Valence bond VB method
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- Introduction to Valence-Bond Theory
- Sigma and Pi Bonds
- Hybridization sp3 hybrid orbitals
- Hybridization Sp2 and Sp Hybrid Orbitals
- Hybridization sp3d and sp3d2 Hybrid Orbitals
- Multiple covalent bonds
- Summary and example of VB theory
- Exercise 1 - part a
- Exercise 1 - part b
- Exercise 1 - part c
- Exercise 2
- Exercise 3 - part a
- Exercise 3 - part b
- Exercise 3 - part c
- Exercise 4
- Exercise 5

Molecular orbital MO theory
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VB and MO description of benzene
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[{"Name":"Valence bond VB method","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction to Valence-Bond Theory","Duration":"4m 57s","ChapterTopicVideoID":23062,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23062.jpeg","UploadDate":"2020-10-12T13:04:18.6800000","DurationForVideoObject":"PT4M57S","Description":null,"MetaTitle":"Introduction to valence-bond theory: Video + Workbook | Proprep","MetaDescription":"More Aspects of Chemical Bonding - Valence bond VB method. 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/more-aspects-of-chemical-bonding/valence-bond-vb-method/vid23904","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.805","Text":"In previous videos, we learned about Lewis theory and VSEPR theory."},{"Start":"00:05.805 ","End":"00:07.110","Text":"In the next few videos,"},{"Start":"00:07.110 ","End":"00:09.375","Text":"we\u0027ll learn about valence bond theory."},{"Start":"00:09.375 ","End":"00:12.075","Text":"So we\u0027re going to talk about valence bond theory."},{"Start":"00:12.075 ","End":"00:15.090","Text":"We can write it for short as VB."},{"Start":"00:15.090 ","End":"00:17.385","Text":"A few facts about valence bond theory."},{"Start":"00:17.385 ","End":"00:19.710","Text":"Valence bond theory was the first description of"},{"Start":"00:19.710 ","End":"00:23.355","Text":"covalent bonding in terms of atomic orbitals."},{"Start":"00:23.355 ","End":"00:27.440","Text":"In a sense, it\u0027s a quantum mechanical version of Lewis theory."},{"Start":"00:27.440 ","End":"00:30.815","Text":"Now, various concepts in valence bond theory,"},{"Start":"00:30.815 ","End":"00:33.970","Text":"such as Sigma and Pi bonds, resonance,"},{"Start":"00:33.970 ","End":"00:37.325","Text":"and hybridization are widely used in chemistry."},{"Start":"00:37.325 ","End":"00:41.240","Text":"We\u0027ve already met resonance and we\u0027re going to talk about Sigma"},{"Start":"00:41.240 ","End":"00:45.590","Text":"and Pi bonds and hybridization in the next few videos."},{"Start":"00:45.590 ","End":"00:51.110","Text":"Now, unlike Lewis theory and VSEPR theory which are qualitative,"},{"Start":"00:51.110 ","End":"00:52.725","Text":"just simple descriptions,"},{"Start":"00:52.725 ","End":"00:57.695","Text":"valence bond theory can be used to actually calculate the properties of molecules."},{"Start":"00:57.695 ","End":"01:00.230","Text":"So it\u0027s a quantitative theory."},{"Start":"01:00.230 ","End":"01:05.090","Text":"Now the motivation for valence bond theory came from the Heitler,"},{"Start":"01:05.090 ","End":"01:10.670","Text":"London description of bonding in the hydrogen molecule and that was in"},{"Start":"01:10.670 ","End":"01:18.590","Text":"1927 only 1 year after Schrodinger wrote his famous equation for the hydrogen atom."},{"Start":"01:18.590 ","End":"01:25.130","Text":"We\u0027re going to plot the potential energy to call V. That\u0027s includes everything"},{"Start":"01:25.130 ","End":"01:27.980","Text":"except the kinetic energy of the nuclei versus"},{"Start":"01:27.980 ","End":"01:31.430","Text":"the distance r between the 2 hydrogen atoms."},{"Start":"01:31.430 ","End":"01:35.555","Text":"Each hydrogen atom is in its ground state where"},{"Start":"01:35.555 ","End":"01:41.675","Text":"each has 1 electron and that electron is in the what 1s orbital."},{"Start":"01:41.675 ","End":"01:48.130","Text":"We have a hydrogen with a 1s orbital and another hydrogen with a 1s orbital."},{"Start":"01:48.130 ","End":"01:49.560","Text":"Here\u0027s the plot,"},{"Start":"01:49.560 ","End":"01:55.130","Text":"we have the energy or potential energy V in kilo joules per mole"},{"Start":"01:55.130 ","End":"02:01.130","Text":"plotted against the distance between 2 hydrogen atoms in picometers."},{"Start":"02:01.130 ","End":"02:04.745","Text":"Now, when the 2 atoms are very far apart,"},{"Start":"02:04.745 ","End":"02:07.310","Text":"the potential energy is 0."},{"Start":"02:07.310 ","End":"02:08.765","Text":"If I went further,"},{"Start":"02:08.765 ","End":"02:12.410","Text":"it would actually get asymptotically to 0."},{"Start":"02:12.410 ","End":"02:16.070","Text":"That means that there\u0027s no interaction between the 2 atoms."},{"Start":"02:16.070 ","End":"02:17.465","Text":"They\u0027re too far apart."},{"Start":"02:17.465 ","End":"02:19.610","Text":"Now as they get closer,"},{"Start":"02:19.610 ","End":"02:22.370","Text":"there is attraction between the 2 atoms."},{"Start":"02:22.370 ","End":"02:27.710","Text":"Essentially, the nucleus of 1 atom attracts the electron in"},{"Start":"02:27.710 ","End":"02:30.380","Text":"the other atom so there\u0027s attraction"},{"Start":"02:30.380 ","End":"02:33.560","Text":"and the potential energy gets more and more negative."},{"Start":"02:33.560 ","End":"02:37.495","Text":"It gets lower until we get to this minimum."},{"Start":"02:37.495 ","End":"02:41.570","Text":"At this point, the attraction between the atoms is"},{"Start":"02:41.570 ","End":"02:47.330","Text":"exactly balanced by the repulsion between the nuclei."},{"Start":"02:47.330 ","End":"02:53.075","Text":"Here\u0027s attraction and here repulsion comes to play as well."},{"Start":"02:53.075 ","End":"02:55.295","Text":"At this point, at this minimum,"},{"Start":"02:55.295 ","End":"02:59.300","Text":"the 2 1s orbitals overlap."},{"Start":"02:59.300 ","End":"03:03.215","Text":"It\u0027s an overlapping region and the 2 electrons,"},{"Start":"03:03.215 ","End":"03:07.100","Text":"one from each hydrogen atom are paired."},{"Start":"03:07.100 ","End":"03:10.475","Text":"This is all summarized in what\u0027s written now."},{"Start":"03:10.475 ","End":"03:12.410","Text":"So when the atoms are far apart,"},{"Start":"03:12.410 ","End":"03:14.060","Text":"V is equal to 0."},{"Start":"03:14.060 ","End":"03:17.675","Text":"As the atoms become closer, V becomes negative."},{"Start":"03:17.675 ","End":"03:20.000","Text":"At the minimum, the attraction between the atoms"},{"Start":"03:20.000 ","End":"03:22.945","Text":"is balanced by the repulsion between the nuclei."},{"Start":"03:22.945 ","End":"03:24.795","Text":"We can see at the minimum,"},{"Start":"03:24.795 ","End":"03:30.770","Text":"that the bond length 74 picometers and the bond dissociation energy,"},{"Start":"03:30.770 ","End":"03:34.430","Text":"that\u0027s the energy required to take us from minus"},{"Start":"03:34.430 ","End":"03:40.460","Text":"436 kilo joules per mole at the minimum to the state of 0 energy,"},{"Start":"03:40.460 ","End":"03:42.560","Text":"where the atoms are far apart."},{"Start":"03:42.560 ","End":"03:45.830","Text":"So that energy is the energy required to"},{"Start":"03:45.830 ","End":"03:50.645","Text":"dissociate the molecule and its 436 kilo joules per mole."},{"Start":"03:50.645 ","End":"03:52.190","Text":"Now, at the minimum,"},{"Start":"03:52.190 ","End":"03:55.820","Text":"the orbitals overlap and the electrons are paired."},{"Start":"03:55.820 ","End":"04:01.140","Text":"Now that\u0027s called a Sigma bond by analogy with the S orbital."},{"Start":"04:01.140 ","End":"04:06.875","Text":"S is a Latin letter and Sigma is a Greek letter."},{"Start":"04:06.875 ","End":"04:09.770","Text":"These have some equivalence."},{"Start":"04:09.770 ","End":"04:12.080","Text":"We can draw it here, it\u0027s 1s,"},{"Start":"04:12.080 ","End":"04:16.890","Text":"on 1 hydrogen and the 1s on the other hydrogen and"},{"Start":"04:16.890 ","End":"04:22.010","Text":"they overlap and this bond is formed is called a Sigma bond."},{"Start":"04:22.010 ","End":"04:25.085","Text":"If we draw this as a z axis,"},{"Start":"04:25.085 ","End":"04:33.035","Text":"we see if we turn this bond route with respect to the z axis,"},{"Start":"04:33.035 ","End":"04:35.465","Text":"nothing happens, it just looks the same,"},{"Start":"04:35.465 ","End":"04:38.185","Text":"so it has no node."},{"Start":"04:38.185 ","End":"04:40.970","Text":"S orbitals also don\u0027t have a node,"},{"Start":"04:40.970 ","End":"04:43.360","Text":"and Sigma bonds also don\u0027t have a node."},{"Start":"04:43.360 ","End":"04:45.085","Text":"Even smaller distances,"},{"Start":"04:45.085 ","End":"04:47.770","Text":"nuclear repulsion takes over."},{"Start":"04:47.770 ","End":"04:52.635","Text":"In this video, we learned about valence bond theory,"},{"Start":"04:52.635 ","End":"04:57.610","Text":"it\u0027s the first of several videos about valence bond theory."}],"ID":23904},{"Watched":false,"Name":"Sigma and Pi Bonds","Duration":"6m 50s","ChapterTopicVideoID":23590,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.590","Text":"In the previous video,"},{"Start":"00:01.590 ","End":"00:04.200","Text":"we introduced valence bond theory."},{"Start":"00:04.200 ","End":"00:07.980","Text":"In this video, we\u0027ll learn about Sigma, and Pi bonds."},{"Start":"00:07.980 ","End":"00:10.560","Text":"Let\u0027s start with Sigma bond."},{"Start":"00:10.560 ","End":"00:12.885","Text":"Sigma bond, we mentioned before,"},{"Start":"00:12.885 ","End":"00:17.730","Text":"that Sigma is the Greek equivalent of the letter S. A Sigma bond"},{"Start":"00:17.730 ","End":"00:23.070","Text":"is formed when orbitals of neighboring atoms overlap head to head."},{"Start":"00:23.070 ","End":"00:26.400","Text":"The overlapping lobes must have the same sign,"},{"Start":"00:26.400 ","End":"00:30.750","Text":"that means the wave functions must have the same phase or the same side"},{"Start":"00:30.750 ","End":"00:36.075","Text":"so that we get constructive interference between the wave functions."},{"Start":"00:36.075 ","End":"00:40.065","Text":"So if we have 1s or the 2 hydrogens,"},{"Start":"00:40.065 ","End":"00:46.505","Text":"this is the area of overlap provided they each have the same side,"},{"Start":"00:46.505 ","End":"00:48.800","Text":"for example, both are positive."},{"Start":"00:48.800 ","End":"00:52.680","Text":"If 1 were positive, and 1 negative,"},{"Start":"00:52.680 ","End":"00:56.060","Text":"there would be destructive interference in"},{"Start":"00:56.060 ","End":"01:00.440","Text":"the middle and there would be no electron density between the 2 atoms,"},{"Start":"01:00.440 ","End":"01:02.300","Text":"and therefore no bond."},{"Start":"01:02.300 ","End":"01:07.220","Text":"Now, a Sigma bond contains 2 paired electrons, 1 from each atom."},{"Start":"01:07.220 ","End":"01:12.695","Text":"Occasionally, we get the situation where 1 atom contributes both the electrons."},{"Start":"01:12.695 ","End":"01:17.000","Text":"You\u0027ll recall this possibility from Lewis theory."},{"Start":"01:17.000 ","End":"01:23.990","Text":"Now, the Sigma bond is cylindrically symmetrical about the z-axis, internuclear axis."},{"Start":"01:23.990 ","End":"01:26.885","Text":"So if this is the z-axis,"},{"Start":"01:26.885 ","End":"01:35.050","Text":"and we rotate this bond about the z-axis, it won\u0027t change."},{"Start":"01:35.540 ","End":"01:41.750","Text":"That\u0027s equivalent to say it has no nodal plane containing the z-axis."},{"Start":"01:41.750 ","End":"01:45.990","Text":"There\u0027s no point at which the bond is equal to 0,"},{"Start":"01:45.990 ","End":"01:48.675","Text":"the electron density of the board is 0."},{"Start":"01:48.675 ","End":"01:51.950","Text":"Now, the Sigma bonds are very important because they define"},{"Start":"01:51.950 ","End":"01:55.580","Text":"the framework or skeleton of a molecule,"},{"Start":"01:55.580 ","End":"01:57.920","Text":"that gives us the shape of the molecule."},{"Start":"01:57.920 ","End":"01:59.840","Text":"Now, let\u0027s take a few examples."},{"Start":"01:59.840 ","End":"02:03.680","Text":"Hydrogen molecule, HCl, and Cl_2."},{"Start":"02:03.680 ","End":"02:05.840","Text":"We\u0027ve already met the hydrogen molecule,"},{"Start":"02:05.840 ","End":"02:07.675","Text":"but we\u0027ll write once again."},{"Start":"02:07.675 ","End":"02:10.785","Text":"The H electron here is in the 1s orbital,"},{"Start":"02:10.785 ","End":"02:12.345","Text":"and we get this situation."},{"Start":"02:12.345 ","End":"02:13.710","Text":"Here\u0027s the 1s orbital,"},{"Start":"02:13.710 ","End":"02:15.030","Text":"the 1s orbital,"},{"Start":"02:15.030 ","End":"02:20.450","Text":"and both the orbitals are positive or both negative,"},{"Start":"02:20.450 ","End":"02:25.055","Text":"but they have to have the same side for overlap to occur."},{"Start":"02:25.055 ","End":"02:28.565","Text":"Now, let\u0027s take the situation of HCl,"},{"Start":"02:28.565 ","End":"02:34.270","Text":"and the unpaired electron in Cl is in the 3p_z orbital."},{"Start":"02:34.270 ","End":"02:36.360","Text":"Here\u0027s a p_z orbital,"},{"Start":"02:36.360 ","End":"02:38.295","Text":"recall that it has a node,"},{"Start":"02:38.295 ","End":"02:40.850","Text":"so one side is negative, for example,"},{"Start":"02:40.850 ","End":"02:42.485","Text":"the other side positive,"},{"Start":"02:42.485 ","End":"02:44.990","Text":"and the 1s orbital also positive."},{"Start":"02:44.990 ","End":"02:52.485","Text":"So here there\u0027s constructive interference between 1 lobe of the 3p_z orbital,"},{"Start":"02:52.485 ","End":"02:55.110","Text":"and the 1s orbital."},{"Start":"02:55.110 ","End":"02:58.140","Text":"That\u0027s why we say that its head to head,"},{"Start":"02:58.140 ","End":"03:00.210","Text":"this is considered head to head."},{"Start":"03:00.210 ","End":"03:02.550","Text":"Now, if we talk about Cl_2,"},{"Start":"03:02.550 ","End":"03:09.405","Text":"then we have 1 electron in the 3p_z orbital,"},{"Start":"03:09.405 ","End":"03:13.520","Text":"and the other chlorine also the 3p_z orbital."},{"Start":"03:13.520 ","End":"03:17.000","Text":"These overlap head to head,"},{"Start":"03:17.000 ","End":"03:19.775","Text":"here\u0027s the area of overlap,"},{"Start":"03:19.775 ","End":"03:22.430","Text":"the positive lobe is here,"},{"Start":"03:22.430 ","End":"03:23.895","Text":"the positive lobe is here,"},{"Start":"03:23.895 ","End":"03:26.120","Text":"this is negative, this is negative."},{"Start":"03:26.120 ","End":"03:28.700","Text":"So this is our Sigma bond,"},{"Start":"03:28.700 ","End":"03:36.860","Text":"overlap between the positive lobes of the 3p_z orbitals on each chlorine atom."},{"Start":"03:36.860 ","End":"03:41.150","Text":"Now, another thing to note is that when there is only a Sigma bond,"},{"Start":"03:41.150 ","End":"03:43.355","Text":"that\u0027s called a single bond."},{"Start":"03:43.355 ","End":"03:48.045","Text":"So a single bond is a Sigma bond."},{"Start":"03:48.045 ","End":"03:51.770","Text":"Now, we\u0027re going to talk about Pi bonds."},{"Start":"03:51.770 ","End":"03:53.330","Text":"Now, a Pi bond,"},{"Start":"03:53.330 ","End":"03:57.400","Text":"and here we note that the Greek equivalent of p is Pi,"},{"Start":"03:57.400 ","End":"04:02.855","Text":"is formed when orbitals of neighboring atoms overlap side-to-side."},{"Start":"04:02.855 ","End":"04:06.015","Text":"Sigma is when they\u0027re head to head,"},{"Start":"04:06.015 ","End":"04:09.075","Text":"and Pi when they\u0027re side-to-side."},{"Start":"04:09.075 ","End":"04:14.225","Text":"The overlapping lobes must have the same side as before,"},{"Start":"04:14.225 ","End":"04:17.090","Text":"as we mentioned at the Sigma bond,"},{"Start":"04:17.090 ","End":"04:21.245","Text":"so that the wave functions interfere constructively."},{"Start":"04:21.245 ","End":"04:23.135","Text":"So here\u0027s an example."},{"Start":"04:23.135 ","End":"04:27.505","Text":"Here are 2p orbitals,"},{"Start":"04:27.505 ","End":"04:30.740","Text":"in this case it\u0027s p_x,"},{"Start":"04:32.480 ","End":"04:35.895","Text":"and supposing this is positive,"},{"Start":"04:35.895 ","End":"04:38.730","Text":"this is negative, this is positive, and negative."},{"Start":"04:38.730 ","End":"04:43.150","Text":"We have overlap between the 2 positive lobes,"},{"Start":"04:43.150 ","End":"04:46.440","Text":"and overlap between the 2 negative lobes,"},{"Start":"04:46.440 ","End":"04:48.935","Text":"and together these 2 parts,"},{"Start":"04:48.935 ","End":"04:50.945","Text":"the positive and the negative,"},{"Start":"04:50.945 ","End":"04:53.185","Text":"are called a Pi bond."},{"Start":"04:53.185 ","End":"04:58.280","Text":"Once again, the bond contains 2 paired electrons,"},{"Start":"04:58.280 ","End":"04:59.745","Text":"one from each atom."},{"Start":"04:59.745 ","End":"05:05.120","Text":"So there are 2 paired electrons in this Pi bond,"},{"Start":"05:05.120 ","End":"05:09.175","Text":"and there\u0027s 1 nodal plane containing the z-axis."},{"Start":"05:09.175 ","End":"05:13.540","Text":"If we draw a plane here containing the z-axis,"},{"Start":"05:13.540 ","End":"05:21.430","Text":"then we\u0027ll see that the bond goes from positive through 0 to negative,"},{"Start":"05:21.430 ","End":"05:23.630","Text":"that\u0027s called a nodal plane."},{"Start":"05:23.630 ","End":"05:28.745","Text":"Now, recall that nitrogen has the electronic structure of 1s^2,"},{"Start":"05:28.745 ","End":"05:30.020","Text":"2 electrons in 1s,"},{"Start":"05:30.020 ","End":"05:31.445","Text":"2 electrons in 2s,"},{"Start":"05:31.445 ","End":"05:33.515","Text":"and 3 electrons in 2p."},{"Start":"05:33.515 ","End":"05:36.925","Text":"So this is 1s, 2s,"},{"Start":"05:36.925 ","End":"05:41.985","Text":"and 3 electrons in 2p."},{"Start":"05:41.985 ","End":"05:45.195","Text":"So we can call this p_x p_y,"},{"Start":"05:45.195 ","End":"05:47.945","Text":"p_z, they all have the same energy."},{"Start":"05:47.945 ","End":"05:51.065","Text":"Now, let\u0027s look at the p_z orbitals."},{"Start":"05:51.065 ","End":"05:54.485","Text":"When they overlap we get the Sigma bond,"},{"Start":"05:54.485 ","End":"05:58.760","Text":"and when the p_x orbitals overlap we get a Pi bond,"},{"Start":"05:58.760 ","End":"06:01.340","Text":"and when the p_y orbitals overlap,"},{"Start":"06:01.340 ","End":"06:03.245","Text":"we get a second Pi bond."},{"Start":"06:03.245 ","End":"06:05.120","Text":"This is illustrated here."},{"Start":"06:05.120 ","End":"06:09.140","Text":"Here overlapping are the p_z orbitals,"},{"Start":"06:09.140 ","End":"06:11.300","Text":"that gives us a Sigma bond."},{"Start":"06:11.300 ","End":"06:16.425","Text":"P_x orbitals on each atom are overlapping,"},{"Start":"06:16.425 ","End":"06:19.710","Text":"that gives us a Pi bond,"},{"Start":"06:19.710 ","End":"06:25.660","Text":"and the p_y overlapping gives us another Pi bond."},{"Start":"06:25.660 ","End":"06:29.075","Text":"Here\u0027s an attempt to draw the whole structure,"},{"Start":"06:29.075 ","End":"06:30.895","Text":"so here is Sigma,"},{"Start":"06:30.895 ","End":"06:33.635","Text":"1 Pi, and another Pi."},{"Start":"06:33.635 ","End":"06:37.580","Text":"So when we have 1 Sigma bond plus 2Pi bonds,"},{"Start":"06:37.580 ","End":"06:39.575","Text":"it\u0027s called a triple bond."},{"Start":"06:39.575 ","End":"06:42.695","Text":"If we have 1 Sigma plus a Pi,"},{"Start":"06:42.695 ","End":"06:45.100","Text":"that\u0027s called a double bond."},{"Start":"06:45.100 ","End":"06:50.440","Text":"In this video we discussed Sigma and Pi bonds."}],"ID":24501},{"Watched":false,"Name":"Hybridization sp3 hybrid orbitals","Duration":"6m 35s","ChapterTopicVideoID":23056,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"In the previous video,"},{"Start":"00:02.040 ","End":"00:05.580","Text":"we learned about valence bond theory and Sigma and Pi bonds."},{"Start":"00:05.580 ","End":"00:10.845","Text":"In this video, we\u0027ll talk about the useful concept of hybridization."},{"Start":"00:10.845 ","End":"00:15.825","Text":"Let\u0027s start by considering the electronic configuration of carbon."},{"Start":"00:15.825 ","End":"00:18.000","Text":"We recall that we have 2 electrons in 1^s,"},{"Start":"00:18.000 ","End":"00:21.525","Text":"2 electrons in 2^s and 2 electrons in 2^p."},{"Start":"00:21.525 ","End":"00:25.815","Text":"Here\u0027s 1^s, 2^s and 2^p."},{"Start":"00:25.815 ","End":"00:29.475","Text":"Say this is px and this is py."},{"Start":"00:29.475 ","End":"00:34.335","Text":"According to this, carbon has 2 unpaired valence electrons."},{"Start":"00:34.335 ","End":"00:37.484","Text":"We expect it to form CH_2 molecules."},{"Start":"00:37.484 ","End":"00:39.015","Text":"Just like sulfur does."},{"Start":"00:39.015 ","End":"00:41.280","Text":"Sulfur has an extra 2 electrons."},{"Start":"00:41.280 ","End":"00:44.660","Text":"We expect the C-H bonds to be perpendicular to each other."},{"Start":"00:44.660 ","End":"00:47.375","Text":"1 hydrogen will overlap with the px"},{"Start":"00:47.375 ","End":"00:52.350","Text":"orbital and 1 hydrogen will overlap with the py orbitals."},{"Start":"00:52.350 ","End":"00:53.745","Text":"We have px,"},{"Start":"00:53.745 ","End":"00:57.560","Text":"the hydrogen, and py another hydrogen."},{"Start":"00:57.560 ","End":"01:01.430","Text":"However, we know that according to experiment in Lewis theory,"},{"Start":"01:01.430 ","End":"01:03.410","Text":"carbon has 4 valence electrons."},{"Start":"01:03.410 ","End":"01:05.135","Text":"We say it\u0027s tetravalent."},{"Start":"01:05.135 ","End":"01:09.050","Text":"It forms CH_4 molecules, not CH_2 molecules."},{"Start":"01:09.050 ","End":"01:13.055","Text":"In addition, we know from VSEPR theory that"},{"Start":"01:13.055 ","End":"01:17.750","Text":"all the C-H bonds are equivalent and the CH_4 is tetrahedral,"},{"Start":"01:17.750 ","End":"01:21.545","Text":"belongs to the AX_4 category of VSEPR theory."},{"Start":"01:21.545 ","End":"01:23.420","Text":"How can we overcome this problem?"},{"Start":"01:23.420 ","End":"01:29.235","Text":"The first thing to do is to promote a 2s electron to a 2p orbital."},{"Start":"01:29.235 ","End":"01:32.045","Text":"We\u0027re going to promote electrons from 2s-2p,"},{"Start":"01:32.045 ","End":"01:35.120","Text":"and that will give us 4 unpaired electrons."},{"Start":"01:35.120 ","End":"01:38.060","Text":"Now we have 1 in 2s and 3 in 2p,"},{"Start":"01:38.060 ","End":"01:42.610","Text":"px, py and pz."},{"Start":"01:42.760 ","End":"01:46.820","Text":"Now, you may think that this is strange because we"},{"Start":"01:46.820 ","End":"01:50.510","Text":"need quite a lot of energy to promote from 2s-2p."},{"Start":"01:50.510 ","End":"01:52.805","Text":"But it\u0027s not as much as you think"},{"Start":"01:52.805 ","End":"01:57.350","Text":"because there\u0027s no less repulsion between the electrons,"},{"Start":"01:57.350 ","End":"02:00.740","Text":"there\u0027s less repulsion where there are different orbitals."},{"Start":"02:00.740 ","End":"02:05.390","Text":"The amount of energy for promotion is not so great as we think."},{"Start":"02:05.390 ","End":"02:07.625","Text":"In addition where the molecules formed,"},{"Start":"02:07.625 ","End":"02:09.650","Text":"we get back energy."},{"Start":"02:09.650 ","End":"02:12.770","Text":"But this is still not an accurate description."},{"Start":"02:12.770 ","End":"02:19.055","Text":"Why? Because it looks now as if we have 3 bonds at right angles to each other,"},{"Start":"02:19.055 ","End":"02:20.360","Text":"one of the x-direction,"},{"Start":"02:20.360 ","End":"02:22.805","Text":"one of the y-direction and one of the z-direction."},{"Start":"02:22.805 ","End":"02:24.685","Text":"These are certainly equivalent."},{"Start":"02:24.685 ","End":"02:25.950","Text":"But in addition,"},{"Start":"02:25.950 ","End":"02:28.625","Text":"we have a hydrogen orbital,"},{"Start":"02:28.625 ","End":"02:34.385","Text":"a hydrogen 1s orbital overlapping with a carbon 2s orbital."},{"Start":"02:34.385 ","End":"02:40.319","Text":"We have hydrogen, 1s overlapping with 2s of carbon."},{"Start":"02:40.319 ","End":"02:44.150","Text":"The problem here is the 1s orbital could attach"},{"Start":"02:44.150 ","End":"02:50.210","Text":"anywhere along the circumference of this carbon 2s orbital."},{"Start":"02:50.210 ","End":"02:53.390","Text":"We don\u0027t know exactly where the bond will be."},{"Start":"02:53.390 ","End":"02:55.760","Text":"Now we have 4 bonds."},{"Start":"02:55.760 ","End":"02:58.175","Text":"It\u0027s an improvement on 2 bonds before,"},{"Start":"02:58.175 ","End":"03:00.900","Text":"but they are not all equivalent."},{"Start":"03:00.900 ","End":"03:06.845","Text":"The way to overcome this is to invoke the concept of hybridization."},{"Start":"03:06.845 ","End":"03:08.970","Text":"We take the 4 atomic orbitals,"},{"Start":"03:08.970 ","End":"03:10.815","Text":"2s, 2px, 2py,"},{"Start":"03:10.815 ","End":"03:15.875","Text":"2pz, and transform them into 4 linear combinations."},{"Start":"03:15.875 ","End":"03:19.370","Text":"These are called sp^3 hybrid orbitals,"},{"Start":"03:19.370 ","End":"03:24.315","Text":"or in this particular case, 2sp^3 hybrid orbitals."},{"Start":"03:24.315 ","End":"03:28.560","Text":"They come from 1s orbital and 3p orbitals."},{"Start":"03:28.560 ","End":"03:31.610","Text":"That\u0027s the media of the sp^3 notation."},{"Start":"03:31.610 ","End":"03:34.190","Text":"Here are the mathematical expressions for them."},{"Start":"03:34.190 ","End":"03:38.990","Text":"You see the only difference between them lies in the signs here,"},{"Start":"03:38.990 ","End":"03:42.811","Text":"all the signs of positive here are the 2 negative and then the others,"},{"Start":"03:42.811 ","End":"03:46.235","Text":"there are 2 positive and 2 negative signs,"},{"Start":"03:46.235 ","End":"03:47.704","Text":"but in different places."},{"Start":"03:47.704 ","End":"03:51.860","Text":"Now, we should note that the hybrid orbitals are still atomic orbitals."},{"Start":"03:51.860 ","End":"03:55.415","Text":"These are not the molecular orbitals that we\u0027ll talk about later."},{"Start":"03:55.415 ","End":"03:58.580","Text":"In addition, one lobe is larger than the other,"},{"Start":"03:58.580 ","End":"04:02.405","Text":"giving better it overlap with the hydrogen orbitals."},{"Start":"04:02.405 ","End":"04:05.440","Text":"We have 1 lobe,"},{"Start":"04:05.440 ","End":"04:08.205","Text":"it\u0027s large and 1 small lobe,"},{"Start":"04:08.205 ","End":"04:09.960","Text":"different signs of course."},{"Start":"04:09.960 ","End":"04:17.510","Text":"The 1s orbital of the hydrogen can overlap with this positive loop."},{"Start":"04:17.510 ","End":"04:21.410","Text":"What goes on in the rest of this talk,"},{"Start":"04:21.410 ","End":"04:27.200","Text":"we\u0027ll ignore the small negative lobe and just talk about the large positive."},{"Start":"04:27.200 ","End":"04:28.775","Text":"What about the energies?"},{"Start":"04:28.775 ","End":"04:30.905","Text":"Well, they all have the same energy."},{"Start":"04:30.905 ","End":"04:36.495","Text":"They have a 1/4 of the energy of 2s and 3/4 of the energy of 2p."},{"Start":"04:36.495 ","End":"04:39.855","Text":"They\u0027re higher than 2s and lower than 2p,"},{"Start":"04:39.855 ","End":"04:42.210","Text":"but closer of course to 2p."},{"Start":"04:42.210 ","End":"04:46.003","Text":"Here\u0027s 2s energy,"},{"Start":"04:46.003 ","End":"04:51.395","Text":"2p energy, and the energy of the 2sp^3 hybrids."},{"Start":"04:51.395 ","End":"04:56.000","Text":"Now important point about these hybrids is that the each point to different vertex,"},{"Start":"04:56.000 ","End":"04:58.700","Text":"a different corner of the tetrahedron."},{"Start":"04:58.700 ","End":"05:00.305","Text":"They\u0027re all equivalent."},{"Start":"05:00.305 ","End":"05:01.880","Text":"Here\u0027s a tetrahedron."},{"Start":"05:01.880 ","End":"05:05.480","Text":"These are the 4 hybrids and with only indicators,"},{"Start":"05:05.480 ","End":"05:09.290","Text":"I said before, the large positive lobes."},{"Start":"05:09.290 ","End":"05:13.055","Text":"Now let\u0027s talk about bonding methane, CH_4."},{"Start":"05:13.055 ","End":"05:19.565","Text":"Each sp^3 orbital on carbon overlaps with the 1s orbital on hydrogen."},{"Start":"05:19.565 ","End":"05:22.966","Text":"We have 1,"},{"Start":"05:22.966 ","End":"05:24.730","Text":"2, 3, 4."},{"Start":"05:24.730 ","End":"05:30.250","Text":"These 1s orbitals, each one contains a pair of electrons."},{"Start":"05:30.250 ","End":"05:33.490","Text":"These are Sigma bonds because they\u0027re head to head,"},{"Start":"05:33.490 ","End":"05:37.830","Text":"and we indicate them as Sigma in brackets,"},{"Start":"05:37.830 ","End":"05:41.920","Text":"carbon 2sp^3, and hydrogen 1s."},{"Start":"05:41.920 ","End":"05:43.900","Text":"That\u0027s how we designate them."},{"Start":"05:43.900 ","End":"05:47.720","Text":"Now we have 4 equivalent bonds."},{"Start":"05:50.480 ","End":"05:55.539","Text":"Now what about ammonia and water,"},{"Start":"05:55.539 ","End":"06:03.930","Text":"which you may recall from VSEPR, AX_3E and AX_2E_2."},{"Start":"06:03.930 ","End":"06:09.245","Text":"An ammonia, the lone pair goes into one of the hybrids."},{"Start":"06:09.245 ","End":"06:15.365","Text":"Then we have overlap with another 3 hydrogens, and in water,"},{"Start":"06:15.365 ","End":"06:17.030","Text":"we have 2 lone pairs,"},{"Start":"06:17.030 ","End":"06:23.645","Text":"it\u0027s AX_2E_2 and overlap with 2 hydrogens."},{"Start":"06:23.645 ","End":"06:30.470","Text":"That\u0027s our explanation of the bonding in ammonia and water."},{"Start":"06:30.470 ","End":"06:35.970","Text":"In this video we talked about sp^3 hybridization."}],"ID":23898},{"Watched":false,"Name":"Hybridization Sp2 and Sp Hybrid Orbitals","Duration":"6m 53s","ChapterTopicVideoID":23735,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.395","Text":"In the previous video,"},{"Start":"00:01.395 ","End":"00:04.290","Text":"we learned about sp^3 hybridization."},{"Start":"00:04.290 ","End":"00:08.625","Text":"This video will talk about sp^2 and sp hybridization."},{"Start":"00:08.625 ","End":"00:11.610","Text":"Let us begin with sp^2 hybridization."},{"Start":"00:11.610 ","End":"00:14.265","Text":"We\u0027ll take the boron atom as an example."},{"Start":"00:14.265 ","End":"00:18.900","Text":"Now the electronic configuration of boron is 2 electrons in 1s,"},{"Start":"00:18.900 ","End":"00:21.135","Text":"2 electrons in 2s,"},{"Start":"00:21.135 ","End":"00:23.940","Text":"and just 1 electron in 2p."},{"Start":"00:23.940 ","End":"00:28.185","Text":"But we know that according to experiment and Lewis theory,"},{"Start":"00:28.185 ","End":"00:30.390","Text":"boron has 3 valence electrons,"},{"Start":"00:30.390 ","End":"00:36.845","Text":"it\u0027s trivalent and it forms BH_3 molecules, BCl_3 molecules."},{"Start":"00:36.845 ","End":"00:41.150","Text":"From VSEPR theory, we know that all the B-H bonds are"},{"Start":"00:41.150 ","End":"00:45.750","Text":"equivalent and BH_3 has trigonal planar geometry."},{"Start":"00:45.750 ","End":"00:48.515","Text":"It\u0027s of the AX_3 category."},{"Start":"00:48.515 ","End":"00:53.030","Text":"So we\u0027re going to do the same thing as we did in the case of sp3."},{"Start":"00:53.030 ","End":"00:57.005","Text":"The first thing to do is to promote 2s electron."},{"Start":"00:57.005 ","End":"01:00.245","Text":"If we take a 2s electron and promote it to 2p,"},{"Start":"01:00.245 ","End":"01:05.940","Text":"then we have only 1 electron in 2s and 2 electrons in 2p."},{"Start":"01:05.940 ","End":"01:08.795","Text":"Now we have 3 unpaired electrons,"},{"Start":"01:08.795 ","End":"01:13.505","Text":"but we still haven\u0027t solved the problem of getting equivalent orbitals."},{"Start":"01:13.505 ","End":"01:20.510","Text":"S is round and 2p 1 point say in the direction of x and 1 in the direction of y,"},{"Start":"01:20.510 ","End":"01:24.545","Text":"so the s orbitals are not equivalent to the p orbitals."},{"Start":"01:24.545 ","End":"01:27.835","Text":"Once again, we can do hybridization."},{"Start":"01:27.835 ","End":"01:29.310","Text":"The 3 atomic orbitals,"},{"Start":"01:29.310 ","End":"01:30.480","Text":"2s, 2ps,"},{"Start":"01:30.480 ","End":"01:37.440","Text":"and 2p_y can be transformed mathematically into 3 linear combinations called sp^2,"},{"Start":"01:37.440 ","End":"01:42.555","Text":"or in this particular case, 2sp^2 hybrid orbitals."},{"Start":"01:42.555 ","End":"01:45.240","Text":"It\u0027s 2 because n is 2,"},{"Start":"01:45.240 ","End":"01:49.635","Text":"we have 2s, 2p_x, and 2p_y orbitals."},{"Start":"01:49.635 ","End":"01:52.355","Text":"Here are the mathematical expressions,"},{"Start":"01:52.355 ","End":"01:56.112","Text":"h_1 consists of s and p_y, h_2 of s,"},{"Start":"01:56.112 ","End":"01:57.440","Text":"p_x, and p_y,"},{"Start":"01:57.440 ","End":"02:01.350","Text":"and h_3 of s, p_x, and p_y again."},{"Start":"02:01.350 ","End":"02:02.820","Text":"But the signs are different."},{"Start":"02:02.820 ","End":"02:08.030","Text":"Here, it\u0027s a positive sign before the p_x and here\u0027s a negative sign before the p_x,"},{"Start":"02:08.030 ","End":"02:10.640","Text":"and here\u0027s a positive sign before the p_y and"},{"Start":"02:10.640 ","End":"02:15.385","Text":"a negative sign in h_2 and h_3 before the p_y."},{"Start":"02:15.385 ","End":"02:17.990","Text":"In addition, they all have the same energy."},{"Start":"02:17.990 ","End":"02:23.575","Text":"The energy is 1/3 of the energy of 2s and 2/3 of the energy of 2p."},{"Start":"02:23.575 ","End":"02:27.515","Text":"The energy is higher than 2s and lower than 2p,"},{"Start":"02:27.515 ","End":"02:30.155","Text":"but more towards the 2p."},{"Start":"02:30.155 ","End":"02:34.295","Text":"Now, since we only have p_x and p_y appearing,"},{"Start":"02:34.295 ","End":"02:36.830","Text":"they all lie in a plane, the x,"},{"Start":"02:36.830 ","End":"02:41.105","Text":"y plane, and they point to the corners of an equilateral triangle."},{"Start":"02:41.105 ","End":"02:42.740","Text":"Here\u0027s 1 of the hybrids."},{"Start":"02:42.740 ","End":"02:48.060","Text":"This is h_1 because you can see it\u0027s pointing in the y-direction."},{"Start":"02:48.160 ","End":"02:50.840","Text":"You can see it has 2 lobes;"},{"Start":"02:50.840 ","End":"02:52.765","Text":"a large 1 and a small 1."},{"Start":"02:52.765 ","End":"02:56.535","Text":"Let\u0027s call that plus and that minus."},{"Start":"02:56.535 ","End":"02:59.615","Text":"Here are the 3 hybrids."},{"Start":"02:59.615 ","End":"03:07.505","Text":"Neglecting the small lobes pointing to the corners of an equilateral triangle."},{"Start":"03:07.505 ","End":"03:12.555","Text":"This is h_1 because p_x is positive in h_2,"},{"Start":"03:12.555 ","End":"03:16.040","Text":"this is h_2, and this is h_3."},{"Start":"03:16.040 ","End":"03:19.385","Text":"Now let\u0027s consider the bonding in BH_3."},{"Start":"03:19.385 ","End":"03:27.140","Text":"We can take a hydrogen 1s orbital and overlap it with each of the hybrids."},{"Start":"03:27.140 ","End":"03:34.990","Text":"These form Sigma bonds and each Sigma bond has a pair of electrons in it."},{"Start":"03:34.990 ","End":"03:37.725","Text":"We can say that each sp^2 orbital,"},{"Start":"03:37.725 ","End":"03:40.890","Text":"B overlaps to the 1s orbital on H to form"},{"Start":"03:40.890 ","End":"03:45.225","Text":"a Sigma bond occupied by 2 electrons with paired spins."},{"Start":"03:45.225 ","End":"03:47.445","Text":"We write this as Sigma,"},{"Start":"03:47.445 ","End":"03:52.800","Text":"Boron 2sp^2, hydrogen 1s."},{"Start":"03:52.800 ","End":"03:58.790","Text":"Of course, each HBH angle is 120 degrees."},{"Start":"03:58.790 ","End":"04:01.990","Text":"Let\u0027s now look at sp hybridization."},{"Start":"04:01.990 ","End":"04:04.865","Text":"We\u0027ll take Beryllium as our example."},{"Start":"04:04.865 ","End":"04:08.870","Text":"The electronic configuration of Beryllium is 1s^2, 2s^2."},{"Start":"04:08.870 ","End":"04:10.925","Text":"2 electrons in 1s,"},{"Start":"04:10.925 ","End":"04:14.275","Text":"2 electrons in 2s."},{"Start":"04:14.275 ","End":"04:19.540","Text":"According to this, it would seem that there are no unpaired electrons"},{"Start":"04:19.540 ","End":"04:25.039","Text":"but we know that beryllium according to Lewis has 2 unpaired electrons."},{"Start":"04:25.039 ","End":"04:27.695","Text":"In order to get 2 unpaired electrons,"},{"Start":"04:27.695 ","End":"04:31.640","Text":"we need to promote an electron from 2s to 2p."},{"Start":"04:31.640 ","End":"04:34.475","Text":"Then we\u0027ll have 2 unpaired electrons;"},{"Start":"04:34.475 ","End":"04:41.135","Text":"1 at 2s and 1 in 2p. We\u0027ll take 2p_z."},{"Start":"04:41.135 ","End":"04:48.640","Text":"However, once again, an s orbital and a 2p_z orbital are not equivalent,"},{"Start":"04:48.640 ","End":"04:53.690","Text":"p_z points in the z-direction and s is round."},{"Start":"04:53.690 ","End":"04:56.825","Text":"Once again, we\u0027ll invoke hybridization."},{"Start":"04:56.825 ","End":"04:59.460","Text":"Returns form the 2 atomic orbitals,"},{"Start":"04:59.460 ","End":"05:08.164","Text":"2s and 2p_z into 2 linear combinations called sp or 2sp in this case, hybrid orbitals."},{"Start":"05:08.164 ","End":"05:14.885","Text":"Note that so it was the same number of atomic orbitals and linear combinations."},{"Start":"05:14.885 ","End":"05:17.605","Text":"Here are the mathematical expressions."},{"Start":"05:17.605 ","End":"05:24.975","Text":"H_1 is proportional to s plus p_z and each 2 to s minus p_z."},{"Start":"05:24.975 ","End":"05:29.329","Text":"The numbers at the beginning are normalization constants,"},{"Start":"05:29.329 ","End":"05:34.430","Text":"they\u0027re not normalization factors and optically important at the moment."},{"Start":"05:34.430 ","End":"05:38.630","Text":"Once again, the 2 hybrids have the same energy,"},{"Start":"05:38.630 ","End":"05:44.495","Text":"which is intermediate halfway between the 2s energy and the 2p energy."},{"Start":"05:44.495 ","End":"05:51.305","Text":"1 hybrid points 1 direction of z-axis and the other direction."},{"Start":"05:51.305 ","End":"05:53.914","Text":"Here\u0027s 1 of them, again,"},{"Start":"05:53.914 ","End":"05:55.970","Text":"with a large lobe and a small lobe,"},{"Start":"05:55.970 ","End":"05:58.730","Text":"let\u0027s call this plus, this minus."},{"Start":"05:58.730 ","End":"06:00.320","Text":"Here are the 2 of them."},{"Start":"06:00.320 ","End":"06:02.930","Text":"Just drawing the large lobes,"},{"Start":"06:02.930 ","End":"06:06.755","Text":"so both plus and the line, a straight line."},{"Start":"06:06.755 ","End":"06:10.670","Text":"The angle here is 180 degrees."},{"Start":"06:10.670 ","End":"06:15.055","Text":"Let\u0027s look at the bonding in Beryllium hydride."},{"Start":"06:15.055 ","End":"06:19.215","Text":"We can take a 1s orbital and overlap it with this lobe,"},{"Start":"06:19.215 ","End":"06:22.840","Text":"and a 1s orbital and overlap with this one."},{"Start":"06:22.840 ","End":"06:26.580","Text":"So we have 2sp orbitals on Beryllium"},{"Start":"06:26.580 ","End":"06:31.358","Text":"overlapping with a 1s orbital in hydrogen to form a Sigma bond,"},{"Start":"06:31.358 ","End":"06:37.755","Text":"and every Sigma bond has 2 electrons in it with paired spins."},{"Start":"06:37.755 ","End":"06:43.800","Text":"We write this as Sigma Beryllium 2sp hydrogen 1s and of course,"},{"Start":"06:43.800 ","End":"06:48.000","Text":"HBH angle is 180 degrees."},{"Start":"06:48.000 ","End":"06:53.530","Text":"In this video, we discussed sp^2 and sp hybridization."}],"ID":24640},{"Watched":false,"Name":"Hybridization sp3d and sp3d2 Hybrid Orbitals","Duration":"7m 43s","ChapterTopicVideoID":23154,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"In previous videos, we learned about sp,"},{"Start":"00:03.210 ","End":"00:06.570","Text":"sp^2 and sp^3 hybridization."},{"Start":"00:06.570 ","End":"00:12.570","Text":"This video, we\u0027ll talk about hybridization schemes that include d orbitals."},{"Start":"00:12.570 ","End":"00:15.315","Text":"Why would we need d orbitals?"},{"Start":"00:15.315 ","End":"00:19.065","Text":"If you recall something like methane,"},{"Start":"00:19.065 ","End":"00:22.575","Text":"for which we use sp^3 hybridization,"},{"Start":"00:22.575 ","End":"00:26.475","Text":"has 4 unpaired electrons around the carbon,"},{"Start":"00:26.475 ","End":"00:32.550","Text":"and a total 8 electrons around carbon once the bonds are formed."},{"Start":"00:32.550 ","End":"00:36.615","Text":"2 electrons in each C-H bond,"},{"Start":"00:36.615 ","End":"00:42.630","Text":"so all together 8 electrons making an octet around the carbon."},{"Start":"00:42.630 ","End":"00:47.435","Text":"If we have more than 8 electrons around the central atom,"},{"Start":"00:47.435 ","End":"00:49.910","Text":"that\u0027s more than an octet of electrons."},{"Start":"00:49.910 ","End":"00:52.565","Text":"We need to use the expanded valence shell,"},{"Start":"00:52.565 ","End":"00:56.839","Text":"and we talked about that when we talked about Lewis theory."},{"Start":"00:56.839 ","End":"00:58.850","Text":"What we said then was that,"},{"Start":"00:58.850 ","End":"01:02.405","Text":"some third-period elements such as phosphorus and sulfur,"},{"Start":"01:02.405 ","End":"01:09.575","Text":"form compounds that can only be explained by having 10 or 12 electrons around them."},{"Start":"01:09.575 ","End":"01:14.300","Text":"For example, PCl_5 which has 10 electrons around the phosphorus,"},{"Start":"01:14.300 ","End":"01:19.615","Text":"and SF_6 which has 12 electrons around the sulfur."},{"Start":"01:19.615 ","End":"01:22.970","Text":"Where we talked about VSEPR theory"},{"Start":"01:22.970 ","End":"01:27.170","Text":"recall the case of 10 electrons around the central atom."},{"Start":"01:27.170 ","End":"01:32.915","Text":"AX_5 and 12 electrodes on the center atom as AX_6,"},{"Start":"01:32.915 ","End":"01:35.250","Text":"that\u0027s 10 and 12."},{"Start":"01:35.250 ","End":"01:37.950","Text":"Now when we have 10 electrons around A,"},{"Start":"01:37.950 ","End":"01:39.675","Text":"that\u0027s the AX_5,"},{"Start":"01:39.675 ","End":"01:43.110","Text":"the shape of the electron groups,"},{"Start":"01:43.110 ","End":"01:47.950","Text":"and of the molecule is trigonal bipyramidal."},{"Start":"01:48.350 ","End":"01:51.060","Text":"We gave it as an example,"},{"Start":"01:51.060 ","End":"01:54.765","Text":"PCl_5 and here is the shape."},{"Start":"01:54.765 ","End":"01:59.355","Text":"We have 3 bonds in the equatorial plane,"},{"Start":"01:59.355 ","End":"02:03.390","Text":"and the bond angle of 120 degrees."},{"Start":"02:03.390 ","End":"02:09.200","Text":"Then we have 2 bonds which are axial to the equatorial plane,"},{"Start":"02:09.200 ","End":"02:16.999","Text":"and the bond angle between an axial bond and an equatorial bond is 90 degrees."},{"Start":"02:16.999 ","End":"02:19.675","Text":"Now when we have 12 electrons around A,"},{"Start":"02:19.675 ","End":"02:24.730","Text":"we call that AX_6 and the shape is octahedral."},{"Start":"02:24.730 ","End":"02:29.330","Text":"We have the examples SF_6 here\u0027s the octahedral shape,"},{"Start":"02:29.330 ","End":"02:32.705","Text":"and all the bond angles are 90 degrees."},{"Start":"02:32.705 ","End":"02:37.130","Text":"Now, we need a hybridization schemes,"},{"Start":"02:37.130 ","End":"02:42.425","Text":"which will explain these 2 geometrical shapes."},{"Start":"02:42.425 ","End":"02:48.620","Text":"Let\u0027s start off with AX_5 and we\u0027re taking it as an example, phosphorus."},{"Start":"02:48.620 ","End":"02:53.750","Text":"Now the electron configuration of phosphorus is same as that of"},{"Start":"02:53.750 ","End":"02:59.190","Text":"neon plus 2 electrons in 3s and 3 electrons in 3p."},{"Start":"02:59.190 ","End":"03:01.845","Text":"We have a pair of electrons in 3s,"},{"Start":"03:01.845 ","End":"03:05.850","Text":"and 3 unpaired electrons in 3p."},{"Start":"03:05.850 ","End":"03:10.420","Text":"The highest number of bonds we can form is 3,"},{"Start":"03:10.420 ","End":"03:13.745","Text":"because we only have 3 unpaired electrons."},{"Start":"03:13.745 ","End":"03:17.435","Text":"This would be fine for PCl_3,"},{"Start":"03:17.435 ","End":"03:20.900","Text":"but can\u0027t accommodate PCl_5."},{"Start":"03:20.900 ","End":"03:22.835","Text":"What we can do is again,"},{"Start":"03:22.835 ","End":"03:26.915","Text":"promote an electron from 3s to 3d."},{"Start":"03:26.915 ","End":"03:29.450","Text":"We take one of the electrons from 3s,"},{"Start":"03:29.450 ","End":"03:31.655","Text":"we only have 1 on the period electron."},{"Start":"03:31.655 ","End":"03:37.920","Text":"We still have the 3 unpaired electrons in 3p, 3s, 3p."},{"Start":"03:37.920 ","End":"03:42.615","Text":"Now we\u0027ve taken 1 of the 3s electrons and put it into 3d."},{"Start":"03:42.615 ","End":"03:45.930","Text":"Now we have 5 unpaired electrons,"},{"Start":"03:45.930 ","End":"03:50.415","Text":"and that would be sufficient to form PCl_5."},{"Start":"03:50.415 ","End":"03:56.610","Text":"Hybridization scheme that we can form from these orbitals,"},{"Start":"03:56.610 ","End":"04:00.690","Text":"remember 5 orbitals gives us 5 hybrids."},{"Start":"04:00.690 ","End":"04:06.975","Text":"Is sp^3d is called sp^3d because there\u0027s 1s orbital,"},{"Start":"04:06.975 ","End":"04:11.295","Text":"3p orbitals, and 1d orbital."},{"Start":"04:11.295 ","End":"04:14.460","Text":"The shape of this is trigonal bipyramidal,"},{"Start":"04:14.460 ","End":"04:15.620","Text":"and here\u0027s the picture."},{"Start":"04:15.620 ","End":"04:21.470","Text":"Here we can see the larger lobes of each of the hybrids."},{"Start":"04:21.470 ","End":"04:24.065","Text":"I haven\u0027t indicated the smaller ones."},{"Start":"04:24.065 ","End":"04:27.605","Text":"Let\u0027s put the sign positive on all of them."},{"Start":"04:27.605 ","End":"04:32.445","Text":"Now if we want to form a molecule such as PCl_5,"},{"Start":"04:32.445 ","End":"04:40.130","Text":"we need to take the p orbital on each chlorine that only has 1 electron in it,"},{"Start":"04:40.130 ","End":"04:45.875","Text":"and overlap them with all these hybrids."},{"Start":"04:45.875 ","End":"04:49.805","Text":"For example, we could have 1 p orbital here."},{"Start":"04:49.805 ","End":"04:56.225","Text":"We need to put the plus sign of the orbital towards the plus of the hybrid,"},{"Start":"04:56.225 ","End":"05:00.530","Text":"and another p orbital here and so on."},{"Start":"05:00.530 ","End":"05:04.445","Text":"That gives us the molecule PCl_5."},{"Start":"05:04.445 ","End":"05:08.000","Text":"We see that according to this hybridization scheme,"},{"Start":"05:08.000 ","End":"05:12.890","Text":"the molecule will have also have the trigonal bipyramidal shape."},{"Start":"05:12.890 ","End":"05:18.380","Text":"Now let\u0027s talk about the case where there are 12 electrons around the central atom."},{"Start":"05:18.380 ","End":"05:20.885","Text":"We take sulfur as a example."},{"Start":"05:20.885 ","End":"05:23.180","Text":"Here\u0027s electron configuration of sulfur."},{"Start":"05:23.180 ","End":"05:24.460","Text":"It\u0027s like neon,"},{"Start":"05:24.460 ","End":"05:28.335","Text":"2 electrons in 3s and 4 electrons in 3p."},{"Start":"05:28.335 ","End":"05:31.620","Text":"We have a pair of electrons in 3s,"},{"Start":"05:31.620 ","End":"05:35.300","Text":"a pair of electrons in this p orbital,"},{"Start":"05:35.300 ","End":"05:39.775","Text":"and 2 unpaired electrons in the other p orbitals."},{"Start":"05:39.775 ","End":"05:43.475","Text":"In total, we only have 2 unpaired electrons."},{"Start":"05:43.475 ","End":"05:47.885","Text":"This is fine for forming a molecule SH_2,"},{"Start":"05:47.885 ","End":"05:51.745","Text":"but can\u0027t help us when we want to form SF_6."},{"Start":"05:51.745 ","End":"05:55.235","Text":"Again, we need to do promotion."},{"Start":"05:55.235 ","End":"06:02.450","Text":"We\u0027re going to promote 1 of the 3s electrons and 1 of the 3p electrons to 3d."},{"Start":"06:02.450 ","End":"06:06.235","Text":"We now have 1 unpaired electron in 3s,"},{"Start":"06:06.235 ","End":"06:09.630","Text":"3 unpaired electrons in 3p,"},{"Start":"06:09.630 ","End":"06:13.065","Text":"and 2 unpaired electrons in 3d."},{"Start":"06:13.065 ","End":"06:15.880","Text":"We have an unpaired electrons in"},{"Start":"06:15.880 ","End":"06:23.070","Text":"6 orbitals and the hybridization scheme is called sp^3d^2,"},{"Start":"06:23.070 ","End":"06:25.830","Text":"1s orbital, 3p orbitals,"},{"Start":"06:25.830 ","End":"06:27.465","Text":"and 2d orbitals,"},{"Start":"06:27.465 ","End":"06:29.630","Text":"and the shape is octahedral."},{"Start":"06:29.630 ","End":"06:33.580","Text":"Here\u0027s a picture here of the large lobes of"},{"Start":"06:33.580 ","End":"06:37.960","Text":"the hybrids indicate plus sign on all of them."},{"Start":"06:37.960 ","End":"06:40.880","Text":"We can see the shape is octahedral,"},{"Start":"06:40.880 ","End":"06:44.900","Text":"and now we can overlap them with the orbital on"},{"Start":"06:44.900 ","End":"06:50.480","Text":"the fluorine that contains only a single unpaired electron."},{"Start":"06:50.480 ","End":"06:51.890","Text":"Here, for example,"},{"Start":"06:51.890 ","End":"06:54.624","Text":"we can have a p orbital."},{"Start":"06:54.624 ","End":"07:00.330","Text":"These pluses and minus or here for example, plus minus."},{"Start":"07:00.330 ","End":"07:05.460","Text":"That way we can form the SF_6 molecule."},{"Start":"07:05.460 ","End":"07:08.840","Text":"Now before closing, we need to note the following."},{"Start":"07:08.840 ","End":"07:12.200","Text":"We need to note that the hybridization schemes"},{"Start":"07:12.200 ","End":"07:16.430","Text":"evolving the s and p orbitals are well established and widely used."},{"Start":"07:16.430 ","End":"07:22.790","Text":"However, those including d orbitals are less well established unless certain,"},{"Start":"07:22.790 ","End":"07:25.790","Text":"but can still be very useful."},{"Start":"07:25.790 ","End":"07:31.640","Text":"The sp types of hybridization schemes are really very,"},{"Start":"07:31.640 ","End":"07:33.080","Text":"very well established,"},{"Start":"07:33.080 ","End":"07:37.054","Text":"but those including d are not at all certain."},{"Start":"07:37.054 ","End":"07:43.830","Text":"In this video, we learned about hybridization schemes that involve d orbitals."}],"ID":23986},{"Watched":false,"Name":"Multiple covalent bonds","Duration":"9m 12s","ChapterTopicVideoID":23152,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.775","Text":"In previous videos, we learned about Sigma and Pi bonds and also about hybridization."},{"Start":"00:05.775 ","End":"00:09.975","Text":"In this video, we\u0027ll apply this knowledge to organic molecules."},{"Start":"00:09.975 ","End":"00:11.940","Text":"Let\u0027s begin with double bonds."},{"Start":"00:11.940 ","End":"00:13.860","Text":"As we learned in the previous video,"},{"Start":"00:13.860 ","End":"00:19.020","Text":"a double bond consists of a Sigma bond that gives a skeleton of the molecule,"},{"Start":"00:19.020 ","End":"00:22.440","Text":"and a Pi bond that gives extra strength."},{"Start":"00:22.440 ","End":"00:26.130","Text":"The Pi bond is usually weaker than the Sigma bond,"},{"Start":"00:26.130 ","End":"00:31.965","Text":"since side-to-side overlap, which gives a Pi bond is smaller than head-to-head overlap."},{"Start":"00:31.965 ","End":"00:36.605","Text":"If we have side-to-side overlap between say, 2p orbitals,"},{"Start":"00:36.605 ","End":"00:38.210","Text":"this should actually overlap,"},{"Start":"00:38.210 ","End":"00:42.355","Text":"but I\u0027m just drawing them for convenience with a space between them."},{"Start":"00:42.355 ","End":"00:46.850","Text":"Here, the plus plus overlaps in the 9, minus, minus overlaps."},{"Start":"00:46.850 ","End":"00:51.095","Text":"That overlap is smaller than the case of a Sigma bond."},{"Start":"00:51.095 ","End":"00:52.984","Text":"We\u0027re here, for example,"},{"Start":"00:52.984 ","End":"00:55.610","Text":"we have 2p orbitals,"},{"Start":"00:55.610 ","End":"00:58.625","Text":"plus plus, minus, minus,"},{"Start":"00:58.625 ","End":"01:00.755","Text":"overlapping head to head."},{"Start":"01:00.755 ","End":"01:02.960","Text":"This is Sigma and this is Pi."},{"Start":"01:02.960 ","End":"01:05.015","Text":"Generally, the overlap."},{"Start":"01:05.015 ","End":"01:09.040","Text":"The Sigma case is greater than that in the Pi case."},{"Start":"01:09.040 ","End":"01:13.135","Text":"Sigma bond is greater than Pi,"},{"Start":"01:13.135 ","End":"01:16.715","Text":"sigma bond is stronger than a Pi bond."},{"Start":"01:16.715 ","End":"01:19.850","Text":"Now, when do we get these double bonds?"},{"Start":"01:19.850 ","End":"01:23.740","Text":"The second period elements like carbon, nitrogen, oxygen."},{"Start":"01:23.740 ","End":"01:28.010","Text":"These are the important elements in organic chemistry,"},{"Start":"01:28.010 ","End":"01:35.840","Text":"often form double bonds either with themselves as we\u0027ll see C C or with one another,"},{"Start":"01:35.840 ","End":"01:38.900","Text":"and also with elements in later periods,"},{"Start":"01:38.900 ","End":"01:43.229","Text":"you get things like P O bonds."},{"Start":"01:43.229 ","End":"01:49.245","Text":"However, the third and higher period elements rarely form double bonds."},{"Start":"01:49.245 ","End":"01:51.650","Text":"That\u0027s because the distance between the atoms is too"},{"Start":"01:51.650 ","End":"01:54.575","Text":"great for side-to-side overlap to occur."},{"Start":"01:54.575 ","End":"01:57.295","Text":"The further away the 2 atoms get,"},{"Start":"01:57.295 ","End":"02:01.010","Text":"the less overlap there is."},{"Start":"02:01.010 ","End":"02:05.165","Text":"Let\u0027s take an example and the example would choose as ethylene,"},{"Start":"02:05.165 ","End":"02:08.140","Text":"that\u0027s called ethene, according to IUPAC,"},{"Start":"02:08.140 ","End":"02:11.395","Text":"that CH_2 a double bond, CH_2."},{"Start":"02:11.395 ","End":"02:14.900","Text":"Now, we know that ethylene is a Planar molecule"},{"Start":"02:14.900 ","End":"02:19.060","Text":"and it has bond angles close to a 120 degrees."},{"Start":"02:19.060 ","End":"02:21.390","Text":"It has 2 carbons,"},{"Start":"02:21.390 ","End":"02:26.295","Text":"4 hydrogens a double bond between the 2 carbons,"},{"Start":"02:26.295 ","End":"02:28.770","Text":"and 4 single bonds,"},{"Start":"02:28.770 ","End":"02:33.185","Text":"4 Sigma bonds between the carbons and hydrogens."},{"Start":"02:33.185 ","End":"02:36.620","Text":"As soon as we hear 120 degrees,"},{"Start":"02:36.620 ","End":"02:43.235","Text":"we should think of a trigonal planar arrangement that\u0027s AX_3 according to VSEPR,"},{"Start":"02:43.235 ","End":"02:46.865","Text":"an sp^2 hybridization for each carbon atom."},{"Start":"02:46.865 ","End":"02:54.885","Text":"These go together, AX_3 and sp^2 and a 120 degrees, all form group."},{"Start":"02:54.885 ","End":"02:57.785","Text":"Whenever say AX_3 is a 120 degrees,"},{"Start":"02:57.785 ","End":"03:01.165","Text":"wherever is a 120 degrees is sp^2."},{"Start":"03:01.165 ","End":"03:06.390","Text":"Now, how can we form sp^2 hybrids in the case of carbon?"},{"Start":"03:06.390 ","End":"03:15.045","Text":"After promotion carbons electron configuration is given by 1 electron into s,"},{"Start":"03:15.045 ","End":"03:18.105","Text":"1 electron into p_x,"},{"Start":"03:18.105 ","End":"03:23.025","Text":"1 into p_y, and 1 into p_z."},{"Start":"03:23.025 ","End":"03:32.070","Text":"We\u0027ve use the s orbital and the p_x and p_y orbitals to form sp^2 hybrids."},{"Start":"03:32.070 ","End":"03:36.615","Text":"Remember, 3 orbitals gives us 3 sp^2 hybrids."},{"Start":"03:36.615 ","End":"03:42.745","Text":"Then we\u0027re left with the p_z orbital to give us the Pi bond."},{"Start":"03:42.745 ","End":"03:45.365","Text":"These will give us Sigma bonds."},{"Start":"03:45.365 ","End":"03:47.600","Text":"This will give us Pi bond."},{"Start":"03:47.600 ","End":"03:52.955","Text":"Here are the sp^2 hybrids on each carbon."},{"Start":"03:52.955 ","End":"03:55.080","Text":"These are in purple."},{"Start":"03:55.080 ","End":"04:00.160","Text":"These can overlap with the hydrogen 1s orbitals."},{"Start":"04:00.160 ","End":"04:03.850","Text":"There\u0027s 4 of them, so we get 4 Sigma bonds here,"},{"Start":"04:03.850 ","End":"04:06.590","Text":"plus a Sigma bond and another one,"},{"Start":"04:06.590 ","End":"04:09.075","Text":"the third one and the fourth one."},{"Start":"04:09.075 ","End":"04:18.525","Text":"In addition, the 2 sp^2s on each carbon can overlap to give us a Sigma bond."},{"Start":"04:18.525 ","End":"04:24.885","Text":"We\u0027ve got 4 sigma bonds between carbon sp^2 and hydrogen 1s,"},{"Start":"04:24.885 ","End":"04:29.830","Text":"and 1 sigma bond between the sp^2 on each carbon."},{"Start":"04:29.830 ","End":"04:32.960","Text":"Now we\u0027re left with the p_z orbital."},{"Start":"04:32.960 ","End":"04:36.450","Text":"Here it is. These can overlap."},{"Start":"04:36.450 ","End":"04:39.830","Text":"In this way positive and this is positive,"},{"Start":"04:39.830 ","End":"04:45.150","Text":"these overlap, and this negative and negative and these overlap."},{"Start":"04:45.150 ","End":"04:52.800","Text":"Together the 2 overlapping p orbitals give us a Pi bond."},{"Start":"04:52.800 ","End":"04:59.915","Text":"It has a positive region on the top and a negative region on the bottom and together,"},{"Start":"04:59.915 ","End":"05:05.915","Text":"these 2 regions are Pi bonds that\u0027s 1 Pi bond."},{"Start":"05:05.915 ","End":"05:10.579","Text":"Now, when we form the Pi bond by overlapping"},{"Start":"05:10.579 ","End":"05:14.885","Text":"the p_z orbital on 1 carbon with a p_z orbital the other carbon,"},{"Start":"05:14.885 ","End":"05:18.350","Text":"we get the greatest overlap when the molecule is planar."},{"Start":"05:18.350 ","End":"05:22.880","Text":"If we were to rotate it out of the plane,"},{"Start":"05:22.880 ","End":"05:24.610","Text":"we would get less overlap."},{"Start":"05:24.610 ","End":"05:27.649","Text":"That\u0027s why this molecule is planar."},{"Start":"05:27.649 ","End":"05:33.830","Text":"Here we have a double bond consisting of a Sigma bond and a Pi bond."},{"Start":"05:33.830 ","End":"05:37.760","Text":"Now let\u0027s talk about triple bonds in organic chemistry."},{"Start":"05:37.760 ","End":"05:41.645","Text":"Let\u0027s take an example of acetylene that\u0027s called ethene,"},{"Start":"05:41.645 ","End":"05:45.335","Text":"ethyne, according to IUPAC."},{"Start":"05:45.335 ","End":"05:48.045","Text":"CH triple bond Ch."},{"Start":"05:48.045 ","End":"05:54.560","Text":"Acetylene is a linear molecule with bond angles of 180 degrees that we can draw"},{"Start":"05:54.560 ","End":"06:01.795","Text":"it to C C H on either side,"},{"Start":"06:01.795 ","End":"06:09.095","Text":"a triple bond between the 2 carbons and a single bond between carbon and hydrogen."},{"Start":"06:09.095 ","End":"06:12.755","Text":"As soon as we see that it\u0027s a linear molecule,"},{"Start":"06:12.755 ","End":"06:15.935","Text":"we know it\u0027s AX_2 on each carbon."},{"Start":"06:15.935 ","End":"06:17.450","Text":"As soon as we know it\u0027s AX_2,"},{"Start":"06:17.450 ","End":"06:19.790","Text":"we know sp hybridization."},{"Start":"06:19.790 ","End":"06:23.900","Text":"How can we get sp hybridization from carbon?"},{"Start":"06:23.900 ","End":"06:32.645","Text":"After we promote the carbon\u0027s electron n configuration is given by 1 electron in 2s,"},{"Start":"06:32.645 ","End":"06:35.460","Text":"1 electron in 2p_z,"},{"Start":"06:35.560 ","End":"06:38.570","Text":"1 electron into p_x,"},{"Start":"06:38.570 ","End":"06:41.580","Text":"and 1 into p_y."},{"Start":"06:41.980 ","End":"06:51.930","Text":"If I use the 2s orbital and the 2p_z orbital to form 2 sp hybrids."},{"Start":"06:51.930 ","End":"06:55.035","Text":"I\u0027m left with p_x and p_y,"},{"Start":"06:55.035 ","End":"06:57.600","Text":"and these can give me the Pi orbitals."},{"Start":"06:57.600 ","End":"06:58.710","Text":"This will give me Pi,"},{"Start":"06:58.710 ","End":"07:00.180","Text":"this will give me Pi."},{"Start":"07:00.180 ","End":"07:03.820","Text":"These together will give me Sigma."},{"Start":"07:03.820 ","End":"07:06.470","Text":"Here\u0027s the molecule."},{"Start":"07:06.470 ","End":"07:10.310","Text":"Here are sp hybrids on this carbon,"},{"Start":"07:10.310 ","End":"07:13.865","Text":"the purple and sp hybrids on the other carbon."},{"Start":"07:13.865 ","End":"07:18.025","Text":"Now these can overlap to give a Sigma bond."},{"Start":"07:18.025 ","End":"07:26.265","Text":"They can also overlap with the 1s on the hydrogen,"},{"Start":"07:26.265 ","End":"07:32.470","Text":"to give another sigma bond and of course on the other side and other Sigma bond."},{"Start":"07:32.470 ","End":"07:39.660","Text":"Then, we\u0027re left with 2p_x and 2p_y orbitals."},{"Start":"07:39.660 ","End":"07:45.020","Text":"If this is supposed to be a 2p_x,"},{"Start":"07:45.020 ","End":"07:51.020","Text":"it can overlap with the other 2p_x on the other carbon."},{"Start":"07:51.020 ","End":"07:53.735","Text":"Say this is 2p_y,"},{"Start":"07:53.735 ","End":"08:01.290","Text":"so that can overlap with the 2p_y on the other carbon."},{"Start":"08:01.290 ","End":"08:07.380","Text":"Then we have this loop overlapping with this one."},{"Start":"08:07.380 ","End":"08:10.679","Text":"Now we have 2 Pi bonds."},{"Start":"08:10.679 ","End":"08:15.670","Text":"1 forms with 2p_x and 1 from 2p_y."},{"Start":"08:16.790 ","End":"08:19.130","Text":"This is our triple bond,"},{"Start":"08:19.130 ","End":"08:23.975","Text":"which consists of a Sigma bond and 2 Pi bonds."},{"Start":"08:23.975 ","End":"08:27.400","Text":"Just a word about the bond energies."},{"Start":"08:27.400 ","End":"08:31.625","Text":"We\u0027ve seen this before, for carbon-carbon bonds,"},{"Start":"08:31.625 ","End":"08:35.390","Text":"the bond energy for a single bond is 347,"},{"Start":"08:35.390 ","End":"08:41.780","Text":"and the bond energy for a double bond 611 and for triple bond 837."},{"Start":"08:41.780 ","End":"08:44.405","Text":"We can see here from the single bond,"},{"Start":"08:44.405 ","End":"08:47.605","Text":"this is the bond energy of a Sigma bond."},{"Start":"08:47.605 ","End":"08:50.090","Text":"From the difference between the two,"},{"Start":"08:50.090 ","End":"08:54.725","Text":"which gives us 264 kilojoules per mole."},{"Start":"08:54.725 ","End":"08:57.470","Text":"That\u0027s the energy bond,"},{"Start":"08:57.470 ","End":"08:59.755","Text":"energy of a Pi bond."},{"Start":"08:59.755 ","End":"09:06.285","Text":"We see that the Sigma bond is much stronger than the Pi bond."},{"Start":"09:06.285 ","End":"09:12.960","Text":"In this video, we discussed double and triple bonds in organic molecules."}],"ID":23984},{"Watched":false,"Name":"Summary and example of VB theory","Duration":"6m 45s","ChapterTopicVideoID":23058,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:04.470","Text":"In this video, we will summarize what we\u0027ve learned"},{"Start":"00:04.470 ","End":"00:07.935","Text":"about valence bond theory in the previous videos,"},{"Start":"00:07.935 ","End":"00:12.630","Text":"and also show an example of how to use it."},{"Start":"00:12.630 ","End":"00:16.560","Text":"Let\u0027s begin with a strategy to describe the structure of"},{"Start":"00:16.560 ","End":"00:19.965","Text":"a molecule in terms of Sigma and pi bonds."},{"Start":"00:19.965 ","End":"00:23.355","Text":"The first step is to write a plausible Lewis structure."},{"Start":"00:23.355 ","End":"00:26.700","Text":"Once we\u0027ve done that, we should note the bond orders,"},{"Start":"00:26.700 ","End":"00:27.930","Text":"whether we have single,"},{"Start":"00:27.930 ","End":"00:29.640","Text":"double, or triple bonds."},{"Start":"00:29.640 ","End":"00:35.940","Text":"Then we can use VSEPR theory to determine the AXE notation,"},{"Start":"00:35.940 ","End":"00:39.629","Text":"and electron group geometry for each of the central atoms."},{"Start":"00:39.629 ","End":"00:41.190","Text":"Once we have done that,"},{"Start":"00:41.190 ","End":"00:47.885","Text":"we can find the corresponding hybridization schemes for each of the central atoms."},{"Start":"00:47.885 ","End":"00:52.580","Text":"Finally, we can form the Sigma bonds and then the pi bonds."},{"Start":"00:52.580 ","End":"00:56.330","Text":"Here\u0027s a table that\u0027s very useful and will help us"},{"Start":"00:56.330 ","End":"01:01.940","Text":"greatly in all that we need to do for valence bond theory."},{"Start":"01:01.940 ","End":"01:04.950","Text":"In it we have the VSEPR notation,"},{"Start":"01:04.950 ","End":"01:06.290","Text":"that\u0027s the first column,"},{"Start":"01:06.290 ","End":"01:12.160","Text":"the hybridization, the electron group geometry, and the angles."},{"Start":"01:12.160 ","End":"01:17.840","Text":"I think this table is so important that every student should learn it by heart."},{"Start":"01:17.840 ","End":"01:20.200","Text":"For example, if we have AX_4,"},{"Start":"01:20.200 ","End":"01:23.760","Text":"we immediately know the hybridization is sp^3,"},{"Start":"01:23.760 ","End":"01:26.660","Text":"the electron group geometry is tetrahedral,"},{"Start":"01:26.660 ","End":"01:30.515","Text":"and the bond angle is 109.5 degrees."},{"Start":"01:30.515 ","End":"01:34.805","Text":"If we\u0027re given any of the other pieces of information,"},{"Start":"01:34.805 ","End":"01:36.470","Text":"we can find the other 3."},{"Start":"01:36.470 ","End":"01:40.340","Text":"For example, if we know the bond angle is 109.5 degrees,"},{"Start":"01:40.340 ","End":"01:44.180","Text":"we should remember that means it\u0027s tetrahedral,"},{"Start":"01:44.180 ","End":"01:46.415","Text":"sp^3, and AX_4."},{"Start":"01:46.415 ","End":"01:50.030","Text":"Of course, this table is also useful when we replace"},{"Start":"01:50.030 ","End":"01:54.020","Text":"an atom X by lone-pair E. For example,"},{"Start":"01:54.020 ","End":"01:57.495","Text":"we have AX_2E_2,"},{"Start":"01:57.495 ","End":"02:02.675","Text":"then we also know that\u0027s sp^3 hybridization,"},{"Start":"02:02.675 ","End":"02:05.839","Text":"the electron group geometry is tetrahedral,"},{"Start":"02:05.839 ","End":"02:09.350","Text":"and the angles are 109.5 degrees."},{"Start":"02:09.350 ","End":"02:14.450","Text":"Let\u0027s take the example of HCOOH, that\u0027s formic acid."},{"Start":"02:14.450 ","End":"02:18.245","Text":"The central carbon is attached to 2 oxygens,"},{"Start":"02:18.245 ","End":"02:21.725","Text":"and 1 of the oxygens is attached to a hydrogen."},{"Start":"02:21.725 ","End":"02:24.365","Text":"Let\u0027s begin by drawing the Lewis structure."},{"Start":"02:24.365 ","End":"02:29.060","Text":"The carbon is attached to an oxygen by a double bond."},{"Start":"02:29.060 ","End":"02:33.455","Text":"It\u0027s also attached to a hydrogen and to another oxygen."},{"Start":"02:33.455 ","End":"02:37.595","Text":"I\u0027ve drawn them in different colors just to distinguish between them."},{"Start":"02:37.595 ","End":"02:43.855","Text":"This oxygen, the black oxygen is attached to 2 atoms and has 2 lone-pairs."},{"Start":"02:43.855 ","End":"02:45.405","Text":"This terminal oxygen,"},{"Start":"02:45.405 ","End":"02:47.315","Text":"we can also discuss it,"},{"Start":"02:47.315 ","End":"02:52.690","Text":"is attached to carbon by double bond and has 2 lone-pairs."},{"Start":"02:52.690 ","End":"03:00.185","Text":"Now, we can work out the VSEPR notation for carbon and oxygen and the other oxygen."},{"Start":"03:00.185 ","End":"03:01.990","Text":"Let\u0027s start off with this carbon."},{"Start":"03:01.990 ","End":"03:06.280","Text":"It\u0027s attached to 3 atoms, so it\u0027s AX_3."},{"Start":"03:06.280 ","End":"03:12.350","Text":"This oxygen is attached to 2 atoms and has 2 lone-pairs, so it\u0027s AX_2E_2."},{"Start":"03:12.350 ","End":"03:20.200","Text":"The terminal oxygen is attached to 1 atom and has 2 lone-pairs, so it\u0027s AX_E_2."},{"Start":"03:20.200 ","End":"03:24.365","Text":"From the table, we know what the bond angles should be."},{"Start":"03:24.365 ","End":"03:27.350","Text":"We know when we\u0027re talking about AX_3,"},{"Start":"03:27.350 ","End":"03:30.155","Text":"we\u0027re talking about 120 degrees."},{"Start":"03:30.155 ","End":"03:32.420","Text":"When we\u0027re talking of AX_2E_2,"},{"Start":"03:32.420 ","End":"03:36.540","Text":"we\u0027re talking about 109.5 degrees."},{"Start":"03:36.540 ","End":"03:39.724","Text":"Now, we can draw the molecule, carbon,"},{"Start":"03:39.724 ","End":"03:43.580","Text":"oxygen, hydrogen, oxygen, and hydrogen."},{"Start":"03:43.580 ","End":"03:46.430","Text":"A double bond between carbon and oxygen,"},{"Start":"03:46.430 ","End":"03:48.894","Text":"a single bond to this hydrogen,"},{"Start":"03:48.894 ","End":"03:50.700","Text":"a single bond to this oxygen,"},{"Start":"03:50.700 ","End":"03:54.225","Text":"a single bond between oxygen and hydrogen."},{"Start":"03:54.225 ","End":"03:56.415","Text":"We can draw the angles."},{"Start":"03:56.415 ","End":"04:00.382","Text":"Here we have 120 degrees, 120 degrees,"},{"Start":"04:00.382 ","End":"04:07.580","Text":"120 degrees, and here we have 109.5 degrees."},{"Start":"04:07.580 ","End":"04:12.605","Text":"Now experimentally, the real angles are very close to these."},{"Start":"04:12.605 ","End":"04:14.990","Text":"This is 118 degrees,"},{"Start":"04:14.990 ","End":"04:20.810","Text":"124 degrees, and 108 degrees."},{"Start":"04:20.810 ","End":"04:24.815","Text":"They\u0027re fairly close to the angles we\u0027ve predicted."},{"Start":"04:24.815 ","End":"04:28.205","Text":"Now we know the VSEPR notation,"},{"Start":"04:28.205 ","End":"04:31.280","Text":"we can work out the hybridization."},{"Start":"04:31.280 ","End":"04:34.490","Text":"If it\u0027s AX_3 or AXE_2,"},{"Start":"04:34.490 ","End":"04:37.495","Text":"we\u0027re talking about sp^2 hybridization."},{"Start":"04:37.495 ","End":"04:42.570","Text":"For AX_2E_2, we\u0027re talking about sp^3 hybridization."},{"Start":"04:42.570 ","End":"04:49.040","Text":"We have sp^2 hybridization for this carbon and terminal oxygen,"},{"Start":"04:49.040 ","End":"04:52.415","Text":"and sp^3 hybridization for this oxygen."},{"Start":"04:52.415 ","End":"04:53.900","Text":"Now, we can form the bonds."},{"Start":"04:53.900 ","End":"04:56.060","Text":"First, we\u0027ll form the Sigma bonds,"},{"Start":"04:56.060 ","End":"04:58.055","Text":"and then will form the pi bond,"},{"Start":"04:58.055 ","End":"05:03.525","Text":"carbon, hydrogen, oxygen, oxygen, and hydrogen."},{"Start":"05:03.525 ","End":"05:06.380","Text":"First, we\u0027re forming the Sigma bonds."},{"Start":"05:06.380 ","End":"05:09.515","Text":"Let\u0027s begin between carbon and hydrogen."},{"Start":"05:09.515 ","End":"05:15.450","Text":"We have sp^2 on this carbon and a 1s orbital on this hydrogen."},{"Start":"05:15.450 ","End":"05:23.905","Text":"Here\u0027s the description of the sigma bond between carbon 2sp^2 and hydrogen 1s."},{"Start":"05:23.905 ","End":"05:27.095","Text":"Then we have a Sigma bond between the carbon,"},{"Start":"05:27.095 ","End":"05:30.140","Text":"which is sp^2 and this terminal oxygen,"},{"Start":"05:30.140 ","End":"05:31.850","Text":"which is also sp^2."},{"Start":"05:31.850 ","End":"05:36.605","Text":"Then we have a single bond between the carbon and this oxygen."},{"Start":"05:36.605 ","End":"05:38.730","Text":"This carbon is sp^2,"},{"Start":"05:38.730 ","End":"05:40.995","Text":"this oxygen is sp^3,"},{"Start":"05:40.995 ","End":"05:42.630","Text":"so here\u0027s the description."},{"Start":"05:42.630 ","End":"05:51.814","Text":"Then we have a bond between the sp^3 on the oxygen and the hydrogen 1s orbital."},{"Start":"05:51.814 ","End":"05:56.030","Text":"This is our description of the 4 Sigma bonds."},{"Start":"05:56.030 ","End":"06:00.560","Text":"Now, we need to add the pi bond in order to get"},{"Start":"06:00.560 ","End":"06:05.320","Text":"a double bond between carbon and this terminal oxygen."},{"Start":"06:05.320 ","End":"06:09.510","Text":"Here is our pi bond between the carbon"},{"Start":"06:09.510 ","End":"06:15.185","Text":"2p orbital on the carbon and the 2p orbital on the oxygen."},{"Start":"06:15.185 ","End":"06:21.425","Text":"This will be perpendicular to the plane of the molecule,"},{"Start":"06:21.425 ","End":"06:25.745","Text":"and this will be perpendicular to the plane of the molecule."},{"Start":"06:25.745 ","End":"06:30.100","Text":"Our pi bond is coming out towards us."},{"Start":"06:30.100 ","End":"06:37.040","Text":"Here\u0027s our description of formic acid in terms of Sigma and pi bonds."},{"Start":"06:37.040 ","End":"06:40.640","Text":"In this video, we talked about the stages in determining"},{"Start":"06:40.640 ","End":"06:46.019","Text":"the structure of a molecule in terms of Sigma and pi bonds."}],"ID":23900},{"Watched":false,"Name":"Exercise 1 - part a","Duration":"5m 40s","ChapterTopicVideoID":23566,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.620 ","End":"00:03.765","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.765 ","End":"00:05.940","Text":"Draw the valence shell orbital diagram,"},{"Start":"00:05.940 ","End":"00:08.785","Text":"which is the hybridization of pure atomic orbitals to"},{"Start":"00:08.785 ","End":"00:12.870","Text":"hybrid atomic orbitals of the central atom in the following molecules."},{"Start":"00:12.870 ","End":"00:18.300","Text":"Let\u0027s begin with a. Again, we"},{"Start":"00:18.300 ","End":"00:19.980","Text":"want to draw the hybridization of"},{"Start":"00:19.980 ","End":"00:23.685","Text":"pure atomic orbitals to hybrid atomic orbitals for carbon."},{"Start":"00:23.685 ","End":"00:30.340","Text":"We\u0027re going to begin with the valence shell electron configuration for carbon."},{"Start":"00:33.740 ","End":"00:37.710","Text":"Now, we have 2 electrons in"},{"Start":"00:37.710 ","End":"00:46.050","Text":"2s and then in our 2p orbitals,"},{"Start":"00:46.050 ","End":"00:48.729","Text":"we have 2 unpaired electrons."},{"Start":"00:49.420 ","End":"00:52.160","Text":"Now electron configuration, just a reminder,"},{"Start":"00:52.160 ","End":"00:57.455","Text":"was explained in Chapter 8 so you can go back if you don\u0027t remember how to do this."},{"Start":"00:57.455 ","End":"01:02.260","Text":"We have the valence shell electron configuration of carbon."},{"Start":"01:02.260 ","End":"01:06.694","Text":"Let\u0027s find the hybridization of carbon."},{"Start":"01:06.694 ","End":"01:11.185","Text":"We\u0027re going to have to write a Lewis structure of our molecule, so let\u0027s begin."},{"Start":"01:11.185 ","End":"01:12.630","Text":"Remember Lewis structure,"},{"Start":"01:12.630 ","End":"01:14.600","Text":"we begin by counting the number of valence electrons."},{"Start":"01:14.600 ","End":"01:16.040","Text":"We have 4 chlorines,"},{"Start":"01:16.040 ","End":"01:20.820","Text":"so that\u0027s 4 times 7 since it\u0027s in Group 17."},{"Start":"01:20.820 ","End":"01:24.300","Text":"Plus we have another 4 valence electrons from"},{"Start":"01:24.300 ","End":"01:28.970","Text":"our carbon so this is going to give us 32 electrons."},{"Start":"01:28.970 ","End":"01:31.370","Text":"Next we\u0027re going to draw the skeletal structure."},{"Start":"01:31.370 ","End":"01:36.720","Text":"Of course we\u0027re going to put the carbon is essential atom and the chlorines around it."},{"Start":"01:41.060 ","End":"01:44.190","Text":"Now we have 4 single bonds,"},{"Start":"01:44.190 ","End":"01:50.610","Text":"meaning we have 8 electrons so we\u0027re going to subtract these 8 electrons from our bonds,"},{"Start":"01:50.610 ","End":"01:54.150","Text":"from the total number of valence electrons."},{"Start":"01:54.150 ","End":"01:57.755","Text":"That\u0027s going to leave us with 24 electrons."},{"Start":"01:57.755 ","End":"01:59.670","Text":"Again, we have 4 chlorines,"},{"Start":"01:59.670 ","End":"02:03.580","Text":"so it\u0027s going to give us 6 electrons for each chlorine."},{"Start":"02:09.530 ","End":"02:12.295","Text":"This is our Lewis structure."},{"Start":"02:12.295 ","End":"02:15.250","Text":"Now the next step after we have Lewis structure is to see"},{"Start":"02:15.250 ","End":"02:18.100","Text":"the number of electron groups we have around our central atom."},{"Start":"02:18.100 ","End":"02:21.650","Text":"Around our carbon, we have 4 single bonds,"},{"Start":"02:21.770 ","End":"02:25.540","Text":"meaning 4 electron groups."},{"Start":"02:25.540 ","End":"02:28.795","Text":"Now remember from the VSEPR theory,"},{"Start":"02:28.795 ","End":"02:31.720","Text":"every single bond is thought of as 1 group."},{"Start":"02:31.720 ","End":"02:35.650","Text":"A double or triple bond is also thought of as"},{"Start":"02:35.650 ","End":"02:39.595","Text":"1 group and the lone pair of electrons is also thought of as 1 group."},{"Start":"02:39.595 ","End":"02:41.590","Text":"In this case again, we have 4 single bonds,"},{"Start":"02:41.590 ","End":"02:43.895","Text":"meaning 4 electron groups."},{"Start":"02:43.895 ","End":"02:46.870","Text":"That\u0027s 4 groups."},{"Start":"02:48.600 ","End":"02:53.440","Text":"Now since we know that our carbon is with 4 groups,"},{"Start":"02:53.440 ","End":"02:56.250","Text":"meaning attached to 4 bonds,"},{"Start":"02:56.250 ","End":"03:00.844","Text":"we need an orbital diagram with 4 unpaired electrons."},{"Start":"03:00.844 ","End":"03:04.925","Text":"If we look at our electron configuration in the beginning,"},{"Start":"03:04.925 ","End":"03:07.655","Text":"we don\u0027t have 4 unpaired electrons."},{"Start":"03:07.655 ","End":"03:11.540","Text":"What we\u0027re going to do, we\u0027re going to take 1 of the 2s orbital"},{"Start":"03:11.540 ","End":"03:16.260","Text":"electrons and we\u0027re going to excite it to a 2p orbital."},{"Start":"03:17.270 ","End":"03:24.870","Text":"That\u0027s going to leave us with 1 unpaired electron in our 2s orbital and our 2p orbital."},{"Start":"03:28.790 ","End":"03:32.950","Text":"We\u0027re going to have 3 unpaired electrons."},{"Start":"03:35.090 ","End":"03:39.110","Text":"These are the unhybridized atomic orbitals."},{"Start":"03:39.110 ","End":"03:43.470","Text":"Now we have to hybridize them. Let\u0027s take a look."},{"Start":"03:46.540 ","End":"03:54.380","Text":"In the beginning, we have our 2s orbital with 1 unpaired electron, and we have 3,"},{"Start":"03:54.380 ","End":"04:02.680","Text":"2p orbitals, each with 1 unpaired electron."},{"Start":"04:10.430 ","End":"04:17.280","Text":"Now these 1s orbital and 3p orbitals"},{"Start":"04:17.280 ","End":"04:25.690","Text":"are going to hybridize into 4 sp3 orbitals."},{"Start":"04:38.270 ","End":"04:41.580","Text":"These are hybridized orbitals."},{"Start":"04:41.580 ","End":"04:45.340","Text":"Our orbital diagram for carbon,"},{"Start":"04:45.980 ","End":"04:49.540","Text":"which is connected to 4 groups,"},{"Start":"04:50.480 ","End":"04:59.383","Text":"we begin with 1s"},{"Start":"04:59.383 ","End":"05:07.804","Text":"orbital containing 1 electron and 3p orbitals containing 3 electrons."},{"Start":"05:07.804 ","End":"05:13.745","Text":"These 4 orbitals or 4 electrons hybridize 2,"},{"Start":"05:13.745 ","End":"05:17.680","Text":"4 electrons in 4 sp3 orbitals."},{"Start":"05:17.680 ","End":"05:19.430","Text":"Now remember that the number of"},{"Start":"05:19.430 ","End":"05:23.960","Text":"unhybridized orbitals is going to equal the number of hybridized orbitals."},{"Start":"05:23.960 ","End":"05:26.930","Text":"Hybridized orbitals are going to be in-between."},{"Start":"05:26.930 ","End":"05:31.820","Text":"The sp3 is going to be in-between our s and our p orbitals as you can see."},{"Start":"05:31.820 ","End":"05:35.210","Text":"That\u0027s the orbital diagram for carbon,"},{"Start":"05:35.210 ","End":"05:37.670","Text":"which is connected to 4 groups."},{"Start":"05:37.670 ","End":"05:40.560","Text":"Now we\u0027re going on to b."}],"ID":24475},{"Watched":false,"Name":"Exercise 1 - part b","Duration":"3m 12s","ChapterTopicVideoID":23567,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:03.520","Text":"Now we\u0027re going to go on to b."},{"Start":"00:04.520 ","End":"00:08.385","Text":"Let\u0027s begin by drawing the Lewis structure."},{"Start":"00:08.385 ","End":"00:11.790","Text":"We have 4 valence electrons from carbon,"},{"Start":"00:11.790 ","End":"00:14.130","Text":"6 valence electrons from oxygen,"},{"Start":"00:14.130 ","End":"00:18.689","Text":"and 2 times 1 valence electrons from hydrogen."},{"Start":"00:18.689 ","End":"00:21.299","Text":"That\u0027s going to give us 12 electrons."},{"Start":"00:21.299 ","End":"00:22.950","Text":"We\u0027re going to draw the skeletal structure,"},{"Start":"00:22.950 ","End":"00:25.485","Text":"so we have our carbon oxygen and 2 hydrogens."},{"Start":"00:25.485 ","End":"00:26.790","Text":"Carbon is the central atom,"},{"Start":"00:26.790 ","End":"00:29.325","Text":"oxygen is terminal atom,"},{"Start":"00:29.325 ","End":"00:31.965","Text":"and our hydrogens are terminal atoms."},{"Start":"00:31.965 ","End":"00:34.650","Text":"We have 3 bonds, meaning 6 electrons."},{"Start":"00:34.650 ","End":"00:36.750","Text":"We\u0027re going to subtract 6 electrons from"},{"Start":"00:36.750 ","End":"00:39.120","Text":"12 and that\u0027s going to leave us with another 6 electrons,"},{"Start":"00:39.120 ","End":"00:42.940","Text":"which we\u0027re going to put on our oxygen."},{"Start":"00:43.460 ","End":"00:45.590","Text":"Now of course, as you can see,"},{"Start":"00:45.590 ","End":"00:47.390","Text":"the carbon does not have an octet."},{"Start":"00:47.390 ","End":"00:49.520","Text":"One of these lone pair of electrons is going to be"},{"Start":"00:49.520 ","End":"00:56.010","Text":"shared between the oxygen and carbon and this is going to give us a double bond."},{"Start":"00:59.090 ","End":"01:02.390","Text":"Lewis structure, we have carbon"},{"Start":"01:02.390 ","End":"01:07.120","Text":"connected with single bonds to hydrogens and with a double bond to oxygen."},{"Start":"01:07.120 ","End":"01:10.925","Text":"In all the carbon has 3 electron groups around it."},{"Start":"01:10.925 ","End":"01:14.160","Text":"2 from single bond, 1 from a double bond."},{"Start":"01:14.210 ","End":"01:16.790","Text":"Since we have 3 groups,"},{"Start":"01:16.790 ","End":"01:20.580","Text":"it means that the hybridization is sp2,"},{"Start":"01:21.350 ","End":"01:25.320","Text":"which is combined from 1s orbital and 2p orbitals."},{"Start":"01:25.320 ","End":"01:28.710","Text":"Let\u0027s take a look at our orbital diagram."},{"Start":"01:29.380 ","End":"01:34.505","Text":"Now remember from a that the carbon,"},{"Start":"01:34.505 ","End":"01:40.055","Text":"we begin with 1 unpaired electron in a 2s orbital,"},{"Start":"01:40.055 ","End":"01:43.734","Text":"and then we have 3 2p orbitals,"},{"Start":"01:43.734 ","End":"01:47.240","Text":"which each contain 1 unpaired electrons,"},{"Start":"01:47.240 ","End":"01:49.980","Text":"meaning we have 4 unpaired electrons."},{"Start":"01:51.920 ","End":"01:54.000","Text":"Now in this case,"},{"Start":"01:54.000 ","End":"02:04.060","Text":"3 of these orbitals hybridize to give us 3 sp2 orbitals."},{"Start":"02:05.930 ","End":"02:09.730","Text":"Again, since we have 3 electron groups,"},{"Start":"02:09.730 ","End":"02:14.690","Text":"each is going to have an unpaired electron and we\u0027re left with"},{"Start":"02:14.690 ","End":"02:20.974","Text":"1 2p orbital with 1 unpaired electron."},{"Start":"02:20.974 ","End":"02:26.195","Text":"Now the sp2 hybridized orbitals are the ones who form the Sigma bonds,"},{"Start":"02:26.195 ","End":"02:30.515","Text":"which are the bonds to the hydrogen and the single bond to the oxygen."},{"Start":"02:30.515 ","End":"02:33.110","Text":"The p orbital, which is not hybridized,"},{"Start":"02:33.110 ","End":"02:38.400","Text":"is what forms the Pi bond between the carbon and oxygen."},{"Start":"02:38.400 ","End":"02:40.995","Text":"Again, our orbital diagram starts"},{"Start":"02:40.995 ","End":"02:48.375","Text":"with 1 unpaired electron and an s orbital and 3 unpaired electrons and 2p orbitals."},{"Start":"02:48.375 ","End":"02:51.290","Text":"Since we have 3 groups around our carbon,"},{"Start":"02:51.290 ","End":"02:58.650","Text":"this hybridizes 2sp2, 3s p2 orbital."},{"Start":"03:00.680 ","End":"03:07.260","Text":"These account for the Sigma bonds and we\u0027re left with 1 2p orbital,"},{"Start":"03:07.260 ","End":"03:09.440","Text":"which accounts for the Pi bond."},{"Start":"03:09.440 ","End":"03:10.790","Text":"That\u0027s our answer for b. Now,"},{"Start":"03:10.790 ","End":"03:13.230","Text":"we\u0027re going to go on to c."}],"ID":24476},{"Watched":false,"Name":"Exercise 1 - part c","Duration":"4m 27s","ChapterTopicVideoID":23568,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.650","Text":"Now we\u0027re going to take a look at c. In c we have nitrogen."},{"Start":"00:07.580 ","End":"00:12.480","Text":"Before we begin, let\u0027s take a look at the electron configuration for nitrogen,"},{"Start":"00:12.480 ","End":"00:15.450","Text":"the valence electron configuration."},{"Start":"00:15.450 ","End":"00:19.140","Text":"Again, that\u0027s explained in Chapter 8 if you don\u0027t remember how to do that."},{"Start":"00:19.140 ","End":"00:25.290","Text":"We have 2 electrons in our 2s orbital."},{"Start":"00:25.290 ","End":"00:34.720","Text":"Then we have 3 unpaired electrons in our 2p orbitals."},{"Start":"00:36.100 ","End":"00:38.910","Text":"Now let\u0027s take a look at our Lewis structure."},{"Start":"00:38.910 ","End":"00:41.365","Text":"We\u0027re going to draw 1 obviously."},{"Start":"00:41.365 ","End":"00:44.240","Text":"Again, you begin with 2 nitrogens in order"},{"Start":"00:44.240 ","End":"00:47.000","Text":"to count the number of valence electrons we have."},{"Start":"00:47.000 ","End":"00:50.465","Text":"Where this is going to multiply 2 by 5 since"},{"Start":"00:50.465 ","End":"00:54.699","Text":"nitrogen is in the 15th group and it\u0027s going to give us 10 electrons."},{"Start":"00:54.699 ","End":"00:58.884","Text":"Obviously we\u0027re going to start with a double bond between nitrogen."},{"Start":"00:58.884 ","End":"01:00.870","Text":"We\u0027re going to take off 2 electrons,"},{"Start":"01:00.870 ","End":"01:02.880","Text":"subtract 2 electrons from our 10."},{"Start":"01:02.880 ","End":"01:04.530","Text":"That gives us 8 electrons left."},{"Start":"01:04.530 ","End":"01:06.180","Text":"It\u0027s going to give me 1, 2, 3, 4,"},{"Start":"01:06.180 ","End":"01:08.095","Text":"5, 6, 7, 8."},{"Start":"01:08.095 ","End":"01:11.795","Text":"Obviously, we can see that our nitrogens do not have octets."},{"Start":"01:11.795 ","End":"01:14.900","Text":"Therefore, we need to share electrons."},{"Start":"01:14.900 ","End":"01:20.255","Text":"We\u0027re going to share 1 lone pair of electrons and another lone pair electrons."},{"Start":"01:20.255 ","End":"01:28.040","Text":"That\u0027s going to give us a triple bond between our nitrogens."},{"Start":"01:28.160 ","End":"01:31.175","Text":"That\u0027s going to give us our Lewis structure."},{"Start":"01:31.175 ","End":"01:33.605","Text":"Now both nitrogens have octet."},{"Start":"01:33.605 ","End":"01:36.845","Text":"Now if we look at each nitrogen, they\u0027re equal."},{"Start":"01:36.845 ","End":"01:38.330","Text":"We\u0027re just going to look at 1."},{"Start":"01:38.330 ","End":"01:40.610","Text":"We can see that we have 2 electron groups around"},{"Start":"01:40.610 ","End":"01:42.740","Text":"it because we have 1 lone pair of electrons,"},{"Start":"01:42.740 ","End":"01:44.240","Text":"which is thought of as 1 electron group,"},{"Start":"01:44.240 ","End":"01:47.330","Text":"and we have a triple bond which has another electron group."},{"Start":"01:47.330 ","End":"01:50.960","Text":"Now the hybridization is sp because it\u0027s going to be combined with"},{"Start":"01:50.960 ","End":"01:54.710","Text":"1s orbital and 1p orbital since we have 2 groups."},{"Start":"01:54.710 ","End":"01:58.790","Text":"Now remember that the hybridized orbitals account"},{"Start":"01:58.790 ","End":"02:03.065","Text":"for the lone pair electrons and they also account for the Sigma bonds."},{"Start":"02:03.065 ","End":"02:07.745","Text":"However, the Pi bonds come from unhybridized orbitals."},{"Start":"02:07.745 ","End":"02:10.475","Text":"Let\u0027s take a look at our orbital diagram."},{"Start":"02:10.475 ","End":"02:14.880","Text":"Let\u0027s begin with our unhybridized orbitals."},{"Start":"02:20.710 ","End":"02:27.340","Text":"We have 2s orbital with 2 electrons."},{"Start":"02:27.340 ","End":"02:31.245","Text":"Then we have 3 2p orbitals,"},{"Start":"02:31.245 ","End":"02:34.510","Text":"each one with an unpaired electron."},{"Start":"02:34.790 ","End":"02:38.020","Text":"This is going to hybridize."},{"Start":"02:40.370 ","End":"02:44.075","Text":"As we said, the hybridization is going to be 2sp."},{"Start":"02:44.075 ","End":"02:51.330","Text":"Now in sp, we know we have 1 lone pair of electrons and 1 Sigma bond."},{"Start":"02:54.580 ","End":"03:01.490","Text":"Our sp, which has 2 orbitals is going to have"},{"Start":"03:01.490 ","End":"03:08.910","Text":"1 of the orbitals with 2 electrons and 1 with 1 unpaired electron."},{"Start":"03:08.910 ","End":"03:10.760","Text":"Since again one of the groups is"},{"Start":"03:10.760 ","End":"03:13.010","Text":"a lone pair of electrons and the other one is a Sigma bond,"},{"Start":"03:13.010 ","End":"03:14.735","Text":"which is the unpaired electron."},{"Start":"03:14.735 ","End":"03:18.995","Text":"Again, this was a combination of 1s orbital and 1p orbital."},{"Start":"03:18.995 ","End":"03:25.480","Text":"What we were left with is 2p orbitals, which are unhybridized."},{"Start":"03:25.870 ","End":"03:28.610","Text":"These account for the Pi bonds,"},{"Start":"03:28.610 ","End":"03:33.964","Text":"meaning the double and triple bond between the nitrogens."},{"Start":"03:33.964 ","End":"03:36.440","Text":"So if we look at our orbital diagram,"},{"Start":"03:36.440 ","End":"03:41.370","Text":"we started with 1s orbital and 3p orbitals."},{"Start":"03:43.070 ","End":"03:51.015","Text":"The s orbital contains 2 electrons and the p orbital has 3 unpaired electrons."},{"Start":"03:51.015 ","End":"03:56.600","Text":"Since we have 2 electron groups in our nitrogen molecule,"},{"Start":"03:56.600 ","End":"04:02.270","Text":"this hybridizes to 2sp orbitals,"},{"Start":"04:02.270 ","End":"04:06.875","Text":"which are the combination of s and p orbitals."},{"Start":"04:06.875 ","End":"04:09.810","Text":"Again, 1 has 2 electrons,"},{"Start":"04:09.810 ","End":"04:12.590","Text":"it accounts for the lone pair group."},{"Start":"04:12.590 ","End":"04:14.300","Text":"And the other one has 1 unpaired electron,"},{"Start":"04:14.300 ","End":"04:15.980","Text":"which accounts for the Sigma bond."},{"Start":"04:15.980 ","End":"04:19.190","Text":"We\u0027re left with 2 unhybridized 2p orbitals,"},{"Start":"04:19.190 ","End":"04:21.710","Text":"which account for the double and triple bond,"},{"Start":"04:21.710 ","End":"04:22.990","Text":"meaning the Pi bonds."},{"Start":"04:22.990 ","End":"04:25.700","Text":"That is our orbital diagram for nitrogen."},{"Start":"04:25.700 ","End":"04:28.440","Text":"Thank you very much for watching."}],"ID":24477},{"Watched":false,"Name":"Exercise 2","Duration":"4m 3s","ChapterTopicVideoID":23569,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:03.075","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.075 ","End":"00:05.070","Text":"What are the hybridizations sp,"},{"Start":"00:05.070 ","End":"00:09.165","Text":"sp^2 and so on of the central atoms in the following molecule."},{"Start":"00:09.165 ","End":"00:11.160","Text":"The first step into finding"},{"Start":"00:11.160 ","End":"00:15.280","Text":"the hybridizations is drawing a Lewis structure of the molecule,"},{"Start":"00:15.280 ","End":"00:18.190","Text":"and in this case we\u0027re given the Lewis structure."},{"Start":"00:18.500 ","End":"00:25.200","Text":"The second step is to take each atom that we want to find the hybridization for,"},{"Start":"00:25.200 ","End":"00:31.380","Text":"and counting the number of electron groups around that atom based on the VSEPR theory."},{"Start":"00:31.380 ","End":"00:35.685","Text":"Now remember, a lone pair of electrons is treated as 1 group,"},{"Start":"00:35.685 ","End":"00:37.890","Text":"a single bond is treated as 1 group and"},{"Start":"00:37.890 ","End":"00:41.500","Text":"a double bond or triple bond is also treated as 1 group."},{"Start":"00:41.500 ","End":"00:44.300","Text":"We\u0027re going to start with the carbon."},{"Start":"00:44.300 ","End":"00:46.226","Text":"Let\u0027s start with carbons 1,"},{"Start":"00:46.226 ","End":"00:48.750","Text":"2, 3, and 4."},{"Start":"00:53.990 ","End":"00:56.865","Text":"If you take a look at carbon 1,"},{"Start":"00:56.865 ","End":"01:01.870","Text":"you\u0027ll see that it has 4 single bonds around it, meaning 4 groups."},{"Start":"01:05.320 ","End":"01:09.980","Text":"This is also the case for carbon 2,3 and 4."},{"Start":"01:09.980 ","End":"01:13.750","Text":"Each of these carbons have 4 electron groups around them."},{"Start":"01:13.750 ","End":"01:20.350","Text":"The hybridization for 4 electron groups is sp^3."},{"Start":"01:22.240 ","End":"01:29.070","Text":"Since this is a combination of 1s orbital and 3p orbitals,"},{"Start":"01:29.070 ","End":"01:35.755","Text":"therefore, we get 4 hybridized orbitals in order to make 4 bonds."},{"Start":"01:35.755 ","End":"01:38.345","Text":"That\u0027s carbons 1-4."},{"Start":"01:38.345 ","End":"01:42.480","Text":"Now let\u0027s take a look at carbon 5 because it\u0027s different."},{"Start":"01:43.310 ","End":"01:47.610","Text":"As you can see, carbon 5 is attached to 2 single bonds,"},{"Start":"01:47.610 ","End":"01:49.200","Text":"but it also has 1 double bond."},{"Start":"01:49.200 ","End":"01:51.495","Text":"Now remember, a double bond is treated as 1 group."},{"Start":"01:51.495 ","End":"01:56.740","Text":"So in all we have 3 groups around carbon number 5."},{"Start":"02:01.490 ","End":"02:06.980","Text":"3 groups is going to give us the hybridization sp^2,"},{"Start":"02:06.980 ","End":"02:11.155","Text":"since it\u0027s a combination of 1s orbital and 2p orbitals,"},{"Start":"02:11.155 ","End":"02:13.640","Text":"therefore we get 3."},{"Start":"02:15.180 ","End":"02:20.060","Text":"Now let\u0027s look at our sulfur atom."},{"Start":"02:21.440 ","End":"02:29.574","Text":"As you can see, sulfur has 2 single bonds around it and also 2 lone pair of electrons."},{"Start":"02:29.574 ","End":"02:31.540","Text":"So each lone pair is thought of as"},{"Start":"02:31.540 ","End":"02:34.150","Text":"1 group and each single bond is thought of as 1 group,"},{"Start":"02:34.150 ","End":"02:41.890","Text":"so it also has 4 groups, making it sp^3."},{"Start":"02:42.620 ","End":"02:47.405","Text":"Again, the combination of 1s orbital and 3p orbitals,"},{"Start":"02:47.405 ","End":"02:51.215","Text":"giving us 4 sp^3 orbitals."},{"Start":"02:51.215 ","End":"02:54.750","Text":"Now let\u0027s take a look at our nitrogen."},{"Start":"02:55.000 ","End":"02:59.165","Text":"As you can see, the nitrogen has 3 single bonds around it"},{"Start":"02:59.165 ","End":"03:04.830","Text":"and 1 lone pair of electrons giving it also 4 groups."},{"Start":"03:06.410 ","End":"03:10.905","Text":"Again, this is going to give us sp^3 hybridization."},{"Start":"03:10.905 ","End":"03:17.015","Text":"Now let\u0027s take a look at our oxygen, right here."},{"Start":"03:17.015 ","End":"03:19.370","Text":"Now our oxygen, just like our sulfur,"},{"Start":"03:19.370 ","End":"03:26.010","Text":"has 2 lone pairs of electrons and 2 single bonds giving it 4 groups at all,"},{"Start":"03:30.080 ","End":"03:34.300","Text":"giving us an sp^3 hybridization."},{"Start":"03:34.340 ","End":"03:39.415","Text":"Again, carbons 1-4 gave us an sp^3 hybridization."},{"Start":"03:39.415 ","End":"03:41.960","Text":"Carbon 5, which had 3 groups around it,"},{"Start":"03:41.960 ","End":"03:44.645","Text":"gave us an sp^2 hybridization."},{"Start":"03:44.645 ","End":"03:49.520","Text":"The sulfur with 4 groups also gave us an sp^3 hybridization."},{"Start":"03:49.520 ","End":"03:53.660","Text":"The nitrogen with 4 groups also gave us an sp^3 hybridization."},{"Start":"03:53.660 ","End":"03:58.280","Text":"Oxygen with 4 groups also gave us an sp^3 hybridization."},{"Start":"03:58.280 ","End":"04:00.605","Text":"That is our final answer."},{"Start":"04:00.605 ","End":"04:03.330","Text":"Thank you very much for watching."}],"ID":24478},{"Watched":false,"Name":"Exercise 3 - part a","Duration":"4m 22s","ChapterTopicVideoID":23561,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.944","Text":"We\u0027re going to solve the following exercise for the following molecules."},{"Start":"00:03.944 ","End":"00:08.040","Text":"One, predict the electron group geometry by VSEPR theory and 2,"},{"Start":"00:08.040 ","End":"00:10.535","Text":"determine the hybridization of the central atom."},{"Start":"00:10.535 ","End":"00:12.550","Text":"We\u0027re going to begin with a."},{"Start":"00:12.550 ","End":"00:15.840","Text":"Since we want to find the electron group geometry by VSEPR theory,"},{"Start":"00:15.840 ","End":"00:18.495","Text":"we\u0027re going to first write the Lewis structure."},{"Start":"00:18.495 ","End":"00:22.335","Text":"The first step is determine the number of valence electrons in our Lewis structure."},{"Start":"00:22.335 ","End":"00:24.300","Text":"We have 8 sulfur atoms,"},{"Start":"00:24.300 ","End":"00:25.740","Text":"sulfur is in the 16th group,"},{"Start":"00:25.740 ","End":"00:29.205","Text":"therefore we\u0027re going to take 6 valence electrons times 8."},{"Start":"00:29.205 ","End":"00:31.350","Text":"Since we have 8 sulfurs,"},{"Start":"00:31.350 ","End":"00:33.270","Text":"this is going to give us a 48."},{"Start":"00:33.270 ","End":"00:35.730","Text":"We have 48 electrons for Lewis structure."},{"Start":"00:35.730 ","End":"00:38.175","Text":"Now we\u0027re going to draw the skeletal structure."},{"Start":"00:38.175 ","End":"00:40.830","Text":"We\u0027re just going to connect our sulfur atoms."},{"Start":"00:40.830 ","End":"00:42.600","Text":"We have sulfur 1,"},{"Start":"00:42.600 ","End":"00:45.315","Text":"2, 3,"},{"Start":"00:45.315 ","End":"00:48.195","Text":"4, 5,"},{"Start":"00:48.195 ","End":"00:52.720","Text":"6, 7, 8."},{"Start":"00:52.720 ","End":"00:56.885","Text":"The next step is to count the number of bonds and subtract the electrons in the bond."},{"Start":"00:56.885 ","End":"00:58.250","Text":"We have 1, 2,"},{"Start":"00:58.250 ","End":"00:59.705","Text":"3, 4, 5,"},{"Start":"00:59.705 ","End":"01:00.980","Text":"6, 7, 8 bonds,"},{"Start":"01:00.980 ","End":"01:02.960","Text":"meaning we have 16 electrons in these bonds."},{"Start":"01:02.960 ","End":"01:09.070","Text":"We\u0027re going to subtract 16 from the 48 valence electrons."},{"Start":"01:10.270 ","End":"01:13.520","Text":"This is going to leave us with 32 electrons."},{"Start":"01:13.520 ","End":"01:15.560","Text":"Now, remember we have 8 sulfur atoms,"},{"Start":"01:15.560 ","End":"01:19.315","Text":"meaning each sulfur atom is going to get 4 electrons."},{"Start":"01:19.315 ","End":"01:23.725","Text":"This is going to be 4, 8, 12,"},{"Start":"01:23.725 ","End":"01:30.030","Text":"16, 24,"},{"Start":"01:30.030 ","End":"01:35.340","Text":"28, and 32."},{"Start":"01:35.340 ","End":"01:36.890","Text":"Now, if we take a look at our Lewis structure,"},{"Start":"01:36.890 ","End":"01:41.465","Text":"let\u0027s just check for a second the formal charge on each sulfur atoms."},{"Start":"01:41.465 ","End":"01:43.460","Text":"If we take one of the sulfur atoms,"},{"Start":"01:43.460 ","End":"01:45.130","Text":"they\u0027re all identical,"},{"Start":"01:45.130 ","End":"01:48.080","Text":"we\u0027re going to start with a number of valence electrons on the free item,"},{"Start":"01:48.080 ","End":"01:51.500","Text":"which is 6 minus the number of lone pair electrons."},{"Start":"01:51.500 ","End":"01:52.580","Text":"Here we have 1, 2, 3,"},{"Start":"01:52.580 ","End":"01:54.260","Text":"4, so that\u0027s minus 4,"},{"Start":"01:54.260 ","End":"01:58.010","Text":"minus half of the number of bonding electrons,"},{"Start":"01:58.010 ","End":"01:59.645","Text":"meaning electrons and bonds."},{"Start":"01:59.645 ","End":"02:01.250","Text":"Around the sulfur atom,"},{"Start":"02:01.250 ","End":"02:04.720","Text":"we have 2 bonds, meaning 4 electrons."},{"Start":"02:04.720 ","End":"02:08.910","Text":"This gives us a total of 6 minus 4 minus 2,"},{"Start":"02:08.910 ","End":"02:10.965","Text":"so this is going to give us a 0."},{"Start":"02:10.965 ","End":"02:14.285","Text":"The formal charge on the sulfur is 0, so we\u0027re good."},{"Start":"02:14.285 ","End":"02:17.120","Text":"Now we have our Lewis structure and we need to"},{"Start":"02:17.120 ","End":"02:21.235","Text":"predict the electron group geometry by the VSEPR theory."},{"Start":"02:21.235 ","End":"02:25.030","Text":"We\u0027re going to look at the AXE notation."},{"Start":"02:25.030 ","End":"02:27.995","Text":"Again, we have our atom,"},{"Start":"02:27.995 ","End":"02:31.080","Text":"which we\u0027re going to call A."},{"Start":"02:31.610 ","End":"02:36.095","Text":"Then X is the number of atoms that it\u0027s connected to."},{"Start":"02:36.095 ","End":"02:38.090","Text":"Here we have 1, 2."},{"Start":"02:38.090 ","End":"02:44.494","Text":"Remember in the AXE notation,"},{"Start":"02:44.494 ","End":"02:46.730","Text":"a single bond, double bond,"},{"Start":"02:46.730 ","End":"02:47.990","Text":"or a triple bond is counted as"},{"Start":"02:47.990 ","End":"02:52.100","Text":"one group and lone pair electrons are also counted as one group."},{"Start":"02:52.100 ","End":"02:53.915","Text":"Here we have 2 bonds,"},{"Start":"02:53.915 ","End":"02:57.800","Text":"that\u0027s AX_2 but we also have 2 pairs of lone pair electrons,"},{"Start":"02:57.800 ","End":"03:05.630","Text":"so that\u0027s going to be E_2 because E is lone pair electrons."},{"Start":"03:05.630 ","End":"03:13.200","Text":"By VSEPR theory, the electron group geometry for AX_2E_2 is tetrahedral."},{"Start":"03:23.030 ","End":"03:26.630","Text":"If you don\u0027t remember this, you can take a look at Chapter 10"},{"Start":"03:26.630 ","End":"03:30.810","Text":"at the lectures just to remind yourself."},{"Start":"03:32.050 ","End":"03:35.120","Text":"That\u0027s 1."},{"Start":"03:35.120 ","End":"03:38.800","Text":"The electron group geometry is tetrahedral."},{"Start":"03:38.800 ","End":"03:41.960","Text":"In 2, we have to determine the hybridization of the central atom."},{"Start":"03:41.960 ","End":"03:47.100","Text":"Remember, we have here that central atom connected to 4 groups."},{"Start":"03:47.230 ","End":"03:52.570","Text":"We have 4 groups that is connected to."},{"Start":"03:52.570 ","End":"03:56.440","Text":"This is going to give us sp^3 notation."},{"Start":"03:58.100 ","End":"04:04.555","Text":"The hybridization of each sulfur atom is sp^3."},{"Start":"04:04.555 ","End":"04:08.885","Text":"Again, the electron group geometry that we found is"},{"Start":"04:08.885 ","End":"04:15.460","Text":"tetrahedral and the hybridization is sp^3."},{"Start":"04:18.050 ","End":"04:22.090","Text":"That\u0027s our answer for a. Now we\u0027re going to go on to b."}],"ID":24470},{"Watched":false,"Name":"Exercise 3 - part b","Duration":"3m 39s","ChapterTopicVideoID":23562,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:03.555","Text":"Now we\u0027re going to take a look at B."},{"Start":"00:03.555 ","End":"00:06.690","Text":"B again, we\u0027re going to draw the Lewis structure."},{"Start":"00:06.690 ","End":"00:08.460","Text":"We\u0027re going to start with the number of valence electrons."},{"Start":"00:08.460 ","End":"00:10.709","Text":"We have 1 sulfur atom and 2 oxygens."},{"Start":"00:10.709 ","End":"00:13.800","Text":"The sulfur is in the 16 group,"},{"Start":"00:13.800 ","End":"00:16.950","Text":"therefore it starts with 6 valence electrons and so is the oxygen,"},{"Start":"00:16.950 ","End":"00:19.785","Text":"so it\u0027s going to be plus 2 times 6."},{"Start":"00:19.785 ","End":"00:25.440","Text":"This is going to give us 18 electrons for our Lewis structure."},{"Start":"00:25.440 ","End":"00:31.780","Text":"We\u0027re just going to connect the sulfur with oxygens at this point."},{"Start":"00:32.480 ","End":"00:35.340","Text":"We have 2 bonds between the sulfur and the oxygens,"},{"Start":"00:35.340 ","End":"00:38.620","Text":"meaning we have to subtract 4 from 18,"},{"Start":"00:38.660 ","End":"00:42.390","Text":"and this is going to leave us with 14 electrons."},{"Start":"00:42.390 ","End":"00:44.010","Text":"We\u0027re going to start with our oxygen,"},{"Start":"00:44.010 ","End":"00:46.420","Text":"so that\u0027s 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14."},{"Start":"00:51.890 ","End":"00:58.630","Text":"The oxygens each got 6 electrons and we\u0027re left with 2 electrons for our sulfur."},{"Start":"00:58.820 ","End":"01:02.675","Text":"The next step is to look at the Lewis structure and see if it satisfies this."},{"Start":"01:02.675 ","End":"01:05.855","Text":"It doesn\u0027t satisfy since our sulfur as missing an octet,"},{"Start":"01:05.855 ","End":"01:08.675","Text":"it only has 6 electrons around it."},{"Start":"01:08.675 ","End":"01:12.065","Text":"Here you have a number of options,"},{"Start":"01:12.065 ","End":"01:15.035","Text":"but what I\u0027m going to do is I\u0027m just going to take a lone pair from"},{"Start":"01:15.035 ","End":"01:20.710","Text":"each oxygen and turn it into shared electrons,"},{"Start":"01:20.710 ","End":"01:26.460","Text":"and this is going to give me a double bond to each oxygen,"},{"Start":"01:27.590 ","End":"01:31.505","Text":"and leave each oxygen with 4 lone pair electrons,"},{"Start":"01:31.505 ","End":"01:34.960","Text":"and we\u0027re still left with 2 on the sulfur."},{"Start":"01:34.960 ","End":"01:38.640","Text":"Remember the sulfur can have an expanded octet."},{"Start":"01:38.900 ","End":"01:42.830","Text":"This is what I decided to draw,"},{"Start":"01:42.830 ","End":"01:45.650","Text":"since if we take a look at the formal charges there is 0."},{"Start":"01:45.650 ","End":"01:49.110","Text":"If we take a look at the formal charges for the oxygen,"},{"Start":"01:49.490 ","End":"01:54.900","Text":"we begin with 6 minus 4 lone pair electrons minus 1/2"},{"Start":"01:54.900 ","End":"02:01.380","Text":"times 4 electrons which are bonded and that gives us 0,"},{"Start":"02:01.380 ","End":"02:04.475","Text":"and the formal charge of sulfur."},{"Start":"02:04.475 ","End":"02:13.140","Text":"Again, we begin with 6 minus 2 lone pair electrons minus 1/2 times 8,"},{"Start":"02:13.140 ","End":"02:15.600","Text":"since we have 4 bonds connected to the sulfur,"},{"Start":"02:15.600 ","End":"02:17.910","Text":"so that\u0027s 8 electrons in these bonds,"},{"Start":"02:17.910 ","End":"02:21.070","Text":"and it\u0027s going to give us 0."},{"Start":"02:21.170 ","End":"02:24.060","Text":"We\u0027re going to take a look at this structure,"},{"Start":"02:24.060 ","End":"02:28.950","Text":"and we need to predict the electron group geometry by the VSEPR theory."},{"Start":"02:32.660 ","End":"02:36.520","Text":"Again, our sulfur is A,"},{"Start":"02:38.600 ","End":"02:42.380","Text":"and we can see that it has 2 double bonds connected to it,"},{"Start":"02:42.380 ","End":"02:45.840","Text":"meaning AX_2, because remember,"},{"Start":"02:45.840 ","End":"02:49.715","Text":"a double bond is 1 group and it also has lone pair electrons,"},{"Start":"02:49.715 ","End":"02:52.495","Text":"1 pair, so that\u0027s going to be AX_2E."},{"Start":"02:52.495 ","End":"02:54.540","Text":"In the VSEPR theory,"},{"Start":"02:54.540 ","End":"02:57.190","Text":"this is going to be a trigonal planar."},{"Start":"02:58.070 ","End":"03:02.770","Text":"The electron group geometry."},{"Start":"03:08.390 ","End":"03:12.065","Text":"The next step is to decide what the hybridization is going to be."},{"Start":"03:12.065 ","End":"03:16.350","Text":"Since our sulfur is connected to 3 groups,"},{"Start":"03:19.670 ","End":"03:24.750","Text":"our hybridization of this molecule is going to be sp^2."},{"Start":"03:25.400 ","End":"03:31.675","Text":"Again, the electron group geometry by the VSEPR theory is trigonal planar,"},{"Start":"03:31.675 ","End":"03:35.410","Text":"and we have sp^2 hybridization."},{"Start":"03:35.450 ","End":"03:39.700","Text":"That is our answer for B, now we\u0027re going to go on to C."}],"ID":24471},{"Watched":false,"Name":"Exercise 3 - part c","Duration":"5m 42s","ChapterTopicVideoID":23563,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.680 ","End":"00:04.420","Text":"In C we have sulfuric acid."},{"Start":"00:04.640 ","End":"00:07.335","Text":"Again, we\u0027re going to begin with the Lewis structure."},{"Start":"00:07.335 ","End":"00:13.170","Text":"We have 2 hydrogens that\u0027s 2 times 1 valence electron"},{"Start":"00:13.170 ","End":"00:21.105","Text":"plus 6 valence electrons for sulfur plus 4 times 6 valence electrons for our oxygen."},{"Start":"00:21.105 ","End":"00:25.419","Text":"This is going to give us a total of 32 electrons."},{"Start":"00:26.690 ","End":"00:29.260","Text":"Now we\u0027re going to draw the skeletal structure."},{"Start":"00:29.260 ","End":"00:31.815","Text":"We\u0027re going to put our sulfur in the middle."},{"Start":"00:31.815 ","End":"00:35.320","Text":"We\u0027re going to attach the oxygens."},{"Start":"00:37.820 ","End":"00:41.350","Text":"Now we\u0027re going to attach the hydrogens."},{"Start":"00:42.680 ","End":"00:46.480","Text":"Now the number of bonds we have is 1,"},{"Start":"00:46.480 ","End":"00:48.040","Text":"2, 3, 4, 5, 6,"},{"Start":"00:48.040 ","End":"00:50.410","Text":"meaning we\u0027re going to subtract 12 electrons from"},{"Start":"00:50.410 ","End":"00:54.890","Text":"our valence electrons and we\u0027re going to be left with 20 electrons."},{"Start":"00:55.140 ","End":"00:57.760","Text":"We\u0027re going to start with our oxygens."},{"Start":"00:57.760 ","End":"01:03.795","Text":"We\u0027re going to give 6 electrons to each oxygen on the bottom and top."},{"Start":"01:03.795 ","End":"01:06.970","Text":"This is going to be 12 electrons,"},{"Start":"01:06.970 ","End":"01:08.290","Text":"meaning we have 8 left."},{"Start":"01:08.290 ","End":"01:13.045","Text":"We\u0027re going to put 8 on each oxygen to the right and to the left."},{"Start":"01:13.045 ","End":"01:15.670","Text":"That\u0027s our 20 electrons."},{"Start":"01:15.670 ","End":"01:19.900","Text":"If you take a look, you\u0027ll see that also the sulfur and also the oxygens have"},{"Start":"01:19.900 ","End":"01:23.740","Text":"octets and the hydrogens have 1 bond, which is what they need."},{"Start":"01:23.740 ","End":"01:26.080","Text":"However, if we look at our formal charges,"},{"Start":"01:26.080 ","End":"01:33.695","Text":"we can see that the sulfur are going to start with 6."},{"Start":"01:33.695 ","End":"01:35.650","Text":"It has no lone pair electrons,"},{"Start":"01:35.650 ","End":"01:42.915","Text":"so that\u0027s minus 0 minus 1/2 times 8 since we have 4 bonds around the sulfur,"},{"Start":"01:42.915 ","End":"01:44.650","Text":"that\u0027s going to give us 6 minus 4,"},{"Start":"01:44.650 ","End":"01:46.910","Text":"which is going to give us a 2."},{"Start":"01:46.910 ","End":"01:51.355","Text":"Now we have 2 oxygens on the left and on the right."},{"Start":"01:51.355 ","End":"01:54.070","Text":"Whichever formal charge of 0,"},{"Start":"01:54.070 ","End":"01:57.485","Text":"since we start with 6 minus 4 lone pair electrons"},{"Start":"01:57.485 ","End":"02:03.760","Text":"minus 1/2 times 4 since they have 2 bonds around them."},{"Start":"02:03.760 ","End":"02:11.820","Text":"This is going to give us 0. For the top and the bottom oxygens, again,"},{"Start":"02:11.820 ","End":"02:15.850","Text":"we\u0027re going to start with 6 minus 6 lone pair electrons minus"},{"Start":"02:15.850 ","End":"02:22.010","Text":"1/2 times 2 bond pair electrons since we have 1 bond connected."},{"Start":"02:22.010 ","End":"02:23.780","Text":"That\u0027s going to give us 6 minus 6 minus 1,"},{"Start":"02:23.780 ","End":"02:25.265","Text":"that\u0027s going to give us a minus 1."},{"Start":"02:25.265 ","End":"02:28.290","Text":"The formal charges are 2 and minus 1."},{"Start":"02:28.560 ","End":"02:35.725","Text":"We\u0027re going to share electrons in our structure in order to fix these formal charges."},{"Start":"02:35.725 ","End":"02:41.830","Text":"We\u0027re just going to chair a lone pair here and another lone parent here."},{"Start":"02:41.830 ","End":"02:44.395","Text":"We\u0027re going to draw the new structures."},{"Start":"02:44.395 ","End":"02:47.590","Text":"That\u0027s the sulfur connected with a double bond to"},{"Start":"02:47.590 ","End":"02:52.070","Text":"2 oxygens and then 2 single bonds to OH."},{"Start":"03:02.030 ","End":"03:05.885","Text":"Now if we take a look at the formal charge of sulfur,"},{"Start":"03:05.885 ","End":"03:10.405","Text":"we can see we\u0027re starting with 6 electrons minus 0,"},{"Start":"03:10.405 ","End":"03:15.175","Text":"since we have no lone pair electrons minus 1/2 times 12,"},{"Start":"03:15.175 ","End":"03:18.285","Text":"since the sulfur has 1, 2, 3, 4,"},{"Start":"03:18.285 ","End":"03:22.560","Text":"5, 6 bonds here."},{"Start":"03:22.560 ","End":"03:28.425","Text":"This is going to give us 6 minus 6 is going to give us a 0."},{"Start":"03:28.425 ","End":"03:31.490","Text":"The formal charge on all the oxygens equals"},{"Start":"03:31.490 ","End":"03:40.860","Text":"6 minus 4 lone pair electrons minus 1/2 times 2 bonds."},{"Start":"03:40.860 ","End":"03:42.960","Text":"That\u0027s 4 electrons."},{"Start":"03:42.960 ","End":"03:45.320","Text":"That\u0027s 6 minus 4 minus 2."},{"Start":"03:45.320 ","End":"03:50.420","Text":"It\u0027s going to give us a 0. We\u0027re going to stick with this Lewis structure."},{"Start":"03:51.120 ","End":"03:56.840","Text":"The next step is to find the electron group geometry by VSEPR theory."},{"Start":"03:57.360 ","End":"04:03.085","Text":"In this case, we have the sulfur and we have our oxygens,"},{"Start":"04:03.085 ","End":"04:05.660","Text":"which are the central atoms."},{"Start":"04:06.150 ","End":"04:09.004","Text":"For the sulfur,"},{"Start":"04:09.004 ","End":"04:11.455","Text":"we can see that it\u0027s connected to 4 groups,"},{"Start":"04:11.455 ","End":"04:13.465","Text":"since again, a double bond is thought of as 1 group."},{"Start":"04:13.465 ","End":"04:15.085","Text":"We have 2 single bonds 2 double bonds."},{"Start":"04:15.085 ","End":"04:17.180","Text":"That\u0027s going to be AX_4."},{"Start":"04:19.320 ","End":"04:27.110","Text":"That\u0027s going to give us a tetrahedral electron group geometry."},{"Start":"04:31.050 ","End":"04:37.160","Text":"That\u0027s going to give us an sp^3 hybridization since it\u0027s connected to 4 groups."},{"Start":"04:40.470 ","End":"04:44.240","Text":"Now we\u0027re going to take a look at our oxygens."},{"Start":"04:46.890 ","End":"04:49.900","Text":"Our oxygens are connected to,"},{"Start":"04:49.900 ","End":"04:53.170","Text":"we have 2 bonds,"},{"Start":"04:53.170 ","End":"04:55.525","Text":"single bonds is going to be AX_2."},{"Start":"04:55.525 ","End":"05:00.340","Text":"We also have 2 pairs of lone pair electrons,"},{"Start":"05:00.340 ","End":"05:04.100","Text":"so that\u0027s going to be AX_2E_2."},{"Start":"05:04.930 ","End":"05:09.120","Text":"Now remember, we\u0027re looking at our central atoms here."},{"Start":"05:10.450 ","End":"05:16.420","Text":"The electron group geometry for AX_2E_2 is also tetrahedral."},{"Start":"05:20.500 ","End":"05:25.230","Text":"Therefore the hybridization is also going to be sp^3."},{"Start":"05:25.870 ","End":"05:30.590","Text":"Again, for the sulfur and oxygen central atoms,"},{"Start":"05:30.590 ","End":"05:37.400","Text":"we found a tetrahedral electron group geometry and an sp^3 hybridization."},{"Start":"05:37.400 ","End":"05:39.425","Text":"That is our final answer."},{"Start":"05:39.425 ","End":"05:42.090","Text":"Thank you very much for watching."}],"ID":24472},{"Watched":false,"Name":"Exercise 4","Duration":"6m 51s","ChapterTopicVideoID":23564,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.670 ","End":"00:04.440","Text":"Determine the hybridization of each of"},{"Start":"00:04.440 ","End":"00:09.990","Text":"the following molecules. We\u0027re going to start with a."},{"Start":"00:09.990 ","End":"00:15.070","Text":"Now, the first step to determine the hybridization is to draw a Lewis structure."},{"Start":"00:15.560 ","End":"00:17.820","Text":"Now in order to draw a Lewis structure,"},{"Start":"00:17.820 ","End":"00:20.895","Text":"we first have to count the number of valence electrons we have."},{"Start":"00:20.895 ","End":"00:23.130","Text":"We can see here that we have 5 chlorines."},{"Start":"00:23.130 ","End":"00:24.975","Text":"Chlorine is in the 17th group,"},{"Start":"00:24.975 ","End":"00:27.390","Text":"meaning 7 valence electrons."},{"Start":"00:27.390 ","End":"00:30.915","Text":"That\u0027s going to be 5 times 7 since we have 5 chlorines,"},{"Start":"00:30.915 ","End":"00:33.060","Text":"plus for the phosphorus,"},{"Start":"00:33.060 ","End":"00:42.765","Text":"we\u0027re going to have 5 valence electrons since phosphorus is in the 15th group."},{"Start":"00:42.765 ","End":"00:45.660","Text":"This is going to give us 35 plus 5,"},{"Start":"00:45.660 ","End":"00:49.450","Text":"this is going to give us 40 valence electrons."},{"Start":"00:50.230 ","End":"00:54.545","Text":"Now, the next step is to draw the skeletal structure of the molecule."},{"Start":"00:54.545 ","End":"00:58.040","Text":"We\u0027re going to put the phosphorus in the middle,"},{"Start":"00:58.040 ","End":"01:01.250","Text":"and we\u0027re going to connect 5"},{"Start":"01:01.250 ","End":"01:10.525","Text":"chlorines to the phosphorus."},{"Start":"01:10.525 ","End":"01:13.500","Text":"Now, since we connected 5 chlorines,"},{"Start":"01:13.500 ","End":"01:15.660","Text":"we have 5 bonds here, 1, 2, 3,"},{"Start":"01:15.660 ","End":"01:18.060","Text":"4, 5, which accounts for 10 electrons."},{"Start":"01:18.060 ","End":"01:21.800","Text":"We\u0027re going to subtract 10 electrons for the whole number of valence electrons."},{"Start":"01:21.800 ","End":"01:25.265","Text":"This gives us 30, meaning we have 30 electrons left."},{"Start":"01:25.265 ","End":"01:26.780","Text":"Since we have 5 chlorines,"},{"Start":"01:26.780 ","End":"01:29.560","Text":"we\u0027re going to put 6 electrons on each one,"},{"Start":"01:29.560 ","End":"01:31.380","Text":"1, 2, 3, 4, 5, 6, 7,"},{"Start":"01:31.380 ","End":"01:32.520","Text":"8, 9, 10, 11,"},{"Start":"01:32.520 ","End":"01:33.990","Text":"12, 13, 14,"},{"Start":"01:33.990 ","End":"01:35.040","Text":"15, 16, 17,"},{"Start":"01:35.040 ","End":"01:36.060","Text":"18, 19, 20,"},{"Start":"01:36.060 ","End":"01:37.785","Text":"21, 22, 23, 24,"},{"Start":"01:37.785 ","End":"01:39.090","Text":"25, 26, 27,"},{"Start":"01:39.090 ","End":"01:40.230","Text":"28, 29, 30."},{"Start":"01:40.230 ","End":"01:43.115","Text":"That\u0027s going to count for all our valence electrons,"},{"Start":"01:43.115 ","End":"01:45.840","Text":"and this is our Lewis structure."},{"Start":"01:45.920 ","End":"01:48.080","Text":"Now that we have a Lewis structure,"},{"Start":"01:48.080 ","End":"01:51.560","Text":"the next step is to count the number of electron groups around the central atom,"},{"Start":"01:51.560 ","End":"01:53.270","Text":"which is around the phosphorus."},{"Start":"01:53.270 ","End":"01:56.300","Text":"In this case, we have 5 single bonds."},{"Start":"01:56.300 ","End":"01:59.940","Text":"That\u0027s 5 electron groups."},{"Start":"02:05.540 ","End":"02:13.180","Text":"Hybridization for 5 electron groups is going to be sp^3d."},{"Start":"02:14.080 ","End":"02:17.120","Text":"That\u0027s a combination of 1 s-orbital,"},{"Start":"02:17.120 ","End":"02:18.770","Text":"3 p-orbitals, and in this case,"},{"Start":"02:18.770 ","End":"02:24.035","Text":"we also have to include a d-orbital in order to give us 5 groups."},{"Start":"02:24.035 ","End":"02:30.395","Text":"Our hybridization for the phosphorus is going to be sp^3d."},{"Start":"02:30.395 ","End":"02:32.660","Text":"Now, we\u0027re going to go on to b."},{"Start":"02:32.660 ","End":"02:34.975","Text":"Let\u0027s continue with b."},{"Start":"02:34.975 ","End":"02:37.800","Text":"Again, we\u0027re going to start with b,"},{"Start":"02:37.800 ","End":"02:39.950","Text":"and we\u0027re going to draw a Lewis structure."},{"Start":"02:39.950 ","End":"02:42.470","Text":"Again, we\u0027re going to count the number of valence electrons we have."},{"Start":"02:42.470 ","End":"02:45.530","Text":"From carbon, we have 4 since it\u0027s in the 14th group."},{"Start":"02:45.530 ","End":"02:48.930","Text":"From oxygen, we have 6."},{"Start":"02:49.700 ","End":"02:53.190","Text":"From chlorine, as before, we have 7 each."},{"Start":"02:53.190 ","End":"02:55.480","Text":"So that\u0027s 2 times 7."},{"Start":"02:56.200 ","End":"02:59.990","Text":"This is going to give us 24 electrons in all."},{"Start":"02:59.990 ","End":"03:02.120","Text":"Now, I\u0027m going to draw the skeletal structure."},{"Start":"03:02.120 ","End":"03:05.495","Text":"We\u0027re going to put the carbon as the central atom,"},{"Start":"03:05.495 ","End":"03:08.995","Text":"we\u0027re going to connect it to oxygen and chlorine."},{"Start":"03:08.995 ","End":"03:12.370","Text":"That\u0027s 1 and 2 chlorines."},{"Start":"03:13.270 ","End":"03:19.070","Text":"The next step is to subtract the number of electrons that we have in the bond."},{"Start":"03:19.070 ","End":"03:21.200","Text":"That\u0027s 3 bonds, meaning 6 electrons."},{"Start":"03:21.200 ","End":"03:22.960","Text":"So 24 minus 6."},{"Start":"03:22.960 ","End":"03:25.974","Text":"It\u0027s going to leave us with 18 electrons."},{"Start":"03:25.974 ","End":"03:30.355","Text":"Now, we\u0027re going to put 6 around each chlorine atom,"},{"Start":"03:30.355 ","End":"03:33.350","Text":"and 6 around our oxygen atom."},{"Start":"03:33.870 ","End":"03:37.315","Text":"Now, if we take a look at our Lewis structure,"},{"Start":"03:37.315 ","End":"03:40.913","Text":"we\u0027re missing an octet in the carbon,"},{"Start":"03:40.913 ","End":"03:46.855","Text":"meaning that we have to take 1 lone pair of electrons and share them."},{"Start":"03:46.855 ","End":"03:49.150","Text":"We\u0027re going to take a lone pair of electrons in form of"},{"Start":"03:49.150 ","End":"03:52.450","Text":"a double bond in order that the carbon will have an octet."},{"Start":"03:52.450 ","End":"03:57.380","Text":"A double bond is usually formed between carbon and oxygen."},{"Start":"04:07.530 ","End":"04:13.070","Text":"We\u0027re going to have a carbon double-bonded to oxygen and 2 chlorines."},{"Start":"04:18.810 ","End":"04:21.700","Text":"Now, if you look at our central carbon atom,"},{"Start":"04:21.700 ","End":"04:24.565","Text":"we can see that it\u0027s connected to 3 electron groups,"},{"Start":"04:24.565 ","End":"04:27.410","Text":"2 single bonds and 1 double bond."},{"Start":"04:31.120 ","End":"04:36.645","Text":"That\u0027s 3 groups and that\u0027s a hybridization of sp^2,"},{"Start":"04:36.645 ","End":"04:41.989","Text":"which are a combination of 1 s-orbital and 2 p-orbitals."},{"Start":"04:42.230 ","End":"04:47.535","Text":"B is sp^2 hybridization."},{"Start":"04:47.535 ","End":"04:51.230","Text":"Now we\u0027re going to go on to c. C,"},{"Start":"04:51.230 ","End":"04:54.455","Text":"we\u0027re going to start by counting the number of valence electrons again."},{"Start":"04:54.455 ","End":"04:58.985","Text":"For the sulfur, we have 6 valence electrons because it\u0027s in group 16,"},{"Start":"04:58.985 ","End":"05:01.520","Text":"and we have to add 6 fluorines,"},{"Start":"05:01.520 ","End":"05:05.780","Text":"meaning 6 times 7 since we have 7 valence electrons for each fluorine,"},{"Start":"05:05.780 ","End":"05:09.455","Text":"which is in group 17."},{"Start":"05:09.455 ","End":"05:11.600","Text":"This is going to give us 42 plus 6."},{"Start":"05:11.600 ","End":"05:14.240","Text":"It\u0027s going to give us 48 electrons."},{"Start":"05:14.240 ","End":"05:17.809","Text":"Now we\u0027re going to draw the skeletal structure."},{"Start":"05:17.809 ","End":"05:22.460","Text":"The skeletal structure, we\u0027re going to put our sulfur as a central atom."},{"Start":"05:22.460 ","End":"05:25.770","Text":"We\u0027re going to connect 6 fluorines to it."},{"Start":"05:35.210 ","End":"05:38.190","Text":"We have 6 bonds, 1, 2, 3, 4, 5, 6,"},{"Start":"05:38.190 ","End":"05:41.700","Text":"meaning that takes care of 12 electrons."},{"Start":"05:41.700 ","End":"05:45.230","Text":"48 minus 12 is going to leave us with 36 electrons."},{"Start":"05:45.230 ","End":"05:46.820","Text":"We have 6 fluorines,"},{"Start":"05:46.820 ","End":"05:51.000","Text":"meaning that\u0027s going to give us 6 electrons for each fluorine."},{"Start":"06:01.760 ","End":"06:05.180","Text":"This is our Lewis structure."},{"Start":"06:05.180 ","End":"06:08.075","Text":"The next step is to see how many electron groups we have."},{"Start":"06:08.075 ","End":"06:09.890","Text":"Around the sulfur, we have 1, 2,"},{"Start":"06:09.890 ","End":"06:11.950","Text":"3, 4, 5, 6."},{"Start":"06:11.950 ","End":"06:13.980","Text":"6 electron groups."},{"Start":"06:13.980 ","End":"06:16.780","Text":"We have 6 groups."},{"Start":"06:20.080 ","End":"06:28.290","Text":"In this case, the hybridization is sp^3 and we still need 2 orbitals."},{"Start":"06:28.290 ","End":"06:30.730","Text":"We\u0027re going to include our d-orbitals,"},{"Start":"06:30.730 ","End":"06:33.130","Text":"so that\u0027s going to be d^2,"},{"Start":"06:33.140 ","End":"06:36.810","Text":"which are a combination of 1 s-orbital, 3 p-orbitals,"},{"Start":"06:36.810 ","End":"06:40.780","Text":"and 2 d-orbitals giving us 6 groups in all."},{"Start":"06:43.670 ","End":"06:48.260","Text":"Our answer for c is sp^3d^2."},{"Start":"06:48.260 ","End":"06:49.460","Text":"That is our final answer."},{"Start":"06:49.460 ","End":"06:52.110","Text":"Thank you very much for watching."}],"ID":24473},{"Watched":false,"Name":"Exercise 5","Duration":"5m 26s","ChapterTopicVideoID":23565,"CourseChapterTopicPlaylistID":101322,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:04.045","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:04.045 ","End":"00:06.140","Text":"For the following molecule."},{"Start":"00:06.140 ","End":"00:09.705","Text":"A, predict the electron group geometry by VSEPR theory."},{"Start":"00:09.705 ","End":"00:12.750","Text":"B, determine the hybridization of the central atoms."},{"Start":"00:12.750 ","End":"00:14.400","Text":"C, identify the orbitals of"},{"Start":"00:14.400 ","End":"00:18.790","Text":"the central and terminal atoms that are involved in orbital overlap."},{"Start":"00:19.070 ","End":"00:22.844","Text":"We\u0027re going to start with the electron group geometry."},{"Start":"00:22.844 ","End":"00:27.360","Text":"We have 3 central atoms in our molecule."},{"Start":"00:27.360 ","End":"00:28.860","Text":"We have the left carbon,"},{"Start":"00:28.860 ","End":"00:31.420","Text":"and the sulfur, and the right carbon."},{"Start":"00:35.180 ","End":"00:39.495","Text":"Let\u0027s begin with the left carbon."},{"Start":"00:39.495 ","End":"00:45.330","Text":"We can see that this carbon is connected to 4 single bonds."},{"Start":"00:45.500 ","End":"00:48.420","Text":"In VSEPR theory, that\u0027s going to give us"},{"Start":"00:48.420 ","End":"00:57.630","Text":"an AX_4 and that\u0027s going to give us a tetrahedral electron group geometry."},{"Start":"01:03.260 ","End":"01:05.610","Text":"Now let\u0027s take a look at it so far."},{"Start":"01:05.610 ","End":"01:10.830","Text":"We can see that we have 2 pairs of lone pair electrons and 2 single bonds."},{"Start":"01:10.830 ","End":"01:17.435","Text":"By VSEPR theory, this is going to be an A x_2 since we have 2 bonds."},{"Start":"01:17.435 ","End":"01:22.900","Text":"E_2, since we have 2 pairs of lone pair electrons."},{"Start":"01:22.980 ","End":"01:28.580","Text":"The electron group geometry is also going to be tetrahedral in this case."},{"Start":"01:28.580 ","End":"01:31.505","Text":"Now let\u0027s take a look at the right carbon."},{"Start":"01:31.505 ","End":"01:35.285","Text":"As you can see, it\u0027s connected to 1 single bond and 1 triple bond."},{"Start":"01:35.285 ","End":"01:37.535","Text":"In VSEPR theory, that\u0027s going to be"},{"Start":"01:37.535 ","End":"01:47.185","Text":"AX_2 and the electron group geometry is linear in this case."},{"Start":"01:47.185 ","End":"01:51.935","Text":"We have the electron group geometries for our 2 carbons and our sulfur."},{"Start":"01:51.935 ","End":"01:54.665","Text":"Next is to find the hybridization."},{"Start":"01:54.665 ","End":"01:57.080","Text":"If we look at our left carbon,"},{"Start":"01:57.080 ","End":"01:59.210","Text":"so it\u0027s connected to 4 groups."},{"Start":"01:59.210 ","End":"02:02.190","Text":"It\u0027s going to have sp^3 hybridization."},{"Start":"02:02.750 ","End":"02:10.260","Text":"Our sulfur is also connected to 4 groups giving us sp^3 hybridization."},{"Start":"02:10.260 ","End":"02:15.610","Text":"Our carbon is connected to 2 groups giving us sp hybridization."},{"Start":"02:16.480 ","End":"02:18.710","Text":"Now we have to find the orbitals of"},{"Start":"02:18.710 ","End":"02:23.160","Text":"the central and terminal atoms that are involved in orbital overlap."},{"Start":"02:23.160 ","End":"02:28.110","Text":"I\u0027m just going to copy our molecule."},{"Start":"02:43.090 ","End":"02:48.990","Text":"First of all, let\u0027s look at the bonds between the left carbon and hydrogen."},{"Start":"02:49.400 ","End":"02:58.395","Text":"The hydrogen orbital is going to be 1s and we know that our carbon is sp^3."},{"Start":"02:58.395 ","End":"03:03.780","Text":"This is going to be hydrogen 1s to carbon sp^3."},{"Start":"03:04.030 ","End":"03:10.515","Text":"Next, let\u0027s take a look at the bond between the left carbon and the sulfur."},{"Start":"03:10.515 ","End":"03:15.105","Text":"The left carbon we know is"},{"Start":"03:15.105 ","End":"03:22.230","Text":"sp^3 and our sulfur we said is also sp^3."},{"Start":"03:22.230 ","End":"03:26.170","Text":"It\u0027s going to be C_sp^3 to sulfur sp^3."},{"Start":"03:28.700 ","End":"03:32.210","Text":"We\u0027re going to go to the next bond that\u0027s between the sulfur and"},{"Start":"03:32.210 ","End":"03:34.760","Text":"the right carbon is going to be"},{"Start":"03:34.760 ","End":"03:40.190","Text":"sulfur sp^3 to carbon."},{"Start":"03:40.190 ","End":"03:45.000","Text":"However, this time we have an sp hybridization, so carbon sp."},{"Start":"03:46.300 ","End":"03:50.525","Text":"Now we need to take a look at our triple bond here."},{"Start":"03:50.525 ","End":"03:53.990","Text":"In order to find the orbitals which are involved in our triple bond."},{"Start":"03:53.990 ","End":"03:57.089","Text":"First we\u0027re going to take a look at our nitrogen."},{"Start":"03:58.330 ","End":"04:00.720","Text":"We\u0027re going to hybridize it as well."},{"Start":"04:00.720 ","End":"04:02.840","Text":"We know that the nitrogen is connected to 2 groups,"},{"Start":"04:02.840 ","End":"04:04.910","Text":"1 triple bond and 1 lone pair electrons,"},{"Start":"04:04.910 ","End":"04:08.200","Text":"meaning that\u0027s going to give us a hybridization of sp."},{"Start":"04:08.200 ","End":"04:10.710","Text":"Here in our triple bond,"},{"Start":"04:10.710 ","End":"04:13.410","Text":"we have 1 Sigma bond and we have 2Pi bonds."},{"Start":"04:13.410 ","End":"04:17.160","Text":"Now the sigma bond is going to go from"},{"Start":"04:17.160 ","End":"04:24.970","Text":"the C_sp orbital to the nitrogen sp orbital."},{"Start":"04:24.970 ","End":"04:28.400","Text":"Now if we look at our double and triple bond,"},{"Start":"04:28.400 ","End":"04:37.210","Text":"that\u0027s going to be between the C_2p orbital to the N_2p, nitrogen, orbital."},{"Start":"04:37.210 ","End":"04:40.445","Text":"That\u0027s going to be for a double and triple bonds."},{"Start":"04:40.445 ","End":"04:43.190","Text":"Again, the H to C bonds,"},{"Start":"04:43.190 ","End":"04:47.395","Text":"the left carbon are H1s to C_sp^3."},{"Start":"04:47.395 ","End":"04:54.630","Text":"The left carbon to sulfur bond is C_sp^3 to S_sp^3."},{"Start":"04:54.630 ","End":"04:57.510","Text":"Then we have the sulfur to right carbon bonds,"},{"Start":"04:57.510 ","End":"05:00.240","Text":"so that\u0027s S_sp^3 to C_sp."},{"Start":"05:00.240 ","End":"05:05.254","Text":"Then we have the last triple bond which is between our carbon to nitrogen."},{"Start":"05:05.254 ","End":"05:14.090","Text":"Our single bond or sigma bond is going to be between the C_sp to N_sp and our Pi bonds,"},{"Start":"05:14.090 ","End":"05:18.335","Text":"which are the double and triple bond are C_sp to N_2p."},{"Start":"05:18.335 ","End":"05:23.210","Text":"This is orbitals which are involved in this molecule."},{"Start":"05:23.210 ","End":"05:26.910","Text":"That is our final answer. Thank you very much for watching."}],"ID":24474}],"Thumbnail":null,"ID":101322},{"Name":"Molecular orbital MO theory","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction to MO theory","Duration":"8m 4s","ChapterTopicVideoID":23052,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"In previous videos,"},{"Start":"00:02.070 ","End":"00:04.185","Text":"we learned about valence bond theory."},{"Start":"00:04.185 ","End":"00:07.980","Text":"In this video, we\u0027ll learn about molecular orbital theory."},{"Start":"00:07.980 ","End":"00:10.425","Text":"What\u0027s molecular orbital theory?"},{"Start":"00:10.425 ","End":"00:12.465","Text":"Just like valence bond theory,"},{"Start":"00:12.465 ","End":"00:14.070","Text":"molecular orbital theory is"},{"Start":"00:14.070 ","End":"00:17.370","Text":"an approximate quantum mechanical theory that was"},{"Start":"00:17.370 ","End":"00:21.585","Text":"developed by Mulliken who got Nobel Prize for 1966,"},{"Start":"00:21.585 ","End":"00:25.500","Text":"and Hund in the late 1920s."},{"Start":"00:25.500 ","End":"00:27.840","Text":"The advantage is that it gives solutions to"},{"Start":"00:27.840 ","End":"00:30.690","Text":"many problems that valence bond fails to solve."},{"Start":"00:30.690 ","End":"00:34.995","Text":"In addition, it\u0027s much easier to use numerically."},{"Start":"00:34.995 ","End":"00:36.570","Text":"However, in recent years,"},{"Start":"00:36.570 ","End":"00:40.290","Text":"valence bond theory has undergone a resurgence,"},{"Start":"00:40.290 ","End":"00:46.055","Text":"and it\u0027s also used to solve quite a number of numerical problems."},{"Start":"00:46.055 ","End":"00:51.080","Text":"What are the disadvantages or the limitations of valence bond theory?"},{"Start":"00:51.080 ","End":"00:56.780","Text":"For example, it can\u0027t explain why oxygen gas is paramagnetic."},{"Start":"00:56.780 ","End":"00:59.025","Text":"According to valence bond theory,"},{"Start":"00:59.025 ","End":"01:02.765","Text":"all the electrons are paired and it\u0027s diamagnetic."},{"Start":"01:02.765 ","End":"01:07.055","Text":"It also can\u0027t explain why H_2 plus is a stable ion."},{"Start":"01:07.055 ","End":"01:11.240","Text":"H_2 plus only has 1 electron and yet is stable."},{"Start":"01:11.240 ","End":"01:15.965","Text":"It can\u0027t explain the structure of electron deficient compounds."},{"Start":"01:15.965 ","End":"01:19.215","Text":"For example, B_2H_6,"},{"Start":"01:19.215 ","End":"01:22.065","Text":"which we can draw as BB,"},{"Start":"01:22.065 ","End":"01:25.005","Text":"there\u0027s 3 hydrogens on each boron."},{"Start":"01:25.005 ","End":"01:27.470","Text":"Now, boron has 3 valence electrons,"},{"Start":"01:27.470 ","End":"01:30.890","Text":"so we have 6 valence electrons from the 2 borons."},{"Start":"01:30.890 ","End":"01:34.640","Text":"Now, we have 6 valence electrons from 6 hydrogens,"},{"Start":"01:34.640 ","End":"01:37.714","Text":"so altogether, we have 12 electrons."},{"Start":"01:37.714 ","End":"01:39.110","Text":"But look at the number of bonds,"},{"Start":"01:39.110 ","End":"01:40.280","Text":"there\u0027s 1, 2, 3,"},{"Start":"01:40.280 ","End":"01:43.129","Text":"4, 5, 6, 7 bonds."},{"Start":"01:43.129 ","End":"01:44.930","Text":"According to valence bond theory,"},{"Start":"01:44.930 ","End":"01:48.020","Text":"we need 2 electrons for each bond,"},{"Start":"01:48.020 ","End":"01:52.650","Text":"so we need 14 electrons yet we only have 12."},{"Start":"01:52.650 ","End":"01:55.475","Text":"What is molecular orbital theory?"},{"Start":"01:55.475 ","End":"02:00.890","Text":"We form the molecular orbitals from linear combinations of atomic orbitals."},{"Start":"02:00.890 ","End":"02:07.920","Text":"That\u0027s called LCA or linear combinations of atomic orbitals, on different atoms."},{"Start":"02:07.920 ","End":"02:10.430","Text":"We\u0027re going to combine the atomic orbitals on"},{"Start":"02:10.430 ","End":"02:14.275","Text":"different atoms in order to get molecular orbitals."},{"Start":"02:14.275 ","End":"02:16.820","Text":"The number of molecular orbitals we get is"},{"Start":"02:16.820 ","End":"02:20.450","Text":"equal to the number of atomic orbitals we started from."},{"Start":"02:20.450 ","End":"02:24.170","Text":"What we do is we form the molecular orbitals and then fill them up with"},{"Start":"02:24.170 ","End":"02:28.085","Text":"valence electrons starting from the lowest energy."},{"Start":"02:28.085 ","End":"02:33.350","Text":"This is similar to the aufbau process for atomic orbitals in atoms."},{"Start":"02:33.350 ","End":"02:38.315","Text":"Remember, we filled them from the lowest energy atomic orbitals."},{"Start":"02:38.315 ","End":"02:44.540","Text":"Now, we\u0027re going to take some examples from the first period of the periodic table."},{"Start":"02:44.540 ","End":"02:46.535","Text":"That\u0027s hydrogen and helium."},{"Start":"02:46.535 ","End":"02:48.830","Text":"We\u0027re going to consider H_2 plus,"},{"Start":"02:48.830 ","End":"02:51.620","Text":"which only has 1 valence electron."},{"Start":"02:51.620 ","End":"02:53.780","Text":"H_2 which has 2,"},{"Start":"02:53.780 ","End":"02:56.275","Text":"and helium 2 which has 4."},{"Start":"02:56.275 ","End":"02:58.440","Text":"From the 1s orbitals,"},{"Start":"02:58.440 ","End":"03:02.250","Text":"1s_A on atom A and 1s_B on atom B,"},{"Start":"03:02.250 ","End":"03:05.070","Text":"we can form 2 linear combinations,"},{"Start":"03:05.070 ","End":"03:06.300","Text":"a positive 1,"},{"Start":"03:06.300 ","End":"03:12.750","Text":"1s_A+1s_B, and a negative 1, 1s_A-1s_B."},{"Start":"03:12.750 ","End":"03:15.600","Text":"The first we call a Sigma orbital,"},{"Start":"03:15.600 ","End":"03:19.440","Text":"Sigma 1s because it comes from 1s atomic orbitals,"},{"Start":"03:19.440 ","End":"03:21.840","Text":"and Sigma because of the symmetry,"},{"Start":"03:21.840 ","End":"03:23.565","Text":"it has no nodes."},{"Start":"03:23.565 ","End":"03:29.495","Text":"Just like we had single bonds without any nodes in valence bond theory."},{"Start":"03:29.495 ","End":"03:31.958","Text":"Here we have also Sigma 1s,"},{"Start":"03:31.958 ","End":"03:34.885","Text":"we call it, but with an asterisk."},{"Start":"03:34.885 ","End":"03:38.405","Text":"This is to indicate that it\u0027s antibonding."},{"Start":"03:38.405 ","End":"03:40.130","Text":"This one is bonding,"},{"Start":"03:40.130 ","End":"03:42.760","Text":"and this one is antibonding."},{"Start":"03:42.760 ","End":"03:47.315","Text":"We said that Sigma 1s is a bonding molecular orbital."},{"Start":"03:47.315 ","End":"03:53.495","Text":"That means that Sigma 1s has lower energy than 1s orbital."},{"Start":"03:53.495 ","End":"03:58.475","Text":"So it\u0027s 1s_A and 1s_B that was the energy of course,"},{"Start":"03:58.475 ","End":"04:00.740","Text":"and Sigma 1s is bonding."},{"Start":"04:00.740 ","End":"04:05.375","Text":"It has lower energy than the 1s orbitals."},{"Start":"04:05.375 ","End":"04:09.330","Text":"Now, here\u0027s the 1s on A,"},{"Start":"04:09.330 ","End":"04:11.715","Text":"and the 1s on B."},{"Start":"04:11.715 ","End":"04:14.450","Text":"Here we\u0027ve drawn them so that they overlap,"},{"Start":"04:14.450 ","End":"04:16.910","Text":"just like we did in valence bond theory."},{"Start":"04:16.910 ","End":"04:19.320","Text":"Now, they\u0027re overlapping."},{"Start":"04:20.290 ","End":"04:27.570","Text":"We add the wave functions for the atomic orbitals."},{"Start":"04:27.570 ","End":"04:28.745","Text":"Let\u0027s do that."},{"Start":"04:28.745 ","End":"04:31.265","Text":"Let\u0027s draw the wave functions."},{"Start":"04:31.265 ","End":"04:36.425","Text":"Here\u0027s 1s_A and here\u0027s 1s_B."},{"Start":"04:36.425 ","End":"04:42.020","Text":"This one is centered on atom A and this is centered on atom B,"},{"Start":"04:42.020 ","End":"04:45.440","Text":"and they both have the same sign because we are considering"},{"Start":"04:45.440 ","End":"04:50.315","Text":"the bonding molecular orbital."},{"Start":"04:50.315 ","End":"04:53.660","Text":"Now, when we add these 2 together,"},{"Start":"04:53.660 ","End":"04:58.805","Text":"we see that the amplitude is"},{"Start":"04:58.805 ","End":"05:06.420","Text":"greater between the atoms than it would be if we just had 1s orbitals."},{"Start":"05:06.420 ","End":"05:10.970","Text":"We have increased electron density between the atoms,"},{"Start":"05:10.970 ","End":"05:14.015","Text":"and this gives us a bonding situation."},{"Start":"05:14.015 ","End":"05:18.995","Text":"Each nucleus can hold onto this electron density."},{"Start":"05:18.995 ","End":"05:23.075","Text":"Here we have constructive interference."},{"Start":"05:23.075 ","End":"05:29.330","Text":"Sigma 1s with the asterisk is an antibody molecular orbital."},{"Start":"05:29.330 ","End":"05:32.780","Text":"That means it has higher energy the 1s, here it\u0027s indicated."},{"Start":"05:32.780 ","End":"05:40.525","Text":"Sigma 1s star or asterisk has a higher energy than the separate 1s orbitals."},{"Start":"05:40.525 ","End":"05:42.165","Text":"Now let\u0027s add them."},{"Start":"05:42.165 ","End":"05:49.800","Text":"Again, we have the 1s orbital centered on atom A,"},{"Start":"05:49.800 ","End":"05:55.040","Text":"but now we see that we have to take the negative of the 1s_B."},{"Start":"05:55.040 ","End":"05:58.415","Text":"So we can draw it like this."},{"Start":"05:58.415 ","End":"06:01.530","Text":"Now we have to add the 2."},{"Start":"06:02.960 ","End":"06:08.295","Text":"It\u0027s a little less here than the 1s,"},{"Start":"06:08.295 ","End":"06:13.320","Text":"goes down, goes through 0, and then up."},{"Start":"06:13.320 ","End":"06:16.760","Text":"We see it\u0027s positive 0 and then negative."},{"Start":"06:16.760 ","End":"06:18.725","Text":"That means there is a node."},{"Start":"06:18.725 ","End":"06:21.990","Text":"This is antibonding."},{"Start":"06:22.060 ","End":"06:26.150","Text":"There\u0027s a node, it\u0027s the opposite of bonding."},{"Start":"06:26.150 ","End":"06:31.180","Text":"The bonding molecular orbital has a lower energy than the 1s,"},{"Start":"06:31.180 ","End":"06:33.590","Text":"and the antibody has a higher end."},{"Start":"06:33.590 ","End":"06:37.285","Text":"Now we can perform our aufbau process."},{"Start":"06:37.285 ","End":"06:39.440","Text":"Start off with H_2 plus,"},{"Start":"06:39.440 ","End":"06:40.820","Text":"it only has one electron,"},{"Start":"06:40.820 ","End":"06:43.295","Text":"so there has to be in the Sigma 1s."},{"Start":"06:43.295 ","End":"06:45.080","Text":"H_2 has 2 electrons."},{"Start":"06:45.080 ","End":"06:48.175","Text":"We have a pair of electrons in 1s."},{"Start":"06:48.175 ","End":"06:51.150","Text":"Helium 2 has 4 electrons,"},{"Start":"06:51.150 ","End":"06:54.090","Text":"so we have 2 electrons in the Sigma 1s,"},{"Start":"06:54.090 ","End":"06:58.745","Text":"and 2 electrons in Sigma 1s asterisk."},{"Start":"06:58.745 ","End":"07:01.150","Text":"We have 2 electrons here."},{"Start":"07:01.150 ","End":"07:02.850","Text":"2 electrons in the bonding,"},{"Start":"07:02.850 ","End":"07:04.855","Text":"2 electrons in the antibonding."},{"Start":"07:04.855 ","End":"07:07.595","Text":"Now we\u0027re going to define the bond order."},{"Start":"07:07.595 ","End":"07:11.570","Text":"The bond order is the number of electrons in the bonding molecular orbitals,"},{"Start":"07:11.570 ","End":"07:16.130","Text":"minus the number of electrons in the antibonding molecular orbitals."},{"Start":"07:16.130 ","End":"07:18.700","Text":"We then divide that by 2."},{"Start":"07:18.700 ","End":"07:22.860","Text":"For H_2 plus we only have 1 electron bonding orbital,"},{"Start":"07:22.860 ","End":"07:25.185","Text":"so the bond order is a half."},{"Start":"07:25.185 ","End":"07:28.455","Text":"H_2, we have 2 electrons in the bonding orbital,"},{"Start":"07:28.455 ","End":"07:30.910","Text":"so the bond order is 1,"},{"Start":"07:30.910 ","End":"07:35.915","Text":"just the same as it would be in valence bond theory, a single bond."},{"Start":"07:35.915 ","End":"07:39.125","Text":"He_2 has 4 electrons,"},{"Start":"07:39.125 ","End":"07:42.260","Text":"2 bonding, 2 antibonding,"},{"Start":"07:42.260 ","End":"07:45.110","Text":"so the bond order is 0."},{"Start":"07:45.110 ","End":"07:49.280","Text":"That tells us H_2 plus is a stable ion."},{"Start":"07:49.280 ","End":"07:53.180","Text":"H_2 has 2 bonding electrons and a single bond,"},{"Start":"07:53.180 ","End":"07:59.495","Text":"and He_2 simply doesn\u0027t exist because the bond order is 0."},{"Start":"07:59.495 ","End":"08:04.680","Text":"In this video, we introduced molecular orbital theory."}],"ID":23894},{"Watched":false,"Name":"Second-period diatomic molecules 1","Duration":"7m 5s","ChapterTopicVideoID":23150,"CourseChapterTopicPlaylistID":101323,"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":"00:06.030","Text":"we applied molecular orbital theory to first period diatomic molecules."},{"Start":"00:06.030 ","End":"00:10.304","Text":"In this video, we\u0027ll apply it to second period diatomic molecules."},{"Start":"00:10.304 ","End":"00:14.658","Text":"Let\u0027s start off with a molecular orbitals derived from 2s orbitals."},{"Start":"00:14.658 ","End":"00:19.945","Text":"2s orbitals are the lowest atomic orbitals in the second period."},{"Start":"00:19.945 ","End":"00:21.815","Text":"Here\u0027s the diagram."},{"Start":"00:21.815 ","End":"00:24.320","Text":"Is the same as we had for 1s."},{"Start":"00:24.320 ","End":"00:27.560","Text":"We have a 2s orbital on atom A and 2s orbital"},{"Start":"00:27.560 ","End":"00:30.690","Text":"in atom B and they combine to form a molecular"},{"Start":"00:30.690 ","End":"00:37.740","Text":"orbital Sigma 2s which is bonding and a Sigma 2s star which is antibonding."},{"Start":"00:37.740 ","End":"00:39.780","Text":"Let\u0027s take 2 examples."},{"Start":"00:39.780 ","End":"00:42.090","Text":"Lithium 2 and beryllium 2."},{"Start":"00:42.090 ","End":"00:44.190","Text":"Lithium has 1 valence electron,"},{"Start":"00:44.190 ","End":"00:45.810","Text":"so lithium 2 has 2."},{"Start":"00:45.810 ","End":"00:48.170","Text":"Beryllium has 2 valence electrons,"},{"Start":"00:48.170 ","End":"00:51.700","Text":"so beryllium 2 has 4 valence electrons."},{"Start":"00:51.700 ","End":"00:55.406","Text":"We can put the 2 electrons of lithium 2 in the Sigma 2s"},{"Start":"00:55.406 ","End":"01:00.500","Text":"molecular orbital is a pair of electrons in the Sigma 2s."},{"Start":"01:00.500 ","End":"01:03.890","Text":"The bond order is 1 because we have"},{"Start":"01:03.890 ","End":"01:12.275","Text":"2 bonding electrons and no antibonding electrons,"},{"Start":"01:12.275 ","End":"01:15.305","Text":"so 2 divide by 2 is 1."},{"Start":"01:15.305 ","End":"01:18.680","Text":"Beryllium 2, we can have 2 electrons in"},{"Start":"01:18.680 ","End":"01:22.855","Text":"Sigma 2s and another 2 electrons in Sigma 2s star."},{"Start":"01:22.855 ","End":"01:25.280","Text":"We have 2 electrons in the bonding orbital,"},{"Start":"01:25.280 ","End":"01:28.295","Text":"2 electrons in the anti-bonding orbital."},{"Start":"01:28.295 ","End":"01:30.260","Text":"The difference between that is 0,"},{"Start":"01:30.260 ","End":"01:32.165","Text":"divide by 2 is still 0,"},{"Start":"01:32.165 ","End":"01:33.875","Text":"so the bond order is 0."},{"Start":"01:33.875 ","End":"01:39.050","Text":"We have a single bond in lithium 2 and we have no bond at all in beryllium 2."},{"Start":"01:39.050 ","End":"01:42.979","Text":"That means beryllium 2 doesn\u0027t exist."},{"Start":"01:42.979 ","End":"01:47.779","Text":"Now let\u0027s go on to the molecular orbitals derived from the 2p orbitals."},{"Start":"01:47.779 ","End":"01:53.190","Text":"The 2p orbitals are higher in energy than the 2s orbitals."},{"Start":"01:54.200 ","End":"02:00.150","Text":"We know that we have 3 degenerate 2p orbitals on each atom, 2p_x,"},{"Start":"02:00.150 ","End":"02:06.300","Text":"2p_y, and 2pz."},{"Start":"02:06.300 ","End":"02:11.565","Text":"That\u0027s a total of 6 atomic orbitals,"},{"Start":"02:11.565 ","End":"02:16.545","Text":"so 6 atomic orbitals will give us 6 molecular orbitals."},{"Start":"02:16.545 ","End":"02:19.545","Text":"Let\u0027s first talk about 2pz."},{"Start":"02:19.545 ","End":"02:23.970","Text":"These orbitals overlap end -to-end or head-to-head."},{"Start":"02:23.970 ","End":"02:25.950","Text":"We can draw it here."},{"Start":"02:25.950 ","End":"02:28.720","Text":"This is the z direction."},{"Start":"02:28.760 ","End":"02:34.205","Text":"This gives us overlap head-to-head."},{"Start":"02:34.205 ","End":"02:36.805","Text":"This is plus and this is plus."},{"Start":"02:36.805 ","End":"02:40.300","Text":"We get a Sigma bonding orbital."},{"Start":"02:40.300 ","End":"02:43.870","Text":"That Sigma 2p bonding molecular orbital."},{"Start":"02:43.870 ","End":"02:51.820","Text":"If we have positive here and negative here,"},{"Start":"02:51.820 ","End":"02:55.275","Text":"in other words the opposite signs then we get"},{"Start":"02:55.275 ","End":"03:00.660","Text":"the antibonding Sigma 2p star molecular orbital."},{"Start":"03:00.660 ","End":"03:02.895","Text":"This is z direction."},{"Start":"03:02.895 ","End":"03:07.480","Text":"These are 2pz atomic orbitals."},{"Start":"03:07.480 ","End":"03:10.015","Text":"Now let\u0027s look at the nodes."},{"Start":"03:10.015 ","End":"03:15.190","Text":"The Sigma orbitals have no planar nodes containing the z axis."},{"Start":"03:15.190 ","End":"03:18.100","Text":"If we look above and below the z axis,"},{"Start":"03:18.100 ","End":"03:20.170","Text":"we\u0027ll see that we have the same sign,"},{"Start":"03:20.170 ","End":"03:21.880","Text":"so there is no node there."},{"Start":"03:21.880 ","End":"03:26.890","Text":"However, the antibonding molecular orbital has a node between the plus"},{"Start":"03:26.890 ","End":"03:32.050","Text":"and minus here that is perpendicular to the intermolecular axis,"},{"Start":"03:32.050 ","End":"03:34.130","Text":"so there\u0027s a node here."},{"Start":"03:34.130 ","End":"03:36.935","Text":"Now let\u0027s look at the 2p_x orbitals."},{"Start":"03:36.935 ","End":"03:40.720","Text":"These overlap side-to-side rather than the end-to-end."},{"Start":"03:40.720 ","End":"03:44.750","Text":"They give us a Pi 2p bonding molecular orbital and a"},{"Start":"03:44.750 ","End":"03:49.190","Text":"Pi_2p^star antibonding molecular orbital. Let\u0027s look at them."},{"Start":"03:49.190 ","End":"03:55.085","Text":"These 2p_x orbitals overlap side-to-side."},{"Start":"03:55.085 ","End":"04:00.870","Text":"This will be positive at the top and negative at the bottom is the z axis."},{"Start":"04:00.910 ","End":"04:05.480","Text":"If they overlap, positive, negative,"},{"Start":"04:05.480 ","End":"04:10.400","Text":"negative, positive, that gives us a node in the center."},{"Start":"04:10.400 ","End":"04:16.010","Text":"We also have a node in both of them parallel to the z axis."},{"Start":"04:16.010 ","End":"04:18.335","Text":"This is because if I draw a line here,"},{"Start":"04:18.335 ","End":"04:22.680","Text":"I can see it goes from positive to negative and here from positive to negative."},{"Start":"04:22.680 ","End":"04:26.360","Text":"Here from positive to negative, negative to positive."},{"Start":"04:26.360 ","End":"04:36.750","Text":"This is the bonding Pi orbital Pi_2p and this is the antibonding Pi_2p^star."},{"Start":"04:36.750 ","End":"04:39.120","Text":"Because there\u0027s no overlap here,"},{"Start":"04:39.120 ","End":"04:40.955","Text":"the 0 on the center."},{"Start":"04:40.955 ","End":"04:42.575","Text":"Here we\u0027ve written it."},{"Start":"04:42.575 ","End":"04:47.270","Text":"The same thing happens for the 2p_y orbitals."},{"Start":"04:47.270 ","End":"04:50.870","Text":"They overlap side-to-side again to give a Pi_2p bonding"},{"Start":"04:50.870 ","End":"04:55.280","Text":"molecular orbital and a Pi_2p^star antibonding molecular orbital."},{"Start":"04:55.280 ","End":"04:59.210","Text":"Now we need to notice that both these Pi orbitals,"},{"Start":"04:59.210 ","End":"05:02.540","Text":"the ones that came from 2p_x and the ones that came from to"},{"Start":"05:02.540 ","End":"05:06.455","Text":"2p_y degenerate the same energy."},{"Start":"05:06.455 ","End":"05:09.625","Text":"This is just to tell us once again that the Pi orbitals have"},{"Start":"05:09.625 ","End":"05:14.000","Text":"1 planar node continuous at z axis."},{"Start":"05:14.000 ","End":"05:22.440","Text":"Pi is like p. Pi is a Greek and p is Latin and they both have 1 planar node."},{"Start":"05:22.440 ","End":"05:25.095","Text":"We\u0027re going to consider the energy."},{"Start":"05:25.095 ","End":"05:27.900","Text":"For oxygen 2 and fluorine 2,"},{"Start":"05:27.900 ","End":"05:31.050","Text":"the diagram is on the right hand side."},{"Start":"05:31.050 ","End":"05:35.120","Text":"The orbitals that come from 2p are higher than the orbitals that come from"},{"Start":"05:35.120 ","End":"05:40.560","Text":"2s and the order is Sigma,"},{"Start":"05:40.560 ","End":"05:44.210","Text":"then Pi, then Pi^star and Sigma^star."},{"Start":"05:44.210 ","End":"05:47.570","Text":"The reason for the different separations is that"},{"Start":"05:47.570 ","End":"05:51.295","Text":"this greater overlap end-to-end or head-to -head,"},{"Start":"05:51.295 ","End":"05:55.580","Text":"giving the Sigma orbitals then there is side-to-side giving the Pi orbitals."},{"Start":"05:55.580 ","End":"05:59.820","Text":"The interaction is greater for the Sigma than"},{"Start":"05:59.820 ","End":"06:05.060","Text":"the Pi and the separation is greater the Sigma then for the Pi."},{"Start":"06:05.060 ","End":"06:12.310","Text":"What\u0027s on the right hand side is true for these heavier molecules, O2 and F2."},{"Start":"06:12.310 ","End":"06:14.225","Text":"However, for the lighter ones,"},{"Start":"06:14.225 ","End":"06:16.265","Text":"the order is slightly different."},{"Start":"06:16.265 ","End":"06:18.935","Text":"The reason for this is as follows."},{"Start":"06:18.935 ","End":"06:23.485","Text":"Sigma_2s interacts with Sigma_2p."},{"Start":"06:23.485 ","End":"06:28.160","Text":"It\u0027s not very great when we have heavy molecules."},{"Start":"06:28.160 ","End":"06:30.110","Text":"However, when they\u0027re lighter,"},{"Start":"06:30.110 ","End":"06:35.210","Text":"what this does is to push the Sigma_2s down a little and to push"},{"Start":"06:35.210 ","End":"06:40.535","Text":"the Sigma_2p up so that it goes higher than the Pi_2p."},{"Start":"06:40.535 ","End":"06:43.580","Text":"Here we have it. The order is changed."},{"Start":"06:43.580 ","End":"06:45.710","Text":"We have Pi first Pi_2p,"},{"Start":"06:45.710 ","End":"06:49.580","Text":"the 2 molecular orbitals, and then Sigma_2p."},{"Start":"06:49.580 ","End":"06:51.740","Text":"Then the same as before,"},{"Start":"06:51.740 ","End":"06:55.310","Text":"Pi_2p^star and then Sigma_2p^star."},{"Start":"06:55.310 ","End":"07:01.520","Text":"This video, we applied molecular orbital theory to the second period diatomic molecules."},{"Start":"07:01.520 ","End":"07:04.830","Text":"In the next video, we\u0027ll give some examples."}],"ID":23982},{"Watched":false,"Name":"Second-period diatomic molecules 2","Duration":"7m 20s","ChapterTopicVideoID":23054,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"In the previous videos,"},{"Start":"00:02.160 ","End":"00:05.130","Text":"we talked about second period molecular orbitals."},{"Start":"00:05.130 ","End":"00:08.250","Text":"This video, we\u0027ll give some examples."},{"Start":"00:08.250 ","End":"00:14.535","Text":"Let\u0027s recall the energy level diagrams for homonuclear diatomic molecules."},{"Start":"00:14.535 ","End":"00:17.640","Text":"That\u0027s where both atoms are the same."},{"Start":"00:17.640 ","End":"00:21.285","Text":"Here are the diagrams. On the right-hand side,"},{"Start":"00:21.285 ","End":"00:25.890","Text":"we have the diagram that\u0027s appropriate for oxygen 2 and fluorine 2."},{"Start":"00:25.890 ","End":"00:33.015","Text":"On the left-hand side for all the other diatomic molecules in the second period."},{"Start":"00:33.015 ","End":"00:35.955","Text":"We\u0027re going to discuss 4 examples."},{"Start":"00:35.955 ","End":"00:39.420","Text":"N_2, N_2+,"},{"Start":"00:39.420 ","End":"00:41.295","Text":"O_2, and O_2+."},{"Start":"00:41.295 ","End":"00:44.550","Text":"Let us begin with N_2."},{"Start":"00:44.550 ","End":"00:51.585","Text":"Nitrogen has 5 valence electron and nitrogen 2 has 10 valence electron."},{"Start":"00:51.585 ","End":"00:55.085","Text":"We have to put 10 valence electrons"},{"Start":"00:55.085 ","End":"01:01.135","Text":"into these molecular orbitals in ascending order of energy."},{"Start":"01:01.135 ","End":"01:05.380","Text":"2 electrons in each orbital."},{"Start":"01:05.380 ","End":"01:09.200","Text":"Recall this from the atomic orbitals,"},{"Start":"01:09.200 ","End":"01:13.355","Text":"this is called the Aufbau method."},{"Start":"01:13.355 ","End":"01:17.090","Text":"We have 2 electrons in Sigma 2s,"},{"Start":"01:17.090 ","End":"01:20.125","Text":"2 electrons in Sigma 2s star,"},{"Start":"01:20.125 ","End":"01:23.130","Text":"4 electrons in Pi 2p,"},{"Start":"01:23.130 ","End":"01:26.410","Text":"2 in each of the molecular orbitals,"},{"Start":"01:26.410 ","End":"01:28.780","Text":"and 2 in Sigma 2p."},{"Start":"01:28.780 ","End":"01:33.415","Text":"That\u0027s a total of 10 electrons, 2,4,6,8,10."},{"Start":"01:33.415 ","End":"01:38.140","Text":"We can write the electron configuration like beryllium 2 that"},{"Start":"01:38.140 ","End":"01:43.055","Text":"means 2 electrons in Sigma 2s and 2 in Sigma 2 star."},{"Start":"01:43.055 ","End":"01:48.705","Text":"4 electrons in Pi 2p and 2 electrons in Sigma 2p."},{"Start":"01:48.705 ","End":"01:51.975","Text":"Now let\u0027s work out the bond order."},{"Start":"01:51.975 ","End":"01:54.330","Text":"As far as 2S goes,"},{"Start":"01:54.330 ","End":"01:56.745","Text":"we have 2 electrons in bonding orbital,"},{"Start":"01:56.745 ","End":"01:58.825","Text":"2 electrons in antibonding orbital."},{"Start":"01:58.825 ","End":"02:02.450","Text":"The bond order from this is just 0."},{"Start":"02:02.450 ","End":"02:07.740","Text":"Then we have 2,4,6 electrons which are bonding,"},{"Start":"02:07.740 ","End":"02:16.580","Text":"6 bonding, and none in antibonding orbitals and 0 antibonding."},{"Start":"02:16.580 ","End":"02:21.455","Text":"That gives us a bond order of 6 minus 0 divided by 2."},{"Start":"02:21.455 ","End":"02:23.890","Text":"The bond order here is 3."},{"Start":"02:23.890 ","End":"02:26.820","Text":"I\u0027ve written it here, the bond order is 3."},{"Start":"02:26.820 ","End":"02:28.895","Text":"That\u0027s a triple bond,"},{"Start":"02:28.895 ","End":"02:35.135","Text":"which was also the prediction of Lewis theory and valence bond theory."},{"Start":"02:35.135 ","End":"02:37.915","Text":"Now let\u0027s look at N_2+."},{"Start":"02:37.915 ","End":"02:42.525","Text":"That\u0027s 1 electron less than N_2."},{"Start":"02:42.525 ","End":"02:45.285","Text":"We have to take out the electron,"},{"Start":"02:45.285 ","End":"02:50.020","Text":"we take it out of the highest energy orbital."},{"Start":"02:50.020 ","End":"02:53.830","Text":"Instead of the pair of electrons in Sigma 2p,"},{"Start":"02:53.830 ","End":"02:56.860","Text":"we just have 1 electron."},{"Start":"02:56.860 ","End":"03:00.380","Text":"Now we can calculate the bond order."},{"Start":"03:00.380 ","End":"03:07.810","Text":"Now we have 5 electrons in bonding orbitals and 0 in antibonding,"},{"Start":"03:07.810 ","End":"03:11.785","Text":"so we have 5 minus 0 divided by 2."},{"Start":"03:11.785 ","End":"03:18.555","Text":"That\u0027s 2.5. The bond order is 2.5 and here\u0027s the electron configuration."},{"Start":"03:18.555 ","End":"03:24.180","Text":"Beryllium 2, 4 electrons in Pi 2p and 1 electron in Sigma 2p."},{"Start":"03:24.180 ","End":"03:34.815","Text":"We see here that the bond order is greater in nitrogen 2 than it is in nitrogen 2 plus."},{"Start":"03:34.815 ","End":"03:38.345","Text":"Nitrogen 2 is a stronger bond."},{"Start":"03:38.345 ","End":"03:45.145","Text":"Takes more energy to dissociate nitrogen 2 than it takes to dissociate nitrogen 2 plus."},{"Start":"03:45.145 ","End":"03:47.590","Text":"Now let\u0027s look at oxygen 2."},{"Start":"03:47.590 ","End":"03:51.445","Text":"Oxygen has 6 valence electrons."},{"Start":"03:51.445 ","End":"03:55.630","Text":"Oxygen 2 has 12 valence electrons."},{"Start":"03:55.630 ","End":"03:59.335","Text":"We need to use the diagram on the right-hand side."},{"Start":"03:59.335 ","End":"04:01.685","Text":"Let\u0027s fill up the orbitals."},{"Start":"04:01.685 ","End":"04:03.615","Text":"2 in Sigma 2s,"},{"Start":"04:03.615 ","End":"04:06.195","Text":"2 in Sigma 2s bar,"},{"Start":"04:06.195 ","End":"04:07.830","Text":"2 in Sigma 2p,"},{"Start":"04:07.830 ","End":"04:14.385","Text":"4 in Pi 2p."},{"Start":"04:14.385 ","End":"04:18.405","Text":"That\u0027s a total of 2,4,6,8,10."},{"Start":"04:18.405 ","End":"04:20.070","Text":"We have another 2 electrons,"},{"Start":"04:20.070 ","End":"04:21.845","Text":"now where are we going to put them?"},{"Start":"04:21.845 ","End":"04:24.620","Text":"We have to go according to the Hund rule,"},{"Start":"04:24.620 ","End":"04:28.670","Text":"which means that when you have 2 degenerate orbitals,"},{"Start":"04:28.670 ","End":"04:32.000","Text":"as we have here in Pi 2p star,"},{"Start":"04:32.000 ","End":"04:36.400","Text":"1 electron goes in one of the orbitals and 1 electron in the other."},{"Start":"04:36.400 ","End":"04:39.320","Text":"We have 2 unpaired electrons."},{"Start":"04:39.320 ","End":"04:43.475","Text":"Here is the electron configuration like beryllium 2,"},{"Start":"04:43.475 ","End":"04:45.635","Text":"2 electrons in Sigma 2p,"},{"Start":"04:45.635 ","End":"04:47.765","Text":"4 electrons in Pi 2p,"},{"Start":"04:47.765 ","End":"04:50.755","Text":"and 2 electrons in Pi 2p star."},{"Start":"04:50.755 ","End":"04:52.515","Text":"What\u0027s the bond order?"},{"Start":"04:52.515 ","End":"04:55.860","Text":"We have 6 electrons that are bonding,"},{"Start":"04:55.860 ","End":"04:57.855","Text":"2 that are antibonding,"},{"Start":"04:57.855 ","End":"05:00.900","Text":"that 6 minus 2 divided by 2,"},{"Start":"05:00.900 ","End":"05:02.565","Text":"and that gives us 2."},{"Start":"05:02.565 ","End":"05:04.080","Text":"The bond order is 2,"},{"Start":"05:04.080 ","End":"05:07.095","Text":"that\u0027s the double bond of oxygen 2."},{"Start":"05:07.095 ","End":"05:09.450","Text":"The same as we have from Lewis theory,"},{"Start":"05:09.450 ","End":"05:12.020","Text":"the same as they have from valence bond theory."},{"Start":"05:12.020 ","End":"05:14.989","Text":"However, here we have something extra."},{"Start":"05:14.989 ","End":"05:21.185","Text":"Molecular orbital theory tells us that there are 2 unpaired electrons."},{"Start":"05:21.185 ","End":"05:25.800","Text":"Oxygen 2 is paramagnetic,"},{"Start":"05:26.960 ","End":"05:32.255","Text":"whereas valence bond theory tells us that all the electrons are paired."},{"Start":"05:32.255 ","End":"05:34.355","Text":"In valence bond theory,"},{"Start":"05:34.355 ","End":"05:37.940","Text":"oxygen 2 is diamagnetic,"},{"Start":"05:37.940 ","End":"05:41.960","Text":"whereas molecular orbital theory gives us the correct answer,"},{"Start":"05:41.960 ","End":"05:45.365","Text":"which is oxygen 2 is paramagnetic."},{"Start":"05:45.365 ","End":"05:47.855","Text":"Now let\u0027s look at O_2+."},{"Start":"05:47.855 ","End":"05:52.915","Text":"Again, we have to take 1 electron out of the highest orbital."},{"Start":"05:52.915 ","End":"05:56.580","Text":"The highest orbital is Pi 2p star."},{"Start":"05:56.580 ","End":"06:00.970","Text":"We have 1 electron in Pi 2p star."},{"Start":"06:01.060 ","End":"06:05.330","Text":"In other words, we can cancel one of these electrons."},{"Start":"06:05.330 ","End":"06:09.110","Text":"We only have 1 electron in Pi 2p star."},{"Start":"06:09.110 ","End":"06:12.055","Text":"Here\u0027s electron configuration."},{"Start":"06:12.055 ","End":"06:15.480","Text":"Sigma 2p, 2 electrons,"},{"Start":"06:15.480 ","End":"06:19.005","Text":"of course, we should have the same as beryllium 2."},{"Start":"06:19.005 ","End":"06:22.275","Text":"Then 4 electrons in Pi 2p,"},{"Start":"06:22.275 ","End":"06:25.385","Text":"and 1 electron Pi 2p star."},{"Start":"06:25.385 ","End":"06:28.655","Text":"The number of bonding electrons is 6,"},{"Start":"06:28.655 ","End":"06:31.550","Text":"antibonding electrons is 1."},{"Start":"06:31.550 ","End":"06:34.820","Text":"We have 6 minus 1 divided by 2,"},{"Start":"06:34.820 ","End":"06:37.060","Text":"which gives us 2.5."},{"Start":"06:37.060 ","End":"06:40.760","Text":"Here we see the very unusual property."},{"Start":"06:40.760 ","End":"06:44.795","Text":"If we look at oxygen 2 the bond order is 2,"},{"Start":"06:44.795 ","End":"06:48.005","Text":"oxygen 2 plus the bond order is 2.5."},{"Start":"06:48.005 ","End":"06:54.205","Text":"That means the oxygen 2 plus is stronger than oxygen 2."},{"Start":"06:54.205 ","End":"07:01.610","Text":"That means it takes more energy to dissociate O_2+ than it does to dissociate O_2."},{"Start":"07:01.610 ","End":"07:06.185","Text":"This is quite the opposite situation from nitrogen,"},{"Start":"07:06.185 ","End":"07:11.465","Text":"where neutral molecule is stronger than the ion."},{"Start":"07:11.465 ","End":"07:15.080","Text":"Here, the ion is stronger than the neutral molecule."},{"Start":"07:15.080 ","End":"07:21.210","Text":"In this video, we gave some examples of second period diatomic molecule."}],"ID":23896},{"Watched":false,"Name":"Heteronuclear diatomic molecules","Duration":"8m 32s","ChapterTopicVideoID":23055,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"In the previous videos,"},{"Start":"00:02.040 ","End":"00:05.235","Text":"we talked about homonuclear diatomic molecules."},{"Start":"00:05.235 ","End":"00:09.240","Text":"In this video, we\u0027ll learn about heteronuclear diatomic molecules."},{"Start":"00:09.240 ","End":"00:10.845","Text":"Now in the previous video,"},{"Start":"00:10.845 ","End":"00:12.960","Text":"we saw that a molecular orbital is written as"},{"Start":"00:12.960 ","End":"00:16.890","Text":"a linear combination of atomic orbitals, Psi_A and Psi_B."},{"Start":"00:16.890 ","End":"00:23.355","Text":"We abbreviated this as LCA or linear combination of atomic orbitals,"},{"Start":"00:23.355 ","End":"00:24.779","Text":"molecular orbitals."},{"Start":"00:24.779 ","End":"00:26.775","Text":"What\u0027s linear combination?"},{"Start":"00:26.775 ","End":"00:31.395","Text":"Psi, the wave function of the molecular orbital is written"},{"Start":"00:31.395 ","End":"00:36.260","Text":"as a sum of the wave function of the first atom,"},{"Start":"00:36.260 ","End":"00:40.090","Text":"atom A with a coefficient CA,"},{"Start":"00:40.090 ","End":"00:42.220","Text":"and Psi B,"},{"Start":"00:42.220 ","End":"00:46.205","Text":"the wave functions of the atom B with a coefficient CB."},{"Start":"00:46.205 ","End":"00:52.850","Text":"Now, CA squared gives us a probability of finding the electron near at"},{"Start":"00:52.850 ","End":"01:00.245","Text":"A and CB squared gives us a probability of finding the electron near B."},{"Start":"01:00.245 ","End":"01:02.224","Text":"So as these are probabilities,"},{"Start":"01:02.224 ","End":"01:05.900","Text":"their sum CA squared plus CB squared is 1."},{"Start":"01:05.900 ","End":"01:10.210","Text":"Let\u0027s recall what we learned about homonuclear diatomic molecules."},{"Start":"01:10.210 ","End":"01:14.935","Text":"In Lewis language, that\u0027s a non-polar covalent bond."},{"Start":"01:14.935 ","End":"01:18.195","Text":"There, A is the same as B,"},{"Start":"01:18.195 ","End":"01:19.785","Text":"the same sort of atom."},{"Start":"01:19.785 ","End":"01:23.960","Text":"CA squared is equal to CB squared and they\u0027re both equal to a half."},{"Start":"01:23.960 ","End":"01:26.060","Text":"The sum has to be 1."},{"Start":"01:26.060 ","End":"01:29.685","Text":"Now, if CA squared is equal to half,"},{"Start":"01:29.685 ","End":"01:37.635","Text":"then CA is equal to either 1 over root 2 or minus 1 over root 2."},{"Start":"01:37.635 ","End":"01:39.735","Text":"The same goes for CB."},{"Start":"01:39.735 ","End":"01:42.505","Text":"There, for Hydrogen 2,"},{"Start":"01:42.505 ","End":"01:44.750","Text":"we had that the sigma orbital,"},{"Start":"01:44.750 ","End":"01:51.920","Text":"the bonding orbital is proportional to 1SA +1SB whereas"},{"Start":"01:51.920 ","End":"01:59.330","Text":"antibonding orbital sigma_1S* is proportional to 1SA -1 SB."},{"Start":"01:59.330 ","End":"02:03.289","Text":"Now when we come to heteronuclear diatomic molecules,"},{"Start":"02:03.289 ","End":"02:07.190","Text":"which in Lewis language it has a polar covalent bond,"},{"Start":"02:07.190 ","End":"02:11.170","Text":"then CA squared is no longer equal to CB squared."},{"Start":"02:11.170 ","End":"02:14.550","Text":"Now if CA squared is greater than CB squared,"},{"Start":"02:14.550 ","End":"02:18.230","Text":"then Psi A contributes more"},{"Start":"02:18.230 ","End":"02:22.085","Text":"to the molecular orbital and the electron density is greatest nearest A."},{"Start":"02:22.085 ","End":"02:25.820","Text":"The other hand of CB squared is greater than the CA squared,"},{"Start":"02:25.820 ","End":"02:28.745","Text":"then Psi B contributes more to the molecular orbital."},{"Start":"02:28.745 ","End":"02:32.295","Text":"The electron density is greater nearer B."},{"Start":"02:32.295 ","End":"02:37.700","Text":"Now let us recall what we learned about the energy of the orbitals."},{"Start":"02:37.700 ","End":"02:40.250","Text":"Now here\u0027s the periodic table."},{"Start":"02:40.250 ","End":"02:44.480","Text":"The energy of 2s and 2p decreases from left to right."},{"Start":"02:44.480 ","End":"02:50.180","Text":"Here is the energy of the orbitals decreasing from left to right,"},{"Start":"02:50.180 ","End":"02:52.895","Text":"increasing from right to left."},{"Start":"02:52.895 ","End":"02:55.460","Text":"The reason for this, if you recall,"},{"Start":"02:55.460 ","End":"03:00.800","Text":"is the increasing effective nuclear charge as you go from left to right."},{"Start":"03:00.800 ","End":"03:06.985","Text":"On the other hand, the electronegativity increases from left to right."},{"Start":"03:06.985 ","End":"03:09.200","Text":"Here we see a correlation between"},{"Start":"03:09.200 ","End":"03:14.375","Text":"electronegativity and the energy of the 2s and 2p orbitals."},{"Start":"03:14.375 ","End":"03:17.555","Text":"One increases, electronegative increases left to right."},{"Start":"03:17.555 ","End":"03:20.600","Text":"The energy decreases left-to-right."},{"Start":"03:20.600 ","End":"03:23.090","Text":"Here\u0027s our energy-level diagram."},{"Start":"03:23.090 ","End":"03:24.905","Text":"Here\u0027s our atomic orbital,"},{"Start":"03:24.905 ","End":"03:31.295","Text":"say 2SA on the left-hand side and 2SB on the right-hand side."},{"Start":"03:31.295 ","End":"03:36.545","Text":"We\u0027ve taken the case where electronegativity of A is greater than that of B."},{"Start":"03:36.545 ","End":"03:40.805","Text":"The orbital energy is lower for A than B."},{"Start":"03:40.805 ","End":"03:43.400","Text":"Now here\u0027s the bonding orbital."},{"Start":"03:43.400 ","End":"03:49.775","Text":"The molecular orbital is more stable than the separate atomic orbitals."},{"Start":"03:49.775 ","End":"03:52.910","Text":"We see that it\u0027s near to 2SA,"},{"Start":"03:52.910 ","End":"03:56.780","Text":"so it has a greater contribution from A then from B."},{"Start":"03:56.780 ","End":"03:59.540","Text":"On the other hand, the antibonding orbital,"},{"Start":"03:59.540 ","End":"04:03.080","Text":"which is less stable than the separate atomic orbitals,"},{"Start":"04:03.080 ","End":"04:06.350","Text":"is closer to B than to A."},{"Start":"04:06.350 ","End":"04:11.740","Text":"It has a greater contribution from 2SB than it has from 2SA."},{"Start":"04:11.740 ","End":"04:13.965","Text":"This is summarized here."},{"Start":"04:13.965 ","End":"04:15.910","Text":"Atom A is lower in energy,"},{"Start":"04:15.910 ","End":"04:19.400","Text":"so it contributes more to the lowest energy molecular orbital,"},{"Start":"04:19.400 ","End":"04:23.390","Text":"less than the highest energy antibonding molecular orbital."},{"Start":"04:23.390 ","End":"04:24.980","Text":"Let\u0027s take an example."},{"Start":"04:24.980 ","End":"04:26.195","Text":"Hydrogen fluoride."},{"Start":"04:26.195 ","End":"04:30.545","Text":"Recall that F is very electronegative and"},{"Start":"04:30.545 ","End":"04:35.810","Text":"H is less so this is a very polar covalent bond."},{"Start":"04:35.810 ","End":"04:39.045","Text":"Here\u0027s hydrogen, 1 electron in 1S."},{"Start":"04:39.045 ","End":"04:42.120","Text":"Here is fluorine, 2 electrons in 2s,"},{"Start":"04:42.120 ","End":"04:45.220","Text":"and 5 electrons in 2p."},{"Start":"04:46.550 ","End":"04:53.375","Text":"The unpaired electron will say it\u0027s a 2pz with 1 unpaired electron for fluorine,"},{"Start":"04:53.375 ","End":"04:56.465","Text":"1 unpaired electron for hydrogen."},{"Start":"04:56.465 ","End":"05:01.620","Text":"The unpaired electrons and hydrogen is 1s and then fluorine 2pz."},{"Start":"05:02.260 ","End":"05:07.460","Text":"Recall the electronegativity of H and F are 2.1 and 4."},{"Start":"05:07.460 ","End":"05:15.000","Text":"This is very polar bond towards F. If we look at the diagram,"},{"Start":"05:15.000 ","End":"05:22.895","Text":"here we have F on the left-hand side and H on the right-hand side."},{"Start":"05:22.895 ","End":"05:28.978","Text":"We can see that the bonding orbital is nearer"},{"Start":"05:28.978 ","End":"05:31.700","Text":"in energy to F and antibonding order is nearer in"},{"Start":"05:31.700 ","End":"05:34.910","Text":"energy to H. We have a pair of electrons,"},{"Start":"05:34.910 ","End":"05:38.424","Text":"we put it in the sigma bonding orbital."},{"Start":"05:38.424 ","End":"05:45.685","Text":"Then the bond order is just (2- 0)/2, that\u0027s 1."},{"Start":"05:45.685 ","End":"05:47.510","Text":"We have our single bond,"},{"Start":"05:47.510 ","End":"05:54.110","Text":"nearer in energy to F. We have a higher electron density on F than on H,"},{"Start":"05:54.110 ","End":"05:58.010","Text":"which is what we\u0027d expect in a polar bond."},{"Start":"05:58.010 ","End":"06:03.645","Text":"Here it\u0027s written F2pz is lower in energy than"},{"Start":"06:03.645 ","End":"06:10.060","Text":"H1s and the pair of electrons is sigma bonding orbital and the bond order is 1."},{"Start":"06:10.060 ","End":"06:13.955","Text":"Now, when we come to other molecules,"},{"Start":"06:13.955 ","End":"06:17.104","Text":"more complicated molecules like CO and NO,"},{"Start":"06:17.104 ","End":"06:19.910","Text":"then we need to calculate the energy levels scheme."},{"Start":"06:19.910 ","End":"06:23.675","Text":"Is not as simple as the case for HF."},{"Start":"06:23.675 ","End":"06:26.150","Text":"Here\u0027s energy level scheme,"},{"Start":"06:26.150 ","End":"06:30.950","Text":"and because the sigma bonds get a contribution from both the S and the P,"},{"Start":"06:30.950 ","End":"06:35.240","Text":"we just number them in increasing order of energy."},{"Start":"06:35.240 ","End":"06:37.885","Text":"We have 1 sigma,"},{"Start":"06:37.885 ","End":"06:40.710","Text":"2 sigma antibonding,"},{"Start":"06:40.710 ","End":"06:43.665","Text":"2 orbitals, 1 Pi."},{"Start":"06:43.665 ","End":"06:46.485","Text":"There were 3 sigma orbital,"},{"Start":"06:46.485 ","End":"06:50.855","Text":"then 2 antibonding pi orbitals,"},{"Start":"06:50.855 ","End":"06:54.275","Text":"and 1 antibonding sigma orbital."},{"Start":"06:54.275 ","End":"06:56.545","Text":"Here\u0027s the energy scheme."},{"Start":"06:56.545 ","End":"07:05.285","Text":"Let\u0027s start off with CO. CO has 4 valence electrons in carbon, 6 in oxygen."},{"Start":"07:05.285 ","End":"07:08.135","Text":"Total of 10 electrons."},{"Start":"07:08.135 ","End":"07:09.639","Text":"Let\u0027s fill up the orbitals."},{"Start":"07:09.639 ","End":"07:11.910","Text":"We have 2 here,"},{"Start":"07:11.910 ","End":"07:14.616","Text":"2 the antibonding, 2,4,"},{"Start":"07:14.616 ","End":"07:22.790","Text":"and then the final 2 in the sigma bonding orbital."},{"Start":"07:22.790 ","End":"07:24.335","Text":"Here\u0027s 2, 4, 6,"},{"Start":"07:24.335 ","End":"07:26.390","Text":"8, 10 electrons."},{"Start":"07:26.390 ","End":"07:28.715","Text":"Now, what\u0027s the bond order?"},{"Start":"07:28.715 ","End":"07:37.385","Text":"Here from these 2 orbitals is 0 because of 2 bonding electrons, 2 antibonding electrons."},{"Start":"07:37.385 ","End":"07:40.790","Text":"Then we have 6 bonding electrons here."},{"Start":"07:40.790 ","End":"07:47.400","Text":"The bond order is (6-0= 6)/ 2= 3."},{"Start":"07:47.400 ","End":"07:49.455","Text":"The bond order is 3."},{"Start":"07:49.455 ","End":"07:51.825","Text":"Let\u0027s go on to NO."},{"Start":"07:51.825 ","End":"07:54.905","Text":"Now, nitrogen has 5 valence electrons,"},{"Start":"07:54.905 ","End":"07:56.900","Text":"and oxygen has 6."},{"Start":"07:56.900 ","End":"07:59.495","Text":"Now we\u0027re talking about 11 electrons."},{"Start":"07:59.495 ","End":"08:05.570","Text":"It\u0027s the same as before with an extra electron in an antibonding orbital."},{"Start":"08:05.570 ","End":"08:09.845","Text":"Now we have 6 bonding electrons,"},{"Start":"08:09.845 ","End":"08:14.435","Text":"1 antibonding electron that\u0027s 6 -1= 5,"},{"Start":"08:14.435 ","End":"08:19.950","Text":"we divide by 2, 5/2= 2.5."},{"Start":"08:19.950 ","End":"08:26.785","Text":"We can see that the bond order is lower in NO than it is in CO."},{"Start":"08:26.785 ","End":"08:32.490","Text":"In this video we discussed heteronuclear diatomic molecules."}],"ID":23897},{"Watched":false,"Name":"Exercise 1","Duration":"6m 17s","ChapterTopicVideoID":23572,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"Hi. We\u0027re going to solve the following exercise,"},{"Start":"00:03.720 ","End":"00:08.475","Text":"sketch the molecular orbital diagrams for the following species."},{"Start":"00:08.475 ","End":"00:10.305","Text":"In a and b,"},{"Start":"00:10.305 ","End":"00:15.443","Text":"we have homonuclear second period diatomic molecules"},{"Start":"00:15.443 ","End":"00:18.030","Text":"and you want to sketch their molecular orbital diagrams."},{"Start":"00:18.030 ","End":"00:20.550","Text":"Before we sketch the molecular orbital diagrams,"},{"Start":"00:20.550 ","End":"00:25.425","Text":"we first need to determine the order of the molecular orbitals in terms of their energy."},{"Start":"00:25.425 ","End":"00:30.250","Text":"We have 2 cases and we\u0027re going to look at the first case."},{"Start":"00:30.320 ","End":"00:33.740","Text":"The first case is appropriate for"},{"Start":"00:33.740 ","End":"00:39.210","Text":"the fluorine diatomic molecule and also for the oxygen diatomic molecule."},{"Start":"00:40.150 ","End":"00:48.299","Text":"As we said, the first case is appropriate for the fluorine and oxygen diatomic molecules."},{"Start":"00:48.299 ","End":"00:50.060","Text":"If we take a look at our diagram,"},{"Start":"00:50.060 ","End":"00:56.180","Text":"we have 2 atomic 2s orbitals which combine and form 2 molecular orbitals."},{"Start":"00:56.180 ","End":"00:58.310","Text":"We have Sigma 2s and Sigma_2s-star."},{"Start":"00:58.310 ","End":"01:01.820","Text":"Sigma_ 2s is bonding and Sigma_2s-star is antibonding."},{"Start":"01:01.820 ","End":"01:04.610","Text":"Next we have our 2p orbitals."},{"Start":"01:04.610 ","End":"01:08.790","Text":"Now remember we have 3 degenerate orbitals in"},{"Start":"01:08.790 ","End":"01:14.046","Text":"each 2p meaning we have 6 atomic orbitals in all,"},{"Start":"01:14.046 ","End":"01:17.765","Text":"therefore we\u0027re going to also have 6 molecular orbitals."},{"Start":"01:17.765 ","End":"01:18.890","Text":"Let\u0027s take a look at"},{"Start":"01:18.890 ","End":"01:25.685","Text":"the molecular orbitals that we achieve from combining the 2p orbitals."},{"Start":"01:25.685 ","End":"01:29.840","Text":"First, in the case again a fluorine and oxygen."},{"Start":"01:29.840 ","End":"01:32.750","Text":"The lowest molecular orbital;"},{"Start":"01:32.750 ","End":"01:35.510","Text":"the lowest energy molecular orbital, is Sigma_2p."},{"Start":"01:35.510 ","End":"01:39.080","Text":"Then we have p_2p,"},{"Start":"01:39.080 ","End":"01:44.772","Text":"then we have p_2p-star and Sigma_2p-star."},{"Start":"01:44.772 ","End":"01:48.110","Text":"Now, if we look at our p_2p and p_2p-star orbitals,"},{"Start":"01:48.110 ","End":"01:52.830","Text":"you can see that each 1 actually has 2 degenerate orbitals."},{"Start":"01:53.290 ","End":"01:57.530","Text":"Again, if we take a look at the order of the energy of the molecular orbitals,"},{"Start":"01:57.530 ","End":"02:01.925","Text":"we have our Sigma_2s orbital which is the lowest in energy."},{"Start":"02:01.925 ","End":"02:04.865","Text":"As we go out, we actually increase the energy."},{"Start":"02:04.865 ","End":"02:07.270","Text":"Next is the Sigma_2-star,"},{"Start":"02:07.270 ","End":"02:09.755","Text":"and then higher in energy is the Sigma_2p,"},{"Start":"02:09.755 ","End":"02:11.435","Text":"and higher energy is the Pi_2p."},{"Start":"02:11.435 ","End":"02:13.499","Text":"Again, 2 degenerate orbitals,"},{"Start":"02:13.499 ","End":"02:17.109","Text":"then we get Pi_2p-star and Sigma_2p-star."},{"Start":"02:17.109 ","End":"02:19.550","Text":"After we have our molecular orbitals,"},{"Start":"02:19.550 ","End":"02:25.205","Text":"the next step is to take our electrons and fill them in."},{"Start":"02:25.205 ","End":"02:28.715","Text":"Again, this is a fluorine molecule."},{"Start":"02:28.715 ","End":"02:33.129","Text":"We have 2 fluorine molecules and each 1 has 7 valence electrons,"},{"Start":"02:33.129 ","End":"02:36.530","Text":"so that\u0027s going to give us 14 electrons to fill in."},{"Start":"02:36.530 ","End":"02:39.980","Text":"We begin filling in the orbital which is lowest in energy,"},{"Start":"02:39.980 ","End":"02:42.065","Text":"which is the Sigma_2s orbital."},{"Start":"02:42.065 ","End":"02:45.050","Text":"In each orbital remember that we can only fill in 2 electrons,"},{"Start":"02:45.050 ","End":"02:47.220","Text":"so that\u0027s going to be 1, 2."},{"Start":"02:47.770 ","End":"02:53.247","Text":"Again, we begin in the lowest energy orbital and we go up increasing in energy,"},{"Start":"02:53.247 ","End":"02:55.340","Text":"so we have 2 in our Sigma_2s."},{"Start":"02:55.340 ","End":"02:59.680","Text":"Two in our Sigma_2s-star is going to give us 4."},{"Start":"02:59.680 ","End":"03:03.975","Text":"Two in our Sigma_2p, that\u0027s going to give us 6."},{"Start":"03:03.975 ","End":"03:07.070","Text":"Next we reach our Pi_2 p orbitals."},{"Start":"03:07.070 ","End":"03:09.953","Text":"Then remember here we have 2 degenerate orbitals,"},{"Start":"03:09.953 ","End":"03:11.780","Text":"so we fill them in one at a time."},{"Start":"03:11.780 ","End":"03:15.900","Text":"That\u0027s 1 electron, 2 electrons, 3 and 4."},{"Start":"03:15.900 ","End":"03:21.810","Text":"This gives us a total of 10 electrons since we had 6 and we added another force."},{"Start":"03:21.810 ","End":"03:23.415","Text":"That\u0027s going to give us 10 electrons,"},{"Start":"03:23.415 ","End":"03:24.690","Text":"and now we\u0027re missing 4."},{"Start":"03:24.690 ","End":"03:28.100","Text":"We\u0027re going to go to our Pi_2p -star and we\u0027re going to fill in 4."},{"Start":"03:28.100 ","End":"03:32.970","Text":"We\u0027re going to fill in 1, 2, 3, 4."},{"Start":"03:33.820 ","End":"03:36.650","Text":"We filled in all our valence electrons,"},{"Start":"03:36.650 ","End":"03:41.850","Text":"and that\u0027s actually our molecular orbital for the fluorine."},{"Start":"03:42.430 ","End":"03:45.665","Text":"We\u0027re going to take a look at b."},{"Start":"03:45.665 ","End":"03:50.587","Text":"As I said, the order of the molecular orbitals are in terms of energy."},{"Start":"03:50.587 ","End":"03:52.790","Text":"In our first case it was appropriate for"},{"Start":"03:52.790 ","End":"03:59.343","Text":"the fluorine and oxygen molecules, diatomic molecules."},{"Start":"03:59.343 ","End":"04:04.280","Text":"We\u0027re going to take a look at another case of molecular orbitals,"},{"Start":"04:04.280 ","End":"04:07.805","Text":"and this is appropriate for the lighter elements."},{"Start":"04:07.805 ","End":"04:11.975","Text":"For example, the nitrogen and the carbon diatomic molecules."},{"Start":"04:11.975 ","End":"04:20.205","Text":"In b we have to sketch a molecular orbital diagram for carbon_2 plus ion."},{"Start":"04:20.205 ","End":"04:25.915","Text":"We\u0027re going to take a look at the order of the molecular orbitals for our second case."},{"Start":"04:25.915 ","End":"04:31.360","Text":"As you can see here, we also have 2 2s atomic orbitals which"},{"Start":"04:31.360 ","End":"04:37.075","Text":"combine to give us 2 molecular orbitals; Sigma_2s and Sigma_2s-star."},{"Start":"04:37.075 ","End":"04:42.594","Text":"Next we have the 2p orbitals which again are 6 orbitals,"},{"Start":"04:42.594 ","End":"04:47.020","Text":"and they combine to form 6 molecular orbitals."},{"Start":"04:47.020 ","End":"04:48.040","Text":"However, in this case,"},{"Start":"04:48.040 ","End":"04:51.490","Text":"the Sigma_2p and the Pi_2p orbitals are reversed."},{"Start":"04:51.490 ","End":"04:57.595","Text":"The lower in energy orbitals are the Pi_2p and then we have the Sigma_2p."},{"Start":"04:57.595 ","End":"05:00.700","Text":"Remember in the first case which is appropriate for"},{"Start":"05:00.700 ","End":"05:03.740","Text":"the fluorine and oxygen molecules, this was reversed."},{"Start":"05:03.740 ","End":"05:08.080","Text":"We first had our Sigma_2p orbitals and then the Pi_2p."},{"Start":"05:08.080 ","End":"05:10.860","Text":"We continue regularly,"},{"Start":"05:10.860 ","End":"05:13.950","Text":"we have our Pi_2p-star and Sigma_2p-star."},{"Start":"05:13.950 ","End":"05:20.000","Text":"Remember that the Pi_2p and Pi_2p-star are both formed from 2 degenerate orbitals."},{"Start":"05:20.380 ","End":"05:26.075","Text":"As we said the second step is to determine how many electrons we have."},{"Start":"05:26.075 ","End":"05:28.220","Text":"We have 2 carbon molecules,"},{"Start":"05:28.220 ","End":"05:33.485","Text":"meaning this is going to be 2 times 4 since carbon has 4 valence electrons,"},{"Start":"05:33.485 ","End":"05:39.115","Text":"and we\u0027re going to subtract 1 since this is a plus ion."},{"Start":"05:39.115 ","End":"05:42.435","Text":"2 times 4 minus 1 is going to give us 7,"},{"Start":"05:42.435 ","End":"05:44.775","Text":"meaning we have 7 electrons to fill in."},{"Start":"05:44.775 ","End":"05:47.055","Text":"We\u0027re going to start with the a Sigma_2s orbitals,"},{"Start":"05:47.055 ","End":"05:50.000","Text":"so we\u0027re going to fill into and then we\u0027re going to"},{"Start":"05:50.000 ","End":"05:53.690","Text":"fill in our Sigma_2s-star another 2 giving us 4."},{"Start":"05:53.690 ","End":"05:55.655","Text":"We have only 3 left."},{"Start":"05:55.655 ","End":"06:00.464","Text":"This we\u0027re going to fill in for that Pi_2p degenerate orbital,"},{"Start":"06:00.464 ","End":"06:03.950","Text":"so we\u0027re going to fill them in one at a time,"},{"Start":"06:03.950 ","End":"06:06.060","Text":"so that\u0027s 1, 2, 3."},{"Start":"06:06.500 ","End":"06:11.345","Text":"That\u0027s all our electrons for our C_2 plus ion,"},{"Start":"06:11.345 ","End":"06:13.835","Text":"and this is our orbital diagram."},{"Start":"06:13.835 ","End":"06:15.410","Text":"That is our final answer."},{"Start":"06:15.410 ","End":"06:17.940","Text":"Thank you very much for watching."}],"ID":24481},{"Watched":false,"Name":"Exercise 2","Duration":"5m 6s","ChapterTopicVideoID":23573,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.105","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.105 ","End":"00:07.905","Text":"A. Calculate the bond order of O2 using the molecular orbital diagram."},{"Start":"00:07.905 ","End":"00:09.390","Text":"B, determine the number of"},{"Start":"00:09.390 ","End":"00:14.740","Text":"unpaired electrons and C is the molecule paramagnetic or diamagnetic."},{"Start":"00:15.560 ","End":"00:20.470","Text":"We\u0027re going to start by sketching the molecular orbital diagram."},{"Start":"00:21.650 ","End":"00:25.245","Text":"Now, we have our molecular orbitals."},{"Start":"00:25.245 ","End":"00:32.205","Text":"Now, remember that for oxygen and for the fluorine diatomic molecules,"},{"Start":"00:32.205 ","End":"00:39.015","Text":"the sigma 2p molecular orbital is lower in energy than the Pi 2p molecular orbital."},{"Start":"00:39.015 ","End":"00:41.890","Text":"This is the diagram we\u0027re going to use."},{"Start":"00:41.960 ","End":"00:44.210","Text":"Now the next step,"},{"Start":"00:44.210 ","End":"00:46.820","Text":"remember, is to fill in the electrons."},{"Start":"00:46.820 ","End":"00:49.055","Text":"First, we need to determine the number of electrons."},{"Start":"00:49.055 ","End":"00:51.679","Text":"We have 2 oxygens."},{"Start":"00:51.679 ","End":"00:55.400","Text":"We have 2 times 6 valence electrons for each oxygen,"},{"Start":"00:55.400 ","End":"00:58.370","Text":"which gives us 12 valence electrons to fill in."},{"Start":"00:58.370 ","End":"01:02.300","Text":"We\u0027re going to start filling in with the lowest energy molecular orbital."},{"Start":"01:02.300 ","End":"01:06.180","Text":"That\u0027s 2 electrons in sigma 2s,"},{"Start":"01:06.180 ","End":"01:07.875","Text":"another 2 in sigma 2s star,"},{"Start":"01:07.875 ","End":"01:10.140","Text":"that\u0027s going to give us 4."},{"Start":"01:10.140 ","End":"01:13.570","Text":"Then we\u0027re going to fill in 2 in sigma 2p."},{"Start":"01:13.640 ","End":"01:16.545","Text":"Which is going to give us a 6."},{"Start":"01:16.545 ","End":"01:18.375","Text":"Then in the Pi 2p,"},{"Start":"01:18.375 ","End":"01:19.965","Text":"we\u0027re going to fill them in 1 at a time."},{"Start":"01:19.965 ","End":"01:27.375","Text":"That\u0027s 7, 8, 9, 10."},{"Start":"01:27.375 ","End":"01:32.015","Text":"Because remember we have 2 degenerate orbitals and the same as in 2p."},{"Start":"01:32.015 ","End":"01:34.785","Text":"The same is in the Pi 2p star."},{"Start":"01:34.785 ","End":"01:37.500","Text":"We filled in 10, we have another 2 electrons."},{"Start":"01:37.500 ","End":"01:39.210","Text":"We\u0027re going to fill 1 in each orbital,"},{"Start":"01:39.210 ","End":"01:41.950","Text":"so that\u0027s going to be 1, 2."},{"Start":"01:43.030 ","End":"01:49.770","Text":"Now we have our molecular orbital diagram and we have to calculate the bond order."},{"Start":"01:50.050 ","End":"01:55.640","Text":"Now, remember that the bond order is the number of electrons in the bonding"},{"Start":"01:55.640 ","End":"01:58.220","Text":"molecular orbitals minus the number of electrons in"},{"Start":"01:58.220 ","End":"02:01.685","Text":"the antibonding molecular orbitals divided by 2."},{"Start":"02:01.685 ","End":"02:07.670","Text":"If we take a look first of all at our 2s molecular orbitals,"},{"Start":"02:07.670 ","End":"02:09.635","Text":"we have Sigma 2s and sigma 2s star."},{"Start":"02:09.635 ","End":"02:14.540","Text":"Sigma 2s is a bonding orbital and sigma 2s is our anti-bonding orbital."},{"Start":"02:14.540 ","End":"02:16.775","Text":"Each has 2 electrons,"},{"Start":"02:16.775 ","End":"02:19.010","Text":"meaning they\u0027ll cancel each other out."},{"Start":"02:19.010 ","End":"02:23.030","Text":"We can just take a look at our 2p electrons."},{"Start":"02:23.030 ","End":"02:25.340","Text":"Now as you can see, we have 6 electrons 1,"},{"Start":"02:25.340 ","End":"02:26.990","Text":"2, 3, 4, 5,"},{"Start":"02:26.990 ","End":"02:30.020","Text":"6 is in our bonding molecular orbitals,"},{"Start":"02:30.020 ","End":"02:33.635","Text":"meaning in our Sigma 2p and Pi 2p orbitals."},{"Start":"02:33.635 ","End":"02:42.950","Text":"That\u0027s going to be bond order equals 6 because that\u0027s the number"},{"Start":"02:42.950 ","End":"02:45.755","Text":"of electrons in our bonding"},{"Start":"02:45.755 ","End":"02:53.275","Text":"molecular orbitals minus the number of electrons in our antibonding molecular orbitals."},{"Start":"02:53.275 ","End":"02:57.360","Text":"Our antibonding molecular orbitals are sigma 2p star and Pi 2p star."},{"Start":"02:57.360 ","End":"02:58.980","Text":"As we can see, we have 2 electrons here,"},{"Start":"02:58.980 ","End":"03:03.345","Text":"so it\u0027s going to be minus 2 and all of this is divided by 2."},{"Start":"03:03.345 ","End":"03:06.465","Text":"This gives us a 4 divided by 2."},{"Start":"03:06.465 ","End":"03:10.270","Text":"This gives us a 2."},{"Start":"03:11.510 ","End":"03:14.290","Text":"Our bond order comes out to 2,"},{"Start":"03:14.290 ","End":"03:16.520","Text":"meaning that we have"},{"Start":"03:16.520 ","End":"03:26.075","Text":"a double bond in our oxygen diatomic molecule."},{"Start":"03:26.075 ","End":"03:31.170","Text":"That\u0027s the bond order of the oxygen, that\u0027s a."},{"Start":"03:31.550 ","End":"03:36.380","Text":"Now in B, we have to determine the number of unpaired electrons that we have."},{"Start":"03:36.380 ","End":"03:39.860","Text":"If we take a look at our molecular orbital diagram,"},{"Start":"03:39.860 ","End":"03:48.400","Text":"we can see that in the Pi 2p star orbitals we have 2 unpaired electrons."},{"Start":"03:48.590 ","End":"03:52.559","Text":"B is going to be 2 unpaired electrons."},{"Start":"03:52.559 ","End":"03:57.095","Text":"In C the molecule is paramagnetic or diamagnetic."},{"Start":"03:57.095 ","End":"03:59.970","Text":"The answer is paramagnetic."},{"Start":"04:03.850 ","End":"04:08.765","Text":"We know this because the molecules has unpaired electrons."},{"Start":"04:08.765 ","End":"04:13.460","Text":"This is very important since the molecular orbital theory is"},{"Start":"04:13.460 ","End":"04:18.560","Text":"the first theory to predict that the oxygen molecule,"},{"Start":"04:18.560 ","End":"04:21.365","Text":"the diatomic molecule is paramagnetic."},{"Start":"04:21.365 ","End":"04:25.215","Text":"As opposed to the valence bond theory or Lewis theory,"},{"Start":"04:25.215 ","End":"04:26.920","Text":"which did not predict this."},{"Start":"04:26.920 ","End":"04:31.140","Text":"It actually predicted that the molecule was diamagnetic."},{"Start":"04:31.420 ","End":"04:35.210","Text":"Again, our answer for A is that our bond order equals 2,"},{"Start":"04:35.210 ","End":"04:37.970","Text":"meaning that oxygen has"},{"Start":"04:37.970 ","End":"04:44.225","Text":"a double bond as we saw earlier in the Lewis theory and the valence bond theory."},{"Start":"04:44.225 ","End":"04:46.670","Text":"Next, we determine the number of unpaired electrons,"},{"Start":"04:46.670 ","End":"04:53.435","Text":"which is 2 unpaired electrons in the Pi 2p star orbitals and we know that the molecule,"},{"Start":"04:53.435 ","End":"04:58.235","Text":"we predict the molecule is paramagnetic since it has unpaired electrons."},{"Start":"04:58.235 ","End":"05:02.610","Text":"Again, this is the first theory that predicts this."},{"Start":"05:03.010 ","End":"05:05.480","Text":"These are our final answers."},{"Start":"05:05.480 ","End":"05:07.950","Text":"Thank you very much for watching."}],"ID":24482},{"Watched":false,"Name":"Exercise 3","Duration":"6m 50s","ChapterTopicVideoID":23574,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:03.840","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.840 ","End":"00:07.500","Text":"Calculate the bond order of the following ions and determine"},{"Start":"00:07.500 ","End":"00:11.590","Text":"if they are stable using the molecular orbital diagram."},{"Start":"00:11.870 ","End":"00:19.920","Text":"Let\u0027s start with a. In a we have a diatomic fluorine ion."},{"Start":"00:19.920 ","End":"00:24.090","Text":"Now instead of sketching the whole molecular orbital diagram,"},{"Start":"00:24.090 ","End":"00:26.625","Text":"we\u0027re just going to take a look at the actual orbitals."},{"Start":"00:26.625 ","End":"00:31.680","Text":"Remember the order of the molecular orbitals in terms of energy."},{"Start":"00:31.680 ","End":"00:36.082","Text":"For fluorine and oxygen diatomic molecules,"},{"Start":"00:36.082 ","End":"00:40.350","Text":"we begin with our Sigma 2s orbital,"},{"Start":"00:40.350 ","End":"00:43.570","Text":"then we have our Sigma 2s star,"},{"Start":"00:43.570 ","End":"00:47.410","Text":"which is our anti-bonding orbital."},{"Start":"00:47.410 ","End":"00:52.000","Text":"Then next in energy we have Sigma 2p."},{"Start":"00:52.370 ","End":"00:57.100","Text":"Then we have 2 degenerate Pi 2p orbitals."},{"Start":"00:57.950 ","End":"01:02.990","Text":"Then we reach our Pi 2p star orbitals,"},{"Start":"01:02.990 ","End":"01:05.495","Text":"again to degenerate once."},{"Start":"01:05.495 ","End":"01:07.160","Text":"Last but not least,"},{"Start":"01:07.160 ","End":"01:11.370","Text":"we have a Sigma 2p star orbital."},{"Start":"01:12.650 ","End":"01:19.307","Text":"We have 2 fluorines, meaning we have 2 times 7 valence electrons in each fluorine,"},{"Start":"01:19.307 ","End":"01:21.860","Text":"but we have a plus 2 ion,"},{"Start":"01:21.860 ","End":"01:23.825","Text":"meaning we have to take off 2 electrons."},{"Start":"01:23.825 ","End":"01:27.060","Text":"This is going to give us 14 minus 2,"},{"Start":"01:27.060 ","End":"01:30.275","Text":"which is going to give us 12 electrons to fill in."},{"Start":"01:30.275 ","End":"01:33.920","Text":"Let\u0027s fill in. We begin with our Sigma 2s."},{"Start":"01:33.920 ","End":"01:39.955","Text":"We have 2, 4, 6,"},{"Start":"01:39.955 ","End":"01:43.025","Text":"7, 8,"},{"Start":"01:43.025 ","End":"01:47.860","Text":"9, 10, 11, 12."},{"Start":"01:47.860 ","End":"01:49.970","Text":"Now remember in our degenerate orbitals,"},{"Start":"01:49.970 ","End":"01:52.730","Text":"which are in the Pi 2p and a Pi 2p star,"},{"Start":"01:52.730 ","End":"01:57.100","Text":"we fill in 1 electron in each orbital every time."},{"Start":"01:57.310 ","End":"02:01.700","Text":"Now we have to calculate the bond order and we have to"},{"Start":"02:01.700 ","End":"02:06.140","Text":"determine if a molecule or ion is stable."},{"Start":"02:06.140 ","End":"02:08.410","Text":"In this case our ion of course."},{"Start":"02:08.410 ","End":"02:15.140","Text":"Again, remember that the bond order equals the number of electrons"},{"Start":"02:15.140 ","End":"02:19.310","Text":"in a bonding molecular orbitals minus"},{"Start":"02:19.310 ","End":"02:20.540","Text":"the number of electrons in"},{"Start":"02:20.540 ","End":"02:24.830","Text":"the anti-bonding molecular orbitals and all this is divided by 2."},{"Start":"02:24.830 ","End":"02:28.415","Text":"We can leave our Sigma 2s as Sigma 2s star"},{"Start":"02:28.415 ","End":"02:32.490","Text":"orbitals since we have 2 in each and they cancel each other out."},{"Start":"02:32.490 ","End":"02:36.645","Text":"We can look at just at our 2p molecular orbitals."},{"Start":"02:36.645 ","End":"02:41.265","Text":"We have our Sigma 2p and our Pi 2p,"},{"Start":"02:41.265 ","End":"02:43.080","Text":"which are bonding orbitals."},{"Start":"02:43.080 ","End":"02:45.435","Text":"We have 6 electrons there."},{"Start":"02:45.435 ","End":"02:48.300","Text":"In our Pi 2p star and Sigma 2p star,"},{"Start":"02:48.300 ","End":"02:50.610","Text":"which are anti-bonding molecular orbitals,"},{"Start":"02:50.610 ","End":"02:52.065","Text":"we have only 2 electrons."},{"Start":"02:52.065 ","End":"02:54.495","Text":"This is going to be 6."},{"Start":"02:54.495 ","End":"03:01.850","Text":"Again, which is the number of electrons in our bonding molecular orbitals minus 2,"},{"Start":"03:01.850 ","End":"03:06.680","Text":"which is the number of electrons in our anti-bonding molecular orbitals."},{"Start":"03:06.680 ","End":"03:09.035","Text":"This is going to be divided by 2."},{"Start":"03:09.035 ","End":"03:12.965","Text":"6 minus 2 is going to give us 4 divided by 2 equals 2,"},{"Start":"03:12.965 ","End":"03:17.130","Text":"meaning this is a double bond."},{"Start":"03:20.450 ","End":"03:26.800","Text":"Our bond order for our fluorine ion is 2."},{"Start":"03:26.820 ","End":"03:32.290","Text":"This molecule is stable since the number of electrons in the bonding orbitals"},{"Start":"03:32.290 ","End":"03:37.015","Text":"is greater than number of electrons in the anti-bonding molecular orbitals."},{"Start":"03:37.015 ","End":"03:39.710","Text":"Now we can take a look at b."},{"Start":"03:55.640 ","End":"04:01.970","Text":"In b, if we take a look at our molecular orbital since we\u0027re dealing with carbon,"},{"Start":"04:03.680 ","End":"04:09.470","Text":"we\u0027re going to start also with our Sigma 2s bonding orbital."},{"Start":"04:09.470 ","End":"04:13.260","Text":"The next in energy is going to be a Sigma 2s star."},{"Start":"04:14.000 ","End":"04:20.800","Text":"However, the next instead of Sigma 2p is going to be our Pi 2p degenerate orbitals."},{"Start":"04:21.050 ","End":"04:24.840","Text":"Then our Sigma 2p orbital,"},{"Start":"04:24.840 ","End":"04:31.740","Text":"meaning that Pi 2p and Sigma 2p orbitals are reversed in this case."},{"Start":"04:31.740 ","End":"04:37.325","Text":"Then we have the Pi 2p star degenerate orbitals"},{"Start":"04:37.325 ","End":"04:43.450","Text":"and a Sigma 2p star orbital."},{"Start":"04:43.450 ","End":"04:48.905","Text":"Now we need to determine the number of electrons and we need to fill them in."},{"Start":"04:48.905 ","End":"04:50.990","Text":"We have 2 carbons,"},{"Start":"04:50.990 ","End":"04:55.380","Text":"meaning 2 times 4 valence electrons each."},{"Start":"04:55.380 ","End":"04:56.805","Text":"That\u0027s going to give us 8."},{"Start":"04:56.805 ","End":"04:59.600","Text":"However, this is ion we have a minus 2 charge,"},{"Start":"04:59.600 ","End":"05:02.495","Text":"meaning we have to add 2 electrons."},{"Start":"05:02.495 ","End":"05:05.240","Text":"8 plus 2 is going to give us 10 electrons,"},{"Start":"05:05.240 ","End":"05:07.520","Text":"and we\u0027re going to start filling them in."},{"Start":"05:07.520 ","End":"05:09.425","Text":"That\u0027s 1, 2,"},{"Start":"05:09.425 ","End":"05:12.740","Text":"3, 4, 5,"},{"Start":"05:12.740 ","End":"05:14.480","Text":"6, 7,"},{"Start":"05:14.480 ","End":"05:19.350","Text":"8, 9, 10."},{"Start":"05:19.660 ","End":"05:22.715","Text":"We filled in all of our electrons."},{"Start":"05:22.715 ","End":"05:25.740","Text":"Now we\u0027re going to find the bond order."},{"Start":"05:29.990 ","End":"05:35.290","Text":"Now remember again, the bond order is the number of electrons in"},{"Start":"05:35.290 ","End":"05:37.600","Text":"our bonding molecular orbitals minus the number of"},{"Start":"05:37.600 ","End":"05:40.940","Text":"electrons in the anti-bonding molecular orbitals."},{"Start":"05:40.970 ","End":"05:49.390","Text":"Now again, we can just look at our 2p orbitals since our Sigma 2s are both full."},{"Start":"05:50.090 ","End":"05:52.530","Text":"We have 1, 2, 3,"},{"Start":"05:52.530 ","End":"05:55.170","Text":"4, 5, 6,"},{"Start":"05:55.170 ","End":"05:59.730","Text":"6 electrons in our bonding orbitals because"},{"Start":"05:59.730 ","End":"06:03.900","Text":"our Pi 2p and Sigma 2p are the bonding orbitals minus,"},{"Start":"06:03.900 ","End":"06:09.940","Text":"as you can see, there are 0 electrons in anti-bonding."},{"Start":"06:10.340 ","End":"06:14.715","Text":"All this is going to be divided by 2."},{"Start":"06:14.715 ","End":"06:17.340","Text":"This is going to equal 3,"},{"Start":"06:17.340 ","End":"06:19.725","Text":"6 minus 0 divided by 2 equals 3."},{"Start":"06:19.725 ","End":"06:23.430","Text":"3 means that there is a triple bond."},{"Start":"06:26.710 ","End":"06:30.650","Text":"In this case, the number of bonding molecular orbitals is"},{"Start":"06:30.650 ","End":"06:34.340","Text":"greater than the number of electrons in the anti-bonding molecular orbitals."},{"Start":"06:34.340 ","End":"06:38.340","Text":"Therefore this ion is also stable."},{"Start":"06:46.120 ","End":"06:48.290","Text":"That is our final answer."},{"Start":"06:48.290 ","End":"06:50.880","Text":"Thank you very much for watching."}],"ID":24483},{"Watched":false,"Name":"Exercise 4","Duration":"10m 25s","ChapterTopicVideoID":23570,"CourseChapterTopicPlaylistID":101323,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.934","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.934 ","End":"00:07.465","Text":"Explain why the oxidation of O_2 decreases the bond distance,"},{"Start":"00:07.465 ","End":"00:11.575","Text":"whereas the oxidation of N_2 increases the bond distance."},{"Start":"00:11.575 ","End":"00:14.515","Text":"First, let\u0027s take a look at oxygen."},{"Start":"00:14.515 ","End":"00:17.781","Text":"We\u0027re going to write our molecular orbitals, again,"},{"Start":"00:17.781 ","End":"00:22.510","Text":"without actually sketching a molecular orbital diagram because it\u0027s not needed."},{"Start":"00:22.510 ","End":"00:28.180","Text":"In the order of increasing energy is to start with our Sigma_2s orbital,"},{"Start":"00:28.180 ","End":"00:31.225","Text":"then we have our Sigma_2s^star."},{"Start":"00:31.225 ","End":"00:35.240","Text":"Next one is our Sigma_2p,"},{"Start":"00:35.360 ","End":"00:40.095","Text":"then our Pi_2p, which is 2 degenerate orbitals."},{"Start":"00:40.095 ","End":"00:47.920","Text":"Then our Pi_2p^star and our Sigma_2p^star."},{"Start":"00:49.320 ","End":"00:51.760","Text":"Now the reason that we\u0027re looking at"},{"Start":"00:51.760 ","End":"00:55.000","Text":"our molecular orbitals and we\u0027re going to fill them in is that we could calculate"},{"Start":"00:55.000 ","End":"01:00.859","Text":"our bond order and then we\u0027ll be able to see what happens to our bond distance."},{"Start":"01:00.960 ","End":"01:03.730","Text":"First, we\u0027re going to look at O_2,"},{"Start":"01:03.730 ","End":"01:08.350","Text":"oxygen. We have 2 oxygens."},{"Start":"01:08.350 ","End":"01:11.620","Text":"Remember that each oxygen has 6 valence electrons,"},{"Start":"01:11.620 ","End":"01:13.360","Text":"meaning we have 12."},{"Start":"01:13.360 ","End":"01:16.405","Text":"We\u0027re going to fill them in 2,"},{"Start":"01:16.405 ","End":"01:19.690","Text":"3, 4, 5,"},{"Start":"01:19.690 ","End":"01:22.720","Text":"6, 7, 8,"},{"Start":"01:22.720 ","End":"01:27.950","Text":"9, 10, 11, 12."},{"Start":"01:29.090 ","End":"01:39.355","Text":"Now let\u0027s calculate our bond order for oxygen."},{"Start":"01:39.355 ","End":"01:44.990","Text":"Again, the bond order is calculated by taking the number of electrons in"},{"Start":"01:44.990 ","End":"01:47.780","Text":"the bonding molecular orbitals"},{"Start":"01:47.780 ","End":"01:53.065","Text":"minus the number of electrons in the anti-bonding molecular orbitals divided by 2."},{"Start":"01:53.065 ","End":"01:56.810","Text":"In the exercises before I told you that"},{"Start":"01:56.810 ","End":"02:03.050","Text":"since we have 2 electrons and Sigma_2s and 2 electrons in Sigma_2s^star,"},{"Start":"02:03.050 ","End":"02:05.390","Text":"you don\u0027t even have to take it into account,"},{"Start":"02:05.390 ","End":"02:06.553","Text":"but you can if you want to,"},{"Start":"02:06.553 ","End":"02:08.330","Text":"so let\u0027s do it this way this time."},{"Start":"02:08.330 ","End":"02:11.300","Text":"We have our bonding molecular orbitals,"},{"Start":"02:11.300 ","End":"02:13.490","Text":"which are Sigma_2s,"},{"Start":"02:13.490 ","End":"02:16.740","Text":"Sigma_2p, and our Pi_2p."},{"Start":"02:16.740 ","End":"02:18.900","Text":"In these orbitals we have 1,"},{"Start":"02:18.900 ","End":"02:20.040","Text":"2, 3,"},{"Start":"02:20.040 ","End":"02:21.315","Text":"4, 5, 6, 7,"},{"Start":"02:21.315 ","End":"02:27.065","Text":"8 electrons minus the number of electrons in our anti-bonding."},{"Start":"02:27.065 ","End":"02:29.540","Text":"Let\u0027s see. We have 2 electrons in"},{"Start":"02:29.540 ","End":"02:34.190","Text":"our Sigma_2s^star and 2 more electrons in our Pi_2p^star,"},{"Start":"02:34.190 ","End":"02:39.350","Text":"so that\u0027s 4 electrons since we have non in our Sigma_2p^star,"},{"Start":"02:39.350 ","End":"02:40.835","Text":"and this is divided by 2."},{"Start":"02:40.835 ","End":"02:43.740","Text":"We have 4 divided by 2 and this is going to give us a 2."},{"Start":"02:43.740 ","End":"02:46.620","Text":"This is a double bond."},{"Start":"02:49.970 ","End":"02:55.970","Text":"That\u0027s our bond order for our O_2 molecule."},{"Start":"02:55.970 ","End":"03:01.370","Text":"Now, we want to see what happens when we oxidize this molecule."},{"Start":"03:01.370 ","End":"03:04.980","Text":"We\u0027re going to oxidize it to O_2^plus."},{"Start":"03:05.180 ","End":"03:09.050","Text":"Meaning we\u0027re taking out an electron."},{"Start":"03:09.050 ","End":"03:11.525","Text":"Now everything is going to be very similar,"},{"Start":"03:11.525 ","End":"03:14.630","Text":"only that we\u0027re going to have less electrons because we have 2 times 6,"},{"Start":"03:14.630 ","End":"03:17.155","Text":"which is the number of valence electrons,"},{"Start":"03:17.155 ","End":"03:19.350","Text":"but minus 1,"},{"Start":"03:19.350 ","End":"03:20.790","Text":"in case we took out an electron."},{"Start":"03:20.790 ","End":"03:23.895","Text":"That\u0027s going to give us 11 electrons in all."},{"Start":"03:23.895 ","End":"03:27.860","Text":"Instead of drawing our molecular orbitals again,"},{"Start":"03:27.860 ","End":"03:33.960","Text":"all we have to do is really just erase one of the electrons that we drew."},{"Start":"03:35.900 ","End":"03:43.510","Text":"Now we can check the bond order for our O_2^plus ion."},{"Start":"03:43.510 ","End":"03:50.020","Text":"Again, the bond order is the number of"},{"Start":"03:50.020 ","End":"03:53.410","Text":"electrons in our bonding molecular orbitals minus"},{"Start":"03:53.410 ","End":"03:57.400","Text":"the number of electrons in our anti-bonding molecular orbitals divided by 2."},{"Start":"03:57.400 ","End":"03:59.440","Text":"In this case, we have 1,"},{"Start":"03:59.440 ","End":"04:00.880","Text":"2, 3, 4,"},{"Start":"04:00.880 ","End":"04:02.530","Text":"5, 6, 7,"},{"Start":"04:02.530 ","End":"04:08.680","Text":"8 electrons in bonding molecular orbitals minus 1,"},{"Start":"04:08.680 ","End":"04:11.065","Text":"2, 3,"},{"Start":"04:11.065 ","End":"04:14.590","Text":"3 electrons in anti-bonding molecular orbitals,"},{"Start":"04:14.590 ","End":"04:16.895","Text":"and then all of this is divided by 2."},{"Start":"04:16.895 ","End":"04:19.590","Text":"This is going to give us 5 divided by 2."},{"Start":"04:19.590 ","End":"04:22.365","Text":"This is going give us a 2.5 bond order."},{"Start":"04:22.365 ","End":"04:28.715","Text":"First of all, we need to explain why the oxidation of O_2 decreases the bond distance."},{"Start":"04:28.715 ","End":"04:30.800","Text":"If we look at our bond order,"},{"Start":"04:30.800 ","End":"04:33.815","Text":"you can see that for O_2 we got a 2."},{"Start":"04:33.815 ","End":"04:39.600","Text":"However, when we oxidize our O_2 into our O_2^plus ion,"},{"Start":"04:39.600 ","End":"04:42.805","Text":"we got a 2.5 bond order."},{"Start":"04:42.805 ","End":"04:45.575","Text":"Meaning we increased the bond order."},{"Start":"04:45.575 ","End":"04:48.595","Text":"But remember, when you increase the bond order,"},{"Start":"04:48.595 ","End":"04:52.519","Text":"you\u0027re actually decreasing the bond distance."},{"Start":"04:52.519 ","End":"04:56.960","Text":"Again, increasing the bond order decreases the bond distance."},{"Start":"04:56.960 ","End":"05:01.265","Text":"The bond is stronger and therefore it\u0027s a shorter bond."},{"Start":"05:01.265 ","End":"05:06.995","Text":"This explains why the oxidation of O_2 decreases the bond distance."},{"Start":"05:06.995 ","End":"05:10.910","Text":"Now let\u0027s take a look at the second part of our exercise."},{"Start":"05:10.910 ","End":"05:18.230","Text":"Now we need to explain why the oxidation of nitrogen increases the bond distance."},{"Start":"05:18.230 ","End":"05:20.810","Text":"Now we\u0027re going to take a look at our nitrogen."},{"Start":"05:20.810 ","End":"05:24.860","Text":"Now, I wrote out the molecular orbitals for us."},{"Start":"05:24.860 ","End":"05:29.885","Text":"Just remember I reversed the order of the energy of the Pi_2p orbital and a Sigma_2p."},{"Start":"05:29.885 ","End":"05:35.825","Text":"In this case, the Pi_2p orbitals are lower in energy than the Sigma_2p."},{"Start":"05:35.825 ","End":"05:39.125","Text":"Therefore, they will fill in first before the Sigma_2p."},{"Start":"05:39.125 ","End":"05:41.795","Text":"Let\u0027s see how many electrons we have to fill in."},{"Start":"05:41.795 ","End":"05:47.485","Text":"For nitrogen, we have to fill in 2 times 5 valence electrons, which equals 10."},{"Start":"05:47.485 ","End":"05:50.460","Text":"We\u0027re going to fill it in 1, 2, 3,"},{"Start":"05:50.460 ","End":"05:52.860","Text":"4, 5,"},{"Start":"05:52.860 ","End":"05:54.675","Text":"6, 7,"},{"Start":"05:54.675 ","End":"05:58.570","Text":"8, 9, 10."},{"Start":"05:59.440 ","End":"06:03.630","Text":"Now we\u0027re going to calculate the bond order."},{"Start":"06:08.770 ","End":"06:13.775","Text":"Again, the number of electrons in our bonding molecular orbitals equals,"},{"Start":"06:13.775 ","End":"06:16.860","Text":"let\u0027s just underline them, that\u0027s Sigma_2s,"},{"Start":"06:16.860 ","End":"06:19.305","Text":"Pi_2p, and Sigma_2p."},{"Start":"06:19.305 ","End":"06:24.735","Text":"That\u0027s going to give us 2 plus 4 plus 2 electrons,"},{"Start":"06:24.735 ","End":"06:27.820","Text":"which gives us a total of 8 electrons,"},{"Start":"06:27.820 ","End":"06:33.920","Text":"minus the number of electrons in our anti-bonding molecular orbitals,"},{"Start":"06:33.920 ","End":"06:37.848","Text":"which we only have 2 because we only have 2 in our Sigma_2s^star,"},{"Start":"06:37.848 ","End":"06:42.630","Text":"so it\u0027s going to be minus 2 divided by 2."},{"Start":"06:42.630 ","End":"06:44.640","Text":"This is going to give us a 6 divided by 2,"},{"Start":"06:44.640 ","End":"06:45.810","Text":"is going to give us a 3,"},{"Start":"06:45.810 ","End":"06:49.030","Text":"meaning we have a triple bond."},{"Start":"06:51.190 ","End":"06:56.670","Text":"Now we want to see what happens when our nitrogen is oxidized."},{"Start":"06:57.760 ","End":"07:03.640","Text":"We\u0027re going to oxidize our nitrogen into N_2^plus ion."},{"Start":"07:03.640 ","End":"07:08.895","Text":"In this case, this changes the number of electrons because we have 2 times 5."},{"Start":"07:08.895 ","End":"07:13.280","Text":"However, we have to subtract 1 since we have a plus 1 charge."},{"Start":"07:13.280 ","End":"07:16.150","Text":"This is going to give us 9 electrons."},{"Start":"07:16.150 ","End":"07:21.830","Text":"Again, instead of just sketching or drawing or writing the molecular orbitals, again,"},{"Start":"07:21.830 ","End":"07:25.742","Text":"we\u0027re just going to erase one of the electrons,"},{"Start":"07:25.742 ","End":"07:28.410","Text":"it\u0027s just easier and quicker."},{"Start":"07:29.350 ","End":"07:35.105","Text":"We\u0027re going to look and calculate our bond order, our new bond order."},{"Start":"07:35.105 ","End":"07:42.990","Text":"This bond order equals,"},{"Start":"07:42.990 ","End":"07:48.260","Text":"now we can see that the number of electrons in our bonding orbitals change."},{"Start":"07:48.260 ","End":"07:49.460","Text":"We have 1, 2,"},{"Start":"07:49.460 ","End":"07:50.825","Text":"3, 4, 5, 6,"},{"Start":"07:50.825 ","End":"07:56.710","Text":"7 minus 2 in our anti-bonding,"},{"Start":"07:56.710 ","End":"07:59.657","Text":"which is still the same as before,"},{"Start":"07:59.657 ","End":"08:02.165","Text":"all this is divided by 2."},{"Start":"08:02.165 ","End":"08:04.880","Text":"This is going to give us 5 divided by 2,"},{"Start":"08:04.880 ","End":"08:08.430","Text":"which is going to give us a 2.5 bond order."},{"Start":"08:09.340 ","End":"08:11.780","Text":"Now if you look at the bond order,"},{"Start":"08:11.780 ","End":"08:15.440","Text":"you can see that the bond order decreased when we"},{"Start":"08:15.440 ","End":"08:20.050","Text":"oxidized our nitrogen diatomic molecule."},{"Start":"08:20.050 ","End":"08:22.370","Text":"If the bond order decreased,"},{"Start":"08:22.370 ","End":"08:25.435","Text":"it means that the bond distance increased."},{"Start":"08:25.435 ","End":"08:29.089","Text":"Again, if the bond order increases,"},{"Start":"08:29.089 ","End":"08:31.625","Text":"it means that the bond distance decreases."},{"Start":"08:31.625 ","End":"08:33.529","Text":"If the bond order decreases,"},{"Start":"08:33.529 ","End":"08:37.535","Text":"it means that the bond distance increases."},{"Start":"08:37.535 ","End":"08:44.400","Text":"In this case, the bond order decreased and therefore the bond distance increased."},{"Start":"08:44.710 ","End":"08:48.515","Text":"I just wrote these out for us again, the bond orders."},{"Start":"08:48.515 ","End":"08:51.110","Text":"In the case of our oxygen molecule,"},{"Start":"08:51.110 ","End":"08:54.710","Text":"we have a bond order of 2."},{"Start":"08:54.710 ","End":"08:56.480","Text":"When we oxidize this molecule,"},{"Start":"08:56.480 ","End":"08:59.460","Text":"we get a bond order of 2.5."},{"Start":"08:59.980 ","End":"09:03.666","Text":"We can see here that the bond order increases."},{"Start":"09:03.666 ","End":"09:04.940","Text":"When"},{"Start":"09:12.970 ","End":"09:15.730","Text":"the bond order increases,"},{"Start":"09:15.730 ","End":"09:24.080","Text":"the bond distance decreases."},{"Start":"09:29.640 ","End":"09:35.170","Text":"This explains why the oxidation of O_2 decreases the bond distance."},{"Start":"09:35.170 ","End":"09:39.925","Text":"Now when we look at our nitrogen molecule,"},{"Start":"09:39.925 ","End":"09:42.170","Text":"we can see that the bond order is 3,"},{"Start":"09:42.170 ","End":"09:45.350","Text":"and when we oxidize this molecule to an ion,"},{"Start":"09:45.350 ","End":"09:51.045","Text":"our bond order decreased to 2.5."},{"Start":"09:51.045 ","End":"10:00.170","Text":"In this case, oxidation causes the bond order to decrease."},{"Start":"10:00.170 ","End":"10:05.930","Text":"The bond order decreases and this means that"},{"Start":"10:05.930 ","End":"10:13.800","Text":"the bond distance increases."},{"Start":"10:15.460 ","End":"10:21.590","Text":"That explains why the oxidation of nitrogen increases the bond distance."},{"Start":"10:21.590 ","End":"10:23.240","Text":"That is our final answer."},{"Start":"10:23.240 ","End":"10:25.980","Text":"Thank you very much for watching."}],"ID":24479},{"Watched":false,"Name":"Exercise 5","Duration":"7m 1s","ChapterTopicVideoID":23571,"CourseChapterTopicPlaylistID":101323,"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":"We\u0027re going to solve the following exercise."},{"Start":"00:02.685 ","End":"00:05.910","Text":"Sketch a molecular orbital diagram for H2 minus."},{"Start":"00:05.910 ","End":"00:11.970","Text":"Does the extra electron make the H-H bonds stronger compared to H2 hydrogen?"},{"Start":"00:11.970 ","End":"00:14.440","Text":"Explain your answer."},{"Start":"00:14.690 ","End":"00:18.630","Text":"We\u0027re going to start with a molecular orbital diagram."},{"Start":"00:18.630 ","End":"00:24.330","Text":"In H2, we have 2 hydrogen atoms and each 1 has 1 electron in a 1s orbital."},{"Start":"00:24.330 ","End":"00:26.190","Text":"Here you can see we have 1Sa,"},{"Start":"00:26.190 ","End":"00:29.898","Text":"which is an atomic orbital of 1 of the hydrogens,"},{"Start":"00:29.898 ","End":"00:32.945","Text":"and 1Sb which is an atomic orbital of the other hydrogen."},{"Start":"00:32.945 ","End":"00:38.090","Text":"Now these 2 atomic orbitals combine and they give us 2 molecular orbitals."},{"Start":"00:38.090 ","End":"00:40.400","Text":"Remember that the number of atomic orbitals"},{"Start":"00:40.400 ","End":"00:42.260","Text":"has to equal the number of molecular orbitals."},{"Start":"00:42.260 ","End":"00:44.900","Text":"Again, we have 2 atomic 1S orbitals,"},{"Start":"00:44.900 ","End":"00:47.750","Text":"which gives us 2 molecular orbitals."},{"Start":"00:47.750 ","End":"00:55.130","Text":"Now 1 of these molecular orbitals is lower in energy than the 1S orbitals,"},{"Start":"00:55.130 ","End":"00:58.070","Text":"and this is our bonding molecular orbital."},{"Start":"00:58.070 ","End":"01:03.468","Text":"Therefore, it\u0027s lower in energy than the 1s atomic orbitals,"},{"Start":"01:03.468 ","End":"01:07.340","Text":"and this orbital is called Sigma 1s and"},{"Start":"01:07.340 ","End":"01:12.010","Text":"we have another molecular orbital which is called Sigma 1S*."},{"Start":"01:12.010 ","End":"01:14.970","Text":"It\u0027s Sigma 1s with an asterisk."},{"Start":"01:14.970 ","End":"01:17.780","Text":"This is an anti-bonding molecular orbital."},{"Start":"01:17.780 ","End":"01:22.690","Text":"Therefore, it\u0027s higher in energy than our 1s atomic orbitals."},{"Start":"01:22.690 ","End":"01:26.310","Text":"Now if we look at our ion H2 minus,"},{"Start":"01:26.310 ","End":"01:30.440","Text":"we know that we have 3 electrons to put in"},{"Start":"01:30.440 ","End":"01:34.730","Text":"our molecular orbital diagram since we have 1 electron from each hydrogen,"},{"Start":"01:34.730 ","End":"01:36.260","Text":"which is 2 times 1."},{"Start":"01:36.260 ","End":"01:45.010","Text":"Then we\u0027re adding another 1 because we have a minus 1 and this gives us 3 electrons."},{"Start":"01:45.160 ","End":"01:47.900","Text":"When you sketch a molecular orbital diagram,"},{"Start":"01:47.900 ","End":"01:50.880","Text":"first you have to write the atomic orbitals,"},{"Start":"01:50.880 ","End":"01:52.155","Text":"write the molecular orbitals,"},{"Start":"01:52.155 ","End":"01:54.020","Text":"and the next step is determine the number of"},{"Start":"01:54.020 ","End":"01:56.285","Text":"electrons you need to fill in and fill them in."},{"Start":"01:56.285 ","End":"01:59.520","Text":"We have 3 electrons we need to fill in."},{"Start":"01:59.600 ","End":"02:02.970","Text":"In each orbital we can have 2 electrons,"},{"Start":"02:02.970 ","End":"02:06.672","Text":"so we\u0027re going to fill Sigma 1S with 2 electrons"},{"Start":"02:06.672 ","End":"02:11.820","Text":"and then we\u0027re just going to have 1 more electron to fill in to Sigma 1S."},{"Start":"02:12.700 ","End":"02:15.800","Text":"We have 3 electrons here."},{"Start":"02:15.800 ","End":"02:21.500","Text":"This is the molecular orbital diagram for H2 minus."},{"Start":"02:21.500 ","End":"02:24.515","Text":"Now let\u0027s go on to the second part of the question."},{"Start":"02:24.515 ","End":"02:30.215","Text":"Does the extra electron make the H-H bonds stronger compared to H2?"},{"Start":"02:30.215 ","End":"02:32.730","Text":"Explain your answer."},{"Start":"02:33.380 ","End":"02:38.524","Text":"In order to find if our bond is stronger,"},{"Start":"02:38.524 ","End":"02:41.785","Text":"what we\u0027re going to do is we\u0027re going to calculate our bond order."},{"Start":"02:41.785 ","End":"02:45.750","Text":"First of all, again,"},{"Start":"02:45.750 ","End":"02:48.360","Text":"we have our H2 minus ion,"},{"Start":"02:48.360 ","End":"02:53.700","Text":"and we can see that in our Sigma 1S orbital we have 2 electrons."},{"Start":"02:53.700 ","End":"02:56.290","Text":"In our Sigma 1S*, again,"},{"Start":"02:56.290 ","End":"02:59.690","Text":"the anti-bonding orbital, we have 1 electron."},{"Start":"03:00.990 ","End":"03:10.635","Text":"Now we\u0027re going to calculate the bond order for H2 minus ion."},{"Start":"03:10.635 ","End":"03:15.965","Text":"Now remember the bond order is the number of electrons in"},{"Start":"03:15.965 ","End":"03:19.970","Text":"our bonding molecular orbitals minus the number of"},{"Start":"03:19.970 ","End":"03:24.185","Text":"electrons in the anti-bonding molecular orbitals divided by 2."},{"Start":"03:24.185 ","End":"03:27.560","Text":"In this case in our binding molecular orbitals,"},{"Start":"03:27.560 ","End":"03:29.630","Text":"which is only Sigma 1S we have 2 electrons,"},{"Start":"03:29.630 ","End":"03:34.430","Text":"so we\u0027re going to start with 2 minus we have 1 electron in our anti-bonding"},{"Start":"03:34.430 ","End":"03:40.620","Text":"molecular orbital so that\u0027s minus 1 divided by 2."},{"Start":"03:40.620 ","End":"03:45.250","Text":"Our bonding order for our H2 minus equals 1/2."},{"Start":"03:46.150 ","End":"03:49.940","Text":"Now we need to compare this bond order to the bond order"},{"Start":"03:49.940 ","End":"03:53.870","Text":"in a hydrogen molecule and in H2 molecule."},{"Start":"03:53.870 ","End":"03:56.010","Text":"If we look at H2,"},{"Start":"03:57.470 ","End":"04:01.975","Text":"we\u0027re going to sketch the same atomic orbitals and the same molecular orbitals."},{"Start":"04:01.975 ","End":"04:04.870","Text":"However, this time we\u0027re only going to have 2 electrons,"},{"Start":"04:04.870 ","End":"04:08.375","Text":"so that\u0027s going to be 2 times 1 so it\u0027s going to give us 2 electrons."},{"Start":"04:08.375 ","End":"04:15.530","Text":"We\u0027re not going to have any electron in our Sigma 1S* in our anti-bonding orbital."},{"Start":"04:16.040 ","End":"04:21.882","Text":"We can write our H2 as Sigma 1S2,"},{"Start":"04:21.882 ","End":"04:30.280","Text":"because again, we\u0027re only going to have 2 electrons in our Sigma 1S molecular orbital,"},{"Start":"04:30.280 ","End":"04:31.300","Text":"which is the bonding orbital."},{"Start":"04:31.300 ","End":"04:33.985","Text":"We\u0027re not going to have any Sigma 1S*."},{"Start":"04:33.985 ","End":"04:36.080","Text":"That\u0027s going to be Sigma 1S2,"},{"Start":"04:36.080 ","End":"04:39.230","Text":"and if we calculate the bond order,"},{"Start":"04:42.830 ","End":"04:45.325","Text":"it\u0027s going to equal, again,"},{"Start":"04:45.325 ","End":"04:49.180","Text":"the number of electrons in our molecular orbitals and our bonding molecular orbitals,"},{"Start":"04:49.180 ","End":"04:54.220","Text":"which is 2 minus the number of electrons in our anti-bonding molecular orbitals,"},{"Start":"04:54.220 ","End":"04:55.645","Text":"which here is 0,"},{"Start":"04:55.645 ","End":"04:57.850","Text":"we have no electrons,"},{"Start":"04:57.850 ","End":"05:02.650","Text":"divided by 2 so there\u0027s going to be 2 divided by 2 and that\u0027s going to give us a 1."},{"Start":"05:03.200 ","End":"05:06.670","Text":"Now remember we have to check which H-H bond is"},{"Start":"05:06.670 ","End":"05:13.090","Text":"stronger in our H2 or in our H2 minus molecule."},{"Start":"05:14.460 ","End":"05:21.280","Text":"We have to check what happens when we add that extra electron to our H2 molecule."},{"Start":"05:21.280 ","End":"05:27.010","Text":"If the H-H bond becomes stronger or less strong."},{"Start":"05:27.380 ","End":"05:29.950","Text":"Again, we\u0027re going to look at our bond order."},{"Start":"05:29.950 ","End":"05:35.530","Text":"For H2, our bond order equals 1."},{"Start":"05:39.210 ","End":"05:42.129","Text":"When we add in an extra electron,"},{"Start":"05:42.129 ","End":"05:49.210","Text":"1 of the electrons went into our anti-bonding orbital and our bond order became 1/2."},{"Start":"05:51.150 ","End":"05:55.000","Text":"We can see that when we added the extra electron from H2 to"},{"Start":"05:55.000 ","End":"05:58.434","Text":"H2 minus our bond order went down,"},{"Start":"05:58.434 ","End":"06:07.660","Text":"meaning our H-H bond became weaker since our bond order is smaller,"},{"Start":"06:18.590 ","End":"06:26.830","Text":"so our bond order decreased and therefore the H-H bond became weaker."},{"Start":"06:32.240 ","End":"06:39.475","Text":"Our bond order decreased from H2 to H2 minus when we added the extra electron."},{"Start":"06:39.475 ","End":"06:44.440","Text":"This extra electron went into our anti-bonding molecular orbital and therefore,"},{"Start":"06:44.440 ","End":"06:47.800","Text":"the H-H bond became weaker."},{"Start":"06:47.800 ","End":"06:52.640","Text":"Again, the bond order decreased the H-H bond became weaker."},{"Start":"06:57.490 ","End":"06:59.660","Text":"That is our final answer."},{"Start":"06:59.660 ","End":"07:02.220","Text":"Thank you very much for watching."}],"ID":24480}],"Thumbnail":null,"ID":101323},{"Name":"VB and MO description of benzene","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"VB description of benzene","Duration":"5m 55s","ChapterTopicVideoID":23050,"CourseChapterTopicPlaylistID":101324,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"In previous videos,"},{"Start":"00:01.920 ","End":"00:04.905","Text":"we learned about valence bond molecular orbital theory."},{"Start":"00:04.905 ","End":"00:09.140","Text":"In this video, we describe benzene using valence bond theory."},{"Start":"00:09.140 ","End":"00:13.320","Text":"In the next video using molecular orbital theory."},{"Start":"00:13.320 ","End":"00:20.955","Text":"Benzene is a planar organic molecule with a formula C_6H_6."},{"Start":"00:20.955 ","End":"00:22.815","Text":"Here\u0027s a picture of it."},{"Start":"00:22.815 ","End":"00:27.360","Text":"The 6 carbon atoms lie at the vertices of"},{"Start":"00:27.360 ","End":"00:33.995","Text":"a regular hexagon and a hydrogen atom is attached to each carbon atom."},{"Start":"00:33.995 ","End":"00:37.610","Text":"Here\u0027s the hydrogen attached to a carbon,"},{"Start":"00:37.610 ","End":"00:40.655","Text":"and so on, hydrogen attached to a carbon."},{"Start":"00:40.655 ","End":"00:45.740","Text":"Now we\u0027ll describe benzene according to valence bond theory."},{"Start":"00:45.740 ","End":"00:50.659","Text":"The first thing to note is that all the angles are 120 degrees."},{"Start":"00:50.659 ","End":"00:56.300","Text":"We have a 120 degrees here, here, and here."},{"Start":"00:56.300 ","End":"01:01.480","Text":"In addition, each carbon is attached to 3 atoms."},{"Start":"01:01.480 ","End":"01:05.730","Text":"Here\u0027s a carbon attached to hydrogen and 2 other carbons."},{"Start":"01:05.730 ","End":"01:12.340","Text":"It can be described as AX_3 in the VSEPR theory."},{"Start":"01:12.740 ","End":"01:15.970","Text":"We have these 2 pieces of knowledge,"},{"Start":"01:15.970 ","End":"01:19.600","Text":"the angle of a 120 degrees and AX_3,"},{"Start":"01:19.600 ","End":"01:26.360","Text":"we can work out that the hybridization of each carbon atom must be sp^2."},{"Start":"01:26.420 ","End":"01:30.765","Text":"Each carbon is sp^2 hybridized."},{"Start":"01:30.765 ","End":"01:33.355","Text":"Let\u0027s look at the molecule again."},{"Start":"01:33.355 ","End":"01:36.760","Text":"We have 6 Sigma bonds between sp^2 on"},{"Start":"01:36.760 ","End":"01:43.454","Text":"1 carbon and sp^2 on another carbon and 6 CH bonds"},{"Start":"01:43.454 ","End":"01:49.465","Text":"between sp^2 on a carbon and a 1s orbital on a hydrogen."},{"Start":"01:49.465 ","End":"01:55.980","Text":"This is written here, 6 Sigma between sp^2 carbon and"},{"Start":"01:55.980 ","End":"02:04.920","Text":"another sp^2 carbon and 6 Sigma bonds between sp^2 on carbon and a 1s on hydrogen."},{"Start":"02:04.920 ","End":"02:10.725","Text":"Now, these 6 Sigma bonds between carbon-carbon,"},{"Start":"02:10.725 ","End":"02:12.225","Text":"carbon and carbon,"},{"Start":"02:12.225 ","End":"02:14.290","Text":"and 6 Sigma bonds between carbon,"},{"Start":"02:14.290 ","End":"02:19.045","Text":"hydrogen give us the backbone the general shape of the molecule."},{"Start":"02:19.045 ","End":"02:21.495","Text":"But in addition to this,"},{"Start":"02:21.495 ","End":"02:28.865","Text":"we have 6 p_z orbitals which are perpendicular to the plane of the molecule."},{"Start":"02:28.865 ","End":"02:33.735","Text":"Recall the same thing we had with ethylene."},{"Start":"02:33.735 ","End":"02:37.140","Text":"We have C-C,"},{"Start":"02:37.700 ","End":"02:46.805","Text":"if I draw it now so that the other bones are coming out of the plane of the screen."},{"Start":"02:46.805 ","End":"02:49.865","Text":"Then perpendicular to this plane,"},{"Start":"02:49.865 ","End":"02:58.280","Text":"we have p_z orbitals and this overlap to give us a Pi bond."},{"Start":"02:58.280 ","End":"03:02.475","Text":"The 2 plus lobes overlap"},{"Start":"03:02.475 ","End":"03:07.680","Text":"the 2 negative lobes overlap and that together gives us a Pi bond."},{"Start":"03:07.680 ","End":"03:11.885","Text":"We can compare this with a valence bond description of ethylene."},{"Start":"03:11.885 ","End":"03:16.485","Text":"The only difference being that here we have"},{"Start":"03:16.485 ","End":"03:23.750","Text":"hydrogens and carbons whereas an ethylene,"},{"Start":"03:23.750 ","End":"03:26.075","Text":"these carbons are hydrogens."},{"Start":"03:26.075 ","End":"03:31.070","Text":"This is even clearer from this picture where we\u0027ve just"},{"Start":"03:31.070 ","End":"03:35.690","Text":"indicated the bones rather than writing specifically carbon,"},{"Start":"03:35.690 ","End":"03:38.060","Text":"this is usually used organic chemistry."},{"Start":"03:38.060 ","End":"03:41.735","Text":"Carbon is most prevalent atom,"},{"Start":"03:41.735 ","End":"03:44.855","Text":"and so it isn\u0027t written specifically."},{"Start":"03:44.855 ","End":"03:48.845","Text":"Now, here we have 1 of the double bonds,"},{"Start":"03:48.845 ","End":"03:52.300","Text":"which is a Sigma bond plus a Pi bond."},{"Start":"03:52.300 ","End":"03:58.450","Text":"Then we have another 1 and the third 1."},{"Start":"03:59.360 ","End":"04:04.510","Text":"Now there\u0027s no specific reason why"},{"Start":"04:04.510 ","End":"04:09.205","Text":"we\u0027ve chosen these particular places to put the Pi bonds."},{"Start":"04:09.205 ","End":"04:13.375","Text":"We could equally well have put them in the other places,"},{"Start":"04:13.375 ","End":"04:15.170","Text":"so here are the 2 versions."},{"Start":"04:15.170 ","End":"04:18.100","Text":"Now the hydrogens are missed out and this"},{"Start":"04:18.100 ","End":"04:22.300","Text":"is often the way it\u0027s written in organic chemistry."},{"Start":"04:22.300 ","End":"04:25.690","Text":"Instead of having a double bond here,"},{"Start":"04:25.690 ","End":"04:30.050","Text":"here and here, here we have it in the other positions."},{"Start":"04:30.050 ","End":"04:33.460","Text":"Neither of them is preferable over the other so there\u0027s"},{"Start":"04:33.460 ","End":"04:36.430","Text":"a resonance between the 2 versions,"},{"Start":"04:36.430 ","End":"04:38.560","Text":"between the 2 structures."},{"Start":"04:38.560 ","End":"04:44.385","Text":"Now this is very important because experimentally,"},{"Start":"04:44.385 ","End":"04:47.540","Text":"we can see that all the C-C bonds are identical,"},{"Start":"04:47.540 ","End":"04:49.850","Text":"they all have the same bond length."},{"Start":"04:49.850 ","End":"04:54.545","Text":"There must be a resonance between these 2 structures."},{"Start":"04:54.545 ","End":"04:58.295","Text":"We can calculate the average bond order."},{"Start":"04:58.295 ","End":"05:00.150","Text":"We have 3 double bonds,"},{"Start":"05:00.150 ","End":"05:03.030","Text":"that\u0027s 3 times 2 and 3 single bonds,"},{"Start":"05:03.030 ","End":"05:04.395","Text":"3 times 1,"},{"Start":"05:04.395 ","End":"05:07.080","Text":"and a total of 6 bonds."},{"Start":"05:07.080 ","End":"05:13.505","Text":"The average bond order is 6 plus 3 divided by 6, that\u0027s 1.5."},{"Start":"05:13.505 ","End":"05:20.695","Text":"Each bond is intermediate in lens and strength between a single and a double bond."},{"Start":"05:20.695 ","End":"05:24.875","Text":"Now this is an example of a conjugated molecule."},{"Start":"05:24.875 ","End":"05:28.280","Text":"A hydrocarbon such as benzene with alternating"},{"Start":"05:28.280 ","End":"05:31.790","Text":"single and double bonds is said to be conjugated."},{"Start":"05:31.790 ","End":"05:33.440","Text":"Here\u0027s a single bond, double bond,"},{"Start":"05:33.440 ","End":"05:35.665","Text":"single, double, single, double."},{"Start":"05:35.665 ","End":"05:40.415","Text":"The Pi electrons here are delocalized over the whole molecule."},{"Start":"05:40.415 ","End":"05:47.465","Text":"The whole concept of delocalization is much more obvious in molecular orbital theory,"},{"Start":"05:47.465 ","End":"05:50.105","Text":"and we\u0027ll deal with that in the next video."},{"Start":"05:50.105 ","End":"05:54.720","Text":"In this video, we applied valence bond theory to benzene."}],"ID":23892},{"Watched":false,"Name":"MO description of benzene","Duration":"6m 36s","ChapterTopicVideoID":23591,"CourseChapterTopicPlaylistID":101324,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"In the previous video,"},{"Start":"00:02.040 ","End":"00:06.374","Text":"we used valence bond theory to describe the benzene molecule."},{"Start":"00:06.374 ","End":"00:10.950","Text":"In this video, we\u0027ll describe it using molecular orbital theory."},{"Start":"00:10.950 ","End":"00:15.765","Text":"We\u0027re going to talk about molecular orbital description of benzene."},{"Start":"00:15.765 ","End":"00:20.745","Text":"Now, when we\u0027re just describing a benzene qualitatively,"},{"Start":"00:20.745 ","End":"00:24.480","Text":"we can use the valence bond description for the Sigma bonds,"},{"Start":"00:24.480 ","End":"00:26.850","Text":"so the C-C and C-H Sigma bonds,"},{"Start":"00:26.850 ","End":"00:28.890","Text":"just like in the previous video."},{"Start":"00:28.890 ","End":"00:32.330","Text":"Then we take the p_z orbitals,"},{"Start":"00:32.330 ","End":"00:35.315","Text":"which are perpendicular to the plane of the molecule,"},{"Start":"00:35.315 ","End":"00:38.360","Text":"and construct 6 molecular orbitals."},{"Start":"00:38.360 ","End":"00:43.895","Text":"We construct 6 molecular orbitals from 6 2p_z orbitals."},{"Start":"00:43.895 ","End":"00:49.160","Text":"The lowest energy molecular orbital which we will call Psi 1,"},{"Start":"00:49.160 ","End":"00:55.580","Text":"is proportional to the sum of all 6 p_z orbitals."},{"Start":"00:55.580 ","End":"00:57.665","Text":"I\u0027ve labeled them Phi 1,"},{"Start":"00:57.665 ","End":"00:59.435","Text":"Phi 2, etc."},{"Start":"00:59.435 ","End":"01:01.562","Text":"Where this is say,"},{"Start":"01:01.562 ","End":"01:03.895","Text":"benzene 1, benzene 2."},{"Start":"01:03.895 ","End":"01:07.245","Text":"Where this is carbon 1, carbon 2,"},{"Start":"01:07.245 ","End":"01:09.405","Text":"carbon 3, carbon 4,"},{"Start":"01:09.405 ","End":"01:12.420","Text":"carbon 5, carbon 6."},{"Start":"01:12.420 ","End":"01:18.335","Text":"Now, here there are no nodes perpendicular to the plane of the molecule."},{"Start":"01:18.335 ","End":"01:19.700","Text":"There is, of course,"},{"Start":"01:19.700 ","End":"01:22.390","Text":"a node in the plane of the molecule,"},{"Start":"01:22.390 ","End":"01:27.920","Text":"and that\u0027s because p_z is above and below the plane of the molecule."},{"Start":"01:27.920 ","End":"01:30.740","Text":"The highest energy molecular orbital,"},{"Start":"01:30.740 ","End":"01:33.200","Text":"which I have called Psi 6,"},{"Start":"01:33.200 ","End":"01:38.455","Text":"and that\u0027s proportional to the 2p_z orbitals,"},{"Start":"01:38.455 ","End":"01:40.040","Text":"but with different signs."},{"Start":"01:40.040 ","End":"01:43.415","Text":"You see the sign of every second orbital changes."},{"Start":"01:43.415 ","End":"01:46.820","Text":"Here, that\u0027s plus in both cases, here plus-minus,"},{"Start":"01:46.820 ","End":"01:49.400","Text":"plus in both cases, plus-minus,"},{"Start":"01:49.400 ","End":"01:52.785","Text":"plus in both cases, plus-minus."},{"Start":"01:52.785 ","End":"01:56.790","Text":"Now, here it\u0027s drawn."},{"Start":"01:56.790 ","End":"02:00.710","Text":"I\u0027ve just written the signs plus,"},{"Start":"02:00.710 ","End":"02:02.645","Text":"minus, plus, minus, plus,"},{"Start":"02:02.645 ","End":"02:06.140","Text":"minus, and every time it goes from plus to minus,"},{"Start":"02:06.140 ","End":"02:07.595","Text":"we must go through 0,"},{"Start":"02:07.595 ","End":"02:09.115","Text":"so this is a node."},{"Start":"02:09.115 ","End":"02:14.705","Text":"In fact, there are 3 nodes perpendicular to the plane of the molecule."},{"Start":"02:14.705 ","End":"02:18.485","Text":"This is antibonding, this one is bonding,"},{"Start":"02:18.485 ","End":"02:20.720","Text":"and this one is antibonding."},{"Start":"02:20.720 ","End":"02:22.070","Text":"Now we\u0027re going to talk about"},{"Start":"02:22.070 ","End":"02:27.770","Text":"the energy level diagram because it\u0027s obvious I\u0027ve called it Psi 1 and Psi 6,"},{"Start":"02:27.770 ","End":"02:32.735","Text":"that there are another 4 molecular orbitals."},{"Start":"02:32.735 ","End":"02:34.190","Text":"The first thing to note,"},{"Start":"02:34.190 ","End":"02:36.755","Text":"and this happens a lot in molecular orbital theory."},{"Start":"02:36.755 ","End":"02:42.835","Text":"The energy increases as the number of nodes increases. Here we have 1."},{"Start":"02:42.835 ","End":"02:45.045","Text":"Here we have 0 nodes."},{"Start":"02:45.045 ","End":"02:47.310","Text":"Here we have 3 nodes."},{"Start":"02:47.310 ","End":"02:50.085","Text":"We\u0027re going to have a molecular orbital was 1 node,"},{"Start":"02:50.085 ","End":"02:51.720","Text":"and with 2 nodes,"},{"Start":"02:51.720 ","End":"02:54.830","Text":"so we have 3 bonding molecular orbitals,"},{"Start":"02:54.830 ","End":"02:59.135","Text":"as you\u0027ll see, and 3 antibonding molecular orbitals."},{"Start":"02:59.135 ","End":"03:01.925","Text":"Here\u0027s the diagram. In green,"},{"Start":"03:01.925 ","End":"03:08.030","Text":"we have the bonding molecular orbitals and in red the antibonding."},{"Start":"03:08.030 ","End":"03:12.260","Text":"Here\u0027s bonding, and here\u0027s anti-bonding."},{"Start":"03:12.260 ","End":"03:19.820","Text":"Here in the dotted line is the energy we would have without the Pi orbitals."},{"Start":"03:19.820 ","End":"03:23.750","Text":"Here, the dotted line is the energy we would have if we didn\u0027t"},{"Start":"03:23.750 ","End":"03:28.565","Text":"make molecular orbitals from the p_z orbitals."},{"Start":"03:28.565 ","End":"03:31.550","Text":"Here is Psi 1."},{"Start":"03:31.550 ","End":"03:33.470","Text":"We drew it before."},{"Start":"03:33.470 ","End":"03:40.960","Text":"Then we have Psi 2 and Psi 3."},{"Start":"03:40.960 ","End":"03:43.070","Text":"These have the same energy,"},{"Start":"03:43.070 ","End":"03:47.870","Text":"the degenerate, and they each have 1 node."},{"Start":"03:47.870 ","End":"03:50.915","Text":"Here it passes between the atoms."},{"Start":"03:50.915 ","End":"03:53.825","Text":"Here it crosses 2 of the atoms,"},{"Start":"03:53.825 ","End":"03:56.375","Text":"but each has 1 node."},{"Start":"03:56.375 ","End":"04:04.450","Text":"Then we have another 2 to degenerate orbitals that are antibonding,"},{"Start":"04:04.450 ","End":"04:07.385","Text":"that\u0027s 4 and 5."},{"Start":"04:07.385 ","End":"04:09.520","Text":"Here\u0027s pictures of them."},{"Start":"04:09.520 ","End":"04:14.725","Text":"We can see that this one has 2 nodes and so does the other one,"},{"Start":"04:14.725 ","End":"04:16.660","Text":"but they\u0027re in the different places."},{"Start":"04:16.660 ","End":"04:18.880","Text":"Here, they are between the atoms,"},{"Start":"04:18.880 ","End":"04:23.990","Text":"and here one of the nodes goes through the 2 atoms."},{"Start":"04:23.990 ","End":"04:26.590","Text":"Then of course, as before,"},{"Start":"04:26.590 ","End":"04:29.645","Text":"we have the highest orbital Psi 6,"},{"Start":"04:29.645 ","End":"04:32.120","Text":"which has 3 nodes."},{"Start":"04:32.120 ","End":"04:35.799","Text":"The energy increases as the number of nodes increases."},{"Start":"04:35.799 ","End":"04:39.535","Text":"We go from 0,1,2,3."},{"Start":"04:39.535 ","End":"04:44.265","Text":"Now we have to fill in the bonding orbitals."},{"Start":"04:44.265 ","End":"04:47.900","Text":"We have 6 electrons,"},{"Start":"04:47.900 ","End":"04:49.280","Text":"so we have a pair here,"},{"Start":"04:49.280 ","End":"04:52.160","Text":"a pair here, and a pair here."},{"Start":"04:52.160 ","End":"04:58.720","Text":"All the bonding orbitals are full and the antibonding orbitals are empty."},{"Start":"04:58.720 ","End":"05:04.240","Text":"Now, we\u0027re going to talk about the important concept of delocalization."},{"Start":"05:04.240 ","End":"05:08.980","Text":"The electrons in the molecular orbital span all 6 carbon atoms,"},{"Start":"05:08.980 ","End":"05:11.440","Text":"so they\u0027re all over the molecule."},{"Start":"05:11.440 ","End":"05:15.750","Text":"The electrons are delocalized all over the molecule."},{"Start":"05:15.750 ","End":"05:18.140","Text":"We say that the Pi electrons,"},{"Start":"05:18.140 ","End":"05:22.420","Text":"we call these Pi electrons because they occupy Pi orbitals,"},{"Start":"05:22.420 ","End":"05:29.020","Text":"are delocalized and this gives extra stability compared to 3 separate Pi bonds,"},{"Start":"05:29.020 ","End":"05:31.840","Text":"as we had at the valence bond theory."},{"Start":"05:31.840 ","End":"05:35.500","Text":"Benzene is said to be an aromatic molecule,"},{"Start":"05:35.500 ","End":"05:38.830","Text":"and it has 4n plus 2 bonding electrons,"},{"Start":"05:38.830 ","End":"05:40.360","Text":"and here is equal to 1,"},{"Start":"05:40.360 ","End":"05:41.740","Text":"so we have 4 times 1,"},{"Start":"05:41.740 ","End":"05:43.525","Text":"that\u0027s 4 plus 2,"},{"Start":"05:43.525 ","End":"05:45.080","Text":"that gives us 6,"},{"Start":"05:45.080 ","End":"05:48.280","Text":"so we have 6 bonding electrons."},{"Start":"05:48.280 ","End":"05:53.215","Text":"If we go on to the next molecule in this series,"},{"Start":"05:53.215 ","End":"05:55.615","Text":"which is called naphthalene,"},{"Start":"05:55.615 ","End":"05:59.935","Text":"it has 10 carbons, 10 bonded electrons,"},{"Start":"05:59.935 ","End":"06:01.900","Text":"and then n is equal to 2,"},{"Start":"06:01.900 ","End":"06:05.255","Text":"4 times 2 plus 2 gives us 10."},{"Start":"06:05.255 ","End":"06:08.660","Text":"Now to emphasize the delocalization,"},{"Start":"06:08.660 ","End":"06:17.120","Text":"benzene is often drawn in this way as a hexagon with a circle inside."},{"Start":"06:17.120 ","End":"06:20.840","Text":"That\u0027s to indicate that area all the carbons are equal."},{"Start":"06:20.840 ","End":"06:25.100","Text":"There is no difference between different carbons,"},{"Start":"06:25.100 ","End":"06:29.885","Text":"and the electrons are delocalized over the whole molecule."},{"Start":"06:29.885 ","End":"06:35.370","Text":"In this video, we describe benzene using molecular orbital theory."}],"ID":24502}],"Thumbnail":null,"ID":101324}]

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