and H_2ONH."},{"Start":"01:37.175 ","End":"01:39.215","Text":"After we\u0027ve done this,"},{"Start":"01:39.215 ","End":"01:46.970","Text":"we\u0027ll see that one of these skeletal structures is more reasonable than the other."},{"Start":"01:46.970 ","End":"01:49.840","Text":"Here we have H, N,"},{"Start":"01:49.840 ","End":"01:54.185","Text":"O, H, and another hydrogen, and a nitrogen."},{"Start":"01:54.185 ","End":"01:58.205","Text":"We have a single bond between the nitrogen and this hydrogen."},{"Start":"01:58.205 ","End":"02:02.980","Text":"Here we have a single bond between the oxygen and the hydrogen."},{"Start":"02:02.980 ","End":"02:07.280","Text":"Let\u0027s calculate the formal charge on each of the atoms."},{"Start":"02:07.280 ","End":"02:09.935","Text":"Let\u0027s start off with the hydrogens."},{"Start":"02:09.935 ","End":"02:13.865","Text":"Each hydrogen has 1 electron."},{"Start":"02:13.865 ","End":"02:16.800","Text":"Here, it has 1 bond pair."},{"Start":"02:16.800 ","End":"02:19.650","Text":"We\u0027ve only to take half of that 1 electron,"},{"Start":"02:19.650 ","End":"02:22.995","Text":"so 1-1 is 0."},{"Start":"02:22.995 ","End":"02:26.839","Text":"The formal charge of hydrogen is 0."},{"Start":"02:26.839 ","End":"02:32.245","Text":"That\u0027s true for all the hydrogens in both a and in b."},{"Start":"02:32.245 ","End":"02:34.395","Text":"Now let\u0027s go to nitrogen."},{"Start":"02:34.395 ","End":"02:39.695","Text":"Nitrogen, we know has 5 electrons in the free atom."},{"Start":"02:39.695 ","End":"02:45.110","Text":"Now let\u0027s look in a and see how many electrons we have to take off."},{"Start":"02:45.110 ","End":"02:49.220","Text":"We take 2 for the lone pair and 1 each,"},{"Start":"02:49.220 ","End":"02:52.015","Text":"for each of these 3 bonds."},{"Start":"02:52.015 ","End":"02:54.360","Text":"In other words, we have 2+3,"},{"Start":"02:54.360 ","End":"02:58.740","Text":"that\u0027s 5, so 5-5 is 0."},{"Start":"02:58.740 ","End":"03:02.975","Text":"In a, the formal charge on nitrogen is 0."},{"Start":"03:02.975 ","End":"03:04.805","Text":"Let\u0027s look at b."},{"Start":"03:04.805 ","End":"03:07.370","Text":"Again, we have 5 electrons."},{"Start":"03:07.370 ","End":"03:11.660","Text":"We have 2 lone pairs so we have 2 electrons for each lone pair,"},{"Start":"03:11.660 ","End":"03:14.644","Text":"that\u0027s 4, and we have 2 bond pairs,"},{"Start":"03:14.644 ","End":"03:16.705","Text":"that\u0027s 5, 6."},{"Start":"03:16.705 ","End":"03:21.195","Text":"We have 5 - 6 = -1."},{"Start":"03:21.195 ","End":"03:25.690","Text":"The formal charge on the nitrogen here is -1."},{"Start":"03:25.690 ","End":"03:27.975","Text":"Now let\u0027s look at oxygen."},{"Start":"03:27.975 ","End":"03:33.800","Text":"We know that oxygen has 6 electrons. Let\u0027s look at a."},{"Start":"03:33.800 ","End":"03:36.605","Text":"We have 6 electrons in the free atom."},{"Start":"03:36.605 ","End":"03:39.500","Text":"We have to take off 2 for each lone pair,"},{"Start":"03:39.500 ","End":"03:43.070","Text":"that\u0027s 4 and 1 for each bond pair."},{"Start":"03:43.070 ","End":"03:45.000","Text":"That gives us another 2,"},{"Start":"03:45.000 ","End":"03:46.410","Text":"a total of 6."},{"Start":"03:46.410 ","End":"03:49.185","Text":"We have 6 - 6 = 0."},{"Start":"03:49.185 ","End":"03:53.390","Text":"The formal charge of oxygen in a is 0."},{"Start":"03:53.390 ","End":"03:55.405","Text":"Now let\u0027s look at b."},{"Start":"03:55.405 ","End":"04:00.075","Text":"We know that oxygen has 6 electrons in free atom."},{"Start":"04:00.075 ","End":"04:07.785","Text":"Now we have 2 electrons for the lone pair and 1 electron for each of the bond pairs."},{"Start":"04:07.785 ","End":"04:09.345","Text":"That\u0027s another 3."},{"Start":"04:09.345 ","End":"04:11.175","Text":"That\u0027s a total of 5."},{"Start":"04:11.175 ","End":"04:15.285","Text":"We have 6 - 5 = 1."},{"Start":"04:15.285 ","End":"04:19.355","Text":"The formal charge on the oxygen is +1."},{"Start":"04:19.355 ","End":"04:24.410","Text":"Now we\u0027re going to learn some rules that allow us to decide which is"},{"Start":"04:24.410 ","End":"04:33.200","Text":"the more plausible skeletal structure HNHOH, or HOHNH."},{"Start":"04:33.200 ","End":"04:38.720","Text":"This is just a summary of how we calculate the formal charge."},{"Start":"04:38.720 ","End":"04:40.760","Text":"Before we go to the rules,"},{"Start":"04:40.760 ","End":"04:44.690","Text":"we should note that the electronegativity of nitrogen is"},{"Start":"04:44.690 ","End":"04:49.880","Text":"3 and the electronegativity of oxygen is 3.5."},{"Start":"04:49.880 ","End":"04:53.345","Text":"This will be relevant when we test the rules."},{"Start":"04:53.345 ","End":"04:57.110","Text":"The first rule is the sum of the formal charges is"},{"Start":"04:57.110 ","End":"05:02.485","Text":"0 or equal to the charge if we\u0027re talking about an ion."},{"Start":"05:02.485 ","End":"05:08.790","Text":"Here we can see that we have 0 on all the atoms in a,"},{"Start":"05:08.790 ","End":"05:10.185","Text":"so the total is 0."},{"Start":"05:10.185 ","End":"05:11.955","Text":"We have here in b,"},{"Start":"05:11.955 ","End":"05:14.115","Text":"0 for the hydrogens,"},{"Start":"05:14.115 ","End":"05:15.855","Text":"+ 1 for the oxygen,"},{"Start":"05:15.855 ","End":"05:17.835","Text":"-1 one for the nitrogen,"},{"Start":"05:17.835 ","End":"05:19.350","Text":"so the sum is 0."},{"Start":"05:19.350 ","End":"05:22.755","Text":"We have 1 - 1= 0."},{"Start":"05:22.755 ","End":"05:24.425","Text":"That is a bead."},{"Start":"05:24.425 ","End":"05:28.750","Text":"The second rule is that the formal charges should be as small as"},{"Start":"05:28.750 ","End":"05:35.005","Text":"possible because the most stable structure has the least reassignment of electrons."},{"Start":"05:35.005 ","End":"05:39.700","Text":"We see that the smallest formal charges are in a,"},{"Start":"05:39.700 ","End":"05:42.320","Text":"where all the formal charges are 0."},{"Start":"05:42.320 ","End":"05:44.110","Text":"Now the third rule is that"},{"Start":"05:44.110 ","End":"05:49.870","Text":"negative formal charges are on the most electronegative atoms,"},{"Start":"05:49.870 ","End":"05:55.455","Text":"and the positive formal charge on the least electronegative atoms."},{"Start":"05:55.455 ","End":"06:00.970","Text":"We see that the most electronegative atom here is O,"},{"Start":"06:00.970 ","End":"06:04.689","Text":"so O should have a negative formal charge."},{"Start":"06:04.689 ","End":"06:07.640","Text":"We see that that\u0027s not the case,"},{"Start":"06:07.640 ","End":"06:12.110","Text":"O has a positive formal charge in b."},{"Start":"06:12.110 ","End":"06:18.320","Text":"Finally, structures with the same sign of formal charge on adjacent atoms are unlikely."},{"Start":"06:18.320 ","End":"06:23.910","Text":"We couldn\u0027t have minus 1 beside minus 1 or plus 1 beside plus 1."},{"Start":"06:23.910 ","End":"06:25.940","Text":"On the basis of all these rules,"},{"Start":"06:25.940 ","End":"06:32.960","Text":"we have to decide which is more probable, H2NOH or H2ONH."},{"Start":"06:32.960 ","End":"06:38.584","Text":"Since H2NOH has all the formal charges is equal to 0,"},{"Start":"06:38.584 ","End":"06:42.040","Text":"we see that that is a more plausible one."},{"Start":"06:42.040 ","End":"06:45.705","Text":"Clearly, the answer is H2NOH."},{"Start":"06:45.705 ","End":"06:50.520","Text":"In this video, we learned about formal charge."}],"ID":21070},{"Watched":false,"Name":"Exercise 7 - part a","Duration":"5m 11s","ChapterTopicVideoID":23604,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:03.779","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.779 ","End":"00:06.420","Text":"State what is wrong with the following structures."},{"Start":"00:06.420 ","End":"00:10.000","Text":"Write acceptable Lewis structures for each structure."},{"Start":"00:10.460 ","End":"00:14.800","Text":"We\u0027re going to start with a and a, we have calcium iodide."},{"Start":"00:14.960 ","End":"00:17.010","Text":"As you can see,"},{"Start":"00:17.010 ","End":"00:19.185","Text":"we have a covalent Lewis structure."},{"Start":"00:19.185 ","End":"00:23.760","Text":"However, remember that calcium is a metal and iodine is a non-metal."},{"Start":"00:23.760 ","End":"00:26.910","Text":"Therefore, we would expect them to have an ionic Lewis structure."},{"Start":"00:26.910 ","End":"00:32.010","Text":"We\u0027re going to write the ionic Lewis structure right now for calcium iodide."},{"Start":"00:32.010 ","End":"00:34.265","Text":"Calcium is in the second group,"},{"Start":"00:34.265 ","End":"00:37.770","Text":"meaning it gives up 2 electrons to the iodine."},{"Start":"00:38.060 ","End":"00:42.240","Text":"We\u0027re going to write the calcium cation as calcium plus 2."},{"Start":"00:42.240 ","End":"00:45.290","Text":"Now, it gives these electrons to the iodine."},{"Start":"00:45.290 ","End":"00:47.300","Text":"Let\u0027s take a look at the iodine."},{"Start":"00:47.300 ","End":"00:49.850","Text":"Remember, iodine is in the 17th group,"},{"Start":"00:49.850 ","End":"00:51.710","Text":"meaning it has 7 valence electrons."},{"Start":"00:51.710 ","End":"00:54.105","Text":"We\u0027re going to draw them here. 1, 2,"},{"Start":"00:54.105 ","End":"00:56.880","Text":"3, 4, 5, 6, 7."},{"Start":"00:56.880 ","End":"00:59.510","Text":"Now again, it has 7 valence electrons,"},{"Start":"00:59.510 ","End":"01:02.375","Text":"meaning it\u0027s missing 1 electron for an octet."},{"Start":"01:02.375 ","End":"01:06.750","Text":"Each iodine is going to accept"},{"Start":"01:06.750 ","End":"01:13.170","Text":"1 electron from calcium and that\u0027s going to make this the iodide anion."},{"Start":"01:13.170 ","End":"01:17.190","Text":"Since each iodine accept only 1 electron,"},{"Start":"01:17.190 ","End":"01:19.660","Text":"we\u0027re going to have 2 of these."},{"Start":"01:20.390 ","End":"01:28.150","Text":"This is the ionic Lewis structure for calcium iodide."},{"Start":"01:28.220 ","End":"01:35.790","Text":"now, just to mention, you can also write this as calcium cation."},{"Start":"01:36.230 ","End":"01:39.120","Text":"Instead of writing the 2 down here,"},{"Start":"01:39.120 ","End":"01:48.190","Text":"you can write 2 times the iodine anion, the iodide."},{"Start":"01:48.520 ","End":"01:51.935","Text":"Sometimes the calcium cation is written in the middle,"},{"Start":"01:51.935 ","End":"01:56.605","Text":"and the iodine counts from right and left."},{"Start":"01:56.605 ","End":"02:00.445","Text":"Now, we\u0027ve finished with a and we\u0027re going to go on to b."},{"Start":"02:00.445 ","End":"02:05.260","Text":"If you look at b, we have to state what is wrong with this Lewis structure."},{"Start":"02:05.260 ","End":"02:11.050","Text":"We can immediately see that we have 1 of the hydrogens with 2 bonds around it,"},{"Start":"02:11.050 ","End":"02:13.200","Text":"and it\u0027s written as a central atom."},{"Start":"02:13.200 ","End":"02:15.700","Text":"Remember, hydrogen can only be"},{"Start":"02:15.700 ","End":"02:20.335","Text":"a terminal atom since it can only have 1 bond, 2 electrons."},{"Start":"02:20.335 ","End":"02:22.750","Text":"Therefore, this hydrogen,"},{"Start":"02:22.750 ","End":"02:25.250","Text":"which is connected to 2 bonds,"},{"Start":"02:25.250 ","End":"02:28.040","Text":"cannot be in this place."},{"Start":"02:28.040 ","End":"02:33.040","Text":"Now, let\u0027s write the Lewis structure for b."},{"Start":"02:33.040 ","End":"02:34.915","Text":"First of all, we have 1 carbon,"},{"Start":"02:34.915 ","End":"02:37.180","Text":"meaning we have 4 valence electrons."},{"Start":"02:37.180 ","End":"02:41.300","Text":"We have 4 valence electrons."},{"Start":"02:41.300 ","End":"02:42.890","Text":"If we look at our hydrogens, we have 1,"},{"Start":"02:42.890 ","End":"02:44.330","Text":"2, 3, 4 hydrogens."},{"Start":"02:44.330 ","End":"02:46.850","Text":"That\u0027s plus 4, times 1,"},{"Start":"02:46.850 ","End":"02:50.630","Text":"since hydrogen is in the first group it has 1 valence electron,"},{"Start":"02:50.630 ","End":"02:56.250","Text":"plus oxygen, which has 6 valence electrons in the 16th group."},{"Start":"02:56.250 ","End":"02:58.520","Text":"Remember, that\u0027s the first step of writing"},{"Start":"02:58.520 ","End":"03:02.150","Text":"Lewis structure to find how many valence electrons we have."},{"Start":"03:02.150 ","End":"03:08.520","Text":"Here we have 4 plus 4 plus 6 giving us a total of 14 electrons."},{"Start":"03:08.520 ","End":"03:14.198","Text":"The second step in writing Lewis structure is to decide which are the central,"},{"Start":"03:14.198 ","End":"03:18.960","Text":"and terminal atoms in writing a skeletal structure."},{"Start":"03:18.960 ","End":"03:23.510","Text":"Of course, our carbon is going to be central and our hydrogens,"},{"Start":"03:23.510 ","End":"03:25.600","Text":"as we said, are going to be terminal."},{"Start":"03:25.600 ","End":"03:30.490","Text":"What we\u0027re going to do is we\u0027re going to connect 3 hydrogens to the carbon."},{"Start":"03:32.200 ","End":"03:35.900","Text":"Here we have to connect our oxygen because if we"},{"Start":"03:35.900 ","End":"03:39.455","Text":"connect another hydrogen we won\u0027t be able to connect the oxygen."},{"Start":"03:39.455 ","End":"03:43.145","Text":"The last hydrogen is going to be connected to the oxygen."},{"Start":"03:43.145 ","End":"03:45.020","Text":"Now that we have our skeletal structure,"},{"Start":"03:45.020 ","End":"03:46.460","Text":"we\u0027re going to count the number of bonds we have."},{"Start":"03:46.460 ","End":"03:47.630","Text":"We have 1, 2,"},{"Start":"03:47.630 ","End":"03:49.475","Text":"3, 4, 5 bonds."},{"Start":"03:49.475 ","End":"03:51.620","Text":"Remember each bond contains 2 electrons,"},{"Start":"03:51.620 ","End":"03:53.600","Text":"so there are 10 electrons in these bonds."},{"Start":"03:53.600 ","End":"03:55.790","Text":"The 3rd step for"},{"Start":"03:55.790 ","End":"03:58.850","Text":"the Lewis structure is to take the total number of valence electrons we have,"},{"Start":"03:58.850 ","End":"03:59.990","Text":"which is 14,"},{"Start":"03:59.990 ","End":"04:03.380","Text":"minus the number of electrons which are placed in the bonds."},{"Start":"04:03.380 ","End":"04:07.360","Text":"That\u0027s minus 10 electrons and we\u0027re left with 4."},{"Start":"04:07.360 ","End":"04:12.530","Text":"Now, at this point we need to take these electrons and we need to deal with"},{"Start":"04:12.530 ","End":"04:17.910","Text":"our terminal atoms first and add electrons where we need,"},{"Start":"04:17.910 ","End":"04:19.880","Text":"and then we have to go to the central atom."},{"Start":"04:19.880 ","End":"04:21.230","Text":"If we look at our terminal atoms,"},{"Start":"04:21.230 ","End":"04:23.300","Text":"we have the hydrogens."},{"Start":"04:23.300 ","End":"04:25.984","Text":"As we said, they only need 2 electrons."},{"Start":"04:25.984 ","End":"04:28.970","Text":"Therefore, we\u0027re not going to add any electrons to them."},{"Start":"04:28.970 ","End":"04:30.590","Text":"If we look at our carbon,"},{"Start":"04:30.590 ","End":"04:33.230","Text":"and we can see that there\u0027s already an octet around the carbon,"},{"Start":"04:33.230 ","End":"04:37.805","Text":"the only atom that\u0027s missing electrons is the oxygen."},{"Start":"04:37.805 ","End":"04:39.215","Text":"It doesn\u0027t have an octet."},{"Start":"04:39.215 ","End":"04:40.730","Text":"We have 4 electrons left,"},{"Start":"04:40.730 ","End":"04:42.730","Text":"so we\u0027re going to add 2,"},{"Start":"04:42.730 ","End":"04:45.660","Text":"4 electrons to the oxygen."},{"Start":"04:45.660 ","End":"04:49.280","Text":"Now, as you can see now the oxygen also has an octet."},{"Start":"04:49.280 ","End":"04:52.250","Text":"This Lewis structure is definitely satisfactory"},{"Start":"04:52.250 ","End":"04:54.515","Text":"because the hydrogens all have 2 electrons."},{"Start":"04:54.515 ","End":"04:57.630","Text":"The carbon and the oxygen have octets."},{"Start":"05:07.510 ","End":"05:11.550","Text":"That\u0027s our answer for b. Now let\u0027s go on to c."}],"ID":24515},{"Watched":false,"Name":"Exercise 7 - part b","Duration":"4m 36s","ChapterTopicVideoID":23605,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.830 ","End":"00:04.110","Text":"Now we\u0027re going to go on to (c)."},{"Start":"00:04.400 ","End":"00:07.395","Text":"If we look at the Lewis structure in c,"},{"Start":"00:07.395 ","End":"00:09.840","Text":"we can see that our carbon has an octet."},{"Start":"00:09.840 ","End":"00:13.230","Text":"However, our oxygens have 6 electrons around them."},{"Start":"00:13.230 ","End":"00:15.449","Text":"1 pair of lone electrons,"},{"Start":"00:15.449 ","End":"00:18.390","Text":"lone pair of electrons, and 2 bonds,"},{"Start":"00:18.390 ","End":"00:21.765","Text":"meaning they have 6 electrons around them instead of an octet."},{"Start":"00:21.765 ","End":"00:23.640","Text":"That\u0027s the problem with our Lewis structure."},{"Start":"00:23.640 ","End":"00:26.655","Text":"Now let\u0027s write a Lewis structure."},{"Start":"00:26.655 ","End":"00:29.190","Text":"Again, our first step for writing Lewis structures is just to"},{"Start":"00:29.190 ","End":"00:31.440","Text":"count the number of valence electrons."},{"Start":"00:31.440 ","End":"00:34.530","Text":"So carbon, which is in the 14th period,"},{"Start":"00:34.530 ","End":"00:37.470","Text":"is going to give us 4 valence electrons and oxygen is in"},{"Start":"00:37.470 ","End":"00:42.290","Text":"the 6th periods so that\u0027s going to give us 6 valence electrons and we have 2 oxygens."},{"Start":"00:42.290 ","End":"00:45.245","Text":"So 4 plus 2 times 6."},{"Start":"00:45.245 ","End":"00:48.849","Text":"This gives us 16 electrons to begin with."},{"Start":"00:48.849 ","End":"00:51.995","Text":"Now, we have to write the skeletal structure."},{"Start":"00:51.995 ","End":"00:55.835","Text":"Of course, the carbon is going to be in the middle, a central atom,"},{"Start":"00:55.835 ","End":"01:00.260","Text":"and both of the oxygens are going to be to the sides of"},{"Start":"01:00.260 ","End":"01:05.760","Text":"the carbon since they\u0027re more electronegative."},{"Start":"01:05.760 ","End":"01:08.840","Text":"We have 2 bonds, so we have to subtract the number of"},{"Start":"01:08.840 ","End":"01:12.450","Text":"electrons from the total number of valence electrons."},{"Start":"01:12.450 ","End":"01:14.150","Text":"We start with the total number of valence electrons,"},{"Start":"01:14.150 ","End":"01:16.820","Text":"which is 16, and we subtract the number of electrons in the bonds."},{"Start":"01:16.820 ","End":"01:19.535","Text":"So we have 2 bonds, that means 4 electrons,"},{"Start":"01:19.535 ","End":"01:23.400","Text":"and we\u0027re left with 12 electrons."},{"Start":"01:23.680 ","End":"01:28.925","Text":"So at this point, we fill in octets for our terminal oxygen."},{"Start":"01:28.925 ","End":"01:33.989","Text":"We\u0027re going to add 6 electrons to each oxygen."},{"Start":"01:34.930 ","End":"01:39.480","Text":"Now if you look, you can see that our oxygens have octets."},{"Start":"01:39.590 ","End":"01:42.030","Text":"However, our carbon does not."},{"Start":"01:42.030 ","End":"01:46.955","Text":"What we\u0027re going to do is we\u0027re going to create multiple bonds here."},{"Start":"01:46.955 ","End":"01:49.540","Text":"We\u0027re going to take 1 pair of electrons and move them in and"},{"Start":"01:49.540 ","End":"01:53.925","Text":"we\u0027re going to write the new Lewis structure here."},{"Start":"01:53.925 ","End":"01:56.915","Text":"We have an oxygen, we have 4 electrons left."},{"Start":"01:56.915 ","End":"01:59.630","Text":"We have a double bond, 2 carbon,"},{"Start":"01:59.630 ","End":"02:01.280","Text":"and as you can see on the other side,"},{"Start":"02:01.280 ","End":"02:03.720","Text":"we still have 1 bond."},{"Start":"02:03.720 ","End":"02:06.050","Text":"What we\u0027re going to do, we\u0027re going to do the same thing on the other side,"},{"Start":"02:06.050 ","End":"02:08.060","Text":"because if we leave it with 1 bond,"},{"Start":"02:08.060 ","End":"02:12.035","Text":"the carbon is only going to have 6 electrons."},{"Start":"02:12.035 ","End":"02:14.600","Text":"We have a double bond on this side too,"},{"Start":"02:14.600 ","End":"02:20.279","Text":"since we moved the electrons and our oxygen has 4 electrons."},{"Start":"02:20.560 ","End":"02:23.150","Text":"Now if we look at a Lewis structure,"},{"Start":"02:23.150 ","End":"02:29.820","Text":"we can see that our carbon has an octet and our oxygens each have an octet."},{"Start":"02:29.820 ","End":"02:33.560","Text":"This is an acceptable Lewis structure."},{"Start":"02:34.280 ","End":"02:37.600","Text":"Now we\u0027re going to go on to (d)."},{"Start":"02:37.600 ","End":"02:41.255","Text":"If we look at our Lewis structure in d,"},{"Start":"02:41.255 ","End":"02:45.845","Text":"we can immediately see the fluorine is placed as a central atom."},{"Start":"02:45.845 ","End":"02:52.990","Text":"The fluorine should be a terminal atom since it has very very high electronegativity."},{"Start":"02:52.990 ","End":"02:56.060","Text":"Then nitrogen should be our central atom."},{"Start":"02:56.060 ","End":"02:58.145","Text":"Now we\u0027re going to write our own Lewis structure."},{"Start":"02:58.145 ","End":"03:01.430","Text":"In the first step of writing Lewis structures,"},{"Start":"03:01.430 ","End":"03:04.040","Text":"we\u0027re going to count the number of valence electrons."},{"Start":"03:04.040 ","End":"03:05.990","Text":"The nitrogen is in the 15th group,"},{"Start":"03:05.990 ","End":"03:08.285","Text":"meaning it has 5 valence electrons,"},{"Start":"03:08.285 ","End":"03:11.450","Text":"and the fluorines are in the 17th group,"},{"Start":"03:11.450 ","End":"03:13.980","Text":"meaning they have 7 valence electrons,"},{"Start":"03:13.980 ","End":"03:15.590","Text":"and we\u0027re going to multiply this by 3,"},{"Start":"03:15.590 ","End":"03:17.930","Text":"since we have 3 fluorines."},{"Start":"03:17.930 ","End":"03:22.640","Text":"This is going to give us 26 electrons in all."},{"Start":"03:22.640 ","End":"03:28.320","Text":"The second step of our Lewis structure is to write our skeletal structure."},{"Start":"03:29.770 ","End":"03:36.575","Text":"Again, we\u0027re going to place the nitrogen in the middle and as the central atom,"},{"Start":"03:36.575 ","End":"03:41.855","Text":"and place the fluorines as terminal atoms on the outside."},{"Start":"03:41.855 ","End":"03:43.460","Text":"Now we have 3 bonds,"},{"Start":"03:43.460 ","End":"03:49.385","Text":"meaning 6 electrons so we\u0027re going to subtract 6 from 26."},{"Start":"03:49.385 ","End":"03:52.565","Text":"This gives us 20 electrons leftover."},{"Start":"03:52.565 ","End":"03:56.930","Text":"At first, we\u0027re going to add 6 electrons, to each fluorine."},{"Start":"03:56.930 ","End":"04:05.880","Text":"They have octets, and we used up 18 electrons,"},{"Start":"04:05.880 ","End":"04:07.575","Text":"6 times 3 equals 18."},{"Start":"04:07.575 ","End":"04:11.045","Text":"We have 2 electrons left since we have a total of 20 electrons,"},{"Start":"04:11.045 ","End":"04:14.920","Text":"and these are going to go all on nitrogen our central atom."},{"Start":"04:14.920 ","End":"04:17.120","Text":"Now, if we look at our Lewis structure,"},{"Start":"04:17.120 ","End":"04:21.620","Text":"we can see that the nitrogen has an octet and the fluorines all have octets also."},{"Start":"04:21.620 ","End":"04:24.800","Text":"This is an acceptable Lewis structure,"},{"Start":"04:24.800 ","End":"04:27.175","Text":"and that\u0027s our answer for (d)."},{"Start":"04:27.175 ","End":"04:30.220","Text":"Now, let\u0027s take a look at (e)."}],"ID":24516},{"Watched":false,"Name":"Exercise 7 - part c","Duration":"3m 45s","ChapterTopicVideoID":23606,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.655","Text":"Now, we\u0027re going on to E. In E,"},{"Start":"00:02.655 ","End":"00:06.660","Text":"we can see immediately that our carbons are missing octets."},{"Start":"00:06.660 ","End":"00:08.520","Text":"They only have 6 electrons around them."},{"Start":"00:08.520 ","End":"00:12.870","Text":"They have 3 bonds each around them containing 6 electrons."},{"Start":"00:12.870 ","End":"00:15.690","Text":"Therefore, they don\u0027t have the octet that they need."},{"Start":"00:15.690 ","End":"00:20.760","Text":"Let\u0027s write an acceptable Lewis structure for E. Again,"},{"Start":"00:20.760 ","End":"00:24.825","Text":"first step is to count the number of valence electrons we have."},{"Start":"00:24.825 ","End":"00:25.890","Text":"We have 2 carbons,"},{"Start":"00:25.890 ","End":"00:28.920","Text":"meaning that\u0027s 2 times 4 valence electrons,"},{"Start":"00:28.920 ","End":"00:32.475","Text":"again, carbon is in the 14th group."},{"Start":"00:32.475 ","End":"00:34.590","Text":"Now, we\u0027re going to add the hydrogen,"},{"Start":"00:34.590 ","End":"00:36.165","Text":"so we have 1, 2, 3, 4."},{"Start":"00:36.165 ","End":"00:40.985","Text":"4 hydrogens and each one has 1 valence electron since they\u0027re in the first group."},{"Start":"00:40.985 ","End":"00:43.450","Text":"This is going to give us 8 plus 4,"},{"Start":"00:43.450 ","End":"00:45.830","Text":"so it\u0027s going to give us 12 electrons."},{"Start":"00:45.830 ","End":"00:49.760","Text":"The second step is to write a skeletal structure."},{"Start":"00:49.760 ","End":"00:53.584","Text":"The skeletal structure that we have in E looks correct."},{"Start":"00:53.584 ","End":"00:58.650","Text":"The carbons are the central atoms and the hydrogens are the terminal atoms."},{"Start":"00:59.110 ","End":"01:03.580","Text":"Let\u0027s look how many bonds we have in our skeletal structure."},{"Start":"01:03.580 ","End":"01:05.730","Text":"We have 1, 2, 3, 4,"},{"Start":"01:05.730 ","End":"01:09.195","Text":"5 bonds, meaning they contain 10 electrons."},{"Start":"01:09.195 ","End":"01:13.460","Text":"The third step of Lewis structure is to take our total number of valence electrons,"},{"Start":"01:13.460 ","End":"01:17.540","Text":"which is 12, and subtract the number of electrons that we have in bonds."},{"Start":"01:17.540 ","End":"01:18.980","Text":"We\u0027re going to subtract 10."},{"Start":"01:18.980 ","End":"01:21.725","Text":"This is going to leave us with 2 electrons."},{"Start":"01:21.725 ","End":"01:26.555","Text":"Since our terminal atoms are hydrogens and they\u0027re all have 1 bond,"},{"Start":"01:26.555 ","End":"01:29.060","Text":"we don\u0027t need to add any electrons to them."},{"Start":"01:29.060 ","End":"01:33.065","Text":"You\u0027re going to add these electrons to one of the carbons."},{"Start":"01:33.065 ","End":"01:35.480","Text":"Now, at this point, if we look at our Lewis structure,"},{"Start":"01:35.480 ","End":"01:38.885","Text":"we can see that 1 carbon has an octet,"},{"Start":"01:38.885 ","End":"01:40.280","Text":"but the other carbon doesn\u0027t."},{"Start":"01:40.280 ","End":"01:41.990","Text":"Therefore, we\u0027re going to take these electrons and we\u0027re"},{"Start":"01:41.990 ","End":"01:44.360","Text":"going to share them between the 2 carbons."},{"Start":"01:44.360 ","End":"01:46.909","Text":"The new Lewis structure is going to look like this;"},{"Start":"01:46.909 ","End":"01:56.520","Text":"a double bond between the carbons and single bonds to the rest of the hydrogens."},{"Start":"01:56.520 ","End":"01:59.320","Text":"Now we can see that all the hydrogens have 1 bond,"},{"Start":"01:59.320 ","End":"02:00.800","Text":"meaning 2 electrons,"},{"Start":"02:00.800 ","End":"02:02.765","Text":"and the carbons have an octet."},{"Start":"02:02.765 ","End":"02:05.510","Text":"This is an acceptable Lewis structure,"},{"Start":"02:05.510 ","End":"02:13.190","Text":"and that\u0027s our answer for E. We\u0027re going onto F. In F,"},{"Start":"02:13.190 ","End":"02:15.275","Text":"we can see that we have hydrofluoric acid."},{"Start":"02:15.275 ","End":"02:18.725","Text":"Now, as you can see, an ionic Lewis structure is given."},{"Start":"02:18.725 ","End":"02:21.440","Text":"However, our Lewis structure needs to be covalent,"},{"Start":"02:21.440 ","End":"02:23.915","Text":"therefore, Lewis structure is not correct."},{"Start":"02:23.915 ","End":"02:27.765","Text":"Now, we\u0027re going to write our own Lewis structure."},{"Start":"02:27.765 ","End":"02:29.345","Text":"Again, first step,"},{"Start":"02:29.345 ","End":"02:31.340","Text":"counting the number of valence electrons."},{"Start":"02:31.340 ","End":"02:34.790","Text":"Hydrogen has one since it\u0027s in Group 1 and"},{"Start":"02:34.790 ","End":"02:39.230","Text":"fluorine being in Group 17 has 7 valence electrons,"},{"Start":"02:39.230 ","End":"02:41.945","Text":"and this gives us a total of 8 electrons."},{"Start":"02:41.945 ","End":"02:47.150","Text":"Next, we\u0027re going to connect the hydrogens to the the fluorine."},{"Start":"02:47.150 ","End":"02:49.910","Text":"We have 1 bond in our skeletal structure."},{"Start":"02:49.910 ","End":"02:56.500","Text":"We\u0027re going to take the 8 valence electrons and subtract 2 since we have 2 in our bond."},{"Start":"02:56.500 ","End":"02:58.939","Text":"This leaves us with 6 electrons."},{"Start":"02:58.939 ","End":"03:03.270","Text":"Now, as we said, the hydrogen needs only 2 electrons,"},{"Start":"03:03.270 ","End":"03:05.405","Text":"so we\u0027re going to add them to the fluorine, 1,"},{"Start":"03:05.405 ","End":"03:09.065","Text":"2, 3, 4, 5, 6."},{"Start":"03:09.065 ","End":"03:11.030","Text":"Now, at this point, again,"},{"Start":"03:11.030 ","End":"03:14.450","Text":"the hydrogen has 1 bond which has 2 electrons,"},{"Start":"03:14.450 ","End":"03:17.510","Text":"and the fluorine, we can see it has an octet, 8 electrons."},{"Start":"03:17.510 ","End":"03:20.940","Text":"Therefore, this is an acceptable Lewis structure."},{"Start":"03:21.470 ","End":"03:26.740","Text":"That\u0027s our answer for F. Thank you very much for watching."}],"ID":24517},{"Watched":false,"Name":"Exercise 8","Duration":"8m 20s","ChapterTopicVideoID":23594,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.685","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.685 ","End":"00:05.640","Text":"Which of the following is the more favorable Lewis structure?"},{"Start":"00:05.640 ","End":"00:08.070","Text":"Let\u0027s begin in a, we have 2 structures."},{"Start":"00:08.070 ","End":"00:11.625","Text":"In order to decide which is the more favorable Lewis structure,"},{"Start":"00:11.625 ","End":"00:13.830","Text":"we\u0027re going to determine the formal charges on"},{"Start":"00:13.830 ","End":"00:16.545","Text":"the atoms and then compare them one to another."},{"Start":"00:16.545 ","End":"00:18.750","Text":"We\u0027re going to start with our left structure,"},{"Start":"00:18.750 ","End":"00:22.095","Text":"and we\u0027re going to start with sulfur."},{"Start":"00:22.095 ","End":"00:26.325","Text":"We\u0027re going to determine the formal charge on the sulfur."},{"Start":"00:26.325 ","End":"00:28.560","Text":"Now, remember the formal charge equals"},{"Start":"00:28.560 ","End":"00:31.140","Text":"the number of valence electrons around the free atom."},{"Start":"00:31.140 ","End":"00:33.810","Text":"Sulfur remember is in group 16,"},{"Start":"00:33.810 ","End":"00:36.135","Text":"meaning it has 6 valence electrons,"},{"Start":"00:36.135 ","End":"00:39.975","Text":"minus the number of lone pair electrons."},{"Start":"00:39.975 ","End":"00:42.330","Text":"Here we have 4 lone pair electrons,"},{"Start":"00:42.330 ","End":"00:43.430","Text":"1, 2, 3, 4,"},{"Start":"00:43.430 ","End":"00:45.095","Text":"so that\u0027s minus 4,"},{"Start":"00:45.095 ","End":"00:49.445","Text":"minus 1/2 times the number of bond pair electrons."},{"Start":"00:49.445 ","End":"00:52.270","Text":"Here the sulfur has 2 bonds next to it,"},{"Start":"00:52.270 ","End":"00:53.906","Text":"meaning it has 1, 2,3,"},{"Start":"00:53.906 ","End":"00:56.960","Text":"4 electrons next to it since in every bond,"},{"Start":"00:56.960 ","End":"01:00.125","Text":"there are 2 electrons, so there\u0027s going to be 1/2 times 4."},{"Start":"01:00.125 ","End":"01:03.455","Text":"This equals 6 minus 4 minus 2,"},{"Start":"01:03.455 ","End":"01:07.070","Text":"and this gives us a 0 for the sulfur."},{"Start":"01:07.070 ","End":"01:10.760","Text":"The left and the right sulfur have the same number of"},{"Start":"01:10.760 ","End":"01:13.900","Text":"lone pair electrons and bonds therefore they both equal 0."},{"Start":"01:13.900 ","End":"01:17.820","Text":"Now we\u0027re going to calculate the formal charge on the carbon."},{"Start":"01:18.140 ","End":"01:21.530","Text":"Again, formal charge of carbon equals the number"},{"Start":"01:21.530 ","End":"01:24.575","Text":"of valence electrons of the free carbon atom,"},{"Start":"01:24.575 ","End":"01:28.430","Text":"which is 4 since the carbon is in the 14th period,"},{"Start":"01:28.430 ","End":"01:30.770","Text":"minus the number of lone pair electrons,"},{"Start":"01:30.770 ","End":"01:38.705","Text":"which in this case is 0 minus 1/2 times the number of bond pair electrons."},{"Start":"01:38.705 ","End":"01:41.000","Text":"Carbon has 4 bonds around it,"},{"Start":"01:41.000 ","End":"01:43.160","Text":"meaning it has 8 electrons around it,"},{"Start":"01:43.160 ","End":"01:45.520","Text":"so that\u0027s going to be a 1/2 times 8."},{"Start":"01:45.520 ","End":"01:50.855","Text":"This equals 4 minus 4, which equals 0."},{"Start":"01:50.855 ","End":"01:55.730","Text":"In the left Lewis structure, we have formal charges which all equal 0."},{"Start":"01:55.730 ","End":"01:58.295","Text":"Let\u0027s look at our right structure."},{"Start":"01:58.295 ","End":"02:01.655","Text":"In this case, you see we have 2 different sulfurs. They\u0027re not the same food."},{"Start":"02:01.655 ","End":"02:03.440","Text":"The formal charge of the left sulfur,"},{"Start":"02:03.440 ","End":"02:06.475","Text":"which we can call it, sulfur number 1,"},{"Start":"02:06.475 ","End":"02:10.310","Text":"equals 6 the number of valence electrons in"},{"Start":"02:10.310 ","End":"02:14.765","Text":"the free atom minus the number of lone pair electrons."},{"Start":"02:14.765 ","End":"02:18.155","Text":"In this case, we can see we have 1,"},{"Start":"02:18.155 ","End":"02:19.617","Text":"2, 3, 4, 5,"},{"Start":"02:19.617 ","End":"02:24.410","Text":"6 lone pair electrons minus 1/2 times the number of bond pair electrons."},{"Start":"02:24.410 ","End":"02:26.420","Text":"Now, we have 1 bond connected to the sulfur,"},{"Start":"02:26.420 ","End":"02:28.145","Text":"meaning we have 2 electrons."},{"Start":"02:28.145 ","End":"02:30.281","Text":"That\u0027s going to give us 6 minus 6 minus 1,"},{"Start":"02:30.281 ","End":"02:32.485","Text":"that\u0027s going to give us a minus 1."},{"Start":"02:32.485 ","End":"02:37.065","Text":"That\u0027s going to be the formal charge on the left sulfur."},{"Start":"02:37.065 ","End":"02:39.140","Text":"Now let\u0027s look at our carbon."},{"Start":"02:39.140 ","End":"02:41.405","Text":"The formal charge on our carbon."},{"Start":"02:41.405 ","End":"02:43.790","Text":"That\u0027s going to be 4, which is the number of valence electrons on"},{"Start":"02:43.790 ","End":"02:48.050","Text":"the free atom minus 0 lone pair electrons"},{"Start":"02:48.050 ","End":"02:50.825","Text":"because there are no lone pair electrons around the carbon minus"},{"Start":"02:50.825 ","End":"02:54.710","Text":"1/2 times the number of bond pair electrons around the carbon."},{"Start":"02:54.710 ","End":"02:57.470","Text":"Now here we have 8 because again we have 4 bonds,"},{"Start":"02:57.470 ","End":"02:58.580","Text":"1, 2, 3, 4,"},{"Start":"02:58.580 ","End":"03:00.484","Text":"and each bond contains 2 electrons,"},{"Start":"03:00.484 ","End":"03:03.935","Text":"so minus 8, and that\u0027s going to give us a 0."},{"Start":"03:03.935 ","End":"03:06.815","Text":"Now we\u0027re going on to the third sulfur,"},{"Start":"03:06.815 ","End":"03:08.940","Text":"which is the right one."},{"Start":"03:08.940 ","End":"03:13.400","Text":"The formal charge of sulfur equals 6,"},{"Start":"03:13.400 ","End":"03:16.130","Text":"which is the number of valence electrons around"},{"Start":"03:16.130 ","End":"03:20.810","Text":"the free sulfur atom minus the number of lone pair electrons,"},{"Start":"03:20.810 ","End":"03:22.175","Text":"which is 2 in our case,"},{"Start":"03:22.175 ","End":"03:25.040","Text":"minus 1/2 times of bond pair electrons,"},{"Start":"03:25.040 ","End":"03:27.380","Text":"which is 6 electrons."},{"Start":"03:27.380 ","End":"03:29.630","Text":"This gives us 6 minus 2 minus 3,"},{"Start":"03:29.630 ","End":"03:33.295","Text":"which equals plus 1."},{"Start":"03:33.295 ","End":"03:36.785","Text":"Now if we compare the left Lewis structure to the right one,"},{"Start":"03:36.785 ","End":"03:38.270","Text":"remember in the left,"},{"Start":"03:38.270 ","End":"03:40.190","Text":"all the formal charges are 0."},{"Start":"03:40.190 ","End":"03:43.980","Text":"However, in the right we have the carbon formal charge is 0"},{"Start":"03:43.980 ","End":"03:48.590","Text":"though one of the sulfur formal charges is minus 1 and the other is plus 1."},{"Start":"03:48.590 ","End":"03:56.360","Text":"Of course, the more favorable Lewis structure is the left with formal charges equaling 0."},{"Start":"03:56.360 ","End":"04:04.345","Text":"Our answer for a is going to be the Lewis structure on the left,"},{"Start":"04:04.345 ","End":"04:07.095","Text":"in which we have 2 double bonds."},{"Start":"04:07.095 ","End":"04:10.430","Text":"That\u0027s our answer for a and now we\u0027re going to go on to b."},{"Start":"04:10.430 ","End":"04:14.870","Text":"Now we\u0027re going to start b, and we\u0027re going to start with our left structure."},{"Start":"04:14.870 ","End":"04:16.580","Text":"Let\u0027s start with our phosphorus,"},{"Start":"04:16.580 ","End":"04:18.530","Text":"which is the central atom."},{"Start":"04:18.530 ","End":"04:23.260","Text":"The formal charge of phosphorous equals 5,"},{"Start":"04:23.260 ","End":"04:25.685","Text":"which is the number of valence electrons in the free atom,"},{"Start":"04:25.685 ","End":"04:29.330","Text":"minus 0, which is the number of lone pair electrons."},{"Start":"04:29.330 ","End":"04:31.685","Text":"There are no lone pair electrons around the phosphorus"},{"Start":"04:31.685 ","End":"04:34.760","Text":"minus 1/2 times the number of bond pair electrons,"},{"Start":"04:34.760 ","End":"04:37.865","Text":"which phosphorus has 4 bonds connected to it,"},{"Start":"04:37.865 ","End":"04:43.195","Text":"meaning it has 8 electrons in its bonds,"},{"Start":"04:43.195 ","End":"04:47.425","Text":"so is going to give us 5 minus 4 and it\u0027s going to give us a plus 1."},{"Start":"04:47.425 ","End":"04:50.915","Text":"Now we\u0027re going to look at the formal charge of the oxygen."},{"Start":"04:50.915 ","End":"04:54.495","Text":"You can see here all the oxygens have the same formal charge."},{"Start":"04:54.495 ","End":"05:00.314","Text":"We\u0027re beginning with 6 valence electrons around the free atoms minus 6, 1, 2, 3,"},{"Start":"05:00.314 ","End":"05:05.720","Text":"4, 5, 6 lone pair electrons around each oxygen minus 1/2 times 2."},{"Start":"05:05.720 ","End":"05:07.385","Text":"Since if you look at our oxygen,"},{"Start":"05:07.385 ","End":"05:09.230","Text":"you can see it\u0027s only connected to one bond,"},{"Start":"05:09.230 ","End":"05:10.835","Text":"so that\u0027s 2 electrons."},{"Start":"05:10.835 ","End":"05:16.240","Text":"This is going to give us a formal charge of minus 1 for each oxygen."},{"Start":"05:16.240 ","End":"05:21.095","Text":"Now just to remind you, the sum of the formal charges has to equal the sum of the charge."},{"Start":"05:21.095 ","End":"05:24.145","Text":"The charge here is minus 3."},{"Start":"05:24.145 ","End":"05:27.650","Text":"This gives us a charge of minus 3 since we have plus 1 on"},{"Start":"05:27.650 ","End":"05:31.580","Text":"the phosphorus and 4 times minus 1 on oxidant,"},{"Start":"05:31.580 ","End":"05:33.425","Text":"so that\u0027s a total of minus 3."},{"Start":"05:33.425 ","End":"05:36.640","Text":"That\u0027s our left structure,"},{"Start":"05:36.640 ","End":"05:38.210","Text":"let\u0027s just write this down."},{"Start":"05:38.210 ","End":"05:45.065","Text":"We have minus 1 for each oxygen and we have a plus 1 for the phosphorus."},{"Start":"05:45.065 ","End":"05:47.240","Text":"That\u0027s our left structure."},{"Start":"05:47.240 ","End":"05:49.865","Text":"Now we\u0027re going to take a look at our right structure."},{"Start":"05:49.865 ","End":"05:53.105","Text":"Again, let\u0027s start with a formal charge of phosphorus,"},{"Start":"05:53.105 ","End":"05:56.870","Text":"which begins with a 5 because we have 5 valence electrons"},{"Start":"05:56.870 ","End":"06:01.100","Text":"minus 0 lone pair electrons minus 1/2 times."},{"Start":"06:01.100 ","End":"06:05.723","Text":"If we look, we have 5 bonds around the phosphorus 1, 2, 3, 4,"},{"Start":"06:05.723 ","End":"06:11.740","Text":"5, and that\u0027s going to be 10 electrons because each one contains 2 electrons."},{"Start":"06:11.740 ","End":"06:17.375","Text":"I just want to remind you that the phosphorus can have an expanded valence shell."},{"Start":"06:17.375 ","End":"06:20.210","Text":"In this case, it has an expanded valence shell."},{"Start":"06:20.210 ","End":"06:21.845","Text":"Let\u0027s continue with our formal charge."},{"Start":"06:21.845 ","End":"06:23.060","Text":"We have 5 minus 5,"},{"Start":"06:23.060 ","End":"06:26.750","Text":"so this is going to give us a 0 formal charge for the phosphorus."},{"Start":"06:26.750 ","End":"06:30.665","Text":"Now we\u0027re continuing with the formal charge on the oxygen."},{"Start":"06:30.665 ","End":"06:33.520","Text":"Now, as you can see, we have 2 different oxygens,"},{"Start":"06:33.520 ","End":"06:37.925","Text":"3 of the oxygens are the same in their formal charge and when it\u0027s going to be different."},{"Start":"06:37.925 ","End":"06:41.945","Text":"Let\u0027s begin with the oxygen, that they are the same."},{"Start":"06:41.945 ","End":"06:46.250","Text":"We\u0027re going to begin with 6 electrons minus the number of lone pair electrons,"},{"Start":"06:46.250 ","End":"06:48.102","Text":"which is 1, 2, 3, 4, 5,"},{"Start":"06:48.102 ","End":"06:51.830","Text":"6 minus 1/2 times bond pair electrons,"},{"Start":"06:51.830 ","End":"06:55.550","Text":"which is 2 since we have 1 bond connected to the oxygen."},{"Start":"06:55.550 ","End":"07:00.690","Text":"That\u0027s going to give us a minus 1 formal charge for these oxygens."},{"Start":"07:00.690 ","End":"07:04.370","Text":"Now if we look at other oxygen,"},{"Start":"07:04.370 ","End":"07:09.560","Text":"again, we begin with the number of valence electrons in the free atom,"},{"Start":"07:09.560 ","End":"07:12.260","Text":"which is 6 minus the number of lone pair electrons,"},{"Start":"07:12.260 ","End":"07:13.880","Text":"which is 1, 2, 3, 4,"},{"Start":"07:13.880 ","End":"07:20.000","Text":"so that\u0027s minus 4 minus 1/2 times the number of electrons in our bonds,"},{"Start":"07:20.000 ","End":"07:22.400","Text":"which is 4 since we have 2 bonds here,"},{"Start":"07:22.400 ","End":"07:24.145","Text":"so that\u0027s times 4."},{"Start":"07:24.145 ","End":"07:27.570","Text":"This equals 6 minus 4 minus 2,"},{"Start":"07:27.570 ","End":"07:30.615","Text":"which gives us 0. Let\u0027s write this down."},{"Start":"07:30.615 ","End":"07:32.535","Text":"This oxygen gives us 0,"},{"Start":"07:32.535 ","End":"07:33.930","Text":"phosphorous gives us 0,"},{"Start":"07:33.930 ","End":"07:35.755","Text":"and then we have 3 minus 1."},{"Start":"07:35.755 ","End":"07:41.330","Text":"In both cases, we can see that the sum of the formal charges equals the formal charge,"},{"Start":"07:41.330 ","End":"07:44.090","Text":"the sum equals the charge, which is okay."},{"Start":"07:44.090 ","End":"07:46.580","Text":"However, we can see that in the right structure we have"},{"Start":"07:46.580 ","End":"07:49.490","Text":"less formal charges than in the left structure therefore,"},{"Start":"07:49.490 ","End":"07:52.680","Text":"the right structure is the more favorable one."},{"Start":"07:52.680 ","End":"07:55.760","Text":"We\u0027re going to write this down. This is the phosphorus has"},{"Start":"07:55.760 ","End":"08:00.660","Text":"one double bond to an oxygen and 3 single bonds, to other oxygens."},{"Start":"08:09.110 ","End":"08:12.650","Text":"Again, the answer to b is the case where we"},{"Start":"08:12.650 ","End":"08:17.135","Text":"have also a double bond between our phosphorus and our oxygen."},{"Start":"08:17.135 ","End":"08:18.710","Text":"That is our final answer."},{"Start":"08:18.710 ","End":"08:21.180","Text":"Thank you very much for watching."}],"ID":24505},{"Watched":false,"Name":"Exercise 9","Duration":"5m 21s","ChapterTopicVideoID":23595,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.895","Text":"Now we\u0027re going to solve the following exercise."},{"Start":"00:02.895 ","End":"00:06.675","Text":"Which of the following is the more favorable Lewis structure?"},{"Start":"00:06.675 ","End":"00:09.795","Text":"Let\u0027s take a look at a. We have 3 structures."},{"Start":"00:09.795 ","End":"00:11.400","Text":"We\u0027re going to begin with determining"},{"Start":"00:11.400 ","End":"00:14.175","Text":"the formal charge on all of the atoms in each structure."},{"Start":"00:14.175 ","End":"00:16.425","Text":"We\u0027re going to start with the left structure."},{"Start":"00:16.425 ","End":"00:18.090","Text":"Now before we begin, just look at"},{"Start":"00:18.090 ","End":"00:22.095","Text":"all the structures and we can see that all they all begin with CH3."},{"Start":"00:22.095 ","End":"00:24.450","Text":"Now if we look at our carbon in"},{"Start":"00:24.450 ","End":"00:29.145","Text":"all three cases and we calculate the formal charge on the carbon."},{"Start":"00:29.145 ","End":"00:31.230","Text":"It\u0027s going to begin with 4 valence,"},{"Start":"00:31.230 ","End":"00:34.020","Text":"which are the valence electrons in the free carbon atom."},{"Start":"00:34.020 ","End":"00:39.590","Text":"Remember, carbon is in group 14 minus the number of lone pair electrons,"},{"Start":"00:39.590 ","End":"00:44.495","Text":"which is 0 minus half times the number of bond pair electrons."},{"Start":"00:44.495 ","End":"00:49.165","Text":"Here we have 4 bonds around the carbon, meaning 8 electrons."},{"Start":"00:49.165 ","End":"00:51.375","Text":"Remember 2 electrons in each bond."},{"Start":"00:51.375 ","End":"00:54.965","Text":"This is going to give us 4 minus 4 and this is going to give us a 0."},{"Start":"00:54.965 ","End":"00:58.330","Text":"In every Lewis structure,"},{"Start":"00:58.330 ","End":"01:02.510","Text":"the carbon in the CH3 is going to be a 0."},{"Start":"01:02.510 ","End":"01:04.850","Text":"We\u0027re going to leave that carbon alone right now and we\u0027re going"},{"Start":"01:04.850 ","End":"01:07.740","Text":"to calculate the rest of the atoms."},{"Start":"01:07.740 ","End":"01:09.935","Text":"Let\u0027s take a look at our sulfur."},{"Start":"01:09.935 ","End":"01:12.440","Text":"The formal charge of sulfur equals,"},{"Start":"01:12.440 ","End":"01:14.360","Text":"remember in our left structure,"},{"Start":"01:14.360 ","End":"01:18.350","Text":"equals 6 because that\u0027s the number of valence electrons in the free atom minus 0."},{"Start":"01:18.350 ","End":"01:22.820","Text":"We have 0 lone pair electrons around the sulfur and we have 4 bonds,"},{"Start":"01:22.820 ","End":"01:26.415","Text":"meaning half times 8 electrons,"},{"Start":"01:26.415 ","End":"01:27.750","Text":"2 electrons per bond."},{"Start":"01:27.750 ","End":"01:30.285","Text":"This is going to give us 6 minus 4."},{"Start":"01:30.285 ","End":"01:32.475","Text":"This is going to give us a 2 or a plus 2."},{"Start":"01:32.475 ","End":"01:37.310","Text":"We\u0027re just going to write that under the sulfur plus 2."},{"Start":"01:37.310 ","End":"01:38.780","Text":"Let\u0027s go to the carbon."},{"Start":"01:38.780 ","End":"01:42.180","Text":"Formal charge of the carbon in this case is 4,"},{"Start":"01:42.180 ","End":"01:45.650","Text":"number of valence electrons in the free atom minus lone pair electrons,"},{"Start":"01:45.650 ","End":"01:50.090","Text":"which is 0 minus half times the number of bond pair electrons,"},{"Start":"01:50.090 ","End":"01:51.260","Text":"which we have 4 bonds here."},{"Start":"01:51.260 ","End":"01:52.820","Text":"This is going to be times 8."},{"Start":"01:52.820 ","End":"01:54.920","Text":"It\u0027s going to give us an overall of 4 minus 4,"},{"Start":"01:54.920 ","End":"01:56.645","Text":"which is going to give us a 0."},{"Start":"01:56.645 ","End":"01:58.430","Text":"The carbon is going to be a 0,"},{"Start":"01:58.430 ","End":"02:01.805","Text":"in this case, 0 formal charge."},{"Start":"02:01.805 ","End":"02:06.140","Text":"Now let\u0027s look at the formal charge of nitrogen in our last structure."},{"Start":"02:06.140 ","End":"02:07.700","Text":"It\u0027s going to begin with a 5,"},{"Start":"02:07.700 ","End":"02:10.340","Text":"because nitrogen is in the 15th group"},{"Start":"02:10.340 ","End":"02:12.860","Text":"minus the number of lone pair electrons who have 1,"},{"Start":"02:12.860 ","End":"02:14.645","Text":"2, 3, 4, 5, 6."},{"Start":"02:14.645 ","End":"02:18.320","Text":"Six lone pair electrons minus half times bond pair electrons,"},{"Start":"02:18.320 ","End":"02:20.630","Text":"we have one bond, so we have 2 electrons."},{"Start":"02:20.630 ","End":"02:23.720","Text":"It\u0027s going to give us 5 minus 6 minus 1,"},{"Start":"02:23.720 ","End":"02:24.890","Text":"meaning 5 minus 7."},{"Start":"02:24.890 ","End":"02:26.375","Text":"That\u0027s going to give us a minus 2,"},{"Start":"02:26.375 ","End":"02:29.620","Text":"so that nitrogen has a minus 2."},{"Start":"02:29.620 ","End":"02:32.375","Text":"Now, even without continuing to the other structures,"},{"Start":"02:32.375 ","End":"02:35.855","Text":"we can see that the formal charges that we get here are pretty high,"},{"Start":"02:35.855 ","End":"02:39.020","Text":"plus 2 and minus 2, so this is probably not a great Lewis structure."},{"Start":"02:39.020 ","End":"02:41.470","Text":"Now we\u0027re going to go on to the middle structure."},{"Start":"02:41.470 ","End":"02:45.080","Text":"We\u0027re going to start with calculating the formal charge of sulfur."},{"Start":"02:45.080 ","End":"02:49.250","Text":"Again, we begin with a 6 minus number of lone pair electrons,"},{"Start":"02:49.250 ","End":"02:56.305","Text":"which is 2 minus half times the number of electrons in three bonds, which is 6."},{"Start":"02:56.305 ","End":"03:01.575","Text":"It\u0027s going to give us an overall of plus 1 for our sulfur."},{"Start":"03:01.575 ","End":"03:06.080","Text":"Look at the carbon, we have 4 valence electrons in the free atom"},{"Start":"03:06.080 ","End":"03:11.335","Text":"minus 0 lone pair electrons minus half times 8."},{"Start":"03:11.335 ","End":"03:13.500","Text":"Since there are 4 bonds around the carbon,"},{"Start":"03:13.500 ","End":"03:17.930","Text":"this is going to give us a 0 because that\u0027s 4 times 4 and the carbon again is 0."},{"Start":"03:17.930 ","End":"03:19.610","Text":"Now we\u0027re going to take a look at our nitrogen."},{"Start":"03:19.610 ","End":"03:25.220","Text":"We\u0027re going to begin with 5 valence electrons minus 4 lone pair electrons, 1, 2, 3,"},{"Start":"03:25.220 ","End":"03:30.140","Text":"4 minus half times number of bond pair electrons,"},{"Start":"03:30.140 ","End":"03:32.665","Text":"which are 4 because we have two bonds."},{"Start":"03:32.665 ","End":"03:35.730","Text":"It\u0027s going to give us a 5 minus 4 minus 2."},{"Start":"03:35.730 ","End":"03:37.755","Text":"It\u0027s going to give us a minus 1."},{"Start":"03:37.755 ","End":"03:40.660","Text":"Sure nitrogen is going to have a minus 1."},{"Start":"03:40.660 ","End":"03:46.235","Text":"Now we\u0027re going to go on to the third structure and see what formal charges we get there."},{"Start":"03:46.235 ","End":"03:50.380","Text":"In B we can see it again we have two structures."},{"Start":"03:50.380 ","End":"03:53.180","Text":"We\u0027re going to calculate the formal charges of the structures."},{"Start":"03:53.180 ","End":"03:57.020","Text":"Let\u0027s begin with a formal charge of the oxygen in our left structure."},{"Start":"03:57.020 ","End":"04:03.860","Text":"This equals 6 number of valence electrons in the free atom minus, we have 1, 2, 3,"},{"Start":"04:03.860 ","End":"04:07.840","Text":"4 lone pair electrons minus half times the bond pair electrons,"},{"Start":"04:07.840 ","End":"04:09.580","Text":"which is 1, 2, 3, 4,"},{"Start":"04:09.580 ","End":"04:12.280","Text":"because we have 4 electrons in two bonds,"},{"Start":"04:12.280 ","End":"04:14.650","Text":"that equals 6 minus 4 minus 2,"},{"Start":"04:14.650 ","End":"04:17.260","Text":"giving us a 0 for the oxygen."},{"Start":"04:17.260 ","End":"04:22.210","Text":"Now the formal charge of nitrogen in the left structure equals 5,"},{"Start":"04:22.210 ","End":"04:27.820","Text":"which is the number of valence electrons for nitrogen minus 1,"},{"Start":"04:27.820 ","End":"04:31.700","Text":"2 lone pair electrons minus half times,"},{"Start":"04:31.700 ","End":"04:34.150","Text":"we have three bonds around the nitrogen,"},{"Start":"04:34.150 ","End":"04:37.435","Text":"meaning 6 electrons in the bonds."},{"Start":"04:37.435 ","End":"04:40.915","Text":"This equals 5 minus 2 minus 3, and this is going to give us a 0."},{"Start":"04:40.915 ","End":"04:45.715","Text":"We have 0 formal charge for the oxygen and 0 for the nitrogen."},{"Start":"04:45.715 ","End":"04:50.195","Text":"Now let\u0027s see what happens with our fluorine."},{"Start":"04:50.195 ","End":"04:52.920","Text":"We begin with 7 electrons in fluorine,"},{"Start":"04:52.920 ","End":"04:56.270","Text":"it\u0027s in the 17th group minus the number of lone pair electrons,"},{"Start":"04:56.270 ","End":"04:57.670","Text":"which is 1, 2, 3, 4, 5,"},{"Start":"04:57.670 ","End":"05:03.050","Text":"6 minus half, times the number of bond pair electrons,"},{"Start":"05:03.050 ","End":"05:06.425","Text":"which is 2 since we have one bond connected here to fluorine."},{"Start":"05:06.425 ","End":"05:10.700","Text":"It\u0027s going to give us 7 minus 6 minus 1. That\u0027s going to give us a 0."},{"Start":"05:10.700 ","End":"05:15.575","Text":"Now the structure we got has formal charge of 0 for all the atoms."},{"Start":"05:15.575 ","End":"05:17.315","Text":"That\u0027s a great Lewis structure."},{"Start":"05:17.315 ","End":"05:19.400","Text":"Now we\u0027re going to go on and see what happens with"},{"Start":"05:19.400 ","End":"05:22.780","Text":"a formal charges in the next structure."}],"ID":24506},{"Watched":false,"Name":"Resonance","Duration":"6m 58s","ChapterTopicVideoID":20280,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"In previous videos, we learned about Lewis structures."},{"Start":"00:03.510 ","End":"00:07.920","Text":"In this video we\u0027ll learn about the important concept of resonance."},{"Start":"00:07.920 ","End":"00:10.425","Text":"So we\u0027re going to talk about resonance."},{"Start":"00:10.425 ","End":"00:14.220","Text":"Sometimes a single Lewis structure does not give"},{"Start":"00:14.220 ","End":"00:19.425","Text":"an adequate description of the molecule and a hybrid of several Lewis structures,"},{"Start":"00:19.425 ","End":"00:22.170","Text":"which we call resonance structures is required."},{"Start":"00:22.170 ","End":"00:24.975","Text":"This is best illustrated by taking an example."},{"Start":"00:24.975 ","End":"00:29.340","Text":"The example I\u0027m going to choose is carbonate CO_3^2 minus,"},{"Start":"00:29.340 ","End":"00:32.040","Text":"we talked about it in a previous video."},{"Start":"00:32.040 ","End":"00:38.060","Text":"In carbonate, all the CO bonds have the same length of 128 picometers."},{"Start":"00:38.060 ","End":"00:41.225","Text":"This is intermediate between a single bond length,"},{"Start":"00:41.225 ","End":"00:43.460","Text":"which is 143 picometers,"},{"Start":"00:43.460 ","End":"00:46.970","Text":"and double bond length which is 120 picometers."},{"Start":"00:46.970 ","End":"00:51.935","Text":"We need some description of the ion CO_3^2 minus"},{"Start":"00:51.935 ","End":"00:57.110","Text":"that allows for the fact that the bond length is intermediate between single,"},{"Start":"00:57.110 ","End":"01:01.260","Text":"double and that all the bonds are equivalent."},{"Start":"01:18.380 ","End":"01:22.775","Text":"Now when we talked about carbonate in a previous video,"},{"Start":"01:22.775 ","End":"01:29.090","Text":"we drew this Lewis structure with a 1 double bond and 2 single bonds."},{"Start":"01:29.090 ","End":"01:32.195","Text":"But there\u0027s really no fundamental reason why"},{"Start":"01:32.195 ","End":"01:35.795","Text":"this should be the double bond and these the single bonds."},{"Start":"01:35.795 ","End":"01:40.100","Text":"Equally well, we could have a double bond between this C and O,"},{"Start":"01:40.100 ","End":"01:43.045","Text":"or a double bond between this C and this O,"},{"Start":"01:43.045 ","End":"01:50.705","Text":"so all 3 are equivalent in the sense that all have 1 double bond and 2 single bonds."},{"Start":"01:50.705 ","End":"01:54.115","Text":"We say that all the structures are in resonance."},{"Start":"01:54.115 ","End":"01:57.320","Text":"This is indicated by the double-sided arrow."},{"Start":"01:57.320 ","End":"02:00.110","Text":"Each structure is called a resonance structure."},{"Start":"02:00.110 ","End":"02:03.139","Text":"The most accurate description of the molecule,"},{"Start":"02:03.139 ","End":"02:06.680","Text":"in this case the ion and the one with the lowest energy"},{"Start":"02:06.680 ","End":"02:10.580","Text":"is a resonance hybrid of the resonance structures."},{"Start":"02:10.580 ","End":"02:13.100","Text":"It doesn\u0027t mean that it jumps from this one to"},{"Start":"02:13.100 ","End":"02:15.860","Text":"this one to this one backwards and forwards."},{"Start":"02:15.860 ","End":"02:21.560","Text":"But that it has the characteristics of all 3 of them, for example,"},{"Start":"02:21.560 ","End":"02:26.840","Text":"if we had a hybrid tea rose very popular in Britain,"},{"Start":"02:26.840 ","End":"02:29.270","Text":"then we may have a pink one,"},{"Start":"02:29.270 ","End":"02:32.240","Text":"that\u0027s a hybrid of a white one and a red one."},{"Start":"02:32.240 ","End":"02:35.225","Text":"It doesn\u0027t jump from being a white rose to red rose,"},{"Start":"02:35.225 ","End":"02:36.695","Text":"but it\u0027s a pink rose,"},{"Start":"02:36.695 ","End":"02:40.120","Text":"which is a hybrid of a white one and a red one."},{"Start":"02:40.120 ","End":"02:44.405","Text":"Now there are few rules about writing resonance structures."},{"Start":"02:44.405 ","End":"02:49.160","Text":"First thing is that in each resonance structure the nuclei are in the same positions."},{"Start":"02:49.160 ","End":"02:50.735","Text":"The nuclei don\u0027t move."},{"Start":"02:50.735 ","End":"02:53.840","Text":"Cs are always here in the same position."},{"Start":"02:53.840 ","End":"02:56.465","Text":"The 3 Os are always in the same position."},{"Start":"02:56.465 ","End":"02:58.805","Text":"The electrons, on the other hand,"},{"Start":"02:58.805 ","End":"03:02.075","Text":"the lone pairs and bond pairs are in different positions."},{"Start":"03:02.075 ","End":"03:04.625","Text":"Here they\u0027re forming a double bond."},{"Start":"03:04.625 ","End":"03:08.095","Text":"Here they\u0027re forming lone pairs and so on."},{"Start":"03:08.095 ","End":"03:10.547","Text":"Now structures with the same energy,"},{"Start":"03:10.547 ","End":"03:12.305","Text":"structures that are equivalent,"},{"Start":"03:12.305 ","End":"03:14.765","Text":"contribute equally to the resonance."},{"Start":"03:14.765 ","End":"03:20.540","Text":"Here we have 3 structures that contribute equally because they\u0027re all equivalent."},{"Start":"03:20.540 ","End":"03:22.375","Text":"They all have the same energy."},{"Start":"03:22.375 ","End":"03:24.859","Text":"If the structures are not all equivalent,"},{"Start":"03:24.859 ","End":"03:28.940","Text":"the low-energy ones contribute more than the high-energy ones."},{"Start":"03:28.940 ","End":"03:32.330","Text":"Now how do we know which ones are the low-energy ones?"},{"Start":"03:32.330 ","End":"03:38.180","Text":"Usually, the structure with lowest formal charge has the lowest energy."},{"Start":"03:38.180 ","End":"03:42.185","Text":"Let\u0027s discuss a non-equivalent resonance structures."},{"Start":"03:42.185 ","End":"03:47.210","Text":"That\u0027s a hybrid which consists of non-equivalent resonance structures."},{"Start":"03:47.210 ","End":"03:48.995","Text":"We\u0027re going to take as an example,"},{"Start":"03:48.995 ","End":"03:51.185","Text":"CO_2, carbon dioxide."},{"Start":"03:51.185 ","End":"03:56.480","Text":"Now, here are 3 possible structures for carbon dioxide."},{"Start":"03:56.480 ","End":"04:00.015","Text":"There\u0027s one with 2 double bonds,"},{"Start":"04:00.015 ","End":"04:05.525","Text":"and then there is one here with a triple bond between the C and O,"},{"Start":"04:05.525 ","End":"04:10.010","Text":"and a single bond between the C and the other oxygen."},{"Start":"04:10.010 ","End":"04:13.650","Text":"We need to add a lone pair here."},{"Start":"04:13.990 ","End":"04:20.715","Text":"Or we can have the triple bond between C and the final oxygen."},{"Start":"04:20.715 ","End":"04:22.700","Text":"These 2 are equivalent."},{"Start":"04:22.700 ","End":"04:24.890","Text":"We have a triple bond here and a single bond,"},{"Start":"04:24.890 ","End":"04:27.005","Text":"and here we have a single bond, and a triple bond."},{"Start":"04:27.005 ","End":"04:29.840","Text":"These 2 Lewis structures are equivalent,"},{"Start":"04:29.840 ","End":"04:33.295","Text":"but the first one is not equivalent to the other 2."},{"Start":"04:33.295 ","End":"04:38.720","Text":"In order to see which structure has the lowest energy,"},{"Start":"04:38.720 ","End":"04:40.940","Text":"we should draw the formal charges,"},{"Start":"04:40.940 ","End":"04:43.369","Text":"we should calculate the formal charges."},{"Start":"04:43.369 ","End":"04:49.115","Text":"Oxygen we know has 6 electrons in the free form."},{"Start":"04:49.115 ","End":"04:55.820","Text":"Here it has 2 electrons from the lone pair and 1 electron for each of the bonds,"},{"Start":"04:55.820 ","End":"04:57.335","Text":"making a total of 5."},{"Start":"04:57.335 ","End":"05:02.090","Text":"So the formal charge is 6 minus 5, giving us 1."},{"Start":"05:02.090 ","End":"05:07.515","Text":"The other oxygen, which should also have 6 in the free form,"},{"Start":"05:07.515 ","End":"05:11.265","Text":"has 1 electron, it\u0027s from the bond,"},{"Start":"05:11.265 ","End":"05:13.295","Text":"and 3 lone pairs,"},{"Start":"05:13.295 ","End":"05:15.580","Text":"that\u0027s making a total of 7."},{"Start":"05:15.580 ","End":"05:18.825","Text":"So 6 minus 7 is equal to minus 1."},{"Start":"05:18.825 ","End":"05:22.040","Text":"This one has a formal charge of minus 1."},{"Start":"05:22.040 ","End":"05:31.380","Text":"We can do the same for the final one and then we get minus 1 here and plus 1 here."},{"Start":"05:31.380 ","End":"05:36.510","Text":"The carbon has 4 electrons in the free form and here"},{"Start":"05:36.510 ","End":"05:41.375","Text":"it has 4 single bonds so it\u0027s 4 minus 4 and that\u0027s 0."},{"Start":"05:41.375 ","End":"05:43.460","Text":"Formal charge on the carbon is 0,"},{"Start":"05:43.460 ","End":"05:44.720","Text":"both here and here."},{"Start":"05:44.720 ","End":"05:47.645","Text":"Now let us look at the first Lewis structure."},{"Start":"05:47.645 ","End":"05:49.429","Text":"Now if we look at the oxygen,"},{"Start":"05:49.429 ","End":"05:53.520","Text":"we should have 6 electrons in the free form, that\u0027s 6,"},{"Start":"05:53.520 ","End":"05:56.600","Text":"and then when it\u0027s in the Lewis structure,"},{"Start":"05:56.600 ","End":"06:02.225","Text":"we have 1 for each bond that\u0027s 2 and 2 lone pairs, that\u0027s another 4."},{"Start":"06:02.225 ","End":"06:05.270","Text":"We have 6 minus 6, that\u0027s 0."},{"Start":"06:05.270 ","End":"06:10.070","Text":"The C also has 0 because it has 4 electrons in the free form,"},{"Start":"06:10.070 ","End":"06:12.275","Text":"and here it has 4 bonds."},{"Start":"06:12.275 ","End":"06:15.540","Text":"That\u0027s 4 minus 4 is 0."},{"Start":"06:16.630 ","End":"06:23.460","Text":"The final oxygen is the same as the first oxygen because the molecule here is symmetric,"},{"Start":"06:23.460 ","End":"06:25.110","Text":"again we have 0."},{"Start":"06:25.110 ","End":"06:29.020","Text":"This Lewis structure has formal charges,"},{"Start":"06:29.020 ","End":"06:34.015","Text":"all 0, whereas the other 2 have plus 1 and minus 1."},{"Start":"06:34.015 ","End":"06:37.720","Text":"It seems that the most stable is the first one with"},{"Start":"06:37.720 ","End":"06:43.255","Text":"2 double bonds and the others contribute only a minor contribution."},{"Start":"06:43.255 ","End":"06:46.450","Text":"We can conclude the structure with the double bonds has"},{"Start":"06:46.450 ","End":"06:49.735","Text":"the greatest contribution to the resonance hybrid."},{"Start":"06:49.735 ","End":"06:53.980","Text":"This is an example of non-equivalent resonance structures."},{"Start":"06:53.980 ","End":"06:59.030","Text":"In this video, we talked about the important concept of resonance."}],"ID":21071},{"Watched":false,"Name":"Exceptions to octet rule","Duration":"6m 13s","ChapterTopicVideoID":20485,"CourseChapterTopicPlaylistID":101312,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:04.860","Text":"In previous videos, we learned about the octet rule."},{"Start":"00:04.860 ","End":"00:09.165","Text":"In this video, we\u0027ll learn about exceptions to this rule."},{"Start":"00:09.165 ","End":"00:13.650","Text":"The first exception are odd electron species."},{"Start":"00:13.650 ","End":"00:16.874","Text":"If a species has an odd number of valence electrons,"},{"Start":"00:16.874 ","End":"00:20.970","Text":"at least 1 atom cannot satisfy the octet rule."},{"Start":"00:20.970 ","End":"00:23.025","Text":"That\u0027s pretty obvious."},{"Start":"00:23.025 ","End":"00:27.660","Text":"Now, odd-electron species are called radical, sometimes free radicals."},{"Start":"00:27.660 ","End":"00:30.420","Text":"They\u0027re paramagnetic, that means weakly attracted to"},{"Start":"00:30.420 ","End":"00:34.605","Text":"an external magnetic field and are usually very reactive."},{"Start":"00:34.605 ","End":"00:38.055","Text":"They don\u0027t stay as radical for very long."},{"Start":"00:38.055 ","End":"00:40.425","Text":"Examples are metal,"},{"Start":"00:40.425 ","End":"00:43.920","Text":"hydroxyl, and nitrogen monoxide."},{"Start":"00:43.920 ","End":"00:46.580","Text":"I have drawn a little dot to the left-hand side to"},{"Start":"00:46.580 ","End":"00:50.315","Text":"indicate that there\u0027s 1 unpaired electron."},{"Start":"00:50.315 ","End":"00:53.585","Text":"Now, here are the structures."},{"Start":"00:53.585 ","End":"00:55.855","Text":"Let\u0027s start with methyl."},{"Start":"00:55.855 ","End":"00:58.865","Text":"Carbon has 4 valence electrons."},{"Start":"00:58.865 ","End":"01:02.030","Text":"Once we\u0027ve used 3 CH bonds,"},{"Start":"01:02.030 ","End":"01:03.845","Text":"we only have 1 leftover,"},{"Start":"01:03.845 ","End":"01:06.155","Text":"and that\u0027s the unpaired electrons."},{"Start":"01:06.155 ","End":"01:12.855","Text":"Oxygen has 6 electrons and refused 1 for the oxygen-hydrogen bond,"},{"Start":"01:12.855 ","End":"01:15.715","Text":"so that leaves 5, and here they are,"},{"Start":"01:15.715 ","End":"01:19.310","Text":"2 lone pairs and 1 unpaired electron."},{"Start":"01:19.310 ","End":"01:21.725","Text":"Now, nitrogen monoxide,"},{"Start":"01:21.725 ","End":"01:24.755","Text":"there\u0027s a double bond between nitrogen and oxygen."},{"Start":"01:24.755 ","End":"01:28.175","Text":"Nitrogen has 5 valence electrons."},{"Start":"01:28.175 ","End":"01:30.575","Text":"We\u0027ve used up 2 for the double bond,"},{"Start":"01:30.575 ","End":"01:32.060","Text":"and that leaves 3,"},{"Start":"01:32.060 ","End":"01:36.725","Text":"so we have lone pair and a single unpaired electron."},{"Start":"01:36.725 ","End":"01:44.375","Text":"Now, you may ask why I\u0027ve put the radical on the nitrogen rather than the oxygen."},{"Start":"01:44.375 ","End":"01:48.529","Text":"If one puts on the oxygen and calculates the formal charges,"},{"Start":"01:48.529 ","End":"01:52.850","Text":"one will see that putting it on nitrogen is preferable."},{"Start":"01:52.850 ","End":"01:56.450","Text":"Now let\u0027s look about at how many electrons"},{"Start":"01:56.450 ","End":"02:00.035","Text":"there are round about each of the central atoms."},{"Start":"02:00.035 ","End":"02:02.690","Text":"In carbon, there are 2, 4, 6,"},{"Start":"02:02.690 ","End":"02:06.200","Text":"7, so has 7 electrons around it."},{"Start":"02:06.200 ","End":"02:09.170","Text":"Oxygen has 2, 4,"},{"Start":"02:09.170 ","End":"02:14.000","Text":"5, and 2 for the single bond, that\u0027s again 7."},{"Start":"02:14.000 ","End":"02:17.705","Text":"Nitrogen has 4 for the double bond,"},{"Start":"02:17.705 ","End":"02:19.470","Text":"another 2 for the lone pair,"},{"Start":"02:19.470 ","End":"02:23.585","Text":"that\u0027s 6 and 1 for the unpaired electron, again 7."},{"Start":"02:23.585 ","End":"02:26.675","Text":"We have an incomplete octet."},{"Start":"02:26.675 ","End":"02:30.755","Text":"The second category are incomplete octets."},{"Start":"02:30.755 ","End":"02:33.050","Text":"Some molecules containing beryllium,"},{"Start":"02:33.050 ","End":"02:38.005","Text":"boron or aluminum have an incomplete octet on the central atom."},{"Start":"02:38.005 ","End":"02:41.445","Text":"Let\u0027s take the example of BF_3."},{"Start":"02:41.445 ","End":"02:47.950","Text":"We have boron and 3 fluorine atoms attached to it."},{"Start":"02:47.950 ","End":"02:52.190","Text":"Now, boron has 3 valence electrons."},{"Start":"02:52.190 ","End":"02:56.420","Text":"We\u0027ve used up three of them to form the BF bonds,"},{"Start":"02:56.420 ","End":"02:59.140","Text":"so we have no electrons left."},{"Start":"02:59.140 ","End":"03:00.870","Text":"If we look around boron,"},{"Start":"03:00.870 ","End":"03:02.490","Text":"we\u0027ll see there are 2, 4,"},{"Start":"03:02.490 ","End":"03:07.280","Text":"6 electrons, that\u0027s an incomplete octet."},{"Start":"03:07.280 ","End":"03:11.765","Text":"Now, since the BF bond is shorter than a single bond,"},{"Start":"03:11.765 ","End":"03:14.600","Text":"some resonance structures have been suggested,"},{"Start":"03:14.600 ","End":"03:17.600","Text":"but the structure with the incomplete octet that we\u0027ve drawn"},{"Start":"03:17.600 ","End":"03:21.259","Text":"above is undoubtedly the dominant structure."},{"Start":"03:21.259 ","End":"03:23.720","Text":"The third category are expanded"},{"Start":"03:23.720 ","End":"03:30.650","Text":"valence shells where there are more than 8 electrons around the central atom."},{"Start":"03:30.650 ","End":"03:33.860","Text":"Now, some third period elements,"},{"Start":"03:33.860 ","End":"03:37.970","Text":"such as sulfur and phosphorus form compounds that can"},{"Start":"03:37.970 ","End":"03:42.370","Text":"only be explained by having 10 or 12 electrons around them;"},{"Start":"03:42.370 ","End":"03:46.940","Text":"10 in the case of phosphorus and 12 in the case of sulfur."},{"Start":"03:46.940 ","End":"03:52.400","Text":"Here are two examples: PCl_5 and SF_6."},{"Start":"03:52.400 ","End":"03:55.150","Text":"Let\u0027s start by drawing them."},{"Start":"03:55.150 ","End":"04:03.820","Text":"Surrounded by 5 chlorine atoms and S surrounded by 6 fluorine atoms."},{"Start":"04:03.820 ","End":"04:06.870","Text":"Now, if we look at phosphorus,"},{"Start":"04:06.870 ","End":"04:08.070","Text":"we\u0027ll see there are 2,"},{"Start":"04:08.070 ","End":"04:09.360","Text":"4, 6, 8,"},{"Start":"04:09.360 ","End":"04:12.245","Text":"10 electrons around the phosphorus."},{"Start":"04:12.245 ","End":"04:15.185","Text":"Now if we look at sulfur, we\u0027ll see there are 2, 4, 6,"},{"Start":"04:15.185 ","End":"04:16.565","Text":"8, 10,"},{"Start":"04:16.565 ","End":"04:19.235","Text":"12 electrons surrounding it."},{"Start":"04:19.235 ","End":"04:24.455","Text":"These are expanded valence shells or expanded octets."},{"Start":"04:24.455 ","End":"04:30.920","Text":"Now, the explanation usually given is that in the third period,"},{"Start":"04:30.920 ","End":"04:33.860","Text":"we can use D electrons."},{"Start":"04:33.860 ","End":"04:38.555","Text":"Electrons can be promoted into the D orbitals and used for bonding."},{"Start":"04:38.555 ","End":"04:44.705","Text":"Now sometimes we can simply get better Lewis structures by using expanded valence shell."},{"Start":"04:44.705 ","End":"04:49.060","Text":"Here\u0027s an example, the sulfate anion."},{"Start":"04:49.060 ","End":"04:52.130","Text":"Now the SO bonds all have the same length,"},{"Start":"04:52.130 ","End":"04:55.865","Text":"but the intermediate between that of single and double bonds,"},{"Start":"04:55.865 ","End":"04:59.710","Text":"and this can be explained by resonance structures."},{"Start":"04:59.710 ","End":"05:03.065","Text":"Here are three of the resonance structures."},{"Start":"05:03.065 ","End":"05:08.750","Text":"One in which 2 double bonds are in the same line,"},{"Start":"05:08.750 ","End":"05:14.630","Text":"one in which the 2 double bonds are perpendicular to each other,"},{"Start":"05:14.630 ","End":"05:18.785","Text":"there are no adjacent oxygen atoms, and one of which,"},{"Start":"05:18.785 ","End":"05:21.860","Text":"again, 2 double bonds are in the same line,"},{"Start":"05:21.860 ","End":"05:24.620","Text":"but instead of being horizontal, they\u0027re vertical."},{"Start":"05:24.620 ","End":"05:28.430","Text":"Now in addition, there are another three structures"},{"Start":"05:28.430 ","End":"05:31.480","Text":"in which the 2 double bonds are adjacent."},{"Start":"05:31.480 ","End":"05:35.525","Text":"Altogether, we have 6 resonance structures."},{"Start":"05:35.525 ","End":"05:38.705","Text":"Now let\u0027s look at the central sulfur."},{"Start":"05:38.705 ","End":"05:45.095","Text":"It has 2 electrons for each single bond and 4 electrons for each double bond."},{"Start":"05:45.095 ","End":"05:51.800","Text":"That makes 12 electrons around the central sulfur and it\u0027s the same for each of them."},{"Start":"05:51.800 ","End":"05:56.240","Text":"Now we\u0027ve used an expanded octet in order to"},{"Start":"05:56.240 ","End":"06:03.605","Text":"explain that the bond lengths of the SO bond are smaller than single bonds."},{"Start":"06:03.605 ","End":"06:08.150","Text":"This is just to say that there are another three structures with adjacent double bonds."},{"Start":"06:08.150 ","End":"06:13.200","Text":"In this video, we learned about exceptions to the octet rule."}],"ID":21360}],"Thumbnail":null,"ID":101312},{"Name":"Shapes of Molecules - VSEPR Theory","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"VSEPR theory AX4","Duration":"7m 20s","ChapterTopicVideoID":20486,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.905","Text":"In previous videos, we learned about the electronic structure of molecules."},{"Start":"00:04.905 ","End":"00:10.230","Text":"In this video, we\u0027ll discuss a simple theory that gives the shape of molecules."},{"Start":"00:10.230 ","End":"00:13.650","Text":"What do we mean by the shape or the geometry of a molecule?"},{"Start":"00:13.650 ","End":"00:15.870","Text":"In order to know the shape of a molecule,"},{"Start":"00:15.870 ","End":"00:18.585","Text":"we need to know the bond lengths and bond angles."},{"Start":"00:18.585 ","End":"00:23.055","Text":"These can be determined experimentally or by calculation."},{"Start":"00:23.055 ","End":"00:25.335","Text":"Let\u0027s take an example of water."},{"Start":"00:25.335 ","End":"00:29.730","Text":"Now we know the electronic structure or the Lewis structure by now."},{"Start":"00:29.730 ","End":"00:35.205","Text":"We know the oxygen has 2 lone pairs and 2 bond pairs pointing to hydrogens."},{"Start":"00:35.205 ","End":"00:39.185","Text":"Now experimentally, it\u0027s known that the bond lengths of the same"},{"Start":"00:39.185 ","End":"00:45.050","Text":"95.8 picometers and the bond angle is 104.45 degrees."},{"Start":"00:45.050 ","End":"00:47.915","Text":"That means the molecule is bent."},{"Start":"00:47.915 ","End":"00:53.495","Text":"Now we\u0027re going to discuss a simple theory that allows us to predict"},{"Start":"00:53.495 ","End":"00:59.660","Text":"the approximate shape of such a molecule without performing detailed calculations."},{"Start":"00:59.660 ","End":"01:06.020","Text":"Now we consider the repulsion of the electron pairs in the valence shell and from that,"},{"Start":"01:06.020 ","End":"01:08.815","Text":"we can derive the approximate shape of a molecule."},{"Start":"01:08.815 ","End":"01:12.370","Text":"The theory is called V-S-E-P-R theory,"},{"Start":"01:12.370 ","End":"01:19.230","Text":"which stands for Valence Shell Electron Pair Repulsion theory."},{"Start":"01:19.230 ","End":"01:23.210","Text":"The basic idea is its electron groups"},{"Start":"01:23.210 ","End":"01:27.920","Text":"repel each other and should therefore be as far apart as possible."},{"Start":"01:27.920 ","End":"01:30.470","Text":"What do we mean by electron groups?"},{"Start":"01:30.470 ","End":"01:33.575","Text":"These can be bond pairs, or lone pairs,"},{"Start":"01:33.575 ","End":"01:37.700","Text":"or double bonds, or triple bonds, or unpaired electrons."},{"Start":"01:37.700 ","End":"01:41.260","Text":"We\u0027re going to treat them all as electron groups,"},{"Start":"01:41.260 ","End":"01:44.675","Text":"and we\u0027re going to use notation like this."},{"Start":"01:44.675 ","End":"01:47.690","Text":"We\u0027re going to denote the central atom as A,"},{"Start":"01:47.690 ","End":"01:51.230","Text":"the atoms bonded to the central atom as X,"},{"Start":"01:51.230 ","End":"01:55.495","Text":"and the lone pairs on the central atom as E. Now,"},{"Start":"01:55.495 ","End":"01:58.715","Text":"we\u0027re going to talk about the repulsion between the electron groups."},{"Start":"01:58.715 ","End":"02:03.545","Text":"We\u0027ll choose the geometry that maximizes the distance between the electron groups,"},{"Start":"02:03.545 ","End":"02:07.580","Text":"and that will minimize the repulsion between the groups."},{"Start":"02:07.580 ","End":"02:12.710","Text":"We want the electron groups to be as far away as possible from each element."},{"Start":"02:12.710 ","End":"02:14.150","Text":"Once we\u0027ve done that,"},{"Start":"02:14.150 ","End":"02:18.905","Text":"we can make a little correction by remembering that"},{"Start":"02:18.905 ","End":"02:26.555","Text":"lone pairs repel each other a little more than a lone pair repels a bond pair,"},{"Start":"02:26.555 ","End":"02:31.595","Text":"and that\u0027s a little more than a bond pair repels another bond pair."},{"Start":"02:31.595 ","End":"02:34.740","Text":"By bond pair here I mean single bonds,"},{"Start":"02:34.740 ","End":"02:37.035","Text":"double and triple bonds."},{"Start":"02:37.035 ","End":"02:39.435","Text":"Let\u0027s start with AX_4,"},{"Start":"02:39.435 ","End":"02:44.560","Text":"that means a central atom A bonded to 4 atoms X."},{"Start":"02:44.560 ","End":"02:49.070","Text":"In order to maximize the distance"},{"Start":"02:49.070 ","End":"02:56.065","Text":"between the bond pairs or the electron groups,"},{"Start":"02:56.065 ","End":"03:01.820","Text":"then we arrange the electron groups to be like a tetrahedral."},{"Start":"03:01.820 ","End":"03:07.430","Text":"Here we have the 4 electron groups and here are the atoms."},{"Start":"03:07.430 ","End":"03:08.570","Text":"This is X,"},{"Start":"03:08.570 ","End":"03:09.980","Text":"X, X,"},{"Start":"03:09.980 ","End":"03:11.930","Text":"X with A in the middle,"},{"Start":"03:11.930 ","End":"03:14.755","Text":"and the shape is tetrahedral."},{"Start":"03:14.755 ","End":"03:21.240","Text":"You may recall that a tetrahedron has 4 faces which are triangular,"},{"Start":"03:21.240 ","End":"03:24.275","Text":"and all 4 faces are identical."},{"Start":"03:24.275 ","End":"03:28.315","Text":"The electron groups have a tetrahedral geometry,"},{"Start":"03:28.315 ","End":"03:30.480","Text":"a tetrahedral arrangement,"},{"Start":"03:30.480 ","End":"03:37.375","Text":"and the geometry by which I mean the actual atoms are also arranged in a tetrahedron,"},{"Start":"03:37.375 ","End":"03:41.460","Text":"and the tetrahedron angle is 109.5 degrees."},{"Start":"03:41.460 ","End":"03:48.870","Text":"The XAX angle is a 109.5 degrees."},{"Start":"03:48.870 ","End":"03:51.765","Text":"Here\u0027s an example of methane."},{"Start":"03:51.765 ","End":"03:54.130","Text":"Here\u0027s the structure of methane,"},{"Start":"03:54.130 ","End":"03:58.735","Text":"a carbon in the center bonded to 4 hydrogens,"},{"Start":"03:58.735 ","End":"04:01.275","Text":"and here\u0027s how we can draw it."},{"Start":"04:01.275 ","End":"04:06.620","Text":"This is called the dash and wedge structure,"},{"Start":"04:06.620 ","End":"04:08.570","Text":"or wedge and dash,"},{"Start":"04:08.570 ","End":"04:10.880","Text":"depending on who you talk to."},{"Start":"04:10.880 ","End":"04:14.960","Text":"In this structure, the 2 bonds,"},{"Start":"04:14.960 ","End":"04:17.720","Text":"CH here and CH here,"},{"Start":"04:17.720 ","End":"04:20.650","Text":"are in the plane of the paper."},{"Start":"04:20.650 ","End":"04:28.070","Text":"This CH bond, where we have the wedge is pointing out of the plane of the paper."},{"Start":"04:28.070 ","End":"04:33.590","Text":"This CH bond which has been denoted by dashes,"},{"Start":"04:33.590 ","End":"04:36.305","Text":"is pointing into the paper."},{"Start":"04:36.305 ","End":"04:41.590","Text":"This denotes a one-dimensional structure on a piece of paper."},{"Start":"04:41.590 ","End":"04:49.670","Text":"Sometimes 1 of the Xs is replaced by a lone pair. Here\u0027s AX3E."},{"Start":"04:49.670 ","End":"04:52.115","Text":"The central atom is bonded to 3 atoms,"},{"Start":"04:52.115 ","End":"04:54.680","Text":"and we have 1 lone pair on A."},{"Start":"04:54.680 ","End":"04:59.375","Text":"The lone pair here is denoted by yellow ball."},{"Start":"04:59.375 ","End":"05:03.845","Text":"Once again, the electron groups are arranged in a tetrahedral arrangement,"},{"Start":"05:03.845 ","End":"05:07.310","Text":"so they\u0027re as far as possible away from each other."},{"Start":"05:07.310 ","End":"05:08.805","Text":"But the geometry,"},{"Start":"05:08.805 ","End":"05:15.420","Text":"that\u0027s the actual shape of the AX3 molecule is trigonal pyramidal."},{"Start":"05:15.420 ","End":"05:18.065","Text":"It looks like a tripod."},{"Start":"05:18.065 ","End":"05:22.310","Text":"Now, because the lone pair repels the bonded pair a"},{"Start":"05:22.310 ","End":"05:26.625","Text":"little more than each bond pair repels each other,"},{"Start":"05:26.625 ","End":"05:30.710","Text":"then these bond pairs are pushed"},{"Start":"05:30.710 ","End":"05:37.545","Text":"together a little so the angle is slightly less than 109.5 degrees."},{"Start":"05:37.545 ","End":"05:39.750","Text":"Here\u0027s an example, ammonia."},{"Start":"05:39.750 ","End":"05:46.340","Text":"We know that ammonia has a lone pair on the nitrogen and 3 bonds to the hydrogens."},{"Start":"05:46.340 ","End":"05:49.085","Text":"Here is its arrangement,"},{"Start":"05:49.085 ","End":"05:50.795","Text":"here is its geometry,"},{"Start":"05:50.795 ","End":"05:56.065","Text":"nitrogen with a lone pair and 3 bonds to hydrogen."},{"Start":"05:56.065 ","End":"06:01.280","Text":"The 3 bonds are arranged like a trigonal pyramid and"},{"Start":"06:01.280 ","End":"06:07.430","Text":"the bond angle is a little less than a 109.5 degrees, it\u0027s 106.8 degrees."},{"Start":"06:07.430 ","End":"06:11.970","Text":"Now, if 2Xs are replaced by 2 lone pairs,"},{"Start":"06:11.970 ","End":"06:14.175","Text":"we get AX2E2,"},{"Start":"06:14.175 ","End":"06:18.245","Text":"that\u0027s the central atom bonded to 2 atoms and 2 lone pairs."},{"Start":"06:18.245 ","End":"06:21.350","Text":"Here\u0027s a picture, a ball-and-stick picture."},{"Start":"06:21.350 ","End":"06:24.600","Text":"We have, on the red is A,"},{"Start":"06:24.600 ","End":"06:26.535","Text":"the white is X,"},{"Start":"06:26.535 ","End":"06:29.420","Text":"and the yellow is E. Again,"},{"Start":"06:29.420 ","End":"06:33.305","Text":"the electron groups are arranged in a tetrahedral arrangement."},{"Start":"06:33.305 ","End":"06:38.180","Text":"The geometry of the AX2 molecule is"},{"Start":"06:38.180 ","End":"06:43.444","Text":"bent and the bond angle has to be less than 109.5 degrees."},{"Start":"06:43.444 ","End":"06:46.235","Text":"Here\u0027s an example of water,"},{"Start":"06:46.235 ","End":"06:53.230","Text":"which we know by now has 2 lone pairs on the oxygen and 2 bonds to the hydrogens."},{"Start":"06:53.230 ","End":"06:59.915","Text":"The bond angle is even less than an ammonia, it\u0027s 104.5 degrees."},{"Start":"06:59.915 ","End":"07:02.600","Text":"We control the molecule just as we did right at"},{"Start":"07:02.600 ","End":"07:06.409","Text":"the beginning of the video has a bent molecule."},{"Start":"07:06.409 ","End":"07:12.230","Text":"In this video we talked about VSEPR theory and AX4."},{"Start":"07:12.230 ","End":"07:13.700","Text":"In the next videos,"},{"Start":"07:13.700 ","End":"07:15.710","Text":"we\u0027ll talk about AX2,"},{"Start":"07:15.710 ","End":"07:20.580","Text":"AX3, AX5, and AX6."}],"ID":21361},{"Watched":false,"Name":"VSEPR theory AX2 AX3","Duration":"4m 16s","ChapterTopicVideoID":23068,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"In the previous video,"},{"Start":"00:01.890 ","End":"00:07.379","Text":"we learned about the VSEPR theory and applied it to AX4 molecules."},{"Start":"00:07.379 ","End":"00:11.985","Text":"In this video, we\u0027ll apply it to AX_2 and AX_3 molecules."},{"Start":"00:11.985 ","End":"00:14.175","Text":"Let us begin with AX_2."},{"Start":"00:14.175 ","End":"00:19.425","Text":"That\u0027s a central atom A bonded to 2x atoms."},{"Start":"00:19.425 ","End":"00:24.435","Text":"In order to keep the electron groups as far away as possible from each other,"},{"Start":"00:24.435 ","End":"00:27.600","Text":"they adopt a linear arrangement."},{"Start":"00:27.600 ","End":"00:31.935","Text":"The electron groups have a linear arrangement and the geometry to the actual molecule"},{"Start":"00:31.935 ","End":"00:36.840","Text":"AX_2 is also linear and the bond angle is 180 degrees."},{"Start":"00:36.840 ","End":"00:42.350","Text":"An example of this is BeH_2, that\u0027s beryllium hydride."},{"Start":"00:42.350 ","End":"00:47.080","Text":"Now you\u0027ll recall that beryllium has just 2 unpaired electrons"},{"Start":"00:47.080 ","End":"00:52.639","Text":"and it uses these 2 unpaired electrons to form 2 beryllium hydrogen bonds."},{"Start":"00:52.639 ","End":"00:55.610","Text":"This is a case of an incomplete octet because"},{"Start":"00:55.610 ","End":"01:00.305","Text":"now the beryllium has only 4 electrons around about it."},{"Start":"01:00.305 ","End":"01:03.065","Text":"Now let\u0027s look at the AX_3 molecule."},{"Start":"01:03.065 ","End":"01:08.570","Text":"That\u0027s a central atom A bonded to 3x atoms in order to"},{"Start":"01:08.570 ","End":"01:14.530","Text":"keep the electron groups as far away as possible from each other,"},{"Start":"01:14.530 ","End":"01:20.670","Text":"the arrangement is in a plane, trigonal planar arrangement."},{"Start":"01:20.670 ","End":"01:23.115","Text":"Here we have x, x,"},{"Start":"01:23.115 ","End":"01:26.805","Text":"x and the angle is a 120 degrees."},{"Start":"01:26.805 ","End":"01:29.375","Text":"The electron groups are trigonal planar."},{"Start":"01:29.375 ","End":"01:35.195","Text":"The jump to that molecule is trigonal planar and the bond angle is 120 degrees."},{"Start":"01:35.195 ","End":"01:38.830","Text":"An example is boron trifluoride."},{"Start":"01:38.830 ","End":"01:45.620","Text":"You recall that boron has only 3 electrons that are unpaired."},{"Start":"01:45.620 ","End":"01:52.204","Text":"They can use them to form 3 boron fluoride bonds."},{"Start":"01:52.204 ","End":"01:56.495","Text":"Now we can draw the boron fluoride molecule,"},{"Start":"01:56.495 ","End":"02:01.420","Text":"3BF bonds and 120 degrees between."},{"Start":"02:01.420 ","End":"02:06.270","Text":"Supposing, we now replace 1 of the x atoms by"},{"Start":"02:06.270 ","End":"02:12.125","Text":"a lone pair so it\u0027s not an AX_2E rather than AX_3."},{"Start":"02:12.125 ","End":"02:15.070","Text":"Let\u0027s take 1 of these x atoms,"},{"Start":"02:15.070 ","End":"02:17.755","Text":"replace it by a lone pair,"},{"Start":"02:17.755 ","End":"02:20.125","Text":"which I\u0027ve indicated by yellow."},{"Start":"02:20.125 ","End":"02:24.070","Text":"The electron groups again have a trigonal planar arrangement,"},{"Start":"02:24.070 ","End":"02:29.770","Text":"but the geometry of the molecule itself is just bent X, A, X."},{"Start":"02:29.770 ","End":"02:32.050","Text":"The bond angle has to be a little less than"},{"Start":"02:32.050 ","End":"02:37.555","Text":"a 120 degrees because the lone pair repels the bond pair."},{"Start":"02:37.555 ","End":"02:41.215","Text":"The examples of this is ozone, O3."},{"Start":"02:41.215 ","End":"02:47.405","Text":"Ozone is a hybrid consisting of 2 resonance structures."},{"Start":"02:47.405 ","End":"02:49.100","Text":"On the left-hand side,"},{"Start":"02:49.100 ","End":"02:52.954","Text":"we have a double bond between the central oxygen"},{"Start":"02:52.954 ","End":"02:58.010","Text":"and the exterior oxygen and a single bond with the other oxygen."},{"Start":"02:58.010 ","End":"03:01.010","Text":"On the right-hand side of the double bond,"},{"Start":"03:01.010 ","End":"03:06.005","Text":"where the single bond was before and a single bond with a double bond was before."},{"Start":"03:06.005 ","End":"03:09.830","Text":"The plus and minus indicate formal charge."},{"Start":"03:09.830 ","End":"03:11.810","Text":"Here we have a central atom,"},{"Start":"03:11.810 ","End":"03:14.429","Text":"oxygen and 2 x\u0027s,"},{"Start":"03:14.429 ","End":"03:16.470","Text":"which are now also oxygen,"},{"Start":"03:16.470 ","End":"03:19.050","Text":"and a lone pair on the central atom."},{"Start":"03:19.050 ","End":"03:22.135","Text":"This is AX_2E."},{"Start":"03:22.135 ","End":"03:28.700","Text":"Now experimentally it\u0027s found that the OOO angle is a 116.8 degrees,"},{"Start":"03:28.700 ","End":"03:32.360","Text":"which is slightly smaller than 120 degrees."},{"Start":"03:32.360 ","End":"03:35.480","Text":"Now we can compare this with water,"},{"Start":"03:35.480 ","End":"03:40.610","Text":"which we learned about in the previous videos is AX_2E_2 and there,"},{"Start":"03:40.610 ","End":"03:45.875","Text":"the HOH angle is a 104.5 degrees."},{"Start":"03:45.875 ","End":"03:49.715","Text":"That\u0027s a little less than the tetrahedral angle,"},{"Start":"03:49.715 ","End":"03:53.375","Text":"which is 109.5 degrees."},{"Start":"03:53.375 ","End":"03:57.215","Text":"We can see the difference now between AX_2E,"},{"Start":"03:57.215 ","End":"04:03.260","Text":"where the bond angle is close to 120 and AX_2E where"},{"Start":"04:03.260 ","End":"04:09.860","Text":"a bond angle is closer a little less than the tetrahedral angle."},{"Start":"04:09.860 ","End":"04:17.040","Text":"In this video, we applied VSEPR theory to AX_2 and AX_3 molecules."}],"ID":23910},{"Watched":false,"Name":"VSEPR theory AX5","Duration":"5m 15s","ChapterTopicVideoID":23069,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.030 ","End":"00:02.230","Text":"In the previous videos,"},{"Start":"00:02.230 ","End":"00:06.565","Text":"we applied VSEPR theory to AX2, AX3,"},{"Start":"00:06.565 ","End":"00:09.640","Text":"and AX4 molecules, in this video,"},{"Start":"00:09.640 ","End":"00:12.020","Text":"we\u0027ll apply it to AX5 molecule,"},{"Start":"00:12.020 ","End":"00:16.120","Text":"so first let us recall what we learned about expanded valence shells."},{"Start":"00:16.120 ","End":"00:21.070","Text":"We learned that some third period elements such as sulfur and phosphorus,"},{"Start":"00:21.070 ","End":"00:25.090","Text":"form compounds that can only be explained by having 10,"},{"Start":"00:25.090 ","End":"00:27.100","Text":"or 12 electrons around them."},{"Start":"00:27.100 ","End":"00:31.405","Text":"In this video, we\u0027re going to talk about AX5 molecules,"},{"Start":"00:31.405 ","End":"00:36.060","Text":"which have 10 electrons around the atom A,"},{"Start":"00:36.060 ","End":"00:42.910","Text":"and the next video we\u0027ll talk about AX6 molecules that have 12 electrons around A."},{"Start":"00:42.910 ","End":"00:45.820","Text":"Let\u0027s discuss AX5 molecules,"},{"Start":"00:45.820 ","End":"00:49.405","Text":"they have a central atom bonded to 5 atoms,"},{"Start":"00:49.405 ","End":"00:53.450","Text":"here\u0027s the arrangement of the electron groups,"},{"Start":"00:53.450 ","End":"00:57.425","Text":"this is a trigonal bipyramidal arrangement."},{"Start":"00:57.425 ","End":"00:58.970","Text":"The electron groups are"},{"Start":"00:58.970 ","End":"01:02.900","Text":"trigonal bipyramidal and the geometry of the actual molecule is also"},{"Start":"01:02.900 ","End":"01:08.600","Text":"trigonal bipyramidal and I have to consider what the bond angles are."},{"Start":"01:08.600 ","End":"01:13.095","Text":"We have 2 bond angles and we have 2 atoms."},{"Start":"01:13.095 ","End":"01:19.900","Text":"The atoms, this 1 and this 1 are in an axial positions whereas"},{"Start":"01:19.900 ","End":"01:27.725","Text":"the atoms that form the base of the pyramid are in equatorial positions."},{"Start":"01:27.725 ","End":"01:31.220","Text":"Now the angle between an axial bond,"},{"Start":"01:31.220 ","End":"01:34.700","Text":"and an equatorial bond is just 90 degrees,"},{"Start":"01:34.700 ","End":"01:40.560","Text":"and the angle between 2 equatorial bonds is 120 degrees."},{"Start":"01:40.560 ","End":"01:43.500","Text":"We really have 2 bonds in"},{"Start":"01:43.500 ","End":"01:47.840","Text":"the AX5 molecules and you can show"},{"Start":"01:47.840 ","End":"01:53.480","Text":"experimentally that the axial bonds are slightly longer than the equatorial."},{"Start":"01:53.480 ","End":"02:00.265","Text":"An example of this is PCL_5 which has 10 electrons around the phosphorus."},{"Start":"02:00.265 ","End":"02:04.650","Text":"Now supposing we replace 1 of the X atoms by lone pair,"},{"Start":"02:04.650 ","End":"02:07.395","Text":"so now we have AX_4E,"},{"Start":"02:07.395 ","End":"02:12.255","Text":"so it\u0027s the central atom bonded to 4 atoms 1 lone pair on A,"},{"Start":"02:12.255 ","End":"02:20.060","Text":"so here\u0027s the arrangement of the electron groups and again, it\u0027s trigonal bipyramidal."},{"Start":"02:20.060 ","End":"02:23.270","Text":"Now, we\u0027re going to put the lone pair,"},{"Start":"02:23.270 ","End":"02:30.545","Text":"we can obviously either put it in an equatorial position or in an axial position."},{"Start":"02:30.545 ","End":"02:37.115","Text":"It turns out that there\u0027s less repulsion when it\u0027s in the equatorial position."},{"Start":"02:37.115 ","End":"02:39.770","Text":"The way we put the lone pair there we see that"},{"Start":"02:39.770 ","End":"02:44.930","Text":"the geometry of the actual molecule is like a seesaw,"},{"Start":"02:44.930 ","End":"02:46.010","Text":"I\u0027ve turned this round,"},{"Start":"02:46.010 ","End":"02:50.165","Text":"this is turned round and so we have a seesaw arrangement."},{"Start":"02:50.165 ","End":"02:52.780","Text":"Now, what about the angles?"},{"Start":"02:52.780 ","End":"02:54.390","Text":"Well, look at this,"},{"Start":"02:54.390 ","End":"03:01.500","Text":"this is a 120 if there\u0027s no extra repulsion and 90,"},{"Start":"03:01.500 ","End":"03:07.240","Text":"now because of the repulsion between the lone pairs and the bond pairs,"},{"Start":"03:07.240 ","End":"03:09.085","Text":"here\u0027s a lone pair,"},{"Start":"03:09.085 ","End":"03:16.300","Text":"these angles are slightly less than 90 and 120 degrees."},{"Start":"03:16.300 ","End":"03:20.235","Text":"The electron groups are trigonal bipyramidal as here,"},{"Start":"03:20.235 ","End":"03:22.254","Text":"the jump is like a seesaw,"},{"Start":"03:22.254 ","End":"03:28.300","Text":"and the bond angles are less than 90 degrees and less than 120 degrees for example,"},{"Start":"03:28.300 ","End":"03:35.160","Text":"SF_4 here the bond angles are 86 degrees which is a little less than 90 degrees,"},{"Start":"03:35.160 ","End":"03:41.185","Text":"and 102 degrees, which is quite a lot less than a 120 degrees."},{"Start":"03:41.185 ","End":"03:46.460","Text":"Now, supposing we replace another of the X atoms by another lone pair,"},{"Start":"03:46.460 ","End":"03:49.510","Text":"so now we have a central atom bonded to 3 atoms and"},{"Start":"03:49.510 ","End":"03:53.575","Text":"2 lone pairs on A, so that\u0027s AX_3E_2."},{"Start":"03:53.575 ","End":"03:56.080","Text":"Again, the electron groups are like"},{"Start":"03:56.080 ","End":"04:01.370","Text":"a trigonal bipyramidal arrangement and the actual molecule,"},{"Start":"04:01.370 ","End":"04:05.545","Text":"well, let\u0027s see first where we should put the lone pairs."},{"Start":"04:05.545 ","End":"04:09.310","Text":"Now if we put the lone pairs on the equatorial arrangement,"},{"Start":"04:09.310 ","End":"04:11.950","Text":"they\u0027ll be quite far away from each other,"},{"Start":"04:11.950 ","End":"04:18.705","Text":"further away than if we put 1 of them in the axial range,"},{"Start":"04:18.705 ","End":"04:20.965","Text":"because as we learned before,"},{"Start":"04:20.965 ","End":"04:27.280","Text":"lone pairs repel each other more than a lone pair repels a bonded pair,"},{"Start":"04:27.280 ","End":"04:32.305","Text":"so the 2 lone pairs are on the equatorial positions."},{"Start":"04:32.305 ","End":"04:34.810","Text":"When we look at the actual molecule,"},{"Start":"04:34.810 ","End":"04:38.280","Text":"we see we have a T arrangement here it is."},{"Start":"04:38.280 ","End":"04:40.495","Text":"Now what about the angles?"},{"Start":"04:40.495 ","End":"04:46.270","Text":"We started off with 90 degrees but now the lone pairs are pushing,"},{"Start":"04:46.270 ","End":"04:48.280","Text":"repelling the bond pairs,"},{"Start":"04:48.280 ","End":"04:52.195","Text":"so we expect to get less than 90 degrees."},{"Start":"04:52.195 ","End":"04:54.895","Text":"To summarize this, the electron groups"},{"Start":"04:54.895 ","End":"04:57.580","Text":"are arranged in a trigonal bipyramidal arrangement,"},{"Start":"04:57.580 ","End":"05:03.025","Text":"the geometry is T-shaped and the bond angles are less than 90 degrees."},{"Start":"05:03.025 ","End":"05:06.000","Text":"For example, ClF_3,"},{"Start":"05:06.000 ","End":"05:09.615","Text":"here the bond angles are 88 degrees,"},{"Start":"05:09.615 ","End":"05:10.895","Text":"so in this video,"},{"Start":"05:10.895 ","End":"05:15.720","Text":"we discussed the geometry of AX5 molecules."}],"ID":23911},{"Watched":false,"Name":"VSEPR theory AX6","Duration":"3m 14s","ChapterTopicVideoID":23070,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"In the previous videos,"},{"Start":"00:01.950 ","End":"00:05.700","Text":"we applied VSEPR theory to AX_2,"},{"Start":"00:05.700 ","End":"00:09.285","Text":"AX_3, AX_4, and AX_5 molecules."},{"Start":"00:09.285 ","End":"00:13.485","Text":"In this video, we\u0027ll apply it to AX_6 molecules."},{"Start":"00:13.485 ","End":"00:18.060","Text":"AX_6 means that there\u0027s a central atom A bonded to"},{"Start":"00:18.060 ","End":"00:23.240","Text":"6 atoms X and A has 12 electrons around it."},{"Start":"00:23.240 ","End":"00:26.595","Text":"We\u0027re talking about an expanded valence shell."},{"Start":"00:26.595 ","End":"00:28.680","Text":"So here\u0027s our molecule."},{"Start":"00:28.680 ","End":"00:32.900","Text":"The electron groups are arranged in an octahedral arrangement."},{"Start":"00:32.900 ","End":"00:34.759","Text":"This is an octahedron."},{"Start":"00:34.759 ","End":"00:38.284","Text":"The geometry of the actual molecule is also octahedral,"},{"Start":"00:38.284 ","End":"00:41.660","Text":"so all the bond angles are 90 degrees."},{"Start":"00:41.660 ","End":"00:46.670","Text":"We should note that all the angles here are equal and all the bond lengths are equal."},{"Start":"00:46.670 ","End":"00:52.690","Text":"This is different from AX_5 that we spoke about in the previous video."},{"Start":"00:52.690 ","End":"00:57.680","Text":"A good example of this is SF_6 which has 12 electrons"},{"Start":"00:57.680 ","End":"01:02.660","Text":"around S. Now suppose that we replace 1 of the X\u0027s by a lone pair,"},{"Start":"01:02.660 ","End":"01:05.423","Text":"so we have AX_5E."},{"Start":"01:05.423 ","End":"01:10.855","Text":"The central atom bonded to 5 atoms and 1 lone pair on A."},{"Start":"01:10.855 ","End":"01:13.489","Text":"Here is our octahedral arrangement."},{"Start":"01:13.489 ","End":"01:18.230","Text":"We have to decide which atom is to be replaced by lone pair."},{"Start":"01:18.230 ","End":"01:21.350","Text":"Since all the atoms are equivalent,"},{"Start":"01:21.350 ","End":"01:22.820","Text":"it doesn\u0027t matter where we put it."},{"Start":"01:22.820 ","End":"01:25.010","Text":"Let\u0027s put it, for example, down here."},{"Start":"01:25.010 ","End":"01:31.970","Text":"Then we get a square pyramidal arrangement for the actual molecule."},{"Start":"01:31.970 ","End":"01:34.480","Text":"The electron groups are octahedral,"},{"Start":"01:34.480 ","End":"01:37.670","Text":"the geometry of the molecule is square pyramidal,"},{"Start":"01:37.670 ","End":"01:40.310","Text":"and the bond angles are less than 90 degrees."},{"Start":"01:40.310 ","End":"01:42.290","Text":"Why are they less than 90 degrees?"},{"Start":"01:42.290 ","End":"01:49.600","Text":"Because the lone pair that\u0027s down here is repelling each of these bond pairs,"},{"Start":"01:49.600 ","End":"01:52.020","Text":"so they turn up a little."},{"Start":"01:52.020 ","End":"01:57.860","Text":"An example of this is BrF_5 where the bond angles are 88 degrees."},{"Start":"01:57.860 ","End":"02:01.370","Text":"Now supposing we replace another of the atoms by lone pair."},{"Start":"02:01.370 ","End":"02:06.550","Text":"Now we have a central atom bonded to 4 atoms and 2 lone pairs on A."},{"Start":"02:06.550 ","End":"02:08.720","Text":"Again, the electron groups form"},{"Start":"02:08.720 ","End":"02:13.055","Text":"an octahedron and we have to decide where to put the 2 lone pairs."},{"Start":"02:13.055 ","End":"02:15.980","Text":"Now we know that a lone pair repels"},{"Start":"02:15.980 ","End":"02:21.190","Text":"another lone pair more than the lone pair repels a bonding pair."},{"Start":"02:21.190 ","End":"02:25.850","Text":"We need to keep the lone pairs as far away as possible from each other."},{"Start":"02:25.850 ","End":"02:29.410","Text":"Let\u0027s put them in opposite positions."},{"Start":"02:29.410 ","End":"02:34.465","Text":"Now the geometry of the actual molecule is square planar."},{"Start":"02:34.465 ","End":"02:38.400","Text":"To summarize that, the electron groups are in octahedral arrangement,"},{"Start":"02:38.400 ","End":"02:40.130","Text":"the geometry is square planar,"},{"Start":"02:40.130 ","End":"02:45.080","Text":"and the bond angles are 90 degrees because the repulsion between"},{"Start":"02:45.080 ","End":"02:50.135","Text":"this lone pair and say this atom will push the atom downwards,"},{"Start":"02:50.135 ","End":"02:52.895","Text":"but the repulsion between this atom"},{"Start":"02:52.895 ","End":"02:56.075","Text":"and the lone pair will push it upwards at an equal amount,"},{"Start":"02:56.075 ","End":"03:01.610","Text":"so they stay in the same place and the bond angles are exactly 90 degrees."},{"Start":"03:01.610 ","End":"03:06.530","Text":"An example would be XeF_4 and there experimentally,"},{"Start":"03:06.530 ","End":"03:09.650","Text":"the bond angles are indeed 90 degrees."},{"Start":"03:09.650 ","End":"03:14.970","Text":"In this video, we discussed the geometry of AX_6 molecules."}],"ID":23912},{"Watched":false,"Name":"Applying VSEPR theory","Duration":"3m 55s","ChapterTopicVideoID":23066,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.065","Text":"In previous videos we discussed VSEPR theory."},{"Start":"00:04.065 ","End":"00:06.900","Text":"In this video, we\u0027ll learn how to apply this theory."},{"Start":"00:06.900 ","End":"00:08.280","Text":"The first thing to do,"},{"Start":"00:08.280 ","End":"00:11.640","Text":"is to draw a loop plausible Lewis structure for the molecule,"},{"Start":"00:11.640 ","End":"00:13.065","Text":"or molecular ion,"},{"Start":"00:13.065 ","End":"00:15.165","Text":"or whatever species we\u0027re talking about."},{"Start":"00:15.165 ","End":"00:17.700","Text":"Then we count the number of bond pairs,"},{"Start":"00:17.700 ","End":"00:20.235","Text":"and that includes double and triple bonds,"},{"Start":"00:20.235 ","End":"00:22.325","Text":"and also the number of lone pairs,"},{"Start":"00:22.325 ","End":"00:24.380","Text":"around each central atom."},{"Start":"00:24.380 ","End":"00:28.890","Text":"Then we assign an AXE notation for each central atom."},{"Start":"00:28.890 ","End":"00:32.040","Text":"A molecule can have more than 1 central atom."},{"Start":"00:32.040 ","End":"00:35.810","Text":"Then we use this AXE notation to"},{"Start":"00:35.810 ","End":"00:39.899","Text":"determine the geometry of atoms bonded to each central atom,"},{"Start":"00:39.899 ","End":"00:41.975","Text":"and finally we draw the molecule."},{"Start":"00:41.975 ","End":"00:46.445","Text":"We\u0027re first going to consider the case where there\u0027s a 1 central atom."},{"Start":"00:46.445 ","End":"00:48.815","Text":"We\u0027re going to take the example of carbonate,"},{"Start":"00:48.815 ","End":"00:53.855","Text":"CO_3^2 minus, which we\u0027ve talked about in previous videos."},{"Start":"00:53.855 ","End":"00:56.780","Text":"Here\u0027s the structure of carbonate."},{"Start":"00:56.780 ","End":"01:02.720","Text":"There are 3 resonance structures that form a resonance hybrid."},{"Start":"01:02.720 ","End":"01:07.325","Text":"In each of them, there are 2 single bonds and 1 double bond."},{"Start":"01:07.325 ","End":"01:13.995","Text":"Now we have to decide to which of the AXE categories this belongs."},{"Start":"01:13.995 ","End":"01:19.115","Text":"Now, we know that you don\u0027t take account of whether we have a single bond,"},{"Start":"01:19.115 ","End":"01:20.525","Text":"or a double bond,"},{"Start":"01:20.525 ","End":"01:22.780","Text":"in deciding the category."},{"Start":"01:22.780 ","End":"01:25.200","Text":"Here we have 3 bonds,"},{"Start":"01:25.200 ","End":"01:26.805","Text":"and no lone pairs,"},{"Start":"01:26.805 ","End":"01:28.710","Text":"so C is A,"},{"Start":"01:28.710 ","End":"01:32.055","Text":"and each O atom is X."},{"Start":"01:32.055 ","End":"01:38.915","Text":"That means that the category that carbonate belongs to is AX3."},{"Start":"01:38.915 ","End":"01:41.090","Text":"Remembering what we learned before,"},{"Start":"01:41.090 ","End":"01:43.660","Text":"we know that this is trigonal planar,"},{"Start":"01:43.660 ","End":"01:46.669","Text":"with bond angles 120 degrees,"},{"Start":"01:46.669 ","End":"01:48.740","Text":"so now we can draw the molecule."},{"Start":"01:48.740 ","End":"01:51.560","Text":"Here\u0027s the molecule, central carbon,"},{"Start":"01:51.560 ","End":"01:53.495","Text":"3 oxygen surrounding it,"},{"Start":"01:53.495 ","End":"02:00.545","Text":"and the angle between each CO bond, is 120 degrees."},{"Start":"02:00.545 ","End":"02:04.055","Text":"This is the structure of carbonate."},{"Start":"02:04.055 ","End":"02:07.685","Text":"Now let\u0027s take a case where there are 2 central atoms."},{"Start":"02:07.685 ","End":"02:10.460","Text":"An example we\u0027re going to choose is once again,"},{"Start":"02:10.460 ","End":"02:14.405","Text":"an example where we\u0027ve already discussed the Lewis structure,"},{"Start":"02:14.405 ","End":"02:20.420","Text":"and this is hydrazine, H2NNH2."},{"Start":"02:20.420 ","End":"02:21.830","Text":"Here\u0027s the Lewis structure."},{"Start":"02:21.830 ","End":"02:27.150","Text":"Each nitrogen is bonded to 3 atoms."},{"Start":"02:27.150 ","End":"02:29.250","Text":"This nitrogen is bond to another nitrogen,"},{"Start":"02:29.250 ","End":"02:32.795","Text":"and 2 hydrogens, and it also has a lone pair."},{"Start":"02:32.795 ","End":"02:35.040","Text":"The second nitrogen is the same,"},{"Start":"02:35.040 ","End":"02:36.410","Text":"has a lone pair,"},{"Start":"02:36.410 ","End":"02:40.685","Text":"and it\u0027s bonded to another nitrogen and 2 hydrogens."},{"Start":"02:40.685 ","End":"02:44.810","Text":"We can work out the AXE configuration."},{"Start":"02:44.810 ","End":"02:49.080","Text":"Each nitrogen is a 3 atoms around about,"},{"Start":"02:49.080 ","End":"02:50.760","Text":"so it\u0027s AX3,"},{"Start":"02:50.760 ","End":"02:52.890","Text":"and 1 lone pair,"},{"Start":"02:52.890 ","End":"02:58.450","Text":"E. We have both nitrogen atoms are AX3E."},{"Start":"02:58.450 ","End":"03:06.350","Text":"Now, we learned before that AX3E means it\u0027s trigonal pyramidal around each N atom."},{"Start":"03:06.350 ","End":"03:09.200","Text":"Here, as opposed to the previous case,"},{"Start":"03:09.200 ","End":"03:14.045","Text":"we have a 3-dimensional structure rather than a planar structure."},{"Start":"03:14.045 ","End":"03:17.060","Text":"We can draw the molecule in 2 ways."},{"Start":"03:17.060 ","End":"03:19.685","Text":"Here we have the ball and stick model,"},{"Start":"03:19.685 ","End":"03:22.643","Text":"usually drawn by a computer,"},{"Start":"03:22.643 ","End":"03:26.570","Text":"and here we have the dash and wedge depiction,"},{"Start":"03:26.570 ","End":"03:29.795","Text":"which is what organic chemists usually use."},{"Start":"03:29.795 ","End":"03:36.045","Text":"Here we have nitrogen with 2 hydrogen atoms pointing out of the plane,"},{"Start":"03:36.045 ","End":"03:38.315","Text":"and this N-N is in the plane."},{"Start":"03:38.315 ","End":"03:45.075","Text":"Then here we have the second nitrogen with 1 hydrogen pointing out of the plane,"},{"Start":"03:45.075 ","End":"03:47.685","Text":"and 1 hydrogen into the plane."},{"Start":"03:47.685 ","End":"03:50.885","Text":"Here we have the structure of the molecule."},{"Start":"03:50.885 ","End":"03:55.950","Text":"In this video, we learned how to apply VSEPR theory."}],"ID":23908},{"Watched":false,"Name":"Exercise 10","Duration":"12m 51s","ChapterTopicVideoID":23608,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:03.405","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.405 ","End":"00:05.940","Text":"Use the VSEPR theory to determine"},{"Start":"00:05.940 ","End":"00:10.575","Text":"the electron group geometry and the molecular geometry of the following molecules."},{"Start":"00:10.575 ","End":"00:14.040","Text":"We\u0027re going to begin with a. In a we have sulfur tetrachloride."},{"Start":"00:14.040 ","End":"00:16.890","Text":"Now the first step that we have to do is to draw"},{"Start":"00:16.890 ","End":"00:20.940","Text":"a plausible Lewis structure for our molecule."},{"Start":"00:20.940 ","End":"00:26.490","Text":"We\u0027re going to put the sulfur as the central atom."},{"Start":"00:26.490 ","End":"00:29.260","Text":"We\u0027re going to connect the chlorines."},{"Start":"00:33.340 ","End":"00:36.860","Text":"Now, remember the first step to write the Lewis structure is"},{"Start":"00:36.860 ","End":"00:39.895","Text":"we have to calculate the number of valence electrons."},{"Start":"00:39.895 ","End":"00:45.750","Text":"Our sulfur has 6 valence electrons because it\u0027s in the 16th group plus 4"},{"Start":"00:45.750 ","End":"00:52.660","Text":"times 7 because we have 4 chlorines and each one has 7 valence electrons,"},{"Start":"00:52.970 ","End":"00:57.690","Text":"and this gives us a total of 34 electrons."},{"Start":"00:57.940 ","End":"01:02.975","Text":"Now we should write the skeleton for our structure."},{"Start":"01:02.975 ","End":"01:04.880","Text":"We put the sulfur in the middle,"},{"Start":"01:04.880 ","End":"01:06.350","Text":"it\u0027s a central atom,"},{"Start":"01:06.350 ","End":"01:08.840","Text":"and we put the chlorines as terminal atoms."},{"Start":"01:08.840 ","End":"01:11.425","Text":"Now we have 4 bonds here."},{"Start":"01:11.425 ","End":"01:14.115","Text":"We\u0027re going to subtract these 4 bonds,"},{"Start":"01:14.115 ","End":"01:19.055","Text":"8 electrons from the total number of valence electrons,"},{"Start":"01:19.055 ","End":"01:24.020","Text":"and that\u0027s going to leave us with 26 electrons to fill in."},{"Start":"01:24.020 ","End":"01:29.270","Text":"We\u0027re going to start by filling octets for the terminal atoms."},{"Start":"01:29.270 ","End":"01:33.254","Text":"We\u0027re going to give 6 electrons to each chlorine atom 1,"},{"Start":"01:33.254 ","End":"01:34.409","Text":"2, 3, 4, 5,"},{"Start":"01:34.409 ","End":"01:35.563","Text":"6, 7, 8, 9,"},{"Start":"01:35.563 ","End":"01:36.862","Text":"10, 11, 12,"},{"Start":"01:36.862 ","End":"01:38.594","Text":"13, 14, 15, 16,"},{"Start":"01:38.594 ","End":"01:39.892","Text":"17, 18, 19,"},{"Start":"01:39.892 ","End":"01:42.285","Text":"20, 21, 22, 23, 24."},{"Start":"01:42.285 ","End":"01:45.755","Text":"We know we have 26 electrons to put in, so we have 2 left."},{"Start":"01:45.755 ","End":"01:48.510","Text":"We\u0027re going to add them to our sulfur."},{"Start":"01:48.970 ","End":"01:51.980","Text":"Now let\u0027s just take a look at our formal charges."},{"Start":"01:51.980 ","End":"01:54.640","Text":"If we look at our formal charges,"},{"Start":"01:54.640 ","End":"01:57.460","Text":"first, we\u0027re going to start with a sulfur."},{"Start":"01:57.460 ","End":"02:02.440","Text":"It begins with a 6 because that\u0027s a 6 valence electrons in the free atom,"},{"Start":"02:02.440 ","End":"02:04.330","Text":"minus the number of lone pair electrons,"},{"Start":"02:04.330 ","End":"02:06.865","Text":"which is 2,"},{"Start":"02:06.865 ","End":"02:09.210","Text":"minus 1/2 times the bond pair electrons,"},{"Start":"02:09.210 ","End":"02:11.520","Text":"and we have 4 bonds connected to the sulfur,"},{"Start":"02:11.520 ","End":"02:13.575","Text":"it\u0027s going to give us an 8."},{"Start":"02:13.575 ","End":"02:16.230","Text":"All in all, we have 6 minus 2 minus 4,"},{"Start":"02:16.230 ","End":"02:19.135","Text":"so that\u0027s going to give us a 0 formal charge for this sulfur."},{"Start":"02:19.135 ","End":"02:22.225","Text":"Let us see the formal charge of our carbon."},{"Start":"02:22.225 ","End":"02:25.120","Text":"Let\u0027s see the formal charge of our chlorine."},{"Start":"02:25.120 ","End":"02:26.830","Text":"We\u0027re going to start with a 7,"},{"Start":"02:26.830 ","End":"02:30.820","Text":"since chlorine is in the 17th group minus the number of lone pair electrons,"},{"Start":"02:30.820 ","End":"02:32.426","Text":"we have 1, 2, 3, 4, 5,"},{"Start":"02:32.426 ","End":"02:39.590","Text":"6 lone pair electrons around each chlorine minus 1/2 times the bond pair electrons,"},{"Start":"02:39.590 ","End":"02:42.235","Text":"we have 1 bond, so that\u0027s 2 electrons."},{"Start":"02:42.235 ","End":"02:44.820","Text":"This is going to give us 7 minus 6 minus 1,"},{"Start":"02:44.820 ","End":"02:46.740","Text":"so this is going to give us a 0."},{"Start":"02:46.740 ","End":"02:48.900","Text":"All the formal charges are 0,"},{"Start":"02:48.900 ","End":"02:51.985","Text":"so we can continue with our Lewis structure."},{"Start":"02:51.985 ","End":"02:56.660","Text":"Now we have to assign an AXE notation for each central atom."},{"Start":"02:56.660 ","End":"02:59.430","Text":"Just going to write that here, AXE."},{"Start":"03:00.950 ","End":"03:03.200","Text":"Now A is the central atom."},{"Start":"03:03.200 ","End":"03:04.880","Text":"Here the sulfur is our only atom,"},{"Start":"03:04.880 ","End":"03:07.240","Text":"so we\u0027re going to write A for sulfur."},{"Start":"03:07.240 ","End":"03:12.030","Text":"X is the number of atoms which are connected to our central atom,"},{"Start":"03:12.030 ","End":"03:14.450","Text":"so we know sulfur is connected to 1,"},{"Start":"03:14.450 ","End":"03:16.370","Text":"2, 3, 4 atoms."},{"Start":"03:16.370 ","End":"03:18.960","Text":"It\u0027s going to be X_4."},{"Start":"03:19.220 ","End":"03:24.865","Text":"We\u0027re left with E. E is the number of pairs of lone pair electrons we have."},{"Start":"03:24.865 ","End":"03:27.820","Text":"If we look at our sulfur, we have 1 pair of lone pair electrons,"},{"Start":"03:27.820 ","End":"03:30.285","Text":"so there\u0027s going to be E_1."},{"Start":"03:30.285 ","End":"03:31.890","Text":"Now we don\u0027t write the 1."},{"Start":"03:31.890 ","End":"03:33.840","Text":"But that\u0027s our AXE notation."},{"Start":"03:33.840 ","End":"03:35.990","Text":"We got AX_4E."},{"Start":"03:35.990 ","End":"03:38.080","Text":"Now if you look back at the lecture,"},{"Start":"03:38.080 ","End":"03:43.330","Text":"you will see that you have a list of AXE notations,"},{"Start":"03:43.330 ","End":"03:47.665","Text":"electron group geometries, and molecular geometries."},{"Start":"03:47.665 ","End":"03:51.265","Text":"If you look at the AX_4E notation,"},{"Start":"03:51.265 ","End":"03:58.029","Text":"you will see that the electron group geometry is trigonal bipyramidal,"},{"Start":"03:58.029 ","End":"04:02.530","Text":"so that\u0027s electron group geometry."},{"Start":"04:02.530 ","End":"04:08.990","Text":"We\u0027re just going to call that EGG is trigonal,"},{"Start":"04:11.760 ","End":"04:20.430","Text":"bipyramidal, and"},{"Start":"04:20.430 ","End":"04:30.960","Text":"the molecular geometry is a seesaw."},{"Start":"04:34.230 ","End":"04:39.895","Text":"Our answer phrase that electron group geometry is trigonal bipyramidal,"},{"Start":"04:39.895 ","End":"04:43.640","Text":"and the molecular geometry is a seesaw."},{"Start":"04:44.090 ","End":"04:46.725","Text":"Now we\u0027re going to go on to b."},{"Start":"04:46.725 ","End":"04:51.155","Text":"In b we have formaldehyde."},{"Start":"04:51.155 ","End":"04:53.950","Text":"We\u0027re going to start with counting the number of"},{"Start":"04:53.950 ","End":"04:57.040","Text":"valence electrons in order to determine our Lewis structure,"},{"Start":"04:57.040 ","End":"05:01.615","Text":"so that\u0027s carbon is going to give us 4 plus oxygen is going to give us 6,"},{"Start":"05:01.615 ","End":"05:03.475","Text":"and hydrogen is going to give us 1,"},{"Start":"05:03.475 ","End":"05:04.720","Text":"but we have 2 hydrogens,"},{"Start":"05:04.720 ","End":"05:06.980","Text":"so that\u0027s going to be 2 times 1,"},{"Start":"05:07.490 ","End":"05:11.245","Text":"and that\u0027s going to give us a total of 12 electrons."},{"Start":"05:11.245 ","End":"05:13.240","Text":"We\u0027re going to write our skeletal structures."},{"Start":"05:13.240 ","End":"05:16.930","Text":"We\u0027re going to write our carbon as our central atom and"},{"Start":"05:16.930 ","End":"05:21.675","Text":"our oxygen and hydrogens as terminal atoms."},{"Start":"05:21.675 ","End":"05:24.490","Text":"Now we have 3 bonds in our skeletal structure,"},{"Start":"05:24.490 ","End":"05:27.160","Text":"so we have to subtract 6 electrons from 12."},{"Start":"05:27.160 ","End":"05:30.075","Text":"That\u0027s going to give us 6 electrons left."},{"Start":"05:30.075 ","End":"05:32.960","Text":"We\u0027re going to begin with our terminal atoms."},{"Start":"05:32.960 ","End":"05:35.030","Text":"Our hydrogens can only form 1 bond."},{"Start":"05:35.030 ","End":"05:37.880","Text":"They can only have 2 electrons, so they\u0027re full."},{"Start":"05:37.880 ","End":"05:39.230","Text":"We\u0027re going to go onto the oxygen,"},{"Start":"05:39.230 ","End":"05:41.663","Text":"so we\u0027re going to add 6 electrons, 1,"},{"Start":"05:41.663 ","End":"05:42.994","Text":"2, 3, 4, 5,"},{"Start":"05:42.994 ","End":"05:44.840","Text":"6 in order to make an octet."},{"Start":"05:44.840 ","End":"05:47.060","Text":"Now at this point, if we look at our Lewis structure,"},{"Start":"05:47.060 ","End":"05:49.040","Text":"we can see that our carbon is missing an octet."},{"Start":"05:49.040 ","End":"05:52.270","Text":"It has 6 electrons around it, 3 bonds."},{"Start":"05:52.270 ","End":"05:54.420","Text":"Our oxygen has an octet,"},{"Start":"05:54.420 ","End":"05:56.730","Text":"and our hydrogens each have 2 electrons."},{"Start":"05:56.730 ","End":"06:02.060","Text":"We\u0027re going to take 2 electrons from the oxygen and share them with the carbon,"},{"Start":"06:02.060 ","End":"06:05.480","Text":"so that\u0027s going to give us C double-bond 2,"},{"Start":"06:05.480 ","End":"06:09.730","Text":"4 electrons, and 2 hydrogens."},{"Start":"06:09.730 ","End":"06:13.340","Text":"Now if we look at our formal charges for the oxygen,"},{"Start":"06:13.340 ","End":"06:20.580","Text":"we can see that we have 6 minus 4 lone pair electrons minus 1/2 times 2 bonds,"},{"Start":"06:20.580 ","End":"06:22.895","Text":"so that\u0027s 4 bond pair electrons."},{"Start":"06:22.895 ","End":"06:24.680","Text":"That\u0027s going to give us 6 minus 4 minus 2,"},{"Start":"06:24.680 ","End":"06:27.575","Text":"and it\u0027s going to give us a 0 for oxygen."},{"Start":"06:27.575 ","End":"06:31.250","Text":"Our formal charge for our carbon is going to start with"},{"Start":"06:31.250 ","End":"06:37.470","Text":"4 minus 0 lone pair electrons minus 1/2 times 8 electrons,"},{"Start":"06:37.470 ","End":"06:39.390","Text":"since we have 4 bonds,"},{"Start":"06:39.390 ","End":"06:41.190","Text":"and each bond contains 2 electrons,"},{"Start":"06:41.190 ","End":"06:42.570","Text":"so that\u0027s going to give us 4 minus 4,"},{"Start":"06:42.570 ","End":"06:44.415","Text":"that\u0027s going to give us a 0."},{"Start":"06:44.415 ","End":"06:47.949","Text":"In our formal charge for hydrogen."},{"Start":"06:48.140 ","End":"06:51.975","Text":"Just going to start with 1 because we\u0027re in Group 1 minus"},{"Start":"06:51.975 ","End":"06:57.735","Text":"0 minus 1/2 times 2 electrons in our bond."},{"Start":"06:57.735 ","End":"07:00.360","Text":"That\u0027s going to give us 1 minus 1, it\u0027s going to give us a 0,"},{"Start":"07:00.360 ","End":"07:02.025","Text":"so all formal charges are 0,"},{"Start":"07:02.025 ","End":"07:05.450","Text":"so we know that this is a good Lewis structure."},{"Start":"07:05.450 ","End":"07:08.630","Text":"Now at this point, we have to assign an AXE notation."},{"Start":"07:08.630 ","End":"07:11.150","Text":"Remember, A is the central atom,"},{"Start":"07:11.150 ","End":"07:12.875","Text":"so that\u0027s our carbon."},{"Start":"07:12.875 ","End":"07:15.650","Text":"X is the number of atoms around the carbon,"},{"Start":"07:15.650 ","End":"07:19.740","Text":"so we have 1, 2, 3 atoms."},{"Start":"07:19.750 ","End":"07:23.750","Text":"Now it doesn\u0027t matter if these atoms are connected through a double bond,"},{"Start":"07:23.750 ","End":"07:26.160","Text":"single bond, triple bond,"},{"Start":"07:26.160 ","End":"07:27.630","Text":"it\u0027s still 3 atoms,"},{"Start":"07:27.630 ","End":"07:31.850","Text":"and E is the number of pairs of lone pair electrons, which we have."},{"Start":"07:31.850 ","End":"07:34.310","Text":"Now around the carbon, we have no lone pair electrons,"},{"Start":"07:34.310 ","End":"07:42.910","Text":"so there\u0027s no E. The AXE notation is going to be AX_3."},{"Start":"07:42.910 ","End":"07:46.520","Text":"Now again, if you look at your lecture or in different tables,"},{"Start":"07:46.520 ","End":"07:53.820","Text":"you will see that the electron group geometry for AX_3 is trigonal planar."},{"Start":"08:00.770 ","End":"08:08.680","Text":"You will see that the molecular geometry"},{"Start":"08:10.460 ","End":"08:14.290","Text":"also equals trigonal planar,"},{"Start":"08:19.250 ","End":"08:22.870","Text":"so our answer for b is the electron group geometry is"},{"Start":"08:22.870 ","End":"08:27.520","Text":"trigonal planar and the molecular geometry is also trigonal planar."},{"Start":"08:27.520 ","End":"08:32.764","Text":"Now we\u0027re going to go on to c. In c we have methanol."},{"Start":"08:32.764 ","End":"08:36.090","Text":"We\u0027re going to start by counting the valence electrons,"},{"Start":"08:36.090 ","End":"08:37.350","Text":"so we have 1 carbon,"},{"Start":"08:37.350 ","End":"08:40.500","Text":"so that\u0027s 4 valence electrons plus 4 hydrogens,"},{"Start":"08:40.500 ","End":"08:42.945","Text":"so that\u0027s 4 times 1 valence electrons,"},{"Start":"08:42.945 ","End":"08:45.460","Text":"plus 1 oxygen, which is 6."},{"Start":"08:45.460 ","End":"08:48.980","Text":"That gives us a total of 14 electrons."},{"Start":"08:48.980 ","End":"08:51.290","Text":"In our skeletal structure, we can see how"},{"Start":"08:51.290 ","End":"08:54.830","Text":"the atoms are connected because of the way this is written."},{"Start":"08:54.830 ","End":"08:56.940","Text":"We have CH_3,"},{"Start":"08:59.860 ","End":"09:05.500","Text":"O-H. Now, the number of bonds here are 1,"},{"Start":"09:05.500 ","End":"09:08.725","Text":"2, 3, 4, 5 meaning we used 10 electrons from our 14,"},{"Start":"09:08.725 ","End":"09:09.995","Text":"so we\u0027re going to subtract them."},{"Start":"09:09.995 ","End":"09:11.870","Text":"We have 4 electrons left."},{"Start":"09:11.870 ","End":"09:13.940","Text":"Our terminal atoms are hydrogens,"},{"Start":"09:13.940 ","End":"09:15.985","Text":"so we\u0027re not going to add them any electrons,"},{"Start":"09:15.985 ","End":"09:19.430","Text":"and we\u0027re going to add these electrons to the oxygen."},{"Start":"09:19.430 ","End":"09:22.430","Text":"Now we can see that the carbon has an octet,"},{"Start":"09:22.430 ","End":"09:23.915","Text":"so does the oxygen,"},{"Start":"09:23.915 ","End":"09:25.700","Text":"and we are settled."},{"Start":"09:25.700 ","End":"09:28.210","Text":"Now, let\u0027s check our formal charges."},{"Start":"09:28.210 ","End":"09:34.475","Text":"The formal charge on our hydrogens equals 1 valence electron minus,"},{"Start":"09:34.475 ","End":"09:36.170","Text":"we don\u0027t have any lone pair electrons,"},{"Start":"09:36.170 ","End":"09:39.515","Text":"it\u0027s 0 minus 1/2 times 1 bond,"},{"Start":"09:39.515 ","End":"09:41.540","Text":"which is 2 bond pair electrons,"},{"Start":"09:41.540 ","End":"09:42.980","Text":"which is going to give us 1 minus 1,"},{"Start":"09:42.980 ","End":"09:45.985","Text":"which gives us is 0 for the hydrogens."},{"Start":"09:45.985 ","End":"09:51.380","Text":"Our formal charge on our carbon equals 4 valence electrons since carbon in"},{"Start":"09:51.380 ","End":"09:58.590","Text":"the 14th group minus 0 lone pair electrons minus 1/2 times 8 bond pair electrons."},{"Start":"09:58.590 ","End":"10:00.380","Text":"Since we have 4 bonds around the carbon,"},{"Start":"10:00.380 ","End":"10:03.390","Text":"it\u0027s going to give us a total of 0 for our formal charge."},{"Start":"10:03.390 ","End":"10:06.560","Text":"The formal charge of oxygen,"},{"Start":"10:06.560 ","End":"10:12.047","Text":"we\u0027re going to begin with 6 since oxygen is at 16th group minus 1,"},{"Start":"10:12.047 ","End":"10:15.415","Text":"2, 3, 4 lone pair electrons minus 1/2."},{"Start":"10:15.415 ","End":"10:17.950","Text":"This bond pair electrons now have 2 bonds,"},{"Start":"10:17.950 ","End":"10:20.270","Text":"so that\u0027s 4 electrons."},{"Start":"10:20.640 ","End":"10:26.050","Text":"Now we have 6 minus 4 minus 2, giving us 0."},{"Start":"10:26.050 ","End":"10:27.470","Text":"Our formal charges are all 0,"},{"Start":"10:27.470 ","End":"10:29.995","Text":"so we\u0027re going to continue with this Lewis structure."},{"Start":"10:29.995 ","End":"10:33.890","Text":"The next step is to assign the AXE notation."},{"Start":"10:33.890 ","End":"10:36.760","Text":"In this case, we have 2 central atoms."},{"Start":"10:36.760 ","End":"10:38.500","Text":"We have carbon and oxygen,"},{"Start":"10:38.500 ","End":"10:40.630","Text":"so let\u0027s start with a carbon."},{"Start":"10:40.630 ","End":"10:43.060","Text":"Our AXE notation will be A,"},{"Start":"10:43.060 ","End":"10:44.590","Text":"which is the central atom."},{"Start":"10:44.590 ","End":"10:47.591","Text":"Carbon as you can see is connected to 1,"},{"Start":"10:47.591 ","End":"10:49.025","Text":"2, 3, 4 atoms,"},{"Start":"10:49.025 ","End":"10:54.020","Text":"meaning AX_4, and it has no lone pair electrons,"},{"Start":"10:54.020 ","End":"10:55.635","Text":"so that\u0027s just going to be AX_4."},{"Start":"10:55.635 ","End":"10:59.460","Text":"Now if we look again at the tables or at the lecture,"},{"Start":"10:59.500 ","End":"11:08.680","Text":"we can see that the electron group geometry for AX_4 equals tetrahedral geometry,"},{"Start":"11:12.830 ","End":"11:17.670","Text":"and the molecular geometry is also tetrahedral."},{"Start":"11:30.380 ","End":"11:38.840","Text":"Now if we look at oxygen as a central atom and we want to assign the AXE notation to it,"},{"Start":"11:38.840 ","End":"11:43.850","Text":"we can see that the oxygen is our A is connected to 2 atoms,"},{"Start":"11:43.850 ","End":"11:45.970","Text":"so that\u0027s going to be X_2."},{"Start":"11:45.970 ","End":"11:48.860","Text":"E, which is the number of pairs of lone pair electrons,"},{"Start":"11:48.860 ","End":"11:51.440","Text":"we can see that we have 2 pairs of lone pair electrons,"},{"Start":"11:51.440 ","End":"11:53.435","Text":"so that\u0027s going to be AX_2E_2."},{"Start":"11:53.435 ","End":"11:57.155","Text":"The electron group geometry for this equals."},{"Start":"11:57.155 ","End":"12:02.580","Text":"The electron group geometry for AX_2E_2 is also tetrahedral."},{"Start":"12:06.080 ","End":"12:08.580","Text":"The molecular geometry,"},{"Start":"12:08.580 ","End":"12:15.680","Text":"for the"},{"Start":"12:15.680 ","End":"12:19.340","Text":"AX_2E_2 notation"},{"Start":"12:19.340 ","End":"12:21.510","Text":"is bent."},{"Start":"12:22.720 ","End":"12:26.720","Text":"Answers for C are around the carbon central atom,"},{"Start":"12:26.720 ","End":"12:28.550","Text":"we have AX_4,"},{"Start":"12:28.550 ","End":"12:31.505","Text":"which gives us an electron group geometry of tetrahedral,"},{"Start":"12:31.505 ","End":"12:33.185","Text":"and the molecular geometry,"},{"Start":"12:33.185 ","End":"12:36.150","Text":"which is also tetrahedral."},{"Start":"12:36.710 ","End":"12:40.790","Text":"The oxygen, which is AX_2E_2 notation"},{"Start":"12:40.790 ","End":"12:43.895","Text":"is going to give us an electron group geometry which is tetrahedral,"},{"Start":"12:43.895 ","End":"12:47.315","Text":"but a molecular geometry which is bent."},{"Start":"12:47.315 ","End":"12:49.340","Text":"That is our final answer."},{"Start":"12:49.340 ","End":"12:52.080","Text":"Thank you very much for watching."}],"ID":24519},{"Watched":false,"Name":"Dipole Moments of Molecules","Duration":"8m 11s","ChapterTopicVideoID":23667,"CourseChapterTopicPlaylistID":101313,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.250","Text":"In a previous video,"},{"Start":"00:02.250 ","End":"00:06.915","Text":"we talked about the dipole moments of individual polar covalent bonds."},{"Start":"00:06.915 ","End":"00:11.850","Text":"In this video, we will determine the dipole moments of molecules."},{"Start":"00:11.850 ","End":"00:14.250","Text":"A polyatomic molecule is polar,"},{"Start":"00:14.250 ","End":"00:17.010","Text":"that means it has a dipole moment if the sum of"},{"Start":"00:17.010 ","End":"00:21.870","Text":"the dipole moments of the individual bonds is non-zero."},{"Start":"00:21.870 ","End":"00:26.235","Text":"In order to find out whether the sum is 0 or non-zero,"},{"Start":"00:26.235 ","End":"00:28.979","Text":"we need to know the geometry of the molecule."},{"Start":"00:28.979 ","End":"00:33.848","Text":"Now we\u0027re going to deal with a number of examples to illustrate this point."},{"Start":"00:33.848 ","End":"00:36.710","Text":"Let\u0027s start off with CO_2."},{"Start":"00:36.710 ","End":"00:42.680","Text":"CO_2 belongs to the AX_2 category of VSEPR."},{"Start":"00:42.680 ","End":"00:46.775","Text":"Now the CO bond itself is a polar covalent bond."},{"Start":"00:46.775 ","End":"00:51.095","Text":"The C is slightly positive and the O is slightly negative."},{"Start":"00:51.095 ","End":"00:54.740","Text":"So the dipole moment points towards O."},{"Start":"00:54.740 ","End":"00:57.830","Text":"Why is O more negative than C?"},{"Start":"00:57.830 ","End":"01:02.215","Text":"That\u0027s because the electronegativity of"},{"Start":"01:02.215 ","End":"01:08.005","Text":"O is greater than the electronegativity of C. C is 2.5 and O is 3.5,"},{"Start":"01:08.005 ","End":"01:12.206","Text":"so the dipole moment of the CO bond points towards O."},{"Start":"01:12.206 ","End":"01:14.855","Text":"Now let\u0027s look at the CO_2 molecule."},{"Start":"01:14.855 ","End":"01:16.285","Text":"It\u0027s linear."},{"Start":"01:16.285 ","End":"01:18.980","Text":"The C is AX_2,"},{"Start":"01:18.980 ","End":"01:22.340","Text":"so the sum of the dipole moments is 0."},{"Start":"01:22.340 ","End":"01:27.425","Text":"We have 1 going this way and 1 going that way, and they cancel."},{"Start":"01:27.425 ","End":"01:31.273","Text":"So CO_2 is a non-polar molecule."},{"Start":"01:31.273 ","End":"01:33.785","Text":"Now let\u0027s look at water, H_2O,"},{"Start":"01:33.785 ","End":"01:39.220","Text":"which is AX_2E_2 in VSEPR language."},{"Start":"01:39.220 ","End":"01:42.765","Text":"Now the HO bond itself is polar."},{"Start":"01:42.765 ","End":"01:45.045","Text":"H is slightly positive,"},{"Start":"01:45.045 ","End":"01:46.985","Text":"O is slightly negative,"},{"Start":"01:46.985 ","End":"01:49.970","Text":"with a dipole moment pointing towards O."},{"Start":"01:49.970 ","End":"01:52.490","Text":"Since O is more electronegative than H,"},{"Start":"01:52.490 ","End":"01:56.615","Text":"H is 2.1 and O, 3.5."},{"Start":"01:56.615 ","End":"02:00.588","Text":"The dipole moment is pointing in this direction."},{"Start":"02:00.588 ","End":"02:03.740","Text":"Now as I\u0027ve said many times in previous videos,"},{"Start":"02:03.740 ","End":"02:10.790","Text":"water is a bent molecule with the angle HOH=104.5 degrees."},{"Start":"02:10.790 ","End":"02:15.575","Text":"The sum of the dipole moments is non-zero. Let\u0027s draw it."},{"Start":"02:15.575 ","End":"02:18.430","Text":"O, H, H,"},{"Start":"02:18.430 ","End":"02:20.540","Text":"1 dipole is pointing this direction,"},{"Start":"02:20.540 ","End":"02:21.950","Text":"1 in this direction,"},{"Start":"02:21.950 ","End":"02:26.828","Text":"and the sum of the dipoles is in this direction,"},{"Start":"02:26.828 ","End":"02:30.405","Text":"so H_2O is a polar molecule."},{"Start":"02:30.405 ","End":"02:32.865","Text":"The third example is BF_3,"},{"Start":"02:32.865 ","End":"02:38.480","Text":"which belongs to AX_3 category of VSEPR."},{"Start":"02:38.480 ","End":"02:43.730","Text":"The BF bond is a polar covalent bond with a dipole pointing"},{"Start":"02:43.730 ","End":"02:49.040","Text":"towards F. That\u0027s because F is more electronegative than B."},{"Start":"02:49.040 ","End":"02:53.465","Text":"B has electronegativity of 2 and F of 4."},{"Start":"02:53.465 ","End":"02:56.765","Text":"Now, what happens when we get to the molecule?"},{"Start":"02:56.765 ","End":"03:04.670","Text":"We\u0027ve said before that BF_3 is a planar molecule with the FBF angle 120 degrees."},{"Start":"03:04.670 ","End":"03:09.700","Text":"B is in the center, 3 Fs."},{"Start":"03:12.140 ","End":"03:18.310","Text":"Now if we add the 2 dipoles pointing down the way,"},{"Start":"03:19.670 ","End":"03:28.700","Text":"we get a dipole of equal value that bisects the FBF angle."},{"Start":"03:28.700 ","End":"03:33.335","Text":"Here is the resultant of these 2 vectors."},{"Start":"03:33.335 ","End":"03:36.310","Text":"The top BF bond,"},{"Start":"03:36.310 ","End":"03:40.025","Text":"for it, the dipole is in the opposite direction."},{"Start":"03:40.025 ","End":"03:45.965","Text":"When we sub the red vector and the black vector, we get 0."},{"Start":"03:45.965 ","End":"03:49.610","Text":"That means BF_3 is a non-polar molecule."},{"Start":"03:49.610 ","End":"03:52.405","Text":"The fourth example, COCL_2,"},{"Start":"03:52.405 ","End":"03:54.830","Text":"which again is AX_3."},{"Start":"03:54.830 ","End":"04:00.515","Text":"But this time the CCL dipole moment is different from the CO dipole moment."},{"Start":"04:00.515 ","End":"04:03.305","Text":"The molecule is polar."},{"Start":"04:03.305 ","End":"04:07.895","Text":"We have CO that dipole in this direction"},{"Start":"04:07.895 ","End":"04:13.910","Text":"and CL resultant of these 2 dipoles."},{"Start":"04:13.910 ","End":"04:17.165","Text":"The same lens and in the downwards direction,"},{"Start":"04:17.165 ","End":"04:18.695","Text":"but it\u0027s a different lens,"},{"Start":"04:18.695 ","End":"04:26.375","Text":"it has a different amplitude than the CO vector pointing towards the top."},{"Start":"04:26.375 ","End":"04:29.330","Text":"The sum is non-zero."},{"Start":"04:29.330 ","End":"04:35.225","Text":"That means that COCL_2 is a polar molecule."},{"Start":"04:35.225 ","End":"04:37.820","Text":"It has a resultant dipole moment,"},{"Start":"04:37.820 ","End":"04:40.280","Text":"a net dipole moment."},{"Start":"04:40.280 ","End":"04:43.960","Text":"The fifth example is CCL_4,"},{"Start":"04:43.960 ","End":"04:49.700","Text":"which belongs to the AX_4 category of VSEPR theory."},{"Start":"04:49.700 ","End":"04:53.734","Text":"Now we\u0027ll see that the sum of the dipole moment is 0."},{"Start":"04:53.734 ","End":"04:57.020","Text":"Each CCl bond has a dipole moment,"},{"Start":"04:57.020 ","End":"04:58.550","Text":"but the sum is 0."},{"Start":"04:58.550 ","End":"05:03.110","Text":"The way to see that is to draw the molecule inside a cube."},{"Start":"05:03.110 ","End":"05:06.620","Text":"We can draw a tetrahedral molecule inside a cube."},{"Start":"05:06.620 ","End":"05:08.950","Text":"We have C in the center,"},{"Start":"05:08.950 ","End":"05:14.615","Text":"a bond pointing towards this vertex."},{"Start":"05:14.615 ","End":"05:16.880","Text":"We take the sum of these two,"},{"Start":"05:16.880 ","End":"05:19.100","Text":"we get a vector pointing down."},{"Start":"05:19.100 ","End":"05:23.544","Text":"Now we have another 2 bonds,"},{"Start":"05:23.544 ","End":"05:25.530","Text":"1 pointing to this vertex,"},{"Start":"05:25.530 ","End":"05:28.040","Text":"and 1 pointing to this vertex."},{"Start":"05:28.040 ","End":"05:30.515","Text":"This makes our tetrahedron."},{"Start":"05:30.515 ","End":"05:34.695","Text":"The sum of these two points upwards."},{"Start":"05:34.695 ","End":"05:39.125","Text":"It\u0027s exactly equal and opposite to the 1 pointing down in the way,"},{"Start":"05:39.125 ","End":"05:41.230","Text":"so the sum is 0."},{"Start":"05:41.230 ","End":"05:50.875","Text":"The sum of bond dipole moments is 0 and that means that CCL_4 is a non-polar molecule."},{"Start":"05:50.875 ","End":"05:55.565","Text":"If 1 of the 4 X atoms is different from the other 3,"},{"Start":"05:55.565 ","End":"05:58.220","Text":"for example, CH_3CL,"},{"Start":"05:58.220 ","End":"06:00.654","Text":"then the molecule will be polar."},{"Start":"06:00.654 ","End":"06:02.265","Text":"Now, why is that so?"},{"Start":"06:02.265 ","End":"06:06.490","Text":"CH has a very small dipole moment,"},{"Start":"06:06.490 ","End":"06:15.920","Text":"0.122 debye, whereas CCL has much greater dipole moment of 1.6 debye."},{"Start":"06:15.920 ","End":"06:17.915","Text":"If we draw the molecule,"},{"Start":"06:17.915 ","End":"06:23.975","Text":"then the dipole moments of the CH bonds will be very small, pointing downwards,"},{"Start":"06:23.975 ","End":"06:29.525","Text":"and the much larger dipole of the CCL bond will be pointing upwards,"},{"Start":"06:29.525 ","End":"06:35.240","Text":"giving us a net dipole moment in the same direction as the CCL bond."},{"Start":"06:35.240 ","End":"06:37.900","Text":"This molecule is polar."},{"Start":"06:37.900 ","End":"06:39.920","Text":"Now if we follow the same reasoning,"},{"Start":"06:39.920 ","End":"06:47.405","Text":"it\u0027s easy to see that AX_5 and AX_6 molecules are non-polar provided of course,"},{"Start":"06:47.405 ","End":"06:51.130","Text":"that all of the X atoms are the same."},{"Start":"06:51.130 ","End":"06:54.960","Text":"That means PCL_5, for example,"},{"Start":"06:54.960 ","End":"07:00.762","Text":"or SF_6 are non-polar molecules."},{"Start":"07:00.762 ","End":"07:07.050","Text":"In this video, we talked about the dipole moments of molecules."}],"ID":24572}],"Thumbnail":null,"ID":101313},{"Name":"Bond lengths","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Bond order and bond length","Duration":"2m 52s","ChapterTopicVideoID":23064,"CourseChapterTopicPlaylistID":101314,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23064.jpeg","UploadDate":"2020-10-27T13:01:34.9900000","DurationForVideoObject":"PT2M52S","Description":null,"MetaTitle":"Bond order and bond length: Video + Workbook | Proprep","MetaDescription":"Introduction to Chemical Bonding - Bond lengths. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/introduction-to-chemical-bonding/bond-lengths/vid23906","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"In previous videos we talked about Lewis structures."},{"Start":"00:03.720 ","End":"00:08.475","Text":"In this video we define the bond order related to bond length."},{"Start":"00:08.475 ","End":"00:12.540","Text":"What\u0027s the bond order which will denote as BO?"},{"Start":"00:12.540 ","End":"00:16.110","Text":"The bond order defines whether covalent bond is single,"},{"Start":"00:16.110 ","End":"00:18.870","Text":"which case BO is 1, double,"},{"Start":"00:18.870 ","End":"00:21.210","Text":"then BO is 2 or triple,"},{"Start":"00:21.210 ","End":"00:22.860","Text":"then B0 is 3."},{"Start":"00:22.860 ","End":"00:27.780","Text":"Sometimes if there is a resonance B0 may even be fractional."},{"Start":"00:27.780 ","End":"00:30.180","Text":"What about the bond length?"},{"Start":"00:30.180 ","End":"00:34.050","Text":"The bond length is the distance between 2 atoms."},{"Start":"00:34.050 ","End":"00:37.290","Text":"Now, a triple bond between 2 atoms is shorter"},{"Start":"00:37.290 ","End":"00:40.865","Text":"than a double bond and that\u0027s shorter than a single bond."},{"Start":"00:40.865 ","End":"00:43.415","Text":"The higher the bond order,"},{"Start":"00:43.415 ","End":"00:50.120","Text":"the shorter the bond length so BO goes up, length goes down."},{"Start":"00:50.120 ","End":"00:55.685","Text":"The reason for this is the issues of electron density."},{"Start":"00:55.685 ","End":"00:59.720","Text":"There is a higher electron density in a double bond than there is in"},{"Start":"00:59.720 ","End":"01:04.175","Text":"a single bond and a even higher electron density in a triple."},{"Start":"01:04.175 ","End":"01:07.235","Text":"Let\u0027s take some examples: CC bonds,"},{"Start":"01:07.235 ","End":"01:10.610","Text":"CN bonds, NN bonds, CO bonds."},{"Start":"01:10.610 ","End":"01:14.045","Text":"We\u0027ll take all the bond length in picometers."},{"Start":"01:14.045 ","End":"01:15.534","Text":"Here\u0027s a table."},{"Start":"01:15.534 ","End":"01:17.955","Text":"Starting off with CC bonds,"},{"Start":"01:17.955 ","End":"01:23.900","Text":"we see that for the case of single bond is a 154 picometers,"},{"Start":"01:23.900 ","End":"01:25.820","Text":"double bond 134,"},{"Start":"01:25.820 ","End":"01:27.665","Text":"and triple bond 120."},{"Start":"01:27.665 ","End":"01:30.225","Text":"That\u0027s the case for all these bonds."},{"Start":"01:30.225 ","End":"01:34.875","Text":"CN bonds: 147,128,116."},{"Start":"01:34.875 ","End":"01:39.480","Text":"NN bonds: 145,123,110,"},{"Start":"01:39.480 ","End":"01:46.365","Text":"and CO bonds: 143,120,113 picometers."},{"Start":"01:46.365 ","End":"01:50.720","Text":"That we can conclude from all this that the triple bond is always"},{"Start":"01:50.720 ","End":"01:55.415","Text":"shorter than the double bond to solve was shorter than the single bond."},{"Start":"01:55.415 ","End":"01:58.690","Text":"Now what happens when we have a resonance hybrid?"},{"Start":"01:58.690 ","End":"02:04.375","Text":"Once again, we\u0027ll take the example of CO_3^2 minus carbonate."},{"Start":"02:04.375 ","End":"02:06.835","Text":"As we said in previous videos,"},{"Start":"02:06.835 ","End":"02:10.640","Text":"there are 3 resonance structures forming a hybrid."},{"Start":"02:10.640 ","End":"02:15.810","Text":"Each of them has 1 double bond and 2 single bonds."},{"Start":"02:15.810 ","End":"02:20.850","Text":"The total bond order is 4 because we have 1 plus 1,"},{"Start":"02:20.850 ","End":"02:23.965","Text":"that\u0027s 2 plus 2 giving 4."},{"Start":"02:23.965 ","End":"02:26.375","Text":"But we have 3 different structures."},{"Start":"02:26.375 ","End":"02:30.955","Text":"The average bond order is 4 divided by 3."},{"Start":"02:30.955 ","End":"02:35.745","Text":"Now 4/3 is intermediate between 1 and 2."},{"Start":"02:35.745 ","End":"02:40.420","Text":"It\u0027s not a surprise that the bond length is intermediate between that of"},{"Start":"02:40.420 ","End":"02:46.665","Text":"a single bond and the double bond and it\u0027s a 129 picometers."},{"Start":"02:46.665 ","End":"02:51.270","Text":"In this video, we talked about bond order and bond length."}],"ID":23906}],"Thumbnail":null,"ID":101314},{"Name":"Bond Energies","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Bond order and bond energy","Duration":"5m 23s","ChapterTopicVideoID":23155,"CourseChapterTopicPlaylistID":101315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:06.195","Text":"In the previous video, we talked about the relation between bond order and bond length."},{"Start":"00:06.195 ","End":"00:11.130","Text":"In this video, we\u0027re talking about bond order and bond energy."},{"Start":"00:11.130 ","End":"00:15.255","Text":"Let\u0027s begin defining the bond dissociation energy."},{"Start":"00:15.255 ","End":"00:18.720","Text":"The bond dissociation energy D is the energy"},{"Start":"00:18.720 ","End":"00:22.935","Text":"required to break 1 mole of covalent bonds in the gas phase."},{"Start":"00:22.935 ","End":"00:24.915","Text":"Let\u0027s take an example."},{"Start":"00:24.915 ","End":"00:30.090","Text":"The bond dissociation or breakage of the Cl_2 bond."},{"Start":"00:30.090 ","End":"00:32.295","Text":"We have Cl_2 in the gas phase,"},{"Start":"00:32.295 ","End":"00:36.215","Text":"breaking up into 2 Cl atoms in the gas phase."},{"Start":"00:36.215 ","End":"00:42.800","Text":"Now, the enthalpy of reaction is the same as the dissociation energy of the bond."},{"Start":"00:42.800 ","End":"00:45.890","Text":"It\u0027s the same as the dissociation energy of the Cl,"},{"Start":"00:45.890 ","End":"00:50.945","Text":"Cl bond and the volume of that is 243 kilojoules per mole."},{"Start":"00:50.945 ","End":"00:54.425","Text":"This is an endothermic process."},{"Start":"00:54.425 ","End":"00:57.440","Text":"It requires energy to break a bond."},{"Start":"00:57.440 ","End":"01:02.345","Text":"What do we call the bond energy or sometimes average bond energy."},{"Start":"01:02.345 ","End":"01:05.300","Text":"The bond energy is the bond dissociation energy of"},{"Start":"01:05.300 ","End":"01:10.115","Text":"a particular bond averaged over a number of species that contain that bond."},{"Start":"01:10.115 ","End":"01:14.200","Text":"The amount of energy required to break a bond is not always the same."},{"Start":"01:14.200 ","End":"01:17.555","Text":"It can vary from molecule to molecule."},{"Start":"01:17.555 ","End":"01:20.405","Text":"We take an average over several molecules."},{"Start":"01:20.405 ","End":"01:24.800","Text":"Sometimes it\u0027s even different for 2 CH bonds,"},{"Start":"01:24.800 ","End":"01:26.885","Text":"for example, in the same molecule."},{"Start":"01:26.885 ","End":"01:29.320","Text":"Now what\u0027s the bond formation energy?"},{"Start":"01:29.320 ","End":"01:32.585","Text":"That\u0027s the opposite of breaking a bond is to form it."},{"Start":"01:32.585 ","End":"01:36.590","Text":"The bond formation energy that\u0027s minus D is the energy"},{"Start":"01:36.590 ","End":"01:40.850","Text":"required to form 1 mole of covalent bonds with the gas phase."},{"Start":"01:40.850 ","End":"01:43.790","Text":"Here\u0027s an example related to the previous one."},{"Start":"01:43.790 ","End":"01:49.940","Text":"Now we have 2 chlorine atoms and they\u0027re going to form a chlorine 2 molecule and"},{"Start":"01:49.940 ","End":"01:56.600","Text":"the enthalpy of reaction is now the minus the dissociation energy of the Cl, Cl bond."},{"Start":"01:56.600 ","End":"02:03.215","Text":"It\u0027s minus 243 kilojoules per mole and is an exothermic process."},{"Start":"02:03.215 ","End":"02:07.925","Text":"Forming a bond is an exothermic process."},{"Start":"02:07.925 ","End":"02:10.880","Text":"Let\u0027s take some examples of the bond energy."},{"Start":"02:10.880 ","End":"02:13.520","Text":"What happens when we increase the bond order?"},{"Start":"02:13.520 ","End":"02:15.260","Text":"As we increase the bond order,"},{"Start":"02:15.260 ","End":"02:17.480","Text":"the bond gets stronger."},{"Start":"02:17.480 ","End":"02:19.775","Text":"The bond energy increases."},{"Start":"02:19.775 ","End":"02:25.130","Text":"With increasing B0 leads to increasing bond energy BE."},{"Start":"02:25.130 ","End":"02:27.295","Text":"Now, what about the bond length?"},{"Start":"02:27.295 ","End":"02:31.590","Text":"As the bond gets stronger then the bond length,"},{"Start":"02:31.590 ","End":"02:34.950","Text":"we\u0027ll call that BL, decreases."},{"Start":"02:34.950 ","End":"02:36.960","Text":"Here we have the relationship."},{"Start":"02:36.960 ","End":"02:39.515","Text":"It\u0027s not a linear relationship."},{"Start":"02:39.515 ","End":"02:42.440","Text":"The energy to break a double bond is not twice"},{"Start":"02:42.440 ","End":"02:45.680","Text":"the energy to break a single bond, and so on."},{"Start":"02:45.680 ","End":"02:47.345","Text":"Let us take some examples."},{"Start":"02:47.345 ","End":"02:50.030","Text":"CC, CN, and NN."},{"Start":"02:50.030 ","End":"02:53.305","Text":"All the volumes are kilojoules per mole."},{"Start":"02:53.305 ","End":"02:55.580","Text":"For example, CC bonds,"},{"Start":"02:55.580 ","End":"02:58.640","Text":"we can have a single bond where the bond order is 1,"},{"Start":"02:58.640 ","End":"03:04.700","Text":"dissociation energy of the bond energy is 347 and where BO is 2,"},{"Start":"03:04.700 ","End":"03:06.905","Text":"that\u0027s the double bond is 611,"},{"Start":"03:06.905 ","End":"03:09.595","Text":"for triple bond, 837."},{"Start":"03:09.595 ","End":"03:13.070","Text":"CN 305 for single bond,"},{"Start":"03:13.070 ","End":"03:15.500","Text":"615 for a double bond,"},{"Start":"03:15.500 ","End":"03:22.235","Text":"and 891 for triple bond and NN 163 for a single bond,"},{"Start":"03:22.235 ","End":"03:24.500","Text":"418 for double bond,"},{"Start":"03:24.500 ","End":"03:28.205","Text":"and enormous 946 for triple bond."},{"Start":"03:28.205 ","End":"03:34.519","Text":"The NN triple bond is extremely energetic, extremely stable."},{"Start":"03:34.519 ","End":"03:37.265","Text":"Now a few other values that might be useful,"},{"Start":"03:37.265 ","End":"03:39.830","Text":"are HH, that\u0027s 436,"},{"Start":"03:39.830 ","End":"03:43.385","Text":"NH 389, CH 414,"},{"Start":"03:43.385 ","End":"03:47.090","Text":"and OH 464 kilojoules per mole."},{"Start":"03:47.090 ","End":"03:52.099","Text":"Now we can use the bond energy to calculate the enthalpy reaction."},{"Start":"03:52.099 ","End":"03:56.780","Text":"This is just an approximate calculation because these bond energies are"},{"Start":"03:56.780 ","End":"04:03.590","Text":"averages and may not relate precisely to the particular reaction we\u0027re talking about."},{"Start":"04:03.590 ","End":"04:07.085","Text":"We get an approximate value of the enthalpy of reaction,"},{"Start":"04:07.085 ","End":"04:09.095","Text":"but still can be very useful."},{"Start":"04:09.095 ","End":"04:14.135","Text":"The enthalpies of reaction in the gas phase can be estimated using the bond energies."},{"Start":"04:14.135 ","End":"04:17.165","Text":"That\u0027s what we\u0027re going to work out a formula for this now."},{"Start":"04:17.165 ","End":"04:19.130","Text":"Supposing we have reactants."},{"Start":"04:19.130 ","End":"04:22.655","Text":"The first thing we do is to break all these bonds."},{"Start":"04:22.655 ","End":"04:24.215","Text":"Now we have atoms."},{"Start":"04:24.215 ","End":"04:27.950","Text":"Then we take the atoms and form them together,"},{"Start":"04:27.950 ","End":"04:30.500","Text":"combine them together to get the products."},{"Start":"04:30.500 ","End":"04:35.870","Text":"The enthalpy of reaction for this process, reactants to products,"},{"Start":"04:35.870 ","End":"04:38.450","Text":"is the enthalpy of reaction for"},{"Start":"04:38.450 ","End":"04:42.620","Text":"bond breakage plus the enthalpy of reaction for bond formation."},{"Start":"04:42.620 ","End":"04:46.340","Text":"The enthalpy of reaction for the bond breakage is"},{"Start":"04:46.340 ","End":"04:50.660","Text":"the sum of all the bond energies of the reactants and that has to be"},{"Start":"04:50.660 ","End":"04:56.000","Text":"positive because it\u0027s an endothermic process"},{"Start":"04:56.000 ","End":"05:00.650","Text":"minus the sum of the bond energies of the products."},{"Start":"05:00.650 ","End":"05:04.545","Text":"It\u0027s minus sign because here it\u0027s exothermic."},{"Start":"05:04.545 ","End":"05:08.985","Text":"This is endothermic and this is exothermic."},{"Start":"05:08.985 ","End":"05:15.245","Text":"In this video, we talked about bond energy and bond length and in the next video,"},{"Start":"05:15.245 ","End":"05:18.650","Text":"we\u0027ll calculate the enthalpy of reaction for"},{"Start":"05:18.650 ","End":"05:23.640","Text":"a particular reaction using these bond energies."}],"ID":23987},{"Watched":false,"Name":"Enthalpy of reaction using bond energies","Duration":"5m 49s","ChapterTopicVideoID":23065,"CourseChapterTopicPlaylistID":101315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:04.290","Text":"In the previous video, we talked about bond energies."},{"Start":"00:04.290 ","End":"00:08.355","Text":"This video, we\u0027ll use them to calculate the enthalpy of reaction."},{"Start":"00:08.355 ","End":"00:12.585","Text":"Here\u0027s the expression we learned in the previous video."},{"Start":"00:12.585 ","End":"00:16.260","Text":"The enthalpy of reaction is approximately equal to the sum of"},{"Start":"00:16.260 ","End":"00:21.690","Text":"all the bond energies of the reactants minus all the bond energies of the products."},{"Start":"00:21.690 ","End":"00:23.610","Text":"Let\u0027s take an example,"},{"Start":"00:23.610 ","End":"00:28.260","Text":"calculate the enthalpy of reaction for the reaction of nitrogen in"},{"Start":"00:28.260 ","End":"00:30.990","Text":"the gas phase plus 3 moles of hydrogen in"},{"Start":"00:30.990 ","End":"00:34.470","Text":"the gas phase to form 2 moles of ammonia in the gas phase."},{"Start":"00:34.470 ","End":"00:37.200","Text":"This is called the Haber process,"},{"Start":"00:37.200 ","End":"00:41.160","Text":"and it\u0027s used in industry to produce ammonia."},{"Start":"00:41.160 ","End":"00:46.085","Text":"The first thing to do is to write Lewis structures for all the reactants and products."},{"Start":"00:46.085 ","End":"00:49.000","Text":"We have nitrogen with this triple bond,"},{"Start":"00:49.000 ","End":"00:54.965","Text":"3 hydrogen molecules with its single hydrogen-hydrogen bond,"},{"Start":"00:54.965 ","End":"00:56.980","Text":"and 2 moles of ammonia,"},{"Start":"00:56.980 ","End":"01:02.685","Text":"each 1 with 3 single nitrogen hydrogen bonds."},{"Start":"01:02.685 ","End":"01:08.865","Text":"Let\u0027s do a little calculation to see how many bonds we have to break and form."},{"Start":"01:08.865 ","End":"01:14.835","Text":"We have 1 N triple bond, 3 H-H bonds."},{"Start":"01:14.835 ","End":"01:20.100","Text":"Here we have 2 moles each with 3 bonds,"},{"Start":"01:20.100 ","End":"01:23.085","Text":"that makes 6 N-H bond."},{"Start":"01:23.085 ","End":"01:25.955","Text":"The sum of the bond energies in the reactants"},{"Start":"01:25.955 ","End":"01:28.625","Text":"is the bond energy of the nitrogen triple bond,"},{"Start":"01:28.625 ","End":"01:34.885","Text":"3 times the bond energy of the hydrogen molecule H-H bonds."},{"Start":"01:34.885 ","End":"01:36.710","Text":"We write the values of this."},{"Start":"01:36.710 ","End":"01:45.785","Text":"We get 946 for nitrogen triple bond plus 3 times 436 for the hydrogen-hydrogen bond."},{"Start":"01:45.785 ","End":"01:50.450","Text":"This gives us 2,254 kilojoules per mole."},{"Start":"01:50.450 ","End":"01:52.865","Text":"Now we need to do the same thing for the products."},{"Start":"01:52.865 ","End":"01:55.500","Text":"We have 6 N-H bonds."},{"Start":"01:55.500 ","End":"01:58.430","Text":"We have the sum of the bond energies for"},{"Start":"01:58.430 ","End":"02:02.050","Text":"the products is 6 times the bond energy of the N-H bond."},{"Start":"02:02.050 ","End":"02:10.250","Text":"That\u0027s 6 times 389 kilojoules per mole and that gives us 2,334 kilojoules per mole."},{"Start":"02:10.250 ","End":"02:12.650","Text":"Now we can combine the 2 parts."},{"Start":"02:12.650 ","End":"02:19.095","Text":"We get Delta H reaction is approximately equal to 2,254."},{"Start":"02:19.095 ","End":"02:25.880","Text":"That\u0027s from the reactants minus 2,334 from the products."},{"Start":"02:25.880 ","End":"02:27.260","Text":"We work that out,"},{"Start":"02:27.260 ","End":"02:30.260","Text":"we get minus 80 kilojoules per mole."},{"Start":"02:30.260 ","End":"02:33.980","Text":"This is an exothermic process."},{"Start":"02:33.980 ","End":"02:38.180","Text":"Now we\u0027re going to do a slightly more complicated reaction."},{"Start":"02:38.180 ","End":"02:41.480","Text":"Calculate the enthalpy of reaction for"},{"Start":"02:41.480 ","End":"02:46.160","Text":"the dehydration reaction of ethanol to ethene and water."},{"Start":"02:46.160 ","End":"02:48.965","Text":"Ethene is the IUPAC name for ethylene."},{"Start":"02:48.965 ","End":"02:52.905","Text":"Let\u0027s write this out in chemical terms."},{"Start":"02:52.905 ","End":"02:56.185","Text":"Ethanol, CH_3CH_2OH,"},{"Start":"02:56.185 ","End":"02:59.985","Text":"the gas phase to give us ethene that CH_2,"},{"Start":"02:59.985 ","End":"03:04.440","Text":"CH_2, the gas phase plus water H_2O in the gas phase."},{"Start":"03:04.440 ","End":"03:08.720","Text":"We start by writing the Lewis structures for all the reactants and products."},{"Start":"03:08.720 ","End":"03:11.240","Text":"Let\u0027s count how many bonds we have."},{"Start":"03:11.240 ","End":"03:13.835","Text":"First, let\u0027s look at the reactants."},{"Start":"03:13.835 ","End":"03:15.545","Text":"You have 1, 2, 3,"},{"Start":"03:15.545 ","End":"03:17.945","Text":"4, 5 C-H bonds,"},{"Start":"03:17.945 ","End":"03:21.065","Text":"1 C-C, 1 C-O,"},{"Start":"03:21.065 ","End":"03:25.060","Text":"and 1 O-H. Now let\u0027s look at the products."},{"Start":"03:25.060 ","End":"03:27.450","Text":"We have 1,"},{"Start":"03:27.450 ","End":"03:28.665","Text":"2, 3, 4,"},{"Start":"03:28.665 ","End":"03:31.470","Text":"4 C-H bonds,"},{"Start":"03:31.470 ","End":"03:34.365","Text":"1 C double bond,"},{"Start":"03:34.365 ","End":"03:38.765","Text":"C-C double bond, and 2 O-H bonds."},{"Start":"03:38.765 ","End":"03:42.745","Text":"Now let\u0027s compare the reactants with the products."},{"Start":"03:42.745 ","End":"03:47.315","Text":"Here we have 5 C-H bonds and here we have 4 C-H bonds."},{"Start":"03:47.315 ","End":"03:53.150","Text":"The net result is that we have just 1 C-H bond."},{"Start":"03:53.150 ","End":"03:56.140","Text":"C-O only appears in the reactants."},{"Start":"03:56.140 ","End":"03:59.070","Text":"C-C only appears in reactants."},{"Start":"03:59.070 ","End":"04:03.210","Text":"Now we look at O-H. O-H appears in the reactants"},{"Start":"04:03.210 ","End":"04:08.340","Text":"1 O-H bond and in the products, 2 O-H bonds."},{"Start":"04:08.340 ","End":"04:13.745","Text":"The net result is just 1 O-H bond."},{"Start":"04:13.745 ","End":"04:16.520","Text":"We could do all the arithmetic and calculate"},{"Start":"04:16.520 ","End":"04:20.705","Text":"5 C-H bonds on the left-hand side and 4 on the right-hand side."},{"Start":"04:20.705 ","End":"04:25.570","Text":"But it\u0027s much easier to do it for the net results."},{"Start":"04:25.570 ","End":"04:29.495","Text":"The sum of the bond energies for the net reactants,"},{"Start":"04:29.495 ","End":"04:36.740","Text":"the bond energy of C-H plus the bond energy of C-C plus the bond energy of C-O."},{"Start":"04:36.740 ","End":"04:38.539","Text":"We insert the values,"},{"Start":"04:38.539 ","End":"04:43.910","Text":"we get 414 plus 347 plus 360 kilojoules per mole."},{"Start":"04:43.910 ","End":"04:48.815","Text":"That gives us a total of 1,121 kilojoules per mole."},{"Start":"04:48.815 ","End":"04:52.575","Text":"Let\u0027s look at the net products."},{"Start":"04:52.575 ","End":"04:58.590","Text":"We have the bond energy of 1 C-C double bond. Here it is."},{"Start":"04:58.590 ","End":"05:03.470","Text":"The bond energy of 1 O-H bond."},{"Start":"05:03.470 ","End":"05:05.180","Text":"When we insert the values,"},{"Start":"05:05.180 ","End":"05:08.750","Text":"we get 611 for the C-C double bond,"},{"Start":"05:08.750 ","End":"05:11.155","Text":"464 for the O-H bond."},{"Start":"05:11.155 ","End":"05:15.455","Text":"That gives us a total of 1,075 kilojoules per mole."},{"Start":"05:15.455 ","End":"05:17.735","Text":"Now we can combine the 2 parts."},{"Start":"05:17.735 ","End":"05:26.030","Text":"We get Delta H of the enthalpy of reaction is approximately equal to 1,121."},{"Start":"05:26.030 ","End":"05:31.430","Text":"That\u0027s from the reactants minus 1,075 from the products."},{"Start":"05:31.430 ","End":"05:33.530","Text":"When we do the arithmetic,"},{"Start":"05:33.530 ","End":"05:36.916","Text":"we get 46 kilojoules per mole,"},{"Start":"05:36.916 ","End":"05:41.105","Text":"so this is an endothermic process."},{"Start":"05:41.105 ","End":"05:44.855","Text":"In this video, we showed how to use bond energies"},{"Start":"05:44.855 ","End":"05:49.470","Text":"to calculate approximate enthalpies of reaction."}],"ID":23907}],"Thumbnail":null,"ID":101315}]

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