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.
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- Do the d orbitals ever get involved in the hybridization. I have watched a couple of videos and have not yet found any f or d hybridizations.(55 votes)
- It will be "sp3d" hybridization. Because, as the electron from s goes to higher energy state to d. We have
One s orbital, Three p orbitals and one new d orbital involved.
Hence sp3d hybridization(19 votes)
- Why is hybridization necessary, why doesn't the carbon just go ahead and form bonds in its original state without hybridizing?(42 votes)
- If carbon does not hybridize then carbon can not form more than 2 bonds as in the last orbital there is only 2 valence electrons if it hybridizes the furthest orbital has 4 valence electrons to bond(66 votes)
- electronegativity play any role in hybridisation(13 votes)
- Well electronegativity is the property of which atoms can attract electrons- it forms a resonance that is the most stable. The more the electron is held tightly to the nucleus of a certain atom in a molecule, the more electronegative that atom is.
I guess you can say the same for hybrid orbitals- an sp orbital has the electron more tightly held together by the nucleus, considering it has a higher s-character than sp2 or sp3, and thus more stable of a hybrid. This means that sp is more electronegative than sp2 or sp3. I hope I helped you some.(34 votes)
- So, does only Carbon have hybridized orbitals?(8 votes)
- How come molecules and substances seem so stable, when the bonds are made up of electrons which just "might" be in the bond?(12 votes)
- because nature tends to only form compounds which are stable. Stable compounds are lower in energy than their separate atoms. So that means, for example H2O, the hydrogen and oxygen atoms are higher in energy than the compound H2O. It is impossible to know at which point an electron is at a certain time. The hybrid orbitals(sp3, sp2, sp) are just figures that tell you where electrons are abundant and where they would most probably be.(6 votes)
- How is sp3 hybridization energetically favorable for the atom if it involves the promotion of an electron from a lower energy s orbital to a higher energy sp3 hybridized orbital?
It can be argued that the energy required for this excitation is compensated by the release of energy due to subsequent formation of bonds but then how does the energetically unfavorable excitation happen before the formation of bonds?(8 votes)
- I am not absolutely certain but I can provide some clarifications. Hybridisation has largely been replaced by the molecular orbital theory which talks about steady-state orbitals and very little about what happens while the bond is forming. Why, because it is extremely difficult to find out what happens and beyond mathematical/computing capacity. The hybridization theory also suffers from the same drawback. It does not describe the temporal evolution of the bond formation well but just attempts to explain the relative bond-angles and energies once the bonds are formed.
Secondly, as a pure imagination exercise, think like this. As the other atom nears, the electronic fields around our atom keep changing, all the while changing the shape and the energies of the allowable orbitals for its electron until it finally reaches the final state which is maintained in a bond. Some of these intervening orbitals might be energetically unfavourable but the overall reaction is favourable.
Okay, to the main part. The jump across the energy is inevitable for most reactions. This is where things get weird. Since the entire quantum mechanics, and to a great extent thermodynamics, is statistically described, it turns out, that there is almost always a non-zero probability of finding the electron in every energy state, even the very seemingly unfavourable ones (see Quantum Tunnelling). This is what makes reactions involving an unfavourable transition state possible. Randomly, the electron will at some point in time be in that transition state allowing the proceeding of the reaction. The more unfavourable the state, lesser is the chance of the electron of being there and slower is the reaction. These fluctuations are extremely common under normal reaction conditions, where energy keeps fluctuating temporally and spatially.(11 votes)
- 1:41he said u can take out one of the electrons and move it up to the P orbital why is that so?
Super confused.....please help(5 votes)
- Some energy was added to the atom to promote that electron to the 2p orbital. When we hybridise the 2s and 2p orbitals we need an unpaired electron in each orbital to form bonds.(8 votes)
- Is a single bond ALWAYS a sigma bond, or are there ever any cases of a pi bond being a single bond ?(2 votes)
- A single bond is always a σ bond.
A double bond is always a σ and a π bond.
However, some theoretical chemists argue that C₂ has the structure :C=C:, in which the double bond consists of two π bonds and no σ bond.(11 votes)
- Please help about how to identify kind of hybridization for a given molecule?(2 votes)
- Count the number of lone pairs + the number of atoms that are directly attached to the central atom.
This is the steric number (SN) of the central atom.
SN = 4 → sp³
SN = 3 → sp²
SN = 2 → sp
For example, the O atom in water (H₂O) has 2 lone pairs and 2 directly attached atoms.
∴ SN = 2 + 2 = 4, and hybridization is sp³.(9 votes)
- Can anyone please explain this percentage character thing? What does it mean if the s character is greater?
I'm confused beacuse I read that the sp character in ethyne makes its hydrogen acidic.(1 vote)
- When atoms combine their atomic orbitals to create hybrid orbitals, they add different amounts of p orbitals. An sp3 hybridized atom combines one s and three p orbitals (and creates four sp3 hybrid orbitals) while an sp2 hybridized atom only combines one s and two orbitals (creating three sp2 hybrid orbitals and a leftover unhybridized p orbital). Each individual hybrid orbital is a combination of multiple atomic orbitals and has different s and p character affecting their shape, length, and acidic properties.
An sp3 hybrid orbital is composed of four atomic orbitals, one s and three p, so the s character is ¼ or 25% (making the p character ¾ or 75%). This means an sp3 hybrid orbital resembles an s orbital only 25%, while it resembles a p orbital 75%. The sp3 orbital looks mostly like a p orbital with two lobes, but the lobes are asymmetrical and smaller because of the s character since s orbitals are spheres and are close to the nucleus.
An sp2 hybrid orbital is composed of three orbitals, one s and two p, so the s character is 1/3 or 33.3% s while the p character is 2/3 or 66.7%. The shape is similar to an sp3 orbital, but the increased s character makes an sp2 orbital even smaller than an sp3 orbital. The same logic follows for sp orbitals having ½ s (50%) and ½ p character.
If we consider hydrocarbons with hybridized carbons such as ethane (sp3), ethene (sp2), and ethyne (sp), they all possess hydrogens capable of being acidic. That is, they have hydrogens which can be removed by bases leaving behind a lone pair on the carbon and making the hydrocarbon a conjugate base. But the acidic strength varies with hybridization with sp hybridized atoms being the most acidic. The acidic trend is sp >> sp2 > sp3, meaning that sp hybridized atoms are much more acidic sp2 hybrid (more so than sp2 is more acidic than sp3 hybridization). The pka values for these hydrocarbons are 50 for ethane, 44 for ethene, and 25 for ethyne where a lower pka indicates a stronger acid. It should be noted that even though ethyne is the strongest acid of the group, it’s still an abysmally weak acid. The pka of water, itself a weak acid too, is much lower at 14 making it a stronger acid than ethyne.
The stability of the conjugate base, and thus the strength of the acid, is determined by the stability of the lone pair leftover on the hydrocarbon after the hydrogen has been removed. We can explain this hybridization difference in acidity by considering the relative stability of the s and p components of hybrid orbitals. An s orbital is lower in energy than a p orbital. Therefore, a lone pair remaining in a hybrid orbital is more stable in a hybrid orbital using more of the s orbital and less of the p (essentially having higher s character). The increased s character of the sp2 hybrid orbital compared to an sp3 hybrid orbital makes lower in energy and the electrons residing in them more stable. The same logic applies to an sp orbital too.
Hope that helps.(10 votes)
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.