Hybridization: Videos & Practice Problems

The Aftbaugh principle explains electron configuration, where electrons fill orbitals in increasing energy order. For carbon, hybridization occurs to form four equivalent sp³ orbitals, allowing for stable bonding. The number of bond sites—counting atoms and lone pairs—determines hybridization: 4 bond sites yield sp³, 3 yield sp², and 2 yield sp. Understanding these concepts is crucial for predicting molecular geometry and reactivity in organic chemistry, particularly with reactive intermediates like carbocations and carbanions.

How carbon creates 4 partially-filled orbitals.

So the first thing I want to talk about is just the theory of why this is even possible, why this is a big deal. The Aufbau principle states that electrons can fill orbitals in order of increasing energy. Okay? Remember that my orbitals in order of increasing energy were 1s, 2s, and then the three 2ps.

If we go ahead and draw the diagram for carbon, let's say. Let's just use carbon as an example. What we're going to find is that the 1s gets completely full at the helium stage. Then the 2s gets completely full at the beryllium stage. Then with carbon, we only have 2 electrons in those p orbitals. Remember, there are 3 p orbitals total, and we only have 2 electrons in them. By the way, the notation that I'm referring to here has to do with electron configuration, which is something that you guys are supposed to remember from general chemistry.

So in this case, what that means is I have 2 electrons in the 2p orbitals. So if I were to draw the actual orbital diagrams for both of these, what you would realize or for the carbon, what you would realize is that like I said, 1s is full, 2s is full, and then the ps are just a mess because what I have is basically 2 unfilled orbitals and then one completely empty orbital. So remember that I told you guys earlier when we were talking about molecular orbital theory that unfilled orbitals like to bond with other unfilled orbitals to become more stable.

So when you're looking at carbon, if it only has 2 unfilled, like partially filled orbitals, you would think, well, why wouldn't it just make 2 bonds? Why does it make 4? Remember, this is another way to think about it. We already talked about bonding preferences and stuff, but this actually has to do with orbitals. So why does it make 4? Well, it turns out that this is just a really bad situation for carbon. Carbon does not like to look like this. The reason is because these orbitals are all messed up. The p orbitals, some are partially filled. Some are not filled. So that means that it would never really be fully stable.

So what carbon actually ends up doing is it winds up exciting one electron up into one of the p orbitals. So what that does is it actually violates the Aufbau Principle and it takes one electron from my 2s orbital and it excites it to one of the 2p orbitals. So now what I end up getting is that instead of having one full orbital of lower energy and then three orbitals that kind of just suck. What ends up happening is that I end up getting 4 orbitals of higher energy that all are partially filled. Why is that better? That even sounds worse. Now we have an excited electron. That means it has more energy, so why is this better?

How carbon makes 4 sp3 orbitals.

The reason is, and this is the crazy thing, it turns out that many of these atoms are going to prefer to blend their second shell orbitals together, and what that's going to do is it's going to make new hybrid orbitals. So, what that means is that instead of the 2s orbital being lower energy and the p orbitals being higher energy, what winds up happening is that the carbon says, "Hey, if I can just take these 4 different orbitals and combine them, kind of mash them up and make them all a little bit higher energy, but all equal, then I could wind up filling them with bonds. If I could fill them with bonds, then I could become more stable because now I could have all of my orbitals filled." So what winds up happening is, check this out: the 2s orbital goes a little bit higher, and because I'm blending the p orbitals with the s, the p orbitals dip down a little bit. So what winds up happening is that all the orbitals hybridize together and they all wind up being the same energy. They all turn into what we call degenerate orbitals. Degenerate just means that they all have the same energy. Once they're degenerate, what that means is that I'm going to follow Hund's rule. So instead of the 2s just having 2 electrons, now I'm going to follow Hund's rule, which is that one each orbital gets one electron. Does that make sense?

So now, here's the interesting part. Since we blended one of the s orbitals and since we had blended 3 of the p orbitals together and we made these 4 degenerate orbitals, these new orbitals are referred to as sp³. Sp³ because one of them is the s and 3 of them are p's. Does that make sense? Now the reason that I have a 2 in front is that because these are all in the second shell. So this would be 2s and 2p's that are blending together to make sp³. So you could either just call it sp³ or sometimes you'll see it called 2sp³. That just means that it's the orbitals from the second shell that are all coming together. All right. So hopefully, that makes sense so far. Now I'm going to go ahead and show you guys how to apply this and how to use this knowledge to figure out hybridization and molecular geometry of atoms. So let's go ahead and get started.

Using bond sites to predict hybridization

Now that we understand the theory behind hybrid orbitals, let's go ahead and just figure out the rules that are going to help us determine what type of orbitals we're dealing with for different atoms. Hybridization, it turns out, instead of looking at just the electrons like we were doing, it can be determined in a much easier way. It can be predicted by the number of what we call bond sites on an atom. Okay? Bond sites, just so you know, some people refer to these as groups. But I think "groups" is a really general word. It's like it's easy to forget what a group is, but a bond site is really way more specific. So a bond site is going to be equal to any atom or lone pair. So basically, any time that an atom is attached to another atom, regardless of what type of bond it is, it could be a single bond, double bond or triple bond, all of those count as just one bond site. Why? Because there's only one place that it's attached to an atom. Then a lone pair counts as another bond site because that's a place that it could form another bond if it wanted to. So what I want to do is I want to go through a hybridization summary and help you guys realize how just by looking at bond sites, we can determine everything about these atoms in terms of their hybridization.

So let's start off with 4 bond sites. What happens if I have a combination of atoms and lone pairs that gives me 4 bond sites? For example, let's say that I just have a carbon with 4 bonds on it. All of these 4 single bonds. All of these bonds represent atoms that it's attached to. Well, if I have a carbon that looks like this, the participating orbitals are going to be that I have the 2s orbital. Remember that I have that 2s orbital that's spherical that is going to hybridize. Then I'm going to blend that with 3 of the p orbitals. Does that make sense so far? Remember that the p orbitals were of higher energy, the s was of lower, but I blend them all together for hybrid orbitals. So what that means is that instead of getting these orbitals of different energies, what I wind up getting is 4 orbitals of the same energy, or those degenerate orbitals. And these orbitals look a little bit different. They're kind of a combination of the p and the s. So what winds up happening is that instead of just looking like a peanut or a sphere, they look kind of like this with one big side and one small side and you get 4 of these. Okay. So basically, oops, that one looks a little bad. I'm going to draw it again. So basically, all of these orbitals combine together and what I wind up getting is 4 of the hybridized orbitals. We're going to talk more about what they are. Unhybridized orbitals actually is none because all of these orbitals participated together and they all blended together to make these 4 degenerate orbitals. Does that make sense so far that they all have the same energy? So now if I was talking about the hybridization, what would we call these orbitals? Would we call them p? Would we call them s? No. We would call them a combination of what they are. So remember what the name would be? It would be sp³. The reason we would call it sp³ is because I have one s and 3 p's combining together to make these orbitals. So what I would get is four sp³ orbitals that are brand new orbitals that I can use. Does that make sense? I hope so.

Well, now let's talk about the bond angle and the s character because these are common things that professors want you to know about the hybridization. Bond angle has to do with how far apart these 4 orbitals can get from each other. And it turns out that if I was on a two-dimensional space, so let's say that this paper I had 4 different orbitals on it and I wanted to get them as far apart from each other as possible, the furthest would be 90 degrees like I have here. Notice that here I drew this at a 90-degree angle. Okay? First would be 90 degrees. But atoms don't exist on a plane. They don't exist on a page. They actually exist in 3D. So in 3D, they can actually move in different directions. So what I could do is I could make 2 of them come out of the plane. If they come out of the plane, what that means is that the bond angles are going to turn into the furthest way they can get from each other is going to be 109.5. This should be an angle that you kind of remember from general chemistry. Okay? The way that you can think of it is that basically 2 of them are going front and back. So imagine that my hands are 2 of the bond sites and then 2 of them are going side to side like my legs. I know you can't see my legs, but I'm just explaining that if my legs were spread apart and if my hands were spread apart this way, that's what a tetrahedral or that's what a 109.5 would look like. I'm going to explain what the shape is in a second. Then finally, what's the s character, percent s character? That doesn't even make sense. Well, all that is, is it's a percentage of s. What part of the entire thing is the s? So I have 4 orbitals total. One of them is s. So overall, what percentage is made out of s? 25%. So let's go ahead and write that in. What I'm trying to say is that if you have 4 and one of them is the s, your percentage of s character is going to be 25%. Why is this important? Because when we get to the acids and bases chapter, we're going to need to know that. We're actually going to need to know the s character to predict some types of acidities.

Now let's move on to the next situation, which is: What if we don't actually have 3 bonds 4 bond sites? How about if we have 3? This is also common. What if I have a molecule that looks like let's say C with a double bond O and 2 carbons on the side. Notice that here, I know that you can't really see the O that well, but notice that this carbon now would have how many bond sites? Only 3. Because remember that I said that any atom counts as a bond site and any lone pair counts as a bond site. Do I have any lone pairs? No. How many atoms do I have attached to that carbon? Only 3. I have a carbon on one side, carbon on the other side and then the oxygen at the top. So what that means is that in this case, I'm only going to get 3 orbitals hybridizing together. The reason is because I only have 3 places where I'm making bonds. Okay? Does that kind of make sense? I only have 3 places where I'm sharing an electron with another atom. So that's the only ones that are going to hybridize. So what I'm going to hybridize is one of the s orbitals, same as before, but now I'm only going to hybridize 2 of the p orbitals. All right. So I have 1 s and 2 p's and what that's going to give me is 3 hybridized orbitals. Does that make sense so far? But now when it comes to unhybridized orbital, is there going to be anything that's not making a bond to anything? Yes, there is. It's going to be one of the p orbitals. One of the p orbitals, I'm just going to put here 1 p or I'm just going to put here p. One of the p orbitals is not going to participate because it's not making a bond to anything, so it's not hybridizing. Okay? So if I were to find a new name for these orbitals, what are they called? What I would call them is just a combination of what they were originally, one of the s's, 2 of the p's, so this would be called sp². sp² is the name of these orbitals and I have 3 of them. So basically, in the first example, I had 4 of the sp³s, but in this next example, I only get 3 of the sp²s and then I have a p orbital that's just left over. So I still have 4 orbitals total, but they're not all hybridizing. Okay? Now in terms of bond angles, how far apart can these guys get? They can get 120 degrees apart. Okay? Because of the fact that now I only have 3 orbitals that I have to worry about I only have 3 bonds that I have to worry about moving apart from each other, repelling each other. The furthest away that three things can get from each other is 120 degrees. So what that means is that it's going to be a slightly different bond angle. Then finally, if I were to analyze the s character for this, I would say, What percentage of the total is the s? And the answer is well, the total is sp². There's only 1 s and there are 3 orbitals total, so it's going to be 33% s character. That shouldn't be so bad. You guys should understand where I'm going with that. It's kind of a pattern.

So now let's do this last one. What if I have a situation where I only have 2 bond sites? So what that means is that, for example, the one that I had with resonance where I had an N on one side and an O on the other. In this case, I have 2 double bonds, but both of them only count as one bond site because it's only attached to one atom. So if I had a situation like this and I'm looking at this carbon right there, how many bonds am I making? 2. So how many orbitals am I going to hybridize? Only 2 because only 2 of these are accepting electrons from other atoms. So I'm going to take an s and I'm going to take a p. Okay? What that means is that when I go ahead and combine these together, these are also going to make those weird looking orbitals, but I'm going to get even less of them. In this case, I'm only going to get 2. Okay? That means in terms of unhybridized orbitals, now I'm going to have 2 p's just left over. Crazy, right? So now I'm going to put here 2 p's. Okay. If I were to go ahead and say what type of hybrid orbitals are these? Well, it would just be a combination of what's coming together, so this would be called sp. And the furthest that two things can get apart from each other because remember now, I only have 2 things repelling, is 180 degrees. And since I only am combining 1 s and 1 p, the s character is going to be 50%. Does that make sense? This is your summary chart. This is what we're going to use to predict hybridizations. All you really have to do is look at the bond sites and that's going to tell you everything else you need to know.

Introduction to the reactive intermediates.

What I want you guys to do is look at these examples here and I want us to predict the hybridization of these different atoms. Okay? What I'm going to do is I'm going to give you the name of all of these. These are called reactive intermediates, by the way. Okay? I'm going to give you the names of all of them, but then I want you to tell me how many bond sites each has, what the hybridization is, and what the intermediate orbital is. Meaning, where is the negative or the positive charge actually going to reside. So let's go ahead and just talk about the names really quick. This first one has a positive charge on a carbon. This is a very common reactive intermediate. What reactive intermediate means is that it's very unstable, so it reacts with lots of things. The reason it's unstable is that it doesn't fulfill its octet and it also doesn't fulfill its bonding preference. Okay? The name of this is a carbocation. I want you guys to know that we're going to deal with these a lot in organic chemistry. Carbocations are unstable because they don't fulfill their octet and they don't fulfill their bonding preference, so they want to bond with something else really badly. Okay? If I have a negative charge on a carbon, that's called a carbanion. Okay? If I have a single electron on a carbon, that's bad because now that orbital is not fully filled. It only has one electron instead of 2. This is called a radical. A radical is the same thing as free radicals. It's an orbital that is not fully filled; it only has one electron. Then finally, this is a very weird intermediate. Valence? This one has 4. But valence? This one has 4. But the problem is that it has the wrong amount of octet electrons. Remember that carbon wants 8 and this one only has 6. This is a very reactive intermediate called a carbene. These are your 4 guys that you're dealing with. What I want you guys to do is use the rule for bond sites and figure out how many, basically, how many bond sites does each of these carbons have. From there, tell me what's the hybridization. Okay? Once you tell me what the hybridization is, then you'll be able to tell me where does this actually reside? Basically, where does the positive charge reside or where does the negative charge reside? Okay. So I'm going to give you guys an opportunity to solve that really quickly and then you guys go ahead try to do all of these in terms of bond sites, hybridization, and then we'll probably do the intermediate orbital together. So go ahead and try to do that now.

Bond sites, hybridization, and intermediate orbitals

So hopefully this wasn't too tricky. Let's go ahead and start with bond sites. All right? So bond sites for the carbocation, what are those? Well, let's go ahead and scroll up one more time just to see what bond sites were. Remember that I said that bond sites were equal to any atom or any lone pair. And remember that I also mentioned that some people call these groups. Okay? So some textbooks, some online homework might call it groups, some might call it bond sites. There's even another name that I've seen and that name is steric number. Okay? So what I want you guys to know is I really want you guys to know all three because your textbook might use one of these words, but your online homework or maybe a supplemental book that you're using to learn might use a different one. So I just want you guys to realize that if I say bond site, group, steric number, that's all the same exact thing. All it has to do with is things around the atom that are going to repel each other.

Let's go ahead and talk about the carbocation. So the carbocation has three atoms, H, H, H, and then it has a positive charge. Does a positive charge count as a bond site? No, it doesn't. Remember that I said the only thing that counts is atoms or lone pairs. So that means that I have three bond sites. If I have three bond sites, then what is my hybridization? Well, my hybridization, I would look at my summary chart and I would say I have three, so my hybridization must be sp². That must mean that one s is coming together with two ps and they're making three sp² orbitals.

So what I'm going to say is that it's sp². And then we have the tricky one. What is intermediate orbital? What that means is which is the leftover orbital that isn't included in bonding where my positive charge is. Okay. Let me put it this way. We just said that we have three sp² orbitals. Where are they? I'll show you. The sp² orbitals are right here. There's one bonding to that H, there's one bonding to that H, and there's one bonding to that H. Okay? So those are the sp² orbitals. By the way, what is this bond going to look like completely? Remember that. What kind of orbital does an H have? Do you remember? An H is just a 1s orbital. Okay. So what I have here, let me use a different color for that since we have two different orbitals. What I have here is a 1s, a 1s, and a 1s. So those are the 1s's. And those are overlapping with sp². So if I were to describe this bond, if I wanted to say what kind of bond is this? What I would say is that it's a 1s-2sp² bond. Why? Because I have a 1s orbital from the hydrogen overlapping with an sp² orbital from the carbon. Remember that it has three of those. It's overlapping with one of those and that's making a sigma bond that has those properties. Isn't that interesting? So I just wanted to show you guys that notation because in a lot of practice problems, professors will ask it just like that. They'll say, what kind of bond is this? Or can you find all of the bonds that are like this? All right.

So that was a little bit of a detour, but I still think that's a good learning moment. Okay? So the question here though is that how many sp² orbitals do I have? I have three. I have one sp², another sp², and another sp². All right? But how many orbitals does this molecule have total? It has four. So what is the leftover orbital? If you look up here, the leftover orbital is this p orbital that never hybridized. That p orbital is the one that's going to get the intermediate in it where it's going to get the positive charge. The reason is because I just showed you the sp²s are already bonding. All of these are taken up by an H. Okay. So that means that the positive charge must go where? The positive charge must be in an empty p orbital that is not bonded. Okay. I know that drawing is really messy, but that's the point of this question.

Okay. So the positive charge is in the p and the other three sp²s are making three bonds. Does that kind of make sense? Let's move on to the carbanion. For the carbanion, how many bond sites do I have? Well, I have three atoms and then I have a negative charge. Is a negative charge a bond site? Actually, not necessarily. I need to

Here’s what students ask on this topic:

What is hybridization in organic chemistry?

Hybridization in organic chemistry refers to the concept where atomic orbitals mix to form new hybrid orbitals. These hybrid orbitals have different shapes and energies compared to the original atomic orbitals. For example, in carbon, the 2s and 2p orbitals combine to form four equivalent sp³ hybrid orbitals. This allows carbon to form four stable covalent bonds, as seen in methane (CH₄). Hybridization helps explain molecular geometry and bonding properties in organic molecules.

How do you determine the hybridization of an atom in a molecule?

To determine the hybridization of an atom in a molecule, count the number of bond sites around the atom. Bond sites include atoms bonded to the atom and lone pairs of electrons. For example, if an atom has 4 bond sites, it is sp³ hybridized. If it has 3 bond sites, it is sp² hybridized, and if it has 2 bond sites, it is sp hybridized. This method helps predict the molecular geometry and reactivity of the atom.

What is the hybridization of carbon in methane (CH₄)?

In methane (CH₄), the carbon atom is sp³ hybridized. This is because the carbon atom forms four single bonds with hydrogen atoms, resulting in four bond sites. The 2s and three 2p orbitals of carbon mix to form four equivalent sp³ hybrid orbitals, each forming a sigma bond with a hydrogen atom. This hybridization results in a tetrahedral geometry with bond angles of 109.5 degrees.

What is the difference between sp² and sp³ hybridization?

The main difference between sp² and sp³ hybridization lies in the number of hybrid orbitals and the bond angles. In sp² hybridization, one s orbital and two p orbitals mix to form three sp² hybrid orbitals, resulting in a trigonal planar geometry with bond angles of 120 degrees. In sp³ hybridization, one s orbital and three p orbitals mix to form four sp³ hybrid orbitals, resulting in a tetrahedral geometry with bond angles of 109.5 degrees. Sp² hybridization is common in alkenes, while sp³ hybridization is common in alkanes.

Why is hybridization important in organic chemistry?

Hybridization is important in organic chemistry because it helps explain the shapes, bond angles, and bonding properties of molecules. By understanding hybridization, chemists can predict molecular geometry, which is crucial for understanding reactivity, stability, and physical properties of organic compounds. For example, knowing that carbon in methane (CH₄) is sp³ hybridized helps explain its tetrahedral shape and bond angles. Hybridization also aids in understanding the behavior of reactive intermediates like carbocations and carbanions.