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AP.Chem:

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in an earlier video we saw that when carbon is bonded to four atoms we had an sp3 hybridization with a tetrahedral geometry and an ideal bond angle of 109.5 degrees if we look at one of the carbons in ethene let's say this carbon right here we don't see the same geometry so the geometry of the atoms around this carbon happens to be planar and so actually this entire molecule is planar so you can think about all this in a plane here and the bond angles are close to 120 degrees so approximately 120 degree bond angles and this carbon that I've underlined here is bonded to only three atoms so a hydrogen a hydrogen and a carbon and so we must need a different hybridization for each of the carbons presence in the ethylene molecule and so we're going to start with with our electron configurations over here the excited stage right so we have carbons four valence electrons represented so 1 2 3 & 4 and in the video on sp3 hybridization we took all four of these orbitals and combined them to make four sp3 hybrid orbitals in this case we only have a carbon bonded to 3 atoms so we only need 3 of our orbitals so we're going to promote the S orbital so we're going to promote the S orbital up and this time we only need two of the P orbitals so we're going to take one of the P's and then another one of the P's here that is going to leave one of our P orbitals unhybridized so each one of these orbitals has one electron in its like that and this is no longer an S orbital this is an sp2 hybrid orbital alright this is no longer a p orbital this is an sp2 hybrid orbital and same with this one an sp2 hybrid orbital we call this sp2 hybridization so let me go ahead and write this up here I don't see us use a different color here so this is sp2 hybridization because we're using 1s orbital and 2p orbitals to form our new hybrid orbitals and so this carbon right here is sp2 hybridized and the same with this carbon alright so notice that we left a p orbital untouched so we have a p orbital unhybridized that in terms of the the shape of our new hybrid orbital let's go ahead and get some more space down here all right so we're taking 1s orbital we know S orbitals are shaped like spheres we're taking 2p orbitals we know that a p orbital is shaped like a dumbbell alright so we're going to take these these orbitals and hybridize them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that and once again when we draw the pictures we're going to ignore the smaller back lobe all right so this gives us our sp2 hybrid orbitals in terms of what percentage character all right we have three orbitals that we're taking here and one of them one of them is an S orbital so one out of 3 gives us 33% s character and our new hybrid sp2 orbital and then we have two P orbitals right so two out of three gives us 67% P character so 33% s character and 67% P character so there's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an S orbital is closer to the nucleus right so we think about the electron density here being closer to the nucleus that means that we can think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that's going to have an effect on the length of the bonds that we're going to be forming all right so let's go ahead and draw the picture of the F of the ethylene molecule now so we know that each of the carbons in ethylene is just going back up here to to emphasize the point each of these carbons here is sp2 hybridized that means each of those carbons is going to have three sp2 hybrid orbitals around it and one unhybridized p orbital so let's go ahead and draw that so we have we have a carbon right here and this is an sp2 hybridized orbital so we're going to draw in there's one sp2 hybrid orbital here's another sp2 hybrid orbital and here's another one and we go back up to here and we can see that each one of those orbitals let me go ahead and mark this each one of those sp2 hybrid orbitals has one electron in it so each one of these orbitals has one electron I go back down here and I put in the one electron in each one of my orbitals like that all right and I know that each of those carbons is going to have an unhybridized p orbital here so an unhybridized p orbital with one electron - so let me go ahead and draw that in so I'll go ahead and use a different color so we have our unhybridized p orbital like that and there's one electron in our unhybridized p orbital alright so each of the carbons right was sp2 hybridized so let me go ahead and draw the dot structure right here again so we can take a look at it all right the dot structure for ethylene okay so let's do let's do the other carbon now so the carbon on the right is also sp2 hybridized so we can go ahead and draw in an sp2 hybrid orbital and there's one electron in that in that orbital and then there's another one with one electron and then here's another one with one electron alright and this carbon being sp2 hybridized also has an unhybridized p orbital with one electron so go ahead and draw in that that p orbital with its one electron we also have some hydrogen's right so we have we have some hydrogen's to think about here so each carbon is bonded to two hydrogen's let me go ahead and put in the hydrogen's the hydrogen has a valence electron in an unhybridized s orbital so I'm going ahead and putting in the S orbital right and the one valence electron from hydrogen like this all right when we take a look at what we've drawn here all right we can see we can see some head-on overlap of orbitals which we know from our earlier video is called a sigma bond alright so here's a head-on overlap of orbitals so that's a sigma bond here's another head-on overlap of orbitals all right the carbon-carbon bond here's also a head-on overlap of orbitals and then we have these two over here and so we have a total of five sigma bonds in our molecule so let me go ahead and write that over here so there are five sigma bonds all right if i'm trying to find those on my on my dot structure that would be this would be a sigma bond this abuse sigma bond one of these two is a sigma bond and then these over here so a total of five sigma bonds and then we have a new type of bonding all right so these these unhybridized p orbitals can overlap side by side right so up up here and down here we get side by side overlap of our p orbitals and this creates a PI bond so a PI bonds go ahead and write that here so a PI bond is side-by-side overlap alright so there is there is overlap above and below this the Sigma bond here and that's going to prevent free rotation so when we're looking at the example of ethane we had free rotation about the Sigma bond that connected the two carbons but because of this PI bond here this pi bond is going to prevent rotation so we don't get we don't get different confirmations of the ethylene molecule so no no free rotation due to the PI bond alright when you're looking at the dot structure one of these bonds is the PI bonds I'm just going to say it's this one right here and so if you have a double bond one of those bonds this Sigma bond and one of those bonds is a PI bond so we have a total of one PI bond alright in the ethylene molecule if you're thinking about the distance between the two carbons right so let me go ahead and use a different color for that the distance between this carbon and this carbon alright so it turns out to be approximately 1.3 4 angstroms all right which is shorter than the distance between the two carbons in the ethane molecule remember for ethane the distance was approximately 1.5 for ox drums and so a double bond is shorter than a single bond one way to think about that is the increased s character right so this increase s character right means the electron density is closer to the nucleus and so that's going to that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here so 1.34 on strums alright let's uh let's look at the dot structure again and see how we can analyze this using the concept of steric number so let me go ahead and redraw the dot structure so we have our carbon-carbon double bond here and our hydrogens like that if you're approaching this this situation using steric number remember to find the hybridization we can use this concept steric number is equal to the number of Sigma bonds plus number of lone pairs of electrons so if my goal was to find the steric number for this carbon right I count out my number of Sigma bonds so that's one two and then I know in a double bond one of those is sigma and one of those is pi so one of those is a sigma bonds of a total of three sigma bonds I have zero lone pairs of electrons around that carbon so three plus zero gives me a steric number of three so I need three hybrid orbitals and we've just seen in this video that three sp2 hybrid orbitals form if you're dealing with sp2 hybridization so if you get a steric number of three you're going to think about sp2 hybridization 1s orbital and 2p orbitals hybridizing alright so that carbon is sp2 hybridized and of course this one is two so both of them are sp2 hybridized let's do another example let's do let's do boron trifluoride right so bf3 if you want to draw the dot structure for bf3 right you would have boron and then you would surround it with your fluorines here and you would have an octet of electrons around each fluorine so I'll go ahead and put those in on my dot structure if your goal is to figure out the hybridization of this boron here alright so what is the hybridization state of this boron let's use the concept of steric number so once again let's use steric number so we'll find the hybridization of this boron steric number is equal to number of Sigma bonds so that's one two three so three sigma bonds plus lone pairs of electrons that's zero so steric number of three tells us this boron is sp2 hybridized so this boron is going to have three sp2 hybrid orbitals and one p orbital one unhybridized p orbital so let's go ahead and draw that so we have a boron here bonded to three fluorine and also is going to have an unhybridized p orbital now remember when you are dealing with boron it has one less valence electron than carbon-carbon had four valence electrons boron has only three so when you're thinking about the sp2 hybrid orbitals that you create sp2 hybrid orbital sp2 sp2 and then one unhybridized p orbital right here boron only has three valence electrons so let's go ahead and put in those valence electrons 1 2 & 3 so it doesn't have any electrons in its unhybridized p orbital and so over here when you're looking at the picture write this as an empty orbital and so boron can accept a pair of electrons so if we're thinking about its chemical behavior right one of the things that that bf3 can do the boron can accept an electron pair and function as a Lewis acid and and so that's one way in thinking about how hybridization allows you to think about the the structure and how something might react alright so this boron turns out to be sp2 hybridized right so this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule right it's planar all right so around this boron it's planar and so therefore your bond angles are 120 degrees so if you had boron right here and you're thinking about a circle a circle is 360 degrees so if you divide a 360 by 3 you get 120 degrees for all of these bond angles alright in the next video we'll look at SP hybridization

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