### Advanced Bonding Theories

- Valence bond theory can explain for observed bond angles via hybrid orbitals, but doesn’t always account for a compound’s magnetic behavior.
- MO theory is a delocalized bonding model, and views electrons as being distributed throughout the entire compound.
- An MO diagram can be used to predict magnetism and bond order for a molecule.

## Advanced Bonding Theories

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Lesson Overview
- Introduction
- Valence Bond Theory
- Valence Bond Theory Cont'd
- Valence Bond Theory Cont'd (spá¶)
- Valence Bond Theory Cont'd (spÂ²)
- Valence Bond Theory Cont'd (sp)
- Valence Bond Theory Cont'd (spÂ³d and spÂ³dÂ²)
- Molecular Orbital Theory
- Molecular Orbital Theory Cont'd
- Molecular Orbital Theory
- Wavefunctions
- How s-orbitals Can Interact
- Bonding Nature of p-orbitals: Head-on
- Bonding Nature of p-orbitals: Parallel
- Interaction Between s and p-orbital
- Molecular Orbital Diagram For Homonuclear Diatomics: Hâ
- Molecular Orbital Diagram For Homonuclear Diatomics: Heâ
- Molecular Orbital Diagram For Homonuclear Diatomic: Liâ
- Molecular Orbital Diagram For Homonuclear Diatomic: Liââº
- Molecular Orbital Diagram For Homonuclear Diatomic: Bâ
- Molecular Orbital Diagram For Homonuclear Diatomic: Nâ
- Molecular Orbital Diagram: Molecular Oxygen
- Molecular Orbital Diagram For Heteronuclear Diatomics: Hydrochloric Acid
- Sample Problem 1: Determine the Atomic Hybridization
- Sample Problem 2

## General Chemistry

## Transcription: Advanced Bonding Theories

*Hi, welcome back to Educator.com.*0000

*Today's lesson in general chemistry is going to be advanced bonding theories.*0003

*We are going to start off with a brief introduction as to why*0010

*we need these advanced bonding theories from what we already know.*0015

*That was really Lewis structures.*0019

*Pretty much the advanced bonding theories that we are going to be talking about are the following.*0022

*One is what we call valence bond theory.*0026

*The second one is what we call molecular orbital theory.*0030

*We will wrap up the session with a brief summary followed by a pair of sample problems.*0033

*What you and I have already discussed is the representation of valence electrons schematically.*0041

*Remember that is what we called Lewis structures.*0050

*We went through a step by step procedure on how to formulate a valid Lewis structure.*0052

*Lewis structures are sufficient because using VSEPR theory we were able to*0058

*come up with some predicted geometries strictly off of the Lewis structure.*0067

*But there exists really two main problems with Lewis theory that we never talked about before.*0076

*The first main problem has to do with magnetism.*0082

*If you noticed that all of the Lewis structures that we looked at, all electrons are paired.*0087

*They are either in a single bond, double bond, or triple bond for electrons.*0093

*And they are either in a lone pair.*0098

*If all electrons are paired, then we would expect everything to be diamagnetic.*0101

*But as we all know, that is not the case.*0108

*There is definitely more than enough compounds that are paramagnetic,*0110

*meaning that they do have at least one unpaired electron.*0116

*Lewis theory does not account for magnetic behavior.*0119

*A second problem with Lewis theory is the following.*0124

*Consider all of the compounds that we looked at.*0130

*For all of the compounds really, the valence shell has been the p orbital.*0133

*We looked at p orbitals before.*0139

*They were pretty much three p orbitals per energy level--p _{x}, p_{y}, and p_{z}.*0141

*We actually graphed them.*0148

*We saw that because of Cartesian coordinates, the p _{x}, p_{y},*0150

*and p _{z} orbitals are going to be perpendicular to each other.*0154

*If you have nothing but 90 degree angles, how do we get 109.5?*0159

*How do we get 120, etc?*0166

*Lewis theory doesn't really account for all of the bond angles that are expected from VSEPR theory.*0175

*We do need advanced additional bonding theories to make up for the flaws of Lewis theory.*0189

*Valence bond theory is what we call a localized bonding model.*0198

*What localized means is the following.*0203

*It is that an atom's atomic orbitals are going to be*0205

*centered around that specific atom when forming covalent bonds.*0209

*In other words, you can think of a specific atom owning*0213

*those electrons rather than being distributed throughout the entire molecule.*0218

*Again that is what we call localized bonding model.*0224

*In order to achieve bond angles that are other than 90 degrees,*0227

*like 120, like 109.5, what valence bond theory suggests is the following.*0231

*It is that an atom's valence orbitals begin to overlap each other.*0239

*When they begin to overlap, they mix.*0245

*They begin to form these mixtures, these combinations of atomic orbitals which we call hybrids or hybrid orbitals.*0248

*What we are going to do first is we are going to take a look at one type of hybrid orbital.*0260

*The first type is what we call an sp3 hybridized carbon atom.*0264

*We go ahead and look at carbon.*0270

*What this sp3 means is that we have one s orbital mixing with three of carbon's p orbitals.*0272

*We can go ahead and show that; let me do that in a different color.*0286

*If I have a carbon s orbital mixing with a p orbital of carbon, we are going to get a hybrid orbital.*0290

*A hybrid orbital is going to have both s character and p character.*0302

*But it is going to be mostly p character because it is one s orbital versus three p's.*0307

*We are going to get a small lobe representative of the s orbital.*0313

*We are going to get a much larger lobe that is representative of the p orbital.*0319

*This is what we call a hybrid orbital.*0326

*Now that we know what the hybrid orbitals look like, they are going to be centered around the carbon atom.*0329

*Remember it is what we call a localized bonding model.*0335

*We are going to have the s orbital right in the middle, closest to the nucleus remember.*0338

*There is going to be one hybrid here; one hybrid here.*0343

*One hybrid here; one hybrid just in the back.*0347

*As you can see the way we have drawn it,*0352

*if you connect the vertices, we do indeed get a tetrahedron.*0355

*The concept of hybrid formation allows us to get these bond angles that are predicted by VSEPR theory.*0366

*Let's go ahead and put that into practice now.*0375

*Now that we physically have a sense of what a hybrid orbital can quote unquote look like.*0378

*Sp3 first, we are going to first start paying attention to what the superscripts mean.*0385

*The superscript tells us actually a great deal.*0391

*In sp3, there is a superscript of 1 for the s orbital.*0395

*There is a superscript of 3 for the p orbital.*0400

*What that translates to is the following.*0403

*That one s orbital mixes with three p orbitals to form the*0405

*sp3 hybrids just like we have seen in the previous slide.*0412

*But the question is how many hybrids do we form?*0416

*Again we turn to the sum of the superscripts to help us answer that.*0419

*Basically the sum of the superscripts equals to the number of hybrid orbitals that are going to form.*0426

*When you look at sp3, the sum of the superscripts is four.*0435

*That is why we form four hybrid orbitals.*0439

*For example, let's go ahead and take a look at water*0445

*and see how hybridization can help to explain for bonding.*0452

*Water has the following structure--the central oxygen, H, and two lone pair.*0457

*Once again the number of electron groups, it is going to be the same as the sum of the superscripts.*0467

*Here we have four electron groups surrounding oxygen.*0491

*There is going to be a total of four for the sum of the superscripts.*0494

*That is going to be sp3.*0499

*We are going to over other hybridization what we call schemes.*0500

*Sp3 is the first one; this is sp3 hybridization.*0505

*We can show this in an energy level diagram; energy.*0511

*Here the sp3 hybridized atom is going to be the oxygen.*0517

*Oxygen starts off as 2s and then 2p.*0522

*Let's go ahead and fill in the electrons now.*0527

*Oxygen is going to be 2s ^{2} and 2p^{4}; this is just atomic oxygen by itself.*0532

*But as soon as oxygen hooks up with those two hydrogens to form water, its atomic orbitals are going to hybridize.*0542

*Again we are going to form a total of four sp3s.*0550

*Where are those four going to form?*0554

*Are they going to form higher in energy than the 2p's?*0556

*Are they going to be less than the energy of the 2s's?*0559

*Or are they going to be intermediate?*0564

*It turns out that we are going to compromise.*0567

*We are going to have an energy that is going to be intermediate for the entire system.*0570

*We form four sp3 hybrid orbitals now.*0577

*When we go ahead now and fill electrons, let's go ahead and fill it.*0581

*I have a total of six electrons--one, two, three, four, five, and six.*0584

*When we do this successfully, the overall picture should agree with the Lewis structure.*0593

*Guess what?--just like the Lewis structure which has two lone pairs around the oxygen,*0599

*we have one lone pair here and one lone pair here.*0604

*What do you think these two electrons are?*0608

*We have two hydrogens bonded to the oxygen.*0612

*These are going to be bonding electrons with H right here.*0616

*These electrons go on to form a single bond with the hydrogen; form single bond with H.*0629

*We now introduce a new terminology.*0642

*A single bond in terms of valence bond theory is a specific type of bond we call a σ bond; σ bond.*0645

*When we go ahead and redraw water, H, H, and O, we are going to see that*0657

*the lone pairs, they are going to exist in sp3 hybrid orbitals of oxygen.*0667

*We are going to have another sp3 here and another sp3 here, each containing one electron.*0676

*Those sp3 orbitals are going to overlap and mix with the hydrogen 1s orbitals which also have one electron.*0689

*Same thing here; that is what we call a σ bond.*0698

*Again we get the very important term, what we call a σ bond, from valence bond theory.*0706

*Let's go ahead and look at one more example of an sp3 hybridized atom.*0717

*The example I would like to look at is ammonia, NH _{3}.*0727

*Step one is always do the Lewis structure.*0735

*Here the Lewis structure for ammonia is going to be here just like that.*0742

*It is going to be three electron groups plus a lone pair*0751

*giving you a total of four electron groups around the nitrogen atom.*0755

*Remember the number of electron groups equals to the sum of the superscripts so this is also sp3.*0760

*Step two, we are going to now come up with our bonding scheme*0771

*and show the formation of these hybrids; show hybrid formation.*0776

*Hopefully again when we are done with this, we can correlate it with the Lewis structure.*0785

*Draw your energy diagram again; there is energy.*0801

*Nitrogen, just bare atomic nitrogen, is going to be 2s ^{2} and 2p^{3}.*0807

*But as soon as the nitrogen starts to form bonds with hydrogen to form ammonia, it is going to mix.*0818

*It is going to form again four sp3 hybrid orbitals.*0827

*Let's go ahead and put the electrons in.*0833

*I have a total of five electrons--one, two, three, four, and five.*0835

*Let's go ahead and see if we can correlate it with the Lewis structure.*0842

*The Lewis structure has one lone pair.*0846

*The lone pair is right there; that is so far so good.*0849

*You notice that there is three single bonds with hydrogen.*0853

*Remember we call the single bond a σ bond.*0856

*σ here, σ here, and σ here; guess what?*0859

*We have three electrons that are each going to form a σ bond with the hydrogen.*0862

*Once again they each form a σ bond with a hydrogen 1s atomic orbital.*0870

*Good, we have a pretty good understanding now of what we mean by sp3 hybridization.*0885

*The next bonding type, the next hybridization type is what we call sp2.*0893

*Again let's look at the sum of the superscripts; that tells us a great deal.*0900

*Sp2, the sum of the superscripts is a grand total of three which means*0906

*an atom is going to be sp2 hybridized when it has three electron groups around it.*0915

*In addition remember that the sum of the superscripts also tells us how many hybrids we are going to form.*0923

*Here we are going to form exactly three sp2 hybrids.*0931

*One more thing, the superscripts also tells us the number of each atomic orbital that is going to be used.*0940

*Here we are going to be having one s orbital mixing with exactly two p orbitals.*0948

*Let's think about that.*0959

*We know that every energy we said has a p _{x}, p_{y}, and p_{z}, three p orbitals.*0961

*But if only two p orbitals are being used out of the three, that tells us the difference.*0967

*One of the p orbitals remains unused, unhybridized.*0974

*I am going to put that in; one p orbital is going to be unhybridized.*0981

*We need to now account for that in our energy diagram.*0990

*Let's go ahead and do that.*0995

*A nice example of something that is going to be sp2 hybridized, let's go ahead and look at BF _{3}.*0997

*BF _{3} here; here is energy.*1013

*Regular boron itself is going to be 2s ^{2} and then just 2p^{1}.*1019

*When that goes ahead and mix, we are going to form exactly three sp2s intermediate in energy.*1030

*Remember that?--always intermediate in energy approximately.*1037

*This one p orbital that remains unhybridized, if it remains unhybridized, that means*1042

*it is going to be the same energy as the initial p orbitals.*1048

*I am going to put that approximately the same just like that.*1056

*Again we have exactly three sp2s; and we have a 2p orbital that remains unhybridized.*1059

*We are going to go ahead and fill the electrons now.*1066

*In BF _{3}, boron just has three electrons--one, two, and three.*1069

*Guess what?--there is no lone pairs.*1075

*There is going to be exactly three σ bonds.*1077

*When we draw out the Lewis structure for boron trifluoride, that is exactly what we get.*1080

*Boron has zero lone pairs in this structure and exactly three σ bonds.*1087

*That looks pretty good right now.*1098

*But let's go ahead and look at another example.*1100

*Here we are going to learn now a new type of bonding.*1103

*An example I want to look at has the following Lewis structure.*1106

*This is going to be a CH, a double bond, O just like that.*1110

*This is what we call formaldehyde.*1120

*Here formaldehyde, the central carbon atom has three electron groups.*1125

*That is also going to be sp2 hybridized.*1131

*Let's go ahead and set up the energy level diagram now for the carbon here.*1136

*Carbon is 2s and 2p; it is going to be 2s ^{2} 2p^{2}.*1140

*That is going to hybridize.*1148

*Again we are going to form exactly three sp2s--one, two, three.*1150

*We are going to have one of the 2p orbitals remaining unmixed just like that.*1154

*Let's go ahead and fill electrons now; carbon has four valence electrons.*1159

*Let's go ahead and populate the orbitals with four--one, two, and three.*1163

*I am going to put a big asterisk here because this is where we start to deviate.*1169

*Where do we put the fourth electron?*1173

*Going by typical Aufbau principle, we would always populate the lower energy*1175

*orbitals first before proceeding on to the higher atomic orbitals.*1182

*But in valence bond theory, we are going to go left to right regardless of energy.*1187

*We are going to go every orbital one time before doubling up.*1193

*I am going to put my fourth electron here, not in the sp2.*1196

*That is this special case for valence bond theory is really*1202

*a big deviation from what we know from Aufbau principle.*1207

*Please again keep that in mind.*1210

*When we do this see, it is going to work out because look at the Lewis structure.*1214

*The Lewis structure has zero lone pairs for the carbon atom.*1218

*As you can see in our bonding scheme, there is zero lone pairs.*1222

*We have the following; we have a σ bond here and a σ bond here; σ, σ.*1227

*And we have a double bond.*1234

*We have to now discuss what we mean by double bond.*1237

*A double bond we learned from Lewis theory has exactly four valence electrons.*1241

*But what type of bonds?*1245

*It turns out that a double bond is composed of the following.*1248

*It is composed of one σ bond and a new type of bond, what we call a π bond.*1250

*Our σ bond is just going to be the regular σ bond from the hybrid orbitals.*1258

*Guess what?--this electron that we have right here in the unhybridized atomic orbital,*1262

*that is going to overlap with oxygen to form a π bond.*1268

*One σ plus one π is equal to a double bond.*1272

*We can make a pretty big conclusion right now from these two exercises.*1279

*Lone pairs and σ bonds are going to come from hybrid orbitals.*1288

*π bonds are going to come from atomic orbitals that are unhybridized.*1300

*You may be wondering then what type of bond is stronger, is it σ or π?*1311

*The lower the energy, the more stable it is going to be.*1318

*The higher the energy, the more unstable it is going to be.*1322

*We are going to learn that very soon in our next discussion of thermodynamics.*1326

*σ bonds are typically going to be stronger than π bonds.*1333

*The reason why σ bonds are stronger is because once again*1341

*lower in energy but also we have more overlap of the orbitals.*1345

*If we have more overlap, the attraction between the nuclei is going to be much stronger.*1355

*That again is sp2 hybridization.*1363

*The next one I want to go over with you is what we call sp hybridization.*1367

*Let's go ahead and take a look at the sum of the superscripts.*1372

*Here is really s ^{1} p^{1}.*1375

*Remember if you don't see a number, 1 is always implied.*1378

*That means we are going to have two electron groups around the central atom.*1381

*Also this tells us that we are going to form a hybrid here by one s mixing with exactly one p.*1392

*Also we are going to form exactly two hybrid orbitals; two sp hybrids.*1404

*Let's go ahead and take a look; let's look at hydrogen cyanide; hydrogen cyanide.*1415

*Step one, it has the following Lewis structure--H, C, triple bond, N.*1424

*As you can see, carbon is two electron groups; that is going to be sp.*1430

*Also nitrogen also has two electron groups around it.*1436

*That too is going to be sp hybridized.*1439

*Let's first tackle carbon.*1441

*Then we will tackle nitrogen to see if it agrees with the Lewis structure.*1443

*Energy; carbon is again 2s 2p.*1447

*Put in its electrons; carbon is 2s ^{2} 2p^{2}.*1454

*That is going to go ahead and mix.*1461

*Again we are going to form exactly two sp orbitals; two sp hybrids.*1463

*It tells us that to form the two sp's, it is the mixing of one s and one p.*1470

*But remember there is three p orbitals per energy level; p _{x}, p_{y}, and p_{z}.*1476

*If I am only using one of those three, that means I have two remaining unhybridized.*1481

*We are also going to put that in; where do we put it in?*1487

*We put it in that it has exactly the same energy level as the initial two p orbitals.*1491

*Let's fill electrons--one, two, and don't forget this is where we deviate.*1497

*We go fill left to right all the orbitals first before double up regardless of energy.*1502

*One electron, two electrons, and now electron number three, and electron number four.*1508

*So far so good; we don't see any lone pairs at all.*1514

*The Lewis structure has no lone pairs around the carbon.*1518

*Let's see what HCN has around carbon; no lone pairs.*1523

*This bond right here is a σ bond; I am going to label that σ.*1531

*Now the nature of the triple bond.*1535

*We saw that a single bond is what we call a σ.*1539

*We saw that a double bond is one σ and one π.*1542

*A triple bond is going to be one σ but two π this time.*1546

*Again a triple bond is going to be one σ and two π.*1551

*We know that σ comes from hybrid orbitals; there is our σ.*1557

*We know that π bonds come from unhybridized atomic orbitals right here; π here and π there.*1561

*As you can see, we have two σ and two π's around carbon which is agreeing with the Lewis structure.*1569

*Let's see if we can get the same agreement with nitrogen this time.*1577

*Energy; nitrogen, 2s 2p; nitrogen is 2s ^{2} and now 2p^{3}.*1582

*Again as we already discussed, this nitrogen atom is also sp hybridized.*1595

*That is going to be one, two.*1600

*We are going to form two p's right here; exactly two of them unhybridized.*1602

*Let's go ahead and fill in the electrons.*1606

*This time it is going to be five--one, two, three, four...*1607

*OK, now we can double up again with the opposite spin.*1611

*According to the Lewis structure for nitrogen in HCN, there is going to be one lone pair.*1617

*There is going to be a triple bond around nitrogen which is one σ and two π.*1626

*Guess what?--it agrees perfectly.*1632

*Here is our lone pair; here is our one σ; here is the two π's.*1636

*Again we see that valence bond theory helps us to count for the*1645

*nature of a single, a double, and now a triple covalent bond.*1650

*There are two other atomic hybridizations that you should be familiar with.*1658

*So far we saw that sp3 is going to be for four electron groups.*1663

*That is going to be a tetrahedral electron geometry.*1671

*We saw that sp2 is going to be three electron groups.*1679

*That is going to be a trigonal planar geometry; again electron geometry.*1684

*We just did sp; we saw that sp was two electron groups.*1692

*That was a linear electron geometry.*1696

*Now sp 3d; sp 3d, now we start talking about the d block orbitals.*1700

*That is a grand total of five electron groups.*1708

*If you recall, five electron groups was a trigonal bipyramid electron geometry.*1712

*Finally sp3 d2, that is six electron groups.*1722

*That is going to be an octahedral electron geometry.*1728

*Again it depends on your instructor and your textbook.*1734

*But I will just present to you the five basic atomic hybridization schemes.*1739

*That is the formation of hybrid orbitals and valence bond theory.*1746

*Valence bond theory as we just saw is successful in explaining for the observed bond angles predicted by VSEPR theory.*1753

*However we still have that one issue about magnetism.*1764

*Take oxygen for example; oxygen's Lewis structure is right here.*1768

*If you look at this, every electron is paired; we would expect diamagnetism; expected.*1776

*In other words, if we take molecular oxygen, it should be repelled by a magnetic field.*1787

*But experimentally when you pour liquid oxygen through a magnet,*1794

*it actually sticks to the magnet showing that O _{2} is actually paramagnetic.*1801

*Obviously valence bond theory does not always successfully account or predict the actual magnetic behavior of a compound.*1814

*Then we need yet another bonding theory.*1826

*This bonding theory is what we call molecular orbital theory, also known as MO theory.*1829

*Here is the main difference.*1840

*In contrast to Lewis and valence bond theory, MO theory is a delocalized bonding model; not localized but delocalized.*1842

*In other words, the electron density is distributed throughout the*1853

*entire molecule rather than being centered around any individual atom.*1857

*You remember our discussion about wave functions.*1866

*It turns out that MO theory is a highly mathematically based theory.*1870

*If you recall the nature of a wave function, a wave function is representative of atomic orbitals.*1878

*The actual function if looked at them contain cosine and/or sine trigonometric functions.*1883

*If we try to think back to that, your basic cosine or sine wave function is a wave.*1891

*From your early math classes, you learned that these waves have both*1901

*positive and negative what are called phases; positive and negative phases.*1909

*Because atomic orbitals are the visual manifestation of a wave function, we can also infer that*1923

*atomic orbitals also can be represented by different phases, a + phase and a ? phase.*1932

*For general chemistry, we are only going to limit our discussion to s and p orbitals.*1945

*Let's first take a look at how s orbitals can interact.*1952

*I can have an s orbital of this phase mixing with another s orbital of the same phase.*1958

*When that happens, I form what is called a σ bonding orbital or a σ bonding MO.*1970

*How does that look like?*1980

*If here the dots represent the nuclei and if you have exactly two of these s orbitals*1982

*mixing with each other, we get a smeared out electron density throughout both nuclei.*1990

*Here is the nuclei.*1997

*But now we are going to get something that looks like this.*1999

*Again this is what we call a σ bonding MO.*2005

*Instead of two separate s orbitals, they now mix and form one entity.*2009

*As you can see from this diagram, it is completely distributed among both nuclei.*2014

*This is what we call delocalized.*2023

*The other possibility can occur.*2026

*What happens when we have s orbitals mixing that are of opposite phases?*2028

*To represent opposite phases, I am going to have now shading in one of these s orbitals just like that.*2034

*Now when I have s orbitals of opposite phases interacting,*2044

*this is going to be equivalent to destructive interference.*2049

*Remember we can interpret orbitals as wave-like properties.*2056

*When we have destructive interference, we actually get the following this time.*2062

*We are going to get something that looks like this now.*2069

*This is actually a node.*2075

*Remember we talked about nodes where we have zero electron probability.*2077

*This type of MO is not a σ bonding anymore.*2085

*But it is called a σ anti bonding MO; a σ antibonding MO.*2089

*For our purposes right now, something you have to remember is that*2099

*σ bonding will always be lower in energy than σ antibonding.*2102

*Once again σ bonding is going to lower in energy to σ antibonding.*2109

*Remember mother nature is always going to favor lower energy.*2114

*If an orbital is going to result in bonding of the molecule, it is going to be lower in energy.*2117

*Once again we went over σ bonding and σ antibonding.*2124

*σ antibonding will typically be represented as a σ*.*2129

*We talked about s orbitals; let's now talk about p orbitals.*2135

*P orbitals can interact in one of two ways.*2139

*They can interact head on or in a parallel fashion.*2143

*Let's talk about head on first.*2152

*When I mean head on, if we have a molecule AB like that, the bonding axis is going to be z-axis.*2154

*I am going to represent this as x; then represent that as y.*2166

*Once again the z-axis is going to be colinear with the bonding axis.*2169

*When we have a head on overlap of p orbitals, it is going to be the p _{z} orbitals.*2174

*Here is one p _{z}; here is another p_{z} interacting head on.*2182

*We can have the same phases of the p _{z} orbitals overlapping with each other like that.*2189

*When that happens, we form a σ bonding orbital.*2203

*Again we form a σ bonding orbital.*2210

*The MO is going to be distributed throughout the entire molecule around both nuclei.*2214

*We are going to get something that looks like this now*2221

*where the two nuclei are here, here, here, and here.*2230

*Again this is what we call a σ bonding MO.*2238

*You may ask then what is the other possibility?*2242

*The other possibility is for the p _{z} orbitals now to interact with their opposite phases like that.*2244

*When the opposite phases interact, it is just like we saw last time with the s orbitals.*2266

*That is going to be antibonding; this is going to be σ*.*2270

*That is going to be represented the following way where the two nuclei again are here and here.*2276

*Let's go ahead and draw the nodes in, shall we?*2295

*Here you see that there is one node here, one node here.*2298

*You see here for the antibonding one node, one node, and one node.*2302

*I want you to compare and contrast.*2307

*You notice that the antibonding MO for the p orbitals has three nodes*2310

*while the bonding MO for the p orbitals has two nodes.*2316

*The same pattern occurs for the s orbitals.*2320

*You notice that the antibonding MO for the σ has one node*2324

*while the bonding MO for the σ has zero nodes.*2331

*The antibonding MOs typically have more nodes than the bonding MOs.*2338

*Let's go ahead and look at the parallel interaction.*2346

*Parallel interaction is going to occur for the remaining two p orbitals, basically p _{x} and p_{y}.*2349

*If I have parallel interaction this time, I can have p orbitals occurring like that.*2355

*When we have parallel interaction, the overlap is not as great.*2365

*This is not a σ MO anymore.*2370

*Instead it is going to be a π MO.*2372

*Because it is the identical phases interacting with each other,*2375

*this is going to be what is called a π bonding MO.*2378

*We are going to get an MO that looks like this now distributed among both nuclei just like that.*2381

*Here we have exactly one node.*2395

*The other possibility is for parallel interaction but now of opposite phases like that.*2401

*When we have opposite phases again that is what we call an antibonding MO.*2409

*This is going to be π*.*2413

*π* is going to look like this distributed throughout the entire compound.*2415

*Let me go ahead and erase that.*2428

*As you can see once again the antibonding MO has more nodes than the corresponding bonding MO.*2434

*One other possibility is having an s orbital directly combined with a p orbital in a head to head fashion.*2448

*That is going to be an s orbital combining with the p _{z} orbital just like that.*2455

*Again that is going to be σ bonding.*2464

*Again we can also have s and p opposite phases.*2467

*That is going to be σ antibonding.*2472

*We pretty much covered all our bases and the different possibilities of the different types*2477

*of MOs that can form from s and p and how they look like.*2482

*Now that we have the combinations down, let's put it all together.*2490

*Let's go ahead and construct what we call a typical MO diagram.*2494

*The simplest type of compounds we are going to be looking at again are going to be homonuclear diatomics.*2499

*Remember two things; σ bonding is going to be less than σ antibonding.*2507

*Similarly π bonding is going to be less than energy of π antibonding.*2518

*σ versus π, σ bonding versus π bonding, that is going to become a major issue pretty soon.*2527

*But let's just go over the small ones first.*2535

*We will come across this issue here.*2540

*Hydrogen H _{2}, molecular H_{2}, two hydrogen atoms are going to combine.*2542

*Hydrogen is very simple.*2550

*We just have the hydrogen 1s orbital here and a hydrogen 1s orbital here.*2551

*Each of them contains one electron.*2555

*When hydrogens come together, we get the formation of MOs.*2560

*When an s orbital combines with an s orbital, remember what the two possibilities were?*2566

*It is going to be σ bonding.*2571

*The other possibility at higher energy is going to be σ antibonding.*2574

*Again bonding is going to be lower in energy than antibonding; that applies there.*2580

*There is some things I want to point out.*2591

*The number of MOs that you form must always equal to the number of atomic orbitals that you initially start with.*2593

*That is another way you check your work.*2601

*Here I start with a total of two atomic orbitals.*2603

*I wind up with two MOs--one bonding, one antibonding.*2606

*Let's go ahead and fill electrons going by the standard Aufbau principle.*2609

*That is going to be one electron here and then one electron here; spin up, spin down.*2615

*Now so what?--what is the big deal?--the big deal is the following.*2621

*We can now calculate something we call bond order.*2626

*Remember bond order?--it was one of those characteristics of a bond.*2629

*The equation for bond order is going to be the following.*2634

*It is going to be equal to the 1/2 number of bonding electrons minus the number of antibonding electrons.*2638

*When we do it for H _{2}, we get 1.*2654

*Remember what bond order of 1 means?*2657

*A bond order of 1 means we have a single bond formed, expected to form.*2659

*We know H _{2} exists, we know hydrogen exists as a diatomic gas.*2668

*It is one of those several elements that occur diatomically.*2673

*Yes, hydrogen is expected to form.*2678

*Again all of these experiments are done in the gas phase.*2680

*Yes, we know that hydrogen exists as diatomic hydrogen in the gas phase.*2686

*A bond order of 1 is what we expect.*2690

*Remember if we have a bond order of 2, we expect double bond.*2693

*We have a bond order of 3, we expect a triple bond.*2698

*What happens if we have bond order less than 1?*2703

*If we have a bond order of less than 1, that is less than a single bond which means*2706

*we don't expect the molecule to exist at all in the gas phase; should not form.*2709

*You see the power of MO theory already.*2716

*It allows us to predict if a molecule is going to be stable or not in the gas phase.*2718

*Let's talk about helium; helium is a noble gas.*2724

*We know that noble gases exist as monoatomic gases, not diatomically.*2729

*Expectation for He _{2} is it should not exist.*2734

*In this case, we expect a bond order of less than 1.*2741

*BO less than 1 expected.*2746

*Let's go ahead and look at the energy diagram here.*2750

*Once again helium here, helium here, and again 1s, 1s.*2754

*Each helium is 1s ^{2}; now let's go ahead and make the MOs.*2760

*Again anytime we have 1s combining, one possibility is going to be σ bonding.*2766

*The other possibility is going to be σ antibonding.*2772

*Now let's go ahead and fill the electrons; one, two, three, and four.*2777

*When we go ahead and look at the bond order equation for He _{2},*2782

*this is going to equal to 1/2, 2 minus 2, which is going to be 0.*2787

*Again a bond order of less than 1 is what we expected.*2793

*That is what we actually get.*2797

*We are going to keep on moving down.*2800

*Now we are going to look at diatomic lithium.*2802

*Let's see if we expect diatomic lithium to exist in the gas phase.*2805

*Here lithium and lithium; lithium is going to be 2s, 2s.*2810

*Remember we only care about valence electrons; here 2s ^{1} 2s^{1}.*2823

*Once again when we have s orbitals, we can expect formation of*2830

*a σ bonding orbital and then of a σ antibonding orbital.*2835

*Let's go ahead and put the electrons in--one electron here, one electron here.*2841

*Lithium, Li _{2}, this bond order is going to be 1/2 times 2 minus 0 which is going to be 1.*2845

*Yes, Li _{2} is observed experimentally in the gas phase.*2853

*A bond order of 1 does make sense.*2858

*Let's move on to the following.*2864

*It turns out that not only can we do MO diagrams for*2865

*neutral species but we can also do it for cations.*2869

*Now this is Li _{2}^{1+}; Li, Li, Li_{2}^{1+} in the middle.*2874

*Again lithium is a 2s atomic orbital; they are going to mix.*2882

*We are going to form σ bonding and then σ antibonding.*2888

*Lithium is 2s ^{1}.*2897

*Again because this is going to be 1+, that means we lose one of the electrons.*2900

*I am going to leave this one blank.*2904

*I am going to leave this one blank to show that we remove one electron.*2906

*Its MO is just going to be that, just one electron in the bonding MO.*2913

*BO or the bonding order is going to be equal to 1/2 times 1 minus 0.*2917

*We get an order of 1/2.*2922

*This suggests that the Li _{2} cation should not occur in the*2925

*gas phase because here the bond order is less than 1.*2931

*Good; we now move on to B _{2}.*2938

*The reason why this is now going to be slightly different is*2943

*because as soon as we enter boron, we enter the p block.*2946

*We have not have not talked about how the p orbitals are going to be expected to be looking.*2953

*What do you get more overlap with?*2960

*Is it going to be s and s?--or is it going to be p and p?*2963

*The s orbitals are going to be closer to the nucleus.*2967

*We expect them to be lower in energy; we know that from Aufbau.*2974

*For example, 2s is usually less in energy than 2p.*2978

*It turns out that the σ bonding here is going to be less than the π bonding.*2982

*What we expect because typically we have more overlap with the σ's than we do with π's.*2993

*Remember σ is more direct overlap; π is parallel interaction.*3000

*We expect σ bonding to be less than π bonding; let's see what happens.*3005

*Boron, boron; boron is going to be a 2s orbital here, 2s orbital here, and 2p's, 2p's.*3016

*Let's fill the electrons in for boron; each boron is 2s ^{2} 2p^{1}, 2s^{2} 2p^{1}.*3028

*Let's go ahead and tackle what we know so far already.*3035

*We know that the 2s orbitals can mix.*3038

*They are going to form our typical σ bonding and σ antibonding.*3042

*But what about the p orbitals?--remember the p orbitals.*3048

*Remember that when p _{z} interacts with p_{z}, we can get σ bonding and σ antibonding.*3053

*Remember that p _{x} and p_{y} are the same.*3062

*Those guys are going to give us π bonding and π antibonding.*3071

*Notice that there is two sets.*3080

*It turns outs that when we talk about diatomic boron, it turns out that the π orbitals are*3086

*going to be lower in energy than the σ orbitals just like that and*3098

*that the π*s are going to be lower in energy than the σ*.*3116

*That is not what is expected.*3125

*Remember we expect σ bonding to be less than π bonding.*3126

*What happens is the following; the reason why σ is above π.*3132

*It is due to the phenomenon that we call sp mixing.*3139

*What happens is the following.*3145

*That a boron 2s orbital is actually going to mix or interact with the boron 2p orbital.*3147

*What that does is it results in a lowering of the σ bonding MO from 2s*3162

*but a raising in energy of the σ bonding orbitals from the 2p's.*3176

*Essentially one goes down lower in energy; one goes up.*3186

*It makes sense that the 2s goes down because it is more stabilized due to the mixing.*3191

*Remember 2s's are going to be closer to the nucleus.*3195

*They get affected more than 2p's due to this interaction.*3198

*For B _{2}, again the big take-home message is that the π bonding MOs are going to be*3203

*lower in energy than the σ bonding MOs of the 2p's due to sp mixing.*3209

*That is our MO diagram; let's go ahead and fill it up now.*3216

*I have a total of six electrons--one, two, three, four, five, and six.*3220

*When we go ahead and calculate now the bond order, the bond order here is going to be equal to 1/2...*3226

*number of bonding electrons is 4, two antibonding electrons.*3232

*That is going to be a bond order of 1.*3236

*Yes, we do expect diatomic boron to exist in the gas phase.*3238

*Let's look at another example, N _{2}.*3246

*N _{2} is something we have looked at already in terms of Lewis structure.*3249

*Look at that; that is our Lewis structure of N _{2}.*3253

*We expect a bond order of 3; let's see if we get that.*3255

*Nitrogen, nitrogen, this is 2s, 2s; then here is nitrogen's 2p orbitals just like that.*3266

*N _{2} is right in the middle.*3276

*Once again the 2s orbitals are going to mix to form σ and σ*.*3279

*Again the π orbitals from the 2p's are going to be lower in energy than the σ orbitals from the 2p's.*3288

*Then followed by π* and σ*.*3299

*Let's go ahead and populate the orbitals with the electrons.*3307

*Nitrogen has a configuration of 2s ^{2} 2p^{3}, 2s^{2} 2p^{3}.*3311

*Now let's go ahead and fill; that is a total of ten electrons.*3321

*One, two, three, four, five, six, seven, eight, nine, and ten.*3324

*When we go ahead and calculate the bond order, 1/2 number of bonding electron which is going to be 8.*3335

*Number of antibonding electrons is 2.*3342

*Look at that; we get a bond order of 3 which is what we expect.*3345

*The accepted Lewis structure for N _{2} is a triple bond.*3351

*The next compound is going to be molecular oxygen.*3359

*Molecular oxygen, that is the one where we get paramagnetism from experiment.*3363

*Paramagnetism expected which means the MO diagram should have at least one unpaired electron.*3373

*Let's go ahead and look at the MO diagram; oxygen here--2s, 2s, and 2p.*3397

*O _{2} right in the middle; let's go ahead and form our MOs.*3413

*Again the 2s's can form σ bonding and σ antibonding and 2p's.*3420

*Let's just for now go by what we have already done*3433

*for the previous compounds where the π's were less than σ.*3436

*This is π; this is σ; and then this is π* and then σ*.*3445

*Let's put the electrons in--2s ^{2} 2p^{4}, 2s^{2} 2p^{4}.*3455

*That is a grand total of twelve electrons that we have to use.*3467

*Let's fill them--one, two, three, four, five, six, seven, eight, nine, ten, and eleven and twelve.*3470

*Let's go ahead now and see do we have lone pair expected?*3482

*Yes, we have lone pairs right there; paramagnetism is good to go.*3487

*But let's go ahead and look at the bond order.*3490

*The bond order is the following.*3492

*1/2 times the number of bonding electrons minus the number of antibonding electrons.*3496

*That is going to be number of bonding electrons is 8*3501

*minus number of antibonding electrons is going to be 4.*3505

*That is going to be 2; yes, double bond expected.*3510

*So far this sample diagram does work.*3514

*However experimentally it turns out that this is not the correct MO diagram for O _{2}.*3517

*It turns out that in the actual MO diagram, these orbitals here,*3525

*the one we had that little controversy with is flipped.*3537

*It turns out that in the actual MO diagram, the σ is going to be less than π.*3542

*Let's go ahead and draw that.*3547

*I will explain why in a second why that makes sense.*3549

*2s ^{2} and then 2p^{4}; let's go ahead and fill everything.*3552

*Here we have our σ bonding, σ antibonding.*3575

*Here is the actual order; σ on the bottom, π on the bottom, then π* here, then σ* here.*3585

*Let me go ahead and label that; σ, π, π*, and σ*.*3595

*Let's go ahead and fill electrons right now.*3601

*One, two, three, four, five, six, seven, eight, nine, and ten, and then eleven and twelve.*3604

*It turns out that this is going to be the accepted one where σ is less than π.*3618

*The reason is because of the following--the sp mixing that we discussed earlier stops.*3624

*The reason why it stops is because it turns out when we reach oxygen*3635

*in the periodic table, it is a relatively heavier element than say nitrogen.*3638

*What that does is the following.*3645

*It is that the s orbital is now too low in energy to mix with the p orbital.*3647

*As the atom gets heavier and heavier, most of the interaction is going to come with the s orbital.*3662

*That is going to if you will benefit first.*3674

*That is going to be much lower in energy.*3677

*Because of that, the s orbital of oxygen is just too low in energy compared to the p orbital of oxygen.*3682

*Sp mixing stops.*3689

*We return to what is expected where the σ bonding MO orbital from the p orbitals is*3692

*going to be less than the energy of the π bonding MO orbitals of the 2p's.*3698

*Again this is always going to be the case; please make a note.*3710

*This is always the case for O _{2} and higher; F_{2} counts, Cl_{2} counts, etc.*3714

*Really everything up to and including N _{2} is one situation.*3727

*Then O _{2} and after is going to be this situation.*3732

*So far we have looked at only homonuclear diatomics.*3738

*We can do MO diagrams also for heteronuclear diatomics too.*3741

*Let's consider hydrochloric acid; hydrochloric acid has the following Lewis structure.*3746

*Here is hydrogen; here is chlorine; hydrogen is 1s.*3756

*Because the chlorine is more electronegative than hydrogen, its 2s orbital is going to be lower in energy than the 1s.*3765

*Then chlorine's 2p is going to be relatively higher in energy there.*3774

*Hydrogen is 1s; chlorine is 3s 3p.*3780

*Chlorine is going to be 3s ^{2} and then 3p^{5} just like that.*3792

*In the Lewis structure for hydrochloric acid, we have a σ bond right there.*3806

*But chlorine has three lone pairs here which means...*3816

*These three lone pairs means that the six electrons are not involved in bonding, are nonbonding.*3823

*In terms of MO purposes, nonbonding electrons will occur not in a*3832

*bonding MO, not in an antibonding MO, but in a nonbonding MO.*3841

*That is a third type of molecular orbital that we now introduce.*3849

*For our purposes in GChem, nonbonding MOs are going to be at*3855

*the same energy as their atomic orbitals; same energy as atomic orbitals.*3858

*Let's go ahead and see what happens then.*3871

*It turns out that the 3s orbital of fluorine is just too low in energy to interact with anything.*3879

*Here is our first nonbonding MO.*3894

*I am going to go ahead and put those two electrons there just like that.*3898

*The next item is now the 1s orbital interacting with the 3p orbitals.*3907

*When this happens, it turns out we are going to get the following experimentally.*3916

*The 3p and the 1s are going to mix.*3922

*They are going to form a σ.*3925

*Of course if they form σ, that means they must also form a σ*.*3927

*Remember we have to conserve orbitals; we have a total of five atomic orbitals.*3936

*We have only used three so far in the MOs.*3943

*That means I have two remaining.*3946

*Those two remaining are going to be additional nonbonding orbitals just like that.*3948

*Let's go ahead and fill everything up.*3955

*We have a grand total of eight electrons; two is already accounted for.*3957

*One, two, three, four, five, six, seven, and eight.*3964

*Guess what?--our MO diagram agrees with our Lewis structure.*3969

*Here is one lone pair nonbonding; here is another lone pair nonbonding.*3974

*Here is another lone pair nonbonding; there we have our σ bond.*3980

*When you calculate the bond order, the bond order is equal to 1/2 number of bonding electrons which is 2.*3986

*You notice that there is no electrons in antibonding MOs.*3993

*You do not count nonbonding electrons in this equation.*3996

*There we go--a bond order of 1 which agrees with the Lewis structure of*3999

*hydrochloric acid where we just get a single covalent bond.*4004

*That is how we tackle heteronuclear diatomics.*4008

*To summarize, valence bond theory was the first bonding theory we talked about today.*4013

*It explains for observed bond angles but doesn't always account for magnetic behavior.*4018

*MO theory is pretty much the one that is going to supersede everything else.*4022

*That is the delocalized bonding model which views electrons as being distributed throughout*4028

*the entire molecule rather than just being centered around any individual atom.*4033

*Let's go ahead and look at a pair of sample problems.*4041

*In this one, determine the atomic hybridization around the central atom in the following molecules.*4044

*I picked these because these were Lewis structures that we have previously seen in the last lecture.*4050

*Here is the Lewis structure for XeO _{3}, for xenon trioxide.*4057

*It is going to look like that.*4061

*Here we have a total of four electron groups around the xenon.*4063

*That is going to be sp3 hybridization.*4068

*SF _{6}, SF_{6} is going to look like this.*4071

*There are lone pairs around each of the fluorines.*4080

*Remember sulfur is one of those elements that can have more than an octet.*4086

*This is completely valid Lewis structure.*4092

*Here we have six electron groups around the sulfur.*4094

*That is going to be a sp3 d2 hybridization.*4097

*Now I _{3}^{1-}, I_{3}^{1-}, its Lewis structure is going to be this.*4102

*When we go ahead and look at this, this is a total of five bonding groups around the iodine.*4124

*That is going to be sp3 d.*4133

*That is atomic hybridization prediction for a central atom from a Lewis structure.*4137

*Now moving on to the final sample problem, sample problem two.*4145

*Draw a qualitative MO diagram for F _{2}^{1-} and comment on its expected stability.*4149

*Let's draw it--energy; fluorine, fluorine, and then F _{2}^{1-}.*4156

*Fluorine has the 2s orbital and then its 2p's; let's go ahead and populate.*4165

*Fluorine is 2s ^{2} 2p^{5} and again 2s^{2} 2p^{5}.*4177

*Don't forget we have a 1- charge.*4190

*I am just going to go ahead and add that to here.*4192

*It doesn't matter which one; they are the same.*4195

*2s orbitals are going to mix.*4198

*They are going to form a σ MO and a σ* MO.*4202

*Here fluorine is after nitrogen; that is when the sp mixing does not occur anymore.*4208

*In this case, from the 2p orbitals, we are going to get σ first*4218

*followed by the π's and then π* and now σ*.*4224

*σ, π, π*, σ*; let's go ahead and fill the electrons in.*4233

*I am going to have a grand total of fifteen electrons to use.*4239

*One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, and fifteen.*4244

*Let's go ahead and calculate its bond order in order to comment on the stability.*4258

*The bond order is equal to 1/2 number of bonding electrons--that is going to be 8.*4262

*Minus number of antibonding electrons--that is going to be 7.*4270

*In this case, we get a bond order of 1/2 which is less than 1.*4278

*Therefore we do not expect F _{2}^{1-} to occur in the gas phase.*4282

*That is another qualitative MO problem.*4292

*Thank you all for your attention; I will see you all next time on Educator.com.*4297

0 answers

Post by Gaurav Kumar on October 25 at 03:31:09 PM

Great lecture...really helped clear things up. Thanks!

1 answer

Last reply by: Professor Franklin Ow

Sat Apr 26, 2014 5:18 PM

Post by Yay Esme on April 21 at 01:03:39 PM

why do you have 2 sigmas when carbon only has one sigma bond ?