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Chemistry: An Introduction to Hybrid Orbitals

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  • Type: Video Tutorial
  • Length: 14:11
  • Media: Video/mp4
  • Use: Watch Online & Download
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  • Download: MP4 (iPod compatible)
  • Size: 153 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Molecular Geometry and Bonding Theory (11 lessons, $18.81)
Chemistry: Valence Bond & Molecular Orbital Theory (7 lessons, $12.87)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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Thinkwell
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I have a question for you. Based on the explanation of valance bond theory that I've given you so far, what does valance bond theory predict should happen when carbon atoms and hydrogen atoms react with each other? Let's take a look. Hydrogen has one valance electron in a 1s orbital and carbon has two unpaired electrons in 2p orbitals. Remember, valance bond theory is about overlapping of orbitals that each have one unpaired electron in them and that's the nature of the bond. The electrons can get together and pair up and that's what forms the bond. So based on this picture, if we take hydrogen atoms and carbon atoms and allow them to react, what we would predict is something like this where we have two hydrogen atoms reacting with the carbon atom to form a CH[2] molecule where one of the hydrogen 1s orbitals overlaps with, say, the p[x], and the reason why I've talked about p[x] and p[y] is because I do want to emphasize in this lecture the directionality of orbitals. The p[x] and the p[y], you'll remember, are orthogonal. They're perpendicular to each other and so this hydrogen-carbon-hydrogen bond is going to be 90. And again, the bonding results from the overlap of the hydrogen 1s with one of the p orbitals to form one of the bonds, and then with a different carbon p orbital to form the other bond. And we call both of these sigma bonds because, you remember, sigma bonding is when you have overlapped increased electron density between the two nuclei.
All right, what's the problem with this picture? The problem with this picture is, just happens to be not quite right. It turns out that that molecule, where the bond angle really is 90, is called carbene, or a carbene, and it exists but it is very unstable. What we know, based on more information, is that the simplest hydrocarbon or compound composed only of carbon and hydrogen is methane, CH[4], meaning that somehow we have to make four bonds between hydrogen and carbon--not just two bonds as what existed in carbene. And so, imagine this: we're going to take valance bond theory as we know it up until now and we're going to put a twist on it, and the twist is to suggest that we can take an electron here that's in a 2s orbital on carbon and we can actually put some energy in and move that electron over here. And don't worry about the fact that I flipped the electron spin over. Just, here's the picture, now that has four unpaired electrons, and four unpaired electrons means that we can make four bonds because, remember, you need an orbital with an unpaired electron to make a bond. Why would we want to do this? And the answer is, we want to do it because making bonds is always an exothermic process. Remember, when you make bonds you get energy out. That's a good thing. That lowers the energy of the molecule to make a bond. So being able to make four bonds instead of two bonds is a good thing. But if we believe, remember that p orbitals are directional, then one of the bonds would be to a p[x] orbital, one would be to a p[y][]orbital and one would be to a p[z] orbital and then one would be to a 2s orbital.
If you think about the relative orientations of those orbitals, we'd have 90, 90, 90, and this hydrogen would just be out here somewhere. Presumably it would just try to stay away from these hydrogens here based on VSEPR kind of ideas of repulsion of electron clouds. Now, that would be our prediction based on the picture that I just showed you, and the reality is that methane is a perfectly tetrahedral molecule with the bond angles at 109.5 so we have to tweak our model again.
And the way we're going to tweak the model is something we're going to call introducing the concept of hybridization. And for methane, and, this is important, all other models that have steric number 4, so hybridization and steric number are intimately related concepts. First we pay the promotion energy. In other words, we put some energy in to put an electron so that we have one electron in the 2s and one in each of the two p's. And then what we're going to do is we're going to mix and match these four orbitals and get out four new orbitals. Where before we had the 2s below each of the two p's, now think of it as a strawberry and three bananas. We're going to throw them into a blender, turn it on, mix it all up and what we end up getting is four new orbitals that we call sp^3 hybrids because we took one s orbital and three p orbitals and they end up at an energy that is an average of the original four orbitals. So you can see that, because these are higher in energy and this one's lower, I drew these orbitals three-quarters of the way up, so three-quarters of the way up between the s and the three p orbitals. These are all exactly the same energy.
So what do they look like? Well, you're going to have to take my word for it a little bit on this one. Here's what they look like, here's a representation of the 2s, each of the mutually perpendicular 2p orbitals. And when we throw these orbitals into a blender and they come back out, it turns out we have something called a hybrid, which looks, sort of fat on one side and skinny on the other side, so a little tiny nub on this side and there are four of them. Four in, four out, and they point at, conveniently, the corners of a tetrahedron. So we have one that points at this corner, one that points at this corner, one that points at this corner, and one that points at that corner. And we can put them all together and overlap them and it would look something like this picture right here where we've sort of forgotten about the little nub there but we know that the nub is there. We just didn't want to represent it because it would really gum up the picture.
But what does this get us? Now we have four orbitals that are hybrid orbitals that each still have one unpaired electron. Here's that picture. They each have one unpaired electron. So they're prepared to make bonds. And let's look what happens. Okay? So now, here's our carbon. Here are three of the four hydrogens and methane that I'm not going to talk about, but remember - they're all equivalent. We have an sp^3 hybrid indicated by this blue thing and there's that little nub that goes along the back. And then the hydrogen 1s overlaps with one of the sp^3 hybrids. Well, there's another picture that looks exactly the same as this except that it would be involving this hydrogen and another one that involves this hydrogen, and another one that involves this hydrogen, because there are four sp^3 hybrid, and what we do is, basically, we have created the sigma bond structure for methane. Four sigma bonds, four orbitals on the carbon overlapping with 4 different hydrogen 1s orbitals, and again a sigma bond because the overlap is between the two carbons.
All right, so, that was a mouthful. But let's do another example for boron triflouride, BF[3], now steric number 3. We need to form three sigma bonds, one between the boron and each of the three flourines and they ought to be equivalent. And furthermore, we know based on VSEPR they have to be positioned relative to each other at 120. So, we take our boron atom. Its electron configuration looks like this. We pay the promotion energy so we can make three bonds and then we hybridize, but this time we don't care about this orbit. There's nothing in it. It's not important. What we want is we want three orbitals that are positioned at 120 to bond with the fluorine atoms. So we're just going to take the three orbitals that we have with one electron in and we're going to again, do the arithmetic average so it will be sort of 2/3 of the way up. We get three equivalent orbitals that are a mixture of this guy, this guy, this guy, and we don't even mess with the last guy that doesn't even have any electrons in it. But what does this look like? It looks like this. We now have sp^2, so one s orbital and two p orbitals, here's the s, here are the two p's and we mix and match, throw them in the blender, and out they come, and we get three orbitals that are hybrid orbitals, one that points at one corner of an equilateral triangle, one that points at a second corner, and one that points at a third corner. And we can overlap them and show them in this picture. And again, there's that little nub that's in the back. These are now at 120 with respect to each other, perfectly oriented in order to form the sigma bonding structure of boron triflouride. And that looks like, schematically, this.
Well once again, I've only focused on one of the three BF bonds. Here's the boron sp^2 hybrid, of which there are three. Three orbitals in; three orbitals out. Here's a fluorine 2p orbital. Because fluorine has a 2p orbital with one unpaired electron in it, it's prepared to form one bond and it forms a sigma bond with boron sp^2 hybrid.
All right now, here's a thought question for you. Where is the empty orbital? Let's go back, briefly. Think about this picture again. If this orbital is in the plane of the paper, and this orbital is in the plane of the paper, and this orbital is in the plane of the paper, then all of these guys should also be in the plane of the paper. So, where's the third p orbital? Where's the third p orbital if this one and this one... Well, it's the one that's coming out of the paper. It's the one that's up and down. And so, for boron triflouride, the p orbital that's empty, okay -- there's nothing there. It has no influence on the shape because there's nothing there, but it's sticking out of the board, okay, or out of the piece of paper. And that's going to come up and be really important when we talk about pi orbitals.
Well, let's keep going and look at steric number 2. Steric number 2 is going to be sp hybridized, sp meaning one s orbital, one p orbital. Beryllium difluoride is a good example. Here is the electron configuration for fluorine. I left a few things off. Here is fluorine. This is 2s, 2p--and then we look at the beryllium. Here's the beryllium--2s 2p and we're going to promote an electron and now I'm doing the short answer. I'm going to promote and I'm going to hybridize. So I promote an electron up here and hybridize. Two of these come down. These two orbitals are exactly in between the original energies of these two guys. And these two guys that are empty, we don't do anything with them. We don't worry about them.
So, we made two orbitals now that are exactly equivalent. They point in opposite directions, it turns out, which is really convenient and here's a picture of what that looks like. We take one s orbital and one p orbital. We hybridize and we get one purple thing that has a big lobe pointing out to the right, another orbital with a lobe that points out to the left. We can overlap them but the point is, we're going to now take these and we're going to form the bonds to the fluorines. That picture looks like this. Here's the fluorine 2p orbital and here's the beryllium, but we've overlapped it so one of the hybrids comes out here and there's a little bit of nub and that forms the sigma bond between this fluorine and this beryllium. And then there's another sp hybrid that's on the other side. They're oriented 180 with respect to each other so the second one is perfectly oriented to overlap with the p orbital on flourine to give another sigma bond. Okay? I think you got it now.
Let's go on. We're only going to touch on what happens with steric number 5 and 6. Remember, steric number and hybrids are intimately related. You take the number of orbitals and it's exactly equal to the steric numbers. So for instance, with steric number 5 we need orbitals to make up hybrids that count up to 5. So we're going to take an s. We're going to take 3 p's and we're going to take a d orbital now. And if you think about the geometry for steric number 5, and here's our model again for steric number 5, remember, there's the trigonal plane. It turns out that the hybridization on the trigonal plane is sp^2. It makes sense because sp^2 has 120 bond angles and then along the axial direction we have pd hybridization. And so we have three orbitals, three equatorial vertices, and then the other two orbitals are the two axial vertices. You can see that these are not exactly equivalent, whereas when we were talking about sp^3, that gave four equivalent orbitals.
Steric number 6 we have to put in another orbital. We put in another d orbital and it turns out, and you wouldn't necessarily have known this, but it turns out that that gives rise to six equivalent hybrids that point at the corners of an octahedron. Convenient? That's exactly what we're saying had to be the case for steric number 6.
Finally, let's ask the question, what about molecules with lone pairs? And what I want to do is look at water. Water has steric number 4. Remember, that number is the sum of the orbitals over here. So 1 + 3 = 4. And here's O2, the central atom. That's its electronic configuration in the ground state. Let's hybridize. If we hybridize we end up with four orbitals that are each sp^3 hybrids. Two of the four are filled so they can't form any bonds. That's where our lone pairs live. And then two of the orbitals, each have one unpaired electron in them. And these orbitals are prepared for bonding to hydrogen to make water.
So what have we done? What we've said is, in order to get to the geometry that we know is true, what we have to do is a mathematical trick called hybridization where we take the atomic orbitals, mix and match them and do mathematical stuff on them, and turn them into equivalent hybrid orbitals of which you always get the same number of hybrid orbitals out as the number of atomic orbitals you put in, and they point at the corners of the shapes of the molecules that we desire. So in other words, the corners of a tetrahedron or the corners of a trigonal bipyramid. And this allows us to reconcile this notion of bonding being overlap of orbitals with one unpaired electron in each one of them with the known geometry of the molecules.
Molecular Geometry and Bonding Theory
Valence Bond Theory and Molecular Orbital Theory
Introduction to Hybrid Orbitals Page [1 of 3]

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