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sp hybridization

AP.Chem:
SAP‑4 (EU)
,
SAP‑4.C (LO)
,
SAP‑4.C.3 (EK)
,
SAP‑4.C.4 (EK)
In sp hybridization, one s orbital and one p orbital hybridize to form two sp orbitals, each consisting of 50% s character and 50% p character. This type of hybridization is required whenever an atom is surrounded by two groups of electrons. Created by Jay.

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  • duskpin ultimate style avatar for user Paru
    Is there an sp4 hybridization too? If so, when does it occur? If not, why not?
    (15 votes)
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    • piceratops ultimate style avatar for user Just Keith
      No, there is no sp⁴ hybridization because any shell has only three p orbitals to offer, thus p³ is as high as it is possible to go (in terms of how many p orbitals can be involved).

      There are is very complicated hybridization that can take place with the d block elements that involve s, p and d orbitals. For example, though a bit rare, there is a d⁵sp³ hybrid.
      (61 votes)
  • leaf green style avatar for user Elsa Langrené
    in the sp hybridization video, why is the SN number 2+0 for acetylene and not 3+0. I thought there were 3 sigma bonds.
    (22 votes)
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  • piceratops seedling style avatar for user shravya
    in pi bond there is side wall overlapping,. what does it mean?
    (4 votes)
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  • aqualine sapling style avatar for user Jana Goodmanson
    Around the 3 minute mark things start getting confusing. I watched the other videos about the Sp bonds, but I don't understand what's happening with his s% p% dialogue at around . I wish he had just rewrote it instead of mushing everything into the same picture. That's what my prof does on the chalkboard and it's very confusing for people trying to learn/review concepts. :(
    (5 votes)
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    • blobby green style avatar for user Drew Leggett
      You can calculate the percentage without even thinking about the molecule just by looking at the type of hybridization. For example, in the characterization "sp," there is one s and one p, so there is 50% s character (it's one of 2 total). In "sp3," there is one s and 3 p's. So there is 25% s character (1/4). As far as what it actually means in terms of imagining molecular bonding, it just refers to how much the hybridized orbital resembles each atomic orbital that combined to make it. S orbitals hold electrons closer to the atom they surround, so a hybridized orbital with 50% s character (sp) holds electrons closer and thus forms a shorter bond than a hybridized orbital with 25% s character (sp3) because it behaves more like an s atomic orbital. Watching the videos from sp3 all the way through sp makes the concept a lot easier to understand.
      (11 votes)
  • blobby green style avatar for user Nashita Rahman
    Is steric number valid only for the central atom?for example in carbondioxide, the steric no. for O IS 3 so is it sp2 hybridised but how??
    (4 votes)
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  • piceratops sapling style avatar for user Alla Privata
    What if the steric number was 0 or 1, can you give some examples of where we'd see that?
    (3 votes)
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    • piceratops seed style avatar for user RogerP
      The steric number refers the sum of the number of sigma bonds and lone pairs in a bonded atom. So you can't have a steric number of 0 otherwise that would mean there was no sigma bonds in the molecule (and having all pi bonds is not possible). A steric number of 1 is only found for hydrogen (H2).
      (6 votes)
  • leafers seedling style avatar for user Samuel Mng
    What does it mean by 'excited state'?
    (1 vote)
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  • blobby green style avatar for user Kavishka Rathnayaka
    So as you have mentioned,if carbon cannot form 4 covalent bonds in the excited state,without hybridization(sp3),whats the difference between forming 4 bonds of sp3 hybrids and just 4 covalent bonds??
    I have already read your coulombs law part.
    (3 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      If CH₄ forms from four sp³ orbitals, all bonds will be identical, and all HCH bond angles will be 109.5 °.
      If CH₄ forms bonds from three p orbitals and an s orbital, there will be three identical bonds at 90 ° to each other and one at no fixed angle.
      The experimentally observed bond angles and bond lengths agree with the first model.
      (3 votes)
  • starky ultimate style avatar for user Tim
    I didn't quite get the reason carbon undergoes sp hybridisation here. Why not sp2 or sp3? Why not no hybridisation at all? Also why can't the pi bond exist alone?
    (3 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      Each C needs to form bonds to only two other atoms, so only two hybridized orbitals are needed.
      If there were no hybridization and there was just overlap with the 2p orbitals of C, the
      H-C-C bond angle would be 90°. This is not what we observe experimentally.
      (2 votes)
  • aqualine seed style avatar for user Shanthamraju Raghavendra
    is steric number works only for organic compounds
    (2 votes)
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Video transcript

Voiceover: The video on SP3 hybridization, we saw a carbon is bonded to four atoms and in the video in SP2 hybridization, we saw that carbon is bonded to three atoms and in this video, we're gonna look at the type of hybridization that's present when carbon is bonded to two atoms. If I look at this carbon right here and the ethyne or the acetylene molecule. This carbon is bonded to a hydrogen and it's also bonded to another carbon. We have carbon bonded to only two atoms and the shape of the acetylene molecule has been determined to be linear. We have a linear geometry. We also have a bond angle here. These bond angles are 180 degrees and so we must have a different hybridization for this carbon. We have a different geometry, a different bond angle and a different number of atoms that this carbon is bonded to. To find our new type of hybridization, we look at our electronic configuration already in the excited stage. We have carbons, four valence electrons represented here in the excited stage. One, two, three and four and we're looking for two hybrid orbitals since carbon is bonded to two atoms. We're going to take an S orbital. We're gonna promote an S orbital in terms of energy and we're going to demote a P orbital, only one P orbital this time. We have an S orbital with one electron. A P orbital with one electron. That's gonna leave behind two P orbitals. Each one of those P orbitals has one electron in it. We have carbons, four valence electrons but this is no longer an S orbital because we're going to hybridized it with a P orbital to make an SP hybrid orbital. This is no longer a P orbital because we're going to hybridize it to form our SP hybrid orbital. This is called SP hybridization. This is SP hybridization because our new hybrid orbitals came from one S orbital and one P orbital like that. This carbon right here is SP hybridized since it bonded to two atoms and this carbon right here is also SP hybridized. Let's think about the shape of our new SP hybrid orbitals. Let's get a little bit of room down here. Once again, we know an S orbital shaped like a sphere. We took one S orbital and we took one P orbital, which is shaped like a dumbbell and we hybridized these two orbitals together to give us two new hybrid orbitals. Two SP hybrid orbitals. Let me go ahead and draw in an SP hybrid orbital here and once again, we're going to ignore the small lobe. We're going to ignore the small lobe. When we draw our picture, only think about this bigger frontal lobe here. When I think about the percentage of S character. We use one S orbital and one P orbital. That means it's 50% S character and 50% P character and this is more S character than in the previous videos. On the video on SP3 hybridization, we're talking about 25% S character The video on SP2 hybridization, we talked about 33% S character and then for these hybrid orbitals, we have even more S character, up to 50% and since the electron density for an S orbital is increased electron density closer to the nucleus for an S orbital than for a P orbital, that means that this lobe here has an increased electron density closer to the nucleus, which is one way to think about why these bonds get shorter as you increase in S character. In general, as you increase in S character, you're going to get shorter bonds because you have smaller hybrid orbitals here. Let's go back up here to this picture of acetylene. I wanna see if we can draw it. We now know that both of these carbons in a acetylene are SP hybridized. Let's go back down here and let me draw up a dot structure really fast. We have acetylene here. We have carbon triple bonded to another carbon. We know each of those carbons as SP hybridized. If each of those carbons is SP hybridized, each carbon has two SP hybrid orbitals. We go ahead and draw in one SP hybrid orbital. Again, I'm ignoring the smaller back lobe and here's our other SP hybrid orbital on this carbon. Let me go back and look at our diagram again. Each carbon, we use red for this. Each SP hybridized carbon has an SP orbital with one valence electron in it and we put that in here and then there's another one. There's another SP hybrid orbital with one valence electron in it. I'm gonna go ahead and put the other electron over here in this hybrid orbital. Also, notice, if you're dealing with an SP hybridized carbon, you also have two P orbitals. Two unhybridized P orbitals. Each P orbital with a valence electron. Let me go back down here and I'm gonna draw in. here's one P orbital. There's one P orbital with one valence electron and there here's another P orbital right here with another one of those valence electrons. Now, we have our picture of an SP hybridized carbon. Let's say that was this carbon over here on the left. Now, let's go ahead and draw in this carbon on the right. This carbon on the right is also SP hybridized. Therefore, this carbon on the right has an SP hybrid orbital with one valence electron in here and then another SP hybrid orbital with one valence electron here. This carbon, it's SP hybridized. Once again, go back up here to this diagram. This carbon is also going to have a P orbital with a valence electron and another P orbital with another electron in it. Go back down to here and we draw in those P orbital. Here is one P orbital. Here is one P orbital with one valence electron and there here's another P orbital with an electron in here like that. Finally, we have to add in hydrogen. We have a hydrogen on either side here. We know that hydrogen has one valence electron in an unhybridized S orbital. Over here, we have a hydrogen with one valence electron in unhybridized S orbital. Now, we can finally analyze the bonding that's present. We know that if we have head on overlap of orbitals like right in here. That's a sigma bond. There's one sigma bond. Here's a head on overlap of orbitals between our two carbons. That's a sigma bond. Then finally, we have a head on overlap of orbitals here. That's another sigma bond. With a total of three sigma bonds in the acetylene molecule. The video on SP2 hybridization, we saw how to make a pi bond. We had this side by side overlap of orbitals. Here we have one pi bond. We have interaction above and below. That's one pi bonds and then we have another pi bond here. We have side by side overlap of orbitals here as well and so we have two pi bonds present. We have two pi bonds present in the acetylene molecules. Let's look at our dot structure again. We saw the bond between this carbon and this hydrogen was a sigma bond. We saw there was one sigma bond between our two carbons. I'm just gonna pick the one on the middle here. Say that's a signal bond and then this bond over here we said was a sigma bond. There's our three sigma bonds and then we have a triple bond presence. There were two pi bonds also present. Two of these are pi bonds here. A total of two pi bonds and three sigma bonds for the acetylene molecule here. Remember pi bonds prevent free rotation. We can't rotate about the sigma bond between the two carbons because of the pi bonds. There's no free rotation for our triple bond. We have a linear shape. Let me go ahead and draw that line in here. You see there's linear geometry for this molecule like that. Also in terms of bond length. The distance between these two carbons. The distance between this carbon and this carbon, let me circle them. The distance between these two carbons turns out to be approximately 1.20 angstroms. An even shorter bond length than in our previous videos. Once again, that's due to the increased S character. Increased S character gives you these smaller orbitals and that's one way to think about the shorter bond distance and the triple bond compared to a double bond or a single bond. That's a lot that we've covered here. Let's go ahead and draw the dot structure one more time and analyze it using steric number. We have our triple bonds. If we're doing steric number to find out the hybridization state, we know to do steric number, you take the number of sigma bonds. Let's say our goal was to figure out the steric number for this carbon. The number of sigma bonds. I know this is a sigma bond. I know on a triple bond, I have one sigma bond and two pi bonds. There are two sigma bonds here and zero lone pairs of electrons. Two plus zero gives me two. I need two hybrid orbitals. Which you make from one S orbital and one P orbital. If you get a steric number of two, you think SP hybridization. This carbon is SP hybridized and so is this carbon as well. That's how to think about it using steric number. Once again, a linear geometry with a bond angles of 180 degrees. Let's do one more example using steric number to analyze the molecule. Let's do carbon dioxide. If we wanted to figure out the hybridization of the carbon there. Let's go ahead and do that. Using steric number. The hybridization of this carbon. The steric number is equal to the number of sigma bonds. If I focus in on the double bond between one of these oxygens in this carbon, I know that one of these bonds is a sigma bond. from our previous video. I have one sigma bond here and then for this other double bond on the right, I know that one of them is a sigma bond. I have two sigma bonds here and zero lone pairs of electrons around the carbon. Two plus zero gives me a steric number of two. I need two hybrid orbitals for that carbon and of course, that must mean this carbon is SP hybridized. This carbon here is SP hybridized as well and therefore, we know that this is a linear molecule with a bond angle of 180 degrees. Once again, steric number is just a nice way of analyzing the hybridization and also the geometry of the molecule. In the next video, we'll look at a couple of examples of organic molecules in different hybridization states.