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Steric number

One way to determine the hybridization of an atom is to calculate its steric number, which is equal to the number of sigma bonds surrounding the atom plus the number of lone pairs on the atoms. In this video, we focus on atoms with a steric number of 4, which corresponds to sp³ hybridization. Created by Jay.

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  • aqualine tree style avatar for user iam.psm7
    when do we have a pi bond?
    (19 votes)
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  • piceratops sapling style avatar for user prachi
    since lone pairs of electrons don't contribute in the bonding, why do they form hybrid orbitals?
    (30 votes)
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  • leaf green style avatar for user wannabeDoc
    So when the Steric number of an element equals 4, then the element MUST BE sp3 hybridized?
    (31 votes)
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  • blobby green style avatar for user Lizzy Chan
    Shouldn't the bond angle of water be approximately 104.5?
    (7 votes)
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    • old spice man green style avatar for user Matt B
      Yup! And for anyone wondering how to get this:
      A tetrahedral structure (e.g. methane) will have bond angles of 109.5 degrees. For every lone pair of electrons that exists in the central atom, you subtract 2.5 degrees. The oxygen in water has two lone pair of electrons, and so it will have 109.5 - 2*(2.5) = 104.5 degree bond angle.
      (36 votes)
  • male robot johnny style avatar for user Auro soni
    What is the meaning of steric?
    (9 votes)
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  • leaf blue style avatar for user Kolta  Hista
    I am confused why water is sp3 hybridised. I thought there are only 1 sigma bond and 1 pi bond, so shouldn't it be a sp hybridised bond?
    (3 votes)
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    • leafers tree style avatar for user drzulu
      Good guess, but as Jay said in the video, the oxygen in water actually has two sigma bonds (one with each attached hydrogen) and then two lone pairs of electrons.

      If it had a pi bond, we would find a double bond in water, which would change its chemistry quite drastically!

      I hope this helps :)
      (22 votes)
  • blobby green style avatar for user Bishr Elias
    Why Nitrogen needs sp3 Hybridization to form ammonia? Why doesn't the 2s orbital have 2 electrons and the 2p bond with those 3 hydrogens
    (6 votes)
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    • aqualine ultimate style avatar for user famousguy786
      If the 2s orbital bonds with a hydrogen the bond will be shorter than the bonds formed by 3 2p orbitals with hydrogen. If this was the case, ammonia would have 3 long bonds and a short bond. However, when the shape of ammonia was discovered, it was clear that all N-H bond lengths were equal, hence the idea of 2s and 2p orbitals forming individual bonds with the hydrogen was discarded and hybridization was able to explain the shape of ammonia.
      (7 votes)
  • duskpin ultimate style avatar for user Prachi Jain
    Why does nitrogen in ammonia have to be hybridized? It has three halfway filled p orbitals, couldn't it simply form bonds with them without hybridization?
    (2 votes)
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    • blobby green style avatar for user xmanaxman
      This is a great question that goes into the details of molecular orbital theory and chemical bonding.

      When a nitrogen atom forms an ammonia molecule, it doesn't use its three half-filled 2p orbitals separately to form bonds. Instead, it "hybridizes" them with its 2s orbital to form four "sp3" hybrid orbitals.

      Here's why this happens:

      - *Geometry*: The geometry of an ammonia molecule (NH3) is trigonal pyramidal, not trigonal planar. If nitrogen used its three 2p orbitals as they are, we would expect the bonds to be 120 degrees apart, resulting in a planar molecule, which is not the case. However, when nitrogen uses sp3 hybrid orbitals to bond, it can achieve the tetrahedral electron-pair geometry (with bond angles about 109.5 degrees) that matches the observed geometry.

      - *Overlapping*: When it comes to forming covalent bonds, the efficiency of orbital overlap is crucial. Hybrid orbitals (sp3 in this case) can form stronger, more stable overlapping (and thus bonds) than the pure s and p orbitals.

      - *Energy*: In order to have the most stable (lowest energy) configuration, atoms will hybridize their orbitals to allow for maximum bonding. In the case of nitrogen in ammonia, this means hybridizing to have four equivalent sp3 orbitals, three of which are used for bonding with hydrogen atoms, and one of which holds the lone pair of electrons.

      So, while it may seem as if nitrogen could use its unhybridized orbitals to form bonds, the reality is that hybridization allows for stronger bonds, a more stable molecule, and a geometry that fits with what we observe in reality.
      (9 votes)
  • aqualine ultimate style avatar for user Verdhaan
    What is VSCPR theory at
    (3 votes)
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  • blobby green style avatar for user -cricket-
    This may sound silly.
    Don't electrons in a bond repel each other the way they repel other electron bond pairs together?

    What keeps them together?
    (3 votes)
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Video transcript

Voiceover: The concept of steric number is very useful, because it tells us the number of hybridized orbitals that we have. So to find the steric number, you add up the number of stigma bonds, or single-bonds, and to that, you add the number of lone pairs of electrons. So, let's go ahead and do it for methane. So, if I wanted to find the steric number, the steric number is equal to the number of sigma bonds, so I look around my carbon here, and I see one, two, three, and four sigma, or single-bonds. So I have four sigma bonds; I have zero lone pairs of electrons around that carbon, so four plus zero gives me a steric number of four. In the last video, we saw that SP three hybridized situation, we get four hybrid orbitals, and that's how many we need, the steric number tells us we need four hybridized orbitals, so we took one S orbital, and three P orbitals, and that gave us four, SP three hybrid orbitals, so this carbon must be SP three hybridized. So let's go ahead, and draw that in here. So this carbon is SP three hybridized, and in the last video, we also drew everything out, so we drew in those four, SP three hybrid orbitals, for that carbon, and we had one valence electron in each of those four, SP three hybrid orbitals, and then hydrogen had one valence electron, in an un-hybridized S orbital, so we drew in our hydrogens, and the one valence electron, like that. This head-on overlap; this is, of course, a sigma bond, so we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs, so these electron pairs are going to repel each other: like charges repel. And so, the idea of the VSEPR theory, tell us these electron pairs are going to repel, and try to get as far away from each other as they possibly can, in space. And this means that the arrangement of those electron pairs, ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon, like that. When we think about the molecular geometry, so that's like electron group geometry, you wanna think about the geometry of the entire molecule. I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, "in the plane of the page." And so, we can go ahead a draw in our hydrogens, and this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face. And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral, the molecular geometry of methane is tetrahedral, and then we also have a bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen bond angle in here, is approximately 109 point five degrees. All right, let's go ahead and do the same type of analysis for a different molecule, here. So let's do it for ammonia, next. So we have NH three, if I want to find the steric number, the steric number is equal to the number of sigma bonds, so that's one, two, three; so three sigma bonds. Plus number of lone pairs of electrons, so I have one lone pair of electrons here, so three plus one gives me a steric number of four. So I need four hybridized orbitals, and once again, when I need four hybridized orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals, and so let's go ahead and draw those four hybrid orbitals. So we would have nitrogen, and let's go ahead and draw in all four of those. So, one, two, three, and four; those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron, and then you'd have two in this one, like that. And then you'd go ahead, and put in your hydrogens, so, once again, each hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those hydrogens, so our overlap of orbitals, so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here. So the arrangement of these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule, so if I go ahead and draw in another picture over here, to talk about the molecular geometry, and go ahead and draw in the bonding electrons, like that, and then I'll put in my non-bonding electrons, up here: this lone pair right here, housed in an SP three hybridized orbital. So, the arrangement of the atoms turns out not to be tetrahedral, and that has to do with this lone pair of electrons up here, at the top. So, this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example, and because it's going to repel those electrons a little bit more strongly, you're not gonna get a bond angle of 109 point five; it's going to decrease the bond angle. So let me go ahead, and use the same color we used before, so this bond angle is not 109 point five; it goes down a bit, because of the extra repulsion, so it turns out to be approximately 107 degrees. And in terms of the shape of the molecule, we don't say "tetrahedral"; we say "trigonal-pyramidal." So let me go ahead, and write that here, so the geometry of the ammonia molecule is trigonal-pyramidal, and let's analyze that a little bit. So, "trigonal" refers to the fact that nitrogen is bonded to three atoms here, so nitrogen is bonded to three hydrogens, so that takes care of the trigonal part. The "pyramidal" part comes in, because when you're doing molecular geometry, you ignore lone pairs of electrons. So if you ignore that lone pair of electrons, and just look at this nitrogen, at the top of this pyramid right here, so that's where the "pyramidal" term comes in. So bonded to three other atoms, like this, this, and this; for our pyramid. So trigonal-pyramidal is the geometry of the ammonia molecule, but the nitrogen is SP three hybridized. All right, let's do one more example; let's do water. So, first we calculate the steric number, so the steric number is equal to the number of sigma bonds, so that's one, two; so, two sigma bonds. Plus numbers of lone pairs of electrons, so here's a lone pair, here's a lone pair; so we have two plus two, which is equal to four, so we need four hybridized orbitals. As we've seen in the previous two examples, when you need four hybridized orbitals, that's an SP three hybridization situation; you have four SP three hybridized orbitals. So this oxygen is SP three hybridized, so I'll go ahead and write that in here, so oxygen is SP three hybridized. So we can draw that out, showing oxygen with its four SP three hybrid orbitals; so there's four of them. So I'm gonna go ahead and draw in all four. In terms of electrons, this orbital gets one, this orbital gets one, and these orbitals are going to get two, like that; so that takes care of oxygen's six valence electrons. When you're drawing in your hyrdogens, so let's go ahead and put in the hydrogen here, so, once again, each hydrogen with one electron, in a un-hybridized S orbital, like that. So in terms of overlap of bonds, here's one sigma bond, and here's another sigma bond; so that's our two sigma bonds for water. Once again, the arrangement of these electron pairs is tetrahedral, so VSEPR theory says the electrons repel, and so the electron group geometry, you could say, is tetrahedral, but that's not the geometry of the entire molecule, 'cause I was just thinking about electron groups, and these hybrid orbitals. The geometry of the molecule is different, so we'll go ahead and draw that over here. So we have our water molecule, and draw in our bonding electrons, and now let's put in our non-bonding electrons, like that, so we have a different situation than with ammonia. With ammonia, we had one lone pair of electrons repelling these bonding electrons up here; for water, we have two lone pairs of electrons repelling these bonding electrons, and so that's going to change the bond angle; it's going to short it even more than in the previous example. So the bond angle decreases, so this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So, thinking about the molecular geometry, or the shape of the water molecule, so we actually call this "bent," or "angular," so this is, "bent geometry," because you ignore the lone pairs of electrons, and that would just give you this oxygen here, and then this angle; so you could also call this, "angular." So we have this bent molecular geometry, like that, or angular, and once again, for molecular geometry, ignore your lone pairs of electrons. So these are examples of three molecules, and the central atom in all three of these molecules is SP three hybridized, and so, this is one way to figure out your overall molecular geometry, and to think about bond angles, and to think about how those hybrid orbitals affect the structure of these molecules.