In sp³ hybridization, one s orbital and three p orbitals hybridize to form four sp³ orbitals, each consisting of 25% s character and 75% p character. This type of hybridization is required whenever an atom is surrounded by four groups of electrons. Created by Jay.
Voiceover: In this video, we're going to look at the SP three hybridization present in methane and ethane; let's start with methane. So that's CH four, if I want to draw a dot structure for methane, I would start with carbon, and its four valence electrons, and then we would put hydrogen around that; each hydrogen has one valence electron, so we go ahead and draw in our hydrogens with one valence electron, and that gives us the Lewis Dot structure. Usually you see it drawn like this, with carbon with its four bonds to hydrogen around it, like that, and in methane, all of these bonds are equivalent, in terms of things like bond length and energy. And so the four valence electrons that carbon brought to the table over here, let me go ahead and highlight those four valence electrons, those should be equivalent, and if we look at the electron configuration for carbon, let's go ahead and do that right now. It's one S two, so go ahead and put in two electrons in the one S orbital, two S two, go ahead and put in two electrons in the two S orbital, and then two P two, and so, I'm assuming you already know your electron configuration, so it would look something like that. If we look at those four valence electrons on our orbital notation here, that would be these four electrons here, the valence electrons in the outer shell. And if we look at this, this implies that carbon would only form two bonds, because I have these unpaired electrons right here, and everything's of different energies, and so, what we see from the dot structure and experimentally, doesn't quite match up with the electron configuration here, and so to explain this difference, Linus Pauling came up with the idea of hybridization. And so, the first thing that he said was, you could go ahead and take out one of these electrons in the two S, and promote it up to the P orbital here, so let me go ahead and show that, so we've moved one of those electrons up to the two P orbital, so we're in the excited state now. And now we have the opportunity for carbon to form four bonds, however, those electrons are not of equivalent energy, and so Linus Pauling said, "Let's do something else here: "Let's go ahead and promote the two S orbital," so we're gonna take this S orbital, and we're gonna promote it in energy, and we're going to take these P orbitals and demote them in energy, so we're gonna lower those P orbitals, like that, so we have our P orbitals here. And these had one electron in each of them, but we're gonna hybridize them, so this is no longer going to be an S orbital; it's going to be an SP three hybrid orbital; this is no longer going to be a P orbital; it's going to be a SP three hybrid orbital, and same with these. So the idea is, you're taking some of the S character, and some of the P character, and you're hybridizing them together into brand new orbitals, and since you're taking this from one S orbital and three P orbitals, we're doing this using one S orbital and three P orbitals, we call this SP three hybridization, so this is SP three hybridization: We create four new, hybrid orbitals. And now we have what we're looking for, because now we have four unpaired electrons, so carbon can form four bonds now, and they're equal in energy, so that's the idea of hybridization. All right, let's think about the character, or the shape of this new hybrid orbital. Well, we know that an S orbital is shaped like a sphere, so we're taking one of those S orbitals here, so one S orbital, and we know that a P orbital is shaped like a dumbbell, so we're taking three of these P orbitals here. So one of these S orbital, and three of these P orbitals, and we're going to hybridize them together, and when that happens, in turns out the shape of the new hybrid orbital you get, has this large frontal lobe here, like this, and then a smaller back lobe, back here like this; so we're gonna make four of these. All right, so once again we have four SP three hybrid orbitals, and each one of these hybrid orbitals is gonna have an electron in it, so we can see that each one of these SP three hybrid orbitals has one electron in there, like that, and so the final orbital, the final hybrid orbitals here contain 25 percent S character. Let me go ahead and write this down here: So 25 percent S character, and 75 percent P character, in this new hybrid orbital. Once again, that's because we started out with one S orbital and three P orbitals for our hybridization. All right, let's go back to methane, and let's go ahead and draw in a picture, because now we know that this carbon is SP three hybridized, so let's go ahead and draw a picture of that hybridized carbon, here. So we're gonna go ahead and draw in our carbon, and we know that it has four SP three hybrid orbitals, and once again, when we draw the orbitals, we're gonna ignore the smaller back lobe here, so it doesn't confuse us. So we go ahead and draw in, here's one of our orbitals, for carbon, so that's an SP three hybrid orbital. Here's another SP three hybrid orbital, here's another one, and then, finally, a fourth one. So let's go back up here, to this picture, 'cause once again, we need to show that each of these hybrid orbitals has one valence electron in it, so I can go ahead and put in my one valence electron, in each of my hybrid orbitals, like that, so here's our valence electron. If we're talking about methane, so carbon is bonded to four hydrogens, each hydrogen has an un-hybridized S orbital, and each hydrogen has one electron in that, so I'm gonna go ahead and sketch that in; let's go ahead and use blue here. So here's an un-hybridized S orbital, I'm gonna go ahead and draw these in, so an un-hybridized S orbital; each one of these un-hybridized S orbitals for hydrogen has one valence electron, so I'm gonna go ahead and put in those one valence electrons, in here, like that, so same for here, and then, finally, for here. So this is just one picture of the methane molecule, so this is hydrogen, these are all the hydrogens right here, like that. All right, let's think about this bond that we formed right here, so here we have an overlap of orbitals, an overlap of an SP three hybrid orbital form carbon, with a un-hybridized S orbital from hydrogen here, and so this is a head-on overlap, so we're sharing electrons here, in this head-on overlap. And a head-on overlap, in chemistry, is called a sigma bond, so this is a sigma bond, sigma bond here, a head-on overlap, and this happens three more times in the methane molecule. So here's a head-on overlap, here's a head-on overlap, and here's a head-on overlap. And so we have a total of four sigma bonds in the methane molecule, so a single-bond here, instead of saying a single-bond now, we're saying it's also can be called, "a sigma bond," and so this head-on overlap. Let's look at the ethane molecule now, so for ethane, we'll go ahead an draw that in, so ethane would be C two H six, so we have two carbons; let's go ahead and draw in those two carbons, and then six hydrogens, so we put in our six hydrogens around there, like that. All right, when we're thinking about hybridization, we've just seen, with methane, that a carbon atom with four single-bonds will be SP three hybridized. So I go back up to here, this carbon right here, four single-bonds; it's SP three hybridized, we could use that same logic and apply it to ethane, here. Each of the carbons in ethane has four single-bonds, so each carbon in ethane is SP three hybridized, so let me go ahead and put SP three hybridized here, so let's go ahead, and draw the picture with the orbitals. So let's get some more room. If each carbon is SP three hybridized, that means each carbon is gonna have four SP three hybrid orbitals. So I can go ahead and sketch in one carbon, once again, I'm ignoring the back lobe, one carbon with four SP three hybrid orbitals, and we know the other carbon is also SP three hybridized, so I can sketch in four SP three hybrid orbitals for this one too, so here's my four SP three hybrid orbitals for this carbon. All right, in terms of electrons, let's go ahead and put in our electrons here. So, let's see, there's one electron in this orbital, one electron in this orbital, one electron in this orbital, one electron from this carbon. And then, for this other carbon, so there's one electron in this orbital, one electron in this orbital, one electron in this orbital, and one electron in this orbital. And then we can go ahead and put in our hydrogens, so we know each hydrogen has an un-hybridized S orbital, with one valence electron, so I can go ahead and do that. And I'll draw in the rest of our hydrogens, so that's four hydrogens, five hydrogens, and then six hydrogens, like that. All right, we just said that a sigma bond is a head-on overlap of orbitals. So, here we have a head-on overlap of orbitals, the bond between the two carbons, and then, of course, all of these are too. So when we count all of those up, that's four, five, six, and seven; so there are seven sigma bonds in the ethane molecule, so seven sigma bonds here. And we could go up here to this dot structure, and look at them again, so here's one, two, three, four, five, six, and then seven: Seven sigma bonds. Let's focus in on the bond between the two carbons now, so this sigma bond, right in here, so that's a sigma bond, and there's free rotation about this sigma bond, so if you could imageine rotating around this bond, so these carbons can rotate in space, and that's gonna give different conformations, so you could have different confirmations of the ethane molecule, which is in later videos. So you have free rotation about sigma bonds. Let me go ahead and write that, 'cause that's pretty important, so free rotation about sigma bonds. And then, the last thing I wanted to point out, about the ethane molecule here, is the bond length: So the length between this carbon and this carbon, so this bond length, in here, turns out to be approximately one point five four angstroms, so you'll see slightly different values in different textbooks, but we'll say it's approximately this value, and the reason we wanna know that, is we're gonna compare this carbon-carbon bond length to the bond length in some other molecules, in some later videos.